The AM Forum

What's a tell tale sign of RF feedback?

Squealing at low or high audio frequencies, runaway modulator current, irregular 60 cycle hummy background noise, audible on-the-edge-of-oscillation that sounds like ringing on audio peaks. Those are a few of my favorites Bob!

Blisters on your microphone hand and / or lips.

--Shane

If done on purpose (correctly) in a high-level rig, less noise and distortion.

Been there!! ;D  'specially while using loop modulation

If by "RF feedback" you mean tx RF getting into your baseband audio gear and distorting the audio, you can spot it when you monitor yourself off-air and the audio sounds fuzzy and unclear on an antenna but is much less so or not at all on a dummy load.  I could be mistaken about this but I think solid state gear, both audio and RF, is more vulnerable to this.  If this is what you are asking about we can continue with methods for eliminating it.

Rob

What I was thinking was more of RF output getting into RF input and how to recognize it.  The reason I ask, I built a solid state linear amplifier for the Retro75, a Communications Concepts EB63.  It's a broadband design.  My input and output jacks are next to each other but are connected with shielded cable.  I didn't think about separating them until now.   I'm having trouble with linearity.  Input looks nasty as does output.  I'm beginning to think it's the load presented to the Retro75 that is the issue but just wanted to know how to recognize RF feedback.  Maybe I don't have an RF feedback issue but an issue with the Retro75.

That amp has the input and output out of phase, thank goodness....

If linearity is suffering, check the bias.  About .7 to .75 will turn the 454s on enough to not have a problem.  Also, you could have a problem with RF NOT being decoupled enough from the collectors, and getting back into the retro75 (possible) or into the bias of the amplifier, causing problems.

And the thing Motorola forgot was the negative feedback.  Try a .01 ceramic disc and a 68 ohm 2 watt resistor from collector to base on BOTH transistors.  I'd say that's probably the problem...  That amp will be MUCHO unstable (especially on the low end of the operating spectrum, where you happen to be having problems) depending on the ferrite they used because they didn't use any type of feedback.


I run a 16 X 2SC2879 amplifier.  Playing with the values of negative feedback got it a lot more stable and tame on the low end of the spectrum, whereas before it only wanted to play on 20 - 10.

--Shane

Don't know if this is any indication but here's what's happening, Retro75 has a nice carrier when the amp isn't keyed (straight through to 50ohms) but as soon as the amp is keyed input linearity goes out the roof.

That's like asking "How can I tell if I've been shot with a .45" or "How can I tell if she's pregnant". Don't worry, YOU WILL KNOW IT.  ;D ;D

Ah Yeah.. as to the The Retro ;D

Umm,.. what ever any owner of the little bugger does.. DO NOT Pair up the Final in that little thing Umm,.. the Tranny,..it gets a Little Bit Real Hot Real Fast LOL.....Yessir and the Audio chip Lets a little bit of Smoke out..LOL..Careful on that one..
Ahh... Recieve is Good in a strong environment butt could use a stage of Gain.. just a L'ill..just a Tad... 8)

But Over all after the l'ill bugger settles down a bit it's a neat little toy... :)

Oh an Be careful about the Fet's... ;D

undersolderlingly yours 8)

Zed.L.R.

I used a different RF source into the amp and got good results, a nice clean wave form. So I'm ruling out the amplifier as the problem, at least for now.  The Retro  75 seems to be the issue at this point.

Bob,

Try some type 45 ferrite cores on the coax between the driver and linear.

Also, do you have more than enuff drive available so that you can add a 50 ohm non-inductive resistor across the linear amp input to provide a more stable load for the driver?   Another version of this might be a 25 ohm in series with the linear input.  This will provide some isolation but increase the input impedance to 75 ohms or so.

If you're not sure about RF in the audio yet, try a .001 cap across the audio input to the rig - and a 500 ohm in series with the same audio input lead. They can both be put inside the audio connector that plugs into the rig.

T



BTW, here's a Mouser order I recently placed for 0.5" I.D. and 0.25 ID cores. The .5" are good for coax and the .25" for cables.  I used all 40 in the shack as a preventive measure. This is now an RF-in-the-audio-proof shack, caw mawn.

TYPE 45 material clamp-on cores: (snap-on)

1/2" ID:
875-28A2024-0A0
Laird Technologies EMI/RFI Suppressors & Ferrites
280ohms 100MHz
US HTS: 8504.90.0080 ECCN:EAR99  20 20 2.420 48.40
2 1  


1/4 ID:
 
875-28A0640-0A0
Laird Technologies EMI/RFI Suppressors & Ferrites
240ohms 100MHz
US HTS: 8504.90.0080 ECCN:EAR99  20 20 1.070 21.40
 1RoHS: Compliant

Shipping Information Merchandise Total
 69.80
 Handling: 0.00
Freight: 18.67
Tax: 0.00
Order TOTAL: $88.47

  
THIS ORDER IS SUBJECT TO ALL TERMS AND CONDITIONS DISPLAYED AT:
www.mouser.com/saleterms
 

I've also put a 560 ohm load across the audio input connector.
radio shack snap on beads could be your friend

Hi Tom,
Yes, I did the 50ohm non-inductive resistor approach across the input to the amp lastnight. That improved the carrier waveform considerably but still the carrier is not a 'clean' sinewave.   Without the 50ohm resistor the Retro75 is behaving like a noise generator, no discernable waveform at all.  The amp is producing what it sees at its input.  The HW-8 and FT-102 have a good match as it seems with the amp.  Their input waveforms are clean and the output of the amp is clean with these two RF sources. I'm wondering if there is some sort of reflection or impedance issue that is messing with the biasing of the final transistor.

I haven't addressed any potential audio problems yet.  Just trying to get the carrier cleaned up. I suspect that if I get the carrier cleaned up the audio amp will not be a problem.

So I 'm wondering on my approach to clean this up.  Do I mod the Retro 75 final or do I need some sort of matching network between the Retro and amp.

http://smallwonderlabs.com/Retro-75_Instr.pdf    The final PA is on page 10.

Lastly, as far as I can tell the phase relationship of the amp from input to  output seems to be 180 degrees apart so any feed back related to the close proximity of the RF input and RF output jacks will only effect gain potentially depending on the shielding between them. So I don't think I have a detrimental issue here, at least not yet.

Bob,

You could try a pi-network between the driver and amp. That will probably solve it. All you need are some small variable caps, like a 365pf multi-section to ground on the input, (1500pf or so)  a roller inductor in series and a 1500pf cap to ground for the ouput of the tuner.


Even  easier, a 'T' network should work just as well. Experiment with clips leads until you find an effective config to build.

You can build one up easily and get a 1:1 match on any freq when using the roller and var caps.

T

bob,
Try a couple beads on your scope probe lead to make sure it isn't common mode crap getting into the scope.

Bob

I was looking at the documentation on the Communications Concepts web site for the EB63.

Some things are peculiar:

A. The documentation says that the input and output transformers are 16:1 (turns ratio)... but this is clearly a typo. The correct value is 1.6:1. I.e., the 20 ohms of swamping resistance from gate to gate x 1.6 x 1.6 = a 51 ohm input impedance.

What is the turns ratio of the transformers you are using?

B. The capacitance across the primary and the secondary of each transformer appears to be too large. For example, the capacitance across the primary of the input transformer is .001 uF. This capacitor has a reactance of only 40 Ohms on 75 meters (and, of course, less at higher frequencies). Something is "fishy" about those capacitances.

I suggest that you reduce the values of the capacitors across the input and the output of the input transformer by a factor of 10.

Best regards
Stu

Bob,
I agree with Stu .001 is way too high. CCI amplifiers have problems below about 3 MHz. The transformer inductance is a bit low for 160 meter operation. I'm thinking you could get away with around 200 pf. This will degrade low frequency operation a bit. If you are running a 12 volt final the output transformer is usually 1:4 or 1:5 turns ratio. 1:5 for higher power devices. Both inputs and outputs of bipolars are quite low impedance. FETs are easier to drive. A Bipolar without RF feedback will have an input impedance all over the map. Best to try a 3 dB pad on the input to control the load on the source.

Frank

Yes... since these are bipolar transistors (it would have been better to use 11N90s for 75 meter operation) the input impedance is very low... and that would explain the large transformer turns ratio. The pair of 10 Ohm resistors are not swamping resistors... but just biasing resistors.

Nevertheless, as we agree, the capacitance across the primary of the input transformer is too large... and the Retro75 is probably very unhappy about looking directly into this amplifier.

I agree that a power divider/attenuator at the input might help.

Stu   

Stu,

You've probably heard a few of the guys on 75M experimenting with 260's and 11N90's for 75M linear service. How clean do you think these devices are for linears?    It seems commercial  amps go to the great expense of paying $35++ per device for the MRF-150's. Why aren't the 11N90's for $3 each being used there?

No one's run spec analyzer tests on these project amps yet, so I'm curious.

I even had problems getting my eight pill  MRF-150 SS linear amp to be acceptably clean, so this is puzzling.


T

Tom

I agree that it is important to consider how "clean" an amplifier is... and that this may be a problem associated with many of the simple linear amplifier designs that are appearing on the air (both in commericially produced products and in homebrew versions)

As you know... there are two complementary issues in that regard. For clarity... I'll state what these are.

The first issue (which is not what I think you are referring to) is whether the amplifier generates harmonics of its intended rf output frequency.

If one is using a broadband output circuit... then getting the amplifier to meet FCC requirements can be challenging. However, if the amplifier has a resonant tank circuit of sufficient Q (such as the series resonant circuit that the Class E folks use) or a selectable low pass filter at the output (which a lot of these types of simple linear amplifiers employ)... or both (which I use with my 1-FET Class E transmitter)...then keeping harmonics under control is a lot easier.

I assume that anyone who builds an amplifier would take reasonable steps to sufficiently attenuate harmonics.

The second issue is whether or not the amplifier produces "splatter". Splatter can be produced in a collector-modulated or drain modulated amplfier by non-linearity of the modulation characteristic... but it can also be produced in a "linear" amplfier by 3rd order rf intermodulation products associated with the rf output vs rf input characteristics. For example, if I have a linear amplifier that is accepting a 3.5 kHz-wide modulated signal as input, centered at 3.885 MHz... then the third order products will include a 10.5 ( 3 x 3.5) kHz wide signal that is also centered at 3.885 MHz. The output circuit, of course, cannot filter this out.

So... what I think you are referring to is this 3rd order product.

I suspect, as you suggest, that many of the linear amplifiers that are appearing on the air have a "wide" signal that is, in fact, associated with 3rd order intermodulation products.

This is certainly something to keep in mind when building one of these amplifiers.... particularly if the input signal is AM-modulated by a fairly wide audio signal, than doesn't have any sharp bandlimiting filters (e.g. +/-7 kHz x 3 = +/- 21 kHz).

Thanks for pointing this out.

Stu    

I have been following the Retro 75 projects for a while. These rigs will find a nice niche on 75m AM.

I have a couple of comments regarding the Retro-75+EB63 combination.

1.) It seems that the PEP output of the Retro75 (~ 11watts pep) will be at least 2X the input requirement of the EB63 (~ 5W pep). So as others have posted, a 3db (at least) attenuator between the Retro and the amp would be helpful. Alternatively adding negative feedback to the amp as Shane suggested will lower the EB63 stage gain, increase the drive requirement, and clean up some of the intermodulation distortion Stu and Tom discussed. Maybe do some of both an attenuator, and RF feedback. Analysis of the EB63 input impedance is warranted.

2.) Since both the Retro75, and the EB63 are powered from 12V, there might be some sort of regeneration between the two units if the power comes from the same source. If so, additional decoupling, and ground considerations should be looked into. The layout of the parts, wire routing, etc. is important. Tom, K1JJ has a bunch of insight with this kind of stuff from his Class E experience. I think Tom eats ferrite beads for breakfast!   ;D

Jim
WD5JKO

Can't recall seeing good IMD performance (-32 dB 3rd and quickly disappearing higher orders) from devices unless they are specifically rated for linear service or run well below (-3 to -6 dB) their saturated power output levels. This goes for tubes, transistors and FETs.   

Yes, this is what I'm referring to, the 3rd order IMD of a device. Thanks for the info Stu and Jay.

Lord knows I've been thru many many "linear" amp tubes trying to find one that will give me -35db 3rd in the real whirl. It was difficult and drove me crazy for years. I finally did it, but it took elaborate measures.

I see that most riceboxes of today are good for only about 30-32db 3rd. So the IMD willl never be better than the ricebox's IMD figures when driving an amp - and always worse. Many SS linear amps are good for -25db 3rd. Gawd, when someone is S9 + 60 over, they will be splattering up the band heavily using specs at -25db 3rd or lower.

Anyway, that said, using these 11N90 amps mobile might fly under the radar, as the signal is already down 15db with the mobile whip - and the narrowband whip may even offer some near-channel attenuation due to the whip's VERY narrow swr curve.

I would certainly like to build one up myself using 11N90's, though, from my past experience with MRF-150's, it will not be an easy task if the amp is doing 1500w pep and operating into a fullsize dipole near others...  ;D

Though I heard one guy with a band scope say that he thought the project amp (running 260's) was reasonably clean the other day. I would like to see a real two-tone scientific  spectrum analyzer test and some careful efficiency measuements made to put the mystery to bed...

I know Bob/KBW has one fired up and is getting the bugs out now. I'll post whatever results we can glean from that. Hope it works FB. What a savings it would be to run $3 11N90 super 900V Fets in a linear amp!!!

T

I'll make the measurements if someone provides an amplifier.

Some folks fixate on the 3rd order products (because they're the most prominent and the ones normally reported) and don't pay attention to what's going on further out. It's the 5th, 7th, 9th etc that make the signal really wide. An amplifier with a bit higher 3rd order products (within reason) but with a rapid attenuation slope of the higher order components is preferable to the opposite condition.

Device types / amplifiers exhibit widely varying slopes so each design really should be analyzed. Idling current, device matching and proximity to saturated output influence IMD level and attenuation slope.   

Yep, Agreed.  When I say 3rd order, I really mean them all, but it wud be too much to type... ;)  Maybe "IMD products" is a better phrase.

When I test my rigs for IMD, I use my reference FT-1000D driver barefoot to mark the crud levels up the band. It is very clean running at 20 watts pep. Then I add the amplifier and see how much it changes, to the limit of the noise floor up the band. I've found my highly modified Henry running reduced power and voltage - and Dr. Love, generally do not add to the trash level more than a few db. That's only when they are running conservatively.  I did run some two-tone tests, but found by playing taped voice programming through the system it is more revealing  for the infinite variations in dynamic range added distortion vs: a steady tone. A voice, at certain "sour spot" audio frequencies to the rig, will make the rig dirtier from what I see.


I'd be curious as to what Bob/KBW's amp shows for efficiency, cuz he said last night the heatsink was getting rather hot. At 33%, you wud think it would. 300w of heat is no small job to sink.

Maybe he'll swing over to your place with his amp when it's ready.  I know lots of guys are interested in hearing about the specs.  I'd be one of the first to knock one out with 12 pills++ if it tests well..

T

I know that I have to pad the input because of the expected PEP.  Not a problem.  Right now I'm just working on just the carrier and trying to clean things up and make the Retro happy. 

The schematic that came with the kit is different than what is on the website.  According to the assembly instructions some changes have occurred with the amp.   

Not sure how you determined that a .001uf is on the input transformer.  The only .001 in this kit is on the biasing portion of the circuit.
 
The input transformer has an 18pf on the primary and an 1100pf on the secondary.  The output primary has a 910pf  and the secondary has a 24pf. 

Should I still knock down the 18 and 1100 down by a factor of 10?

Bob

Ok... It appears that the noisy schematic on Communication Concepts site makes "C2" look like "C7". I found a clean schematic and associated documentation on another web page of their site:

http://www.communication-concepts.com/appnotes/eb63300sharp.pdf

The values you listed, above, for the capacitors are reasonable.

Try putting a 50 ohm, non-inductive resistor in series with the output of the Retro 75, looking into the input of the amp. I.e. center-pin-to-center-pin, not center-pin-to-ground
.
Stu

Bob,
I have a Motorola app note from 1995 that says 1100 pf on the secondary of the input transformer and 910 pf on the primary of the output transformer. My guess is your transistors and not a matched pair and one is hogging the bias. I have seen this effect playing with bigger 48 volt devices.
So one device may be in class B and the other AB. Usually there is a color dot on the device case. Both should have the same color dot for hfe match.
The efficiency will be around 50% so expect heat. I suspect EB63 isn't great for IMD or input VSWR without feedback. I can scan the app for you after the weekend if you remind me.

That's an interesting point. I wonder if the 11N90 linear amp could be made to work well (acceptable IMD) with a large amount of RF negative feedback?  Maybe a few 11N90's driving a group of twelve (6X6 p-p) with NFB around the whole bunch?

T

Yes Tom, I have simulated a 22 FET linear with source resistors and RF feedback. It will do 5500 watts out on 75 with 130 volts on the drains. About 2400 with 80 volts. I have a 7000 square inch heat sink area all machined waiting for me to assemble it. IMD looks pretty good but need to run simulation in FFT mode that Jay just pointed out.

This is using 11N90's ?  Should be quite the amp if it works well. Better than fils and blowers during hot wx.

T

The transistors are matched according to the assembly instructions and the former packaging they came in.  I don't seem to have power hogging on one transistor.  They seem equal with an infrared thermometer and to feel.  I have the same app note too.  Heat not a  problem to deal with.  The MRFs have a nice hefty heat sink.  On the subject of heat sinks, if anyone is looking for a cheap hefty heatsink go to Allelectronics.com.  They have some 10" x 7" x 2" heavy duty heatsinks for $18.00.  They're used but beefy.  PN: HS-620

I guess what I'm having a hard time understanding is why the amp works fine with 2 other rigs but not the Retro75.  Unless I live in bizarro world I would think the Retro75 is the suspect rather than the amp. 

I'm going to try Stu's suggestion with the resistor and/or pad.  May even take a stab at messing with the network on the Retro75 by putting in trimmer caps.

Bob

The transistors in the amplifier represent a very non-linear load for the Retro 75. I.e. the impedance from base to emitter of each transistor varies a great deal as its collector current varies. The emitters are connected directly to ground (no emitter resistors to provide a more stable resistance looking into the base of each transistor).

The Retro 75 output stage appears to be interacting with this non-linear load (which also has reactance) to produce the instability you are seeing.

The other transmitters you tried are apparently less sensitive to looking into this kind of a load. For example, the high Q tank circuit of the FT-102 is apparently acting to reduce the tendancy toward instability associated with the transistor amplifier. The Retro 75 output circuit is a low pass filter (to remove harmonics) and not a high Q bandpass filter.

The series resistor and/or an attenuator will provide some isolation between the Retro 75 and the amplifier... and will hopefully tame the instability.

Stu

Great analysis Stu...

So is it also possible we're seeing regeneration through the 12v line since the Retro-75 and the EB-63 both run off 12V? If powered from the same supply, and if 12V decoupling is not optimal (daisy chained power versus star power), I see opportunities for instability via the power rail(s). Powering the Retro-75 from a 12V gel cell as a test might verify/eliminate this possibility.

Also, a P-P AM linear does not always have to be biased class AB. Remember the Vacuum Tube class BC P-P linears where each tube is biased at cutoff? This only works with AM (not SSB). The non-linear transfer around cutoff results in the RF output of the AMP to have a higher modulation percentage than the driver, maybe 70% driver, 90% AMP output.

I have run those CB type P-P solid sate amps on 10m AM, and turning off the AB bias seems to make little difference on AM whereas on SSB the results are horrible. No bias requires a bit more drive though for the same RF output. It seems that the thresholding cross-over issue is overcome with the presence of the carrier where it don't really matter.

Jim
WD5JKO

Jim

I think that an interaction associated with a shared power supply is also a possible source of this instability. However, I would expect the instability, in that case, to take the form of a spurious audio frequency modulation of the carrier (either a periodic modulation or a chaotic/noiselike modulation at audio frequencies). In any event, if I were directly trouble-shooting this problem, I would try powering the Retro 75 from a separate power supply... as you suggest. I have a small storage battery in my shack (the size used with a garden tractor)... and that's what I would use for testing.

I agree that, in AM operation, running a linear amplifier close to Class B might work as well as Class AB (maybe even better)... depending upon the specific characteristics of the tube(s), and the peak modulation index being used. As has been discussed earlier in this thread... 3rd order IMD is a key factor in determining how to operate a linear amplifier. I.e. how faithfully does the amplitude of the output rf waveform (at the fundamental frequency) track the amplitude of the input r.f. waveform... over the range of amplitudes that are being employed. As you have pointed out.. the nice thing about AM, in this regard, is that (if you control the negative peaks) you can keep away from the region where the output rf amplitude is a very non-linear function of the input rf amplitude.  

Best regards
Stu

Took some hints and tips from everyone.  

There is indeed an interaction between the Retro 75 and the amp using a single supply (battery).  Before going to dual power supplies I put 100ohms in series with the Retro and amp.  The Retro stablized and had a pretty descent sinewave at the input to the amp however, the sinewave appeared 'rough' as if something was very lightly superimposed on the trace.  Output looked better but not great.  And I think this is where an oscillation comes into play.

Went to dual power supplies and Retro works without the 100ohm resistor however, output sinewave is worse than above.  With the 100ohm resistor only a slight improvement.  Again, probably oscillation issues.

On a long shot I started probing the stages in the amp with a scope.  The amp is oscillating. It's very apparent on the secondary of T2 and more so on the collectors of each transistor.  It occurs both in standby and transmit.

Bob

Does the amp oscillate if you disconnect the output of the Retro75 from the input to the amp... and leave the input to the amp open?

Stu

Patient: I can't sleep
Doctor: Try drinking a glass of warm milk before going to bed
Patient: But doctor, the last time I had this problem you told me not to eat or drink anything at bed time.
Doctor: Well, medical science has come a long way since then.

Hi Stu,
Yes it does oscillate with the Retro disconnected. If I key the relay by shorting the CE junction on the relay driver the amp oscillates as well and does so without any RF at the input. Tried it with the input shorted as well. No change.

Bob

What type of load is the amp looking into?

Stu

Stu,  a 50ohm non-inductive load.  I 've 2 DL's.  One is a cantenna the other is a 2kw dry DL.  Both do the same thing.  Just so you know when the amp is in standby, the relay opens the input and output to the amp bypassing it so the input and output are floating.  However, even in Xmit with the input shorted or connected to the Retro75 (power off) the amp still oscillates.  The frequency of the oscillation does change somewhat.  Oscillation is in the 1.5Mhz range.  Also, I increase the bias temporarily to see what would happen.  Going from .645V to .675Vdc 'cleaned' up the oscillation. 

Bob

Try adding some additional capacitance directly from base to ground for each of the transistors in the amplifier. Use something like .005 uf (5000 pF) from base to ground on each transistor. Keep the leads short.

Stu

I'll have to try that next week when I get a new pair of transistors. Opened a base emitter junction on one.

Looking at the EB63 schematic, the bias is there all the time. This means the AMP has an open input and output and the zitters are idling away trying to amplify something. It can be debated whether an AM P-P linear needs to be biased at all, but the need for bias during idle periods is unnecessary, and likely causing the issue.

Once you rebuild the amp, maybe add a fuse on the +12v power buss.  ???

Also consider jumpering that bias diode to go to pure class B with NO idle current.

Jim
WD5JKO

Jim,
I was thinking of just turning the amp off all together when no RF is applied.  And rather than using a COR, switch it directly. The amp does have a fuse on the B+. Definitely a must.

Bob
et al.

Take a look at the schematic of the biasing circuit of the amplifier:

http://www.communication-concepts.com/appnotes/eb63300sharp.pdf

A few observations (comments?)

A. The bias supply is a forward biased diode (D2), bypassed by a 500uF 3V electrolyic capacitor (C22). This feeds bias current, at 0.7 Volts, through the center tap of the input transformer.

It seems to me that an additional, low impedance (at RF) capacitor... in parallel with the 500uF electrolytic... is appropriate. E.g. .01 uF... which has an impedance of 4 Ohms on 75 meters... or maybe a .01uF capacitor in parallel with a 0.1 uF capacitor).

B. According to the document (referenced above)... the transistors are biased at around 500mA each. Assuming the DC current gain is 100 (i.e. the MRF454 specification sheet says that the DC current gain is a minimum of 40 and a maximum of 150) ...  each transistor has about 5mA of base bias current. In addition, each of the 10 Ohm physical resistors from base-to-ground of their respective transistors will draw around 0.7 Volts / 10 Ohms =70 mA. The total of: 2 x (5mA + 70mA) = 150mA of bias current has to be delivered to the center tap of the input transformer by the 33ohm resistor (R4). This is not problem, because the 33 ohm resistor is delivering a total current of (13.6 Volts - 0.7 Volts)/33 Ohms = 390 mA. So far, so good. [150mA goes to the center tap of the input transformer, and the remaining 240mA flows through diode D2].

When the transistors are delivering (for example) a total of 140 watts of RF output power... corresponding to a total of (approximately)  140/0.66 Watts = 212 Watts of DC input power...  then the total collector current is around 212 Watts/13.6 Volts = 16 Amps. This corresponds to a total average base current (assuming a current gain of 100) of 160 mA. Adding this to the 140 mA of current flowing through the pair of 10 ohm resistors... we get a total of 300 mA of average current that must be delivered to the center tap of the input transformer. This is still less than the 390 mA of current that the 33 Ohm resistor (R4) is delivering.

But, what if the current gain of each of the matched pair of transistors is closer to its minimum data sheet value of 40? Let's say the current gain is 75. Then the total base current corresponding to a total of 16 Amps of collector current will be: 213 mA. Adding this to the 140mA of total current through the pair of 10 Ohm base-to-ground resistors, we get: 353mA... which is getting close to the maximum current that the 33 Ohm resistor (R4) can deliver... while still maintaining a 0.7 volts bias.

So... I'm wondering if the oscillation might me caused by an instability in the base bias voltage.

Stu  

Stu,
I agree on the .01uF.  I thought about that yesterday when I was messing around with the bias and happened to stumble upon the oscillation and looking at the schematic.  The lack of one raise a flag.  I was about to solder one into place when I discovered the blown transistor.

Also, I wouldn't go by the schematic posted on their website or the engineering notes.  The bias circuit has morphed a little bit over the years.  See the current circuit diagram.  I should have posted this way in the beginning.  

D2 is a transistor being used as a diode.  It is heat sinked or sandwiched between the PC board and heat sink.  R4 is now 82 ohms.

BoB

I tried magnifying the schematic... but there is one key value I cannot read.

What is the value of R4 in your amplifier?

Stu

Stu,
R4 is 82 ohms.  Also see my previous post. I've included some of the changed components.

Bob

et al.

Based on the analysis I did in my previous post... I can't see how R4= 82 Ohms could possibly deliver the required average base current when the amplifier is putting out significant RF power. This, by itself, would likely produce a feedback mechanism between the output and the input of the transistors... leading to instability.

Stu

I kind of think the same Stu based on some empirical poking yesterday.  I dropped the R4 value to about 70ohms to turn on the  transistors a little more.  With R4 being 82 ohms they were floating around .64Vdc.  Bumping them up closer to .68 seemed to turn them on better but also changed the oscillation rather than eliminate it..

Bob

I think that this version (with R4=82 Ohms) is not appropriate for AM.

The 470uF capacitor across the bias source (the diode) can act as a charge resevoir for brief periods of time during modulation peaks in SSB mode... allowing the base of each transistor to draw more current than R4 can deliver, for those brief periods of time... and the charge that is removed from the capacitor will be restored (refilled) via the 82 Ohm resistor (R4) during periods of relatively low RF output.

However, the 82 Ohm resistor cannot deliver enough continuous base current to keep the amplfier at carrier in AM mode.

I think you need to revert back to 33 Ohms... and after making this change, you need to check that the resting collector current (no RF input) is not too high.

Stu

Bob,
Try .01 in series with 100 ohms (2W) between base and drain of each transistor.
A little NFB should tame it. I've found diode thermal bias compensation not the greatest. It will still run away if the devices get hot so best to starve then a bit and let heat bring up the current. You can increase the collector to ground caps if you don't plan to use it on 10 meters. Make sure the Bias diode has a good thermal bond to the devices
I played with the MRF429 enough to prefer the mRF150 FET. The PCB layout and NFB are very critical using bipolars if you want a BB flat input Z.

I just looked at my modified 10 meter amp after conversion from 11m done sometime the last solar cycle. I made some changes I don't remember, but that is another thread, under CRS.  I did the NFB just as Francis had suggested (and Shane) to lower the gain, improve stability, and lower IMD. I also put a 510 ohm 2W resistor across the RF amp output so during standby periods the amp at least had some load instead of an open circuit. The EB63 also leaves the AMP unloaded when idle, and biased up too!!  :-[

I see considerable rework with the bias supply done (CRS though), and I ended up with the bias source shorted out (no bias). The amp as I recall worked just fine on AM (3w in 20w out AM).

After reading Stu's analysis of how the EB63 stock bias is unsuitable for AM, I understand why I eliminated the bias on my amp. For AM, NO bias is preferable than a high-Z bias source. With the EB63, changing that bias resistor from 82 to 33 ohms is likely to raise the idle current too high, increasing the 12v draw, heat, and lowering efficiency.

Since we seem to want to preserve the bias with this amp, we need a low - Z bias source. An idea comes to mind and I will outline the concept here. Use a LM-317 adjustable regulator which provides 1.25v when the adjustment pin is grounded. Then in series with the 1.25v low-Z source put a series resistor, maybe 10 ohms to the bias diode D2. Now the threshold is gone, and the Rs of the bias supply is 10 ohms instead of 33, or 82 ohms.

We can vary the idle current as before by playing with the series resistor value, or maybe make the LM317 adjustable over the range of 1.25v to 2.0v.

I was just typing as I was thinking, so hopefully this lower Z bias source concept will work.

BTW the term CRS means, "can't remember sh_t"  ::)

Jim
WD5JKO

Jim

I'm beginning to think that your suggestion to just ground the center tap of the secondary of the input transformer (no fixed bias) is the simplest solution for AM operation, if the modulation index (for negative peaks) is not too much greater than 50%.

The more I think about, the harder is seems to be to bias the transistors, properly, to operate in Class AB or Class B with an AM input signal.  

With (for example) a 2 Watt input (at carrier), you have 14 Volts peak (assuming the input impedance is close to 50 Ohms). The 4:1 (turns ratio) input transformer will reduce this to 3.5 Volts peak (i.e. 1.75 Volts peak from each side of the secondary to the center tap/RF ground). This is probably enough signal to drive the transistors... while still providing reasonable linearity (RF amplitude out v. RF amplitude in) on up to 50% negative modulation peaks. I.e. (1- 0.5) x 1.75 Volts = .875 Volts

What you really want to do (but which is hard to do) is:

 A) to inject a fixed amout of current into the base of each transistor... adjusted to produce the desired resting collector current in each transistor. I.e. you really don't want a fixed voltage... and trying to use a fixed voltage will lead to problems such as the thermal runaway problem that Frank mentioned.

and

B) to provide a low impedance DC path from center tap to ground... in order to provide the additional average base current, as needed, when there is R.F. applied to the input.

As a compromise, one might redesign the biasing circuit to employ a 33 Ohm (or lower value) resistor... but to use a biasing voltage that is closer to 0.5 Volts (A Schottky diode that can handle the current?). That way, the transistors will be operating closer to Class B (but still a little bit into Class BC).

Stu

Changing the bias will also effect the input Z of the amplifier. I would put a shunt resistor load in parallel with the thermal diode so the voltage can't raise high enough to hurt the devices. I would set the bais so there is just a bit of collector current close to class B. This way as the parts heat up the diode will draw additional current off the bias. When diode cools down the resistor sets the maximum voltage. I found this worked best when playing with MRF429s

Frank,
I have no intention of running this amp higher than 75 meters. So if narrow banding it is necessary, I won't lose any sleep over it.

I blew a transistor by messing around with what I thought was NFB.  Not sure that I was doing it right.  If anything I only improved the oscillation to the point of destruction.

As soon as I get replacement transistors I'll tinker using some of the above suggestions.

A current limied power supply is your friend

OK, got new transistors today.  Installed them and started troubleshooting again. 

Put a 0.01uf from R4 (82ohms) to ground.  It did nothing to suppress the oscillation. 

Tried 2  0.005uf from base to emitter on each transistor.  They reduced the amplitude of the oscillation by 10%.

In standby mode when oscillation isn't under load, putting a .005uf across the primary of the input transformer kills the oscillation.  The same with a jumper. 

I quit for now.  Will come back to it tomorrow for a brief stint. 

Need to try to get some rest. Been running full tilt the last 3 days with minimal sleep and don't want to screw up.


 

Yes best to go slow, and deliberate when you are not tired.

So in my opinion, stabilizing the AMP when the relays disconnect the inputs and outputs should be attacked first. Here is what I'd do:

1.) Remove the class AB bias (short out D2) during receive. Could add a relay, or figure a way to use a 2N7000 fet. No since having the transistors sucking current while you are receiving. In fact they might create noise causing poor reception problem.

This should quench it, but if not, then:

2.) Add a 510 ohm resistor (2W) across the RF output (of the AMP) to always have some load on the AMP.

If this is still less than 100%, then:

3.) Add a 510 ohm resistor across the input (of the Amp) to always have an input load on the AMP. Better yet, install a 3 db pad at the input since the amp has too much gain anyway.


Then when you finally get to running the AMP from the retro, to get the gain & linearity right you will likely need NFB to achieve both variables. Shane suggested adding NFB early in this thread. I just looked at my old 10m amplifier, and I did the same. see attached picture. Notice the Amp already had 47 ohm base-emitter resistors. The R-C I added from C-B on each transistor used a 2200pf - 47 ohm 2W. For your AMP on 80m. the cap should be around .01uf, and I might start with ~ 220 ohms, and work my way down till the gain and linearity are right to match the Retro-75.

The AMP pictured would do 20W carrier AM, and ~ 80W PEP. The bias was OFF all the time since it caused more problems than it benefited. I don't recall all the details, but I did a lot of modification in that area, and ended up jumpering out the bias. That meant NO SSB however.

Jim
WD5JKO

Rob

Putting 0.005uF across the primary of the transformer is equivalent to putting .005uF x 16 from base-to-base (but not base-to-ground). 0.080 uF has an impedance of around 2 ohms at 1 MHz. This is forcing the inputs of the two transistors to be closer to being in phase... and since their outputs are connected out-of-phase, it is killing the oscillation.

The question remains: what is the cause of the feedback.

I think you should, next, try reducing R4, the 82 Ohm resistor, to something like 33 Ohms... and also add a 0.1 uF (or larger) r.f. bypass between R4 and ground (in parallel with the existing 470uF charge reservoir capacitor).

Stu

Bob

One other thing...

It is possible that feedback from the output is getting back to the input via the primary of the input winding (and the various things that are connected to the primary of the input winding... like the automatic antenna switch).

Try flipping/transposing the primary leads.

Separately... even if you stop the oscillaton by eliminating or cancelling the feedback... you still have to change R4 to 33 Ohms if you want this amplifier to work on AM (rather than just SSB).

Stu

any idea what the B-E current of the MRF454 is. The data sheet doesn't provide that.  I'm leary of changing the 82 ohm resistor to 33ohms .

Do not increase bias current unless you want to buy more transistors.

Bob

You can check the current gain of the transistor. The Base-to-Emitter current is (1/the current gain) x the average collector current (at carrier). According to the MRF 454 specification sheet, the DC current gain (H_fe) is between 40 (minimum) and 150 (maximum).

If you use a value of 100, for example, the average base-to-emitter current has to be the average collector current /100.

Let's assume that the average collector current for each transistor is (for example) 3 amps at carrier. I.e. 3 amps x 12 volts =  36 watts per transistor of electrical input power, at carrier... corresponding to around 12 watts of r.f. output power, at carrier, in Class AB linear operation (per transistor).

That means that the average base current has to be 3A/100 = 30mA per transistor => 60mA total average base current.

Add, to that, the average current drawn by the two 10 ohm resistors (one from each base to ground) = 2 x 0.7 volts /10 Ohms = 140 mA

So the total current is 200 mA (not including the current through the biasing diode)

200mA x 82 Ohms = 16.5 Volts.. i.e., you can't deliver the average base current to the transistors and the two 10 ohm resistors with an 82 ohm resistor fed by a 13.6 volt supply.

You can do the job in sideband mode, because you don't have to deliver the constant average base current associated with the AM carrier.

This is why they used a 33 Ohm resistor in the original design. The revised design uses an 82 Ohm resistor... but it is specified only for use on SSB (not AM).

Try a gradual reduction of that resistor's value... e.g. put 150 ohms in parallel with it to get a combined resistance of 53 Ohms


Stu

I 1/2'ed the 82ohms to 41ohms. I didn't have a 33 on hand.  Brought the DC up slowly and got to about 11.0V and heard a phfft sound and immediately killed power.  Not sure what it was  but feared the worst.  So I pulled the transistors and checked them including the bias transistor.  All seem OK for the most part but I do question the MRFs.  Even though the BE and BC junctions check good CE readings are all over the place and  in most cases around 50K but waver. Sometimes they're as low as 37K and as high as infinite. I don't get a consistent reading on either transistor and neither behaves the same.  Different meters give different results.  Tried my old VTVM and it shows both CEs being shorted.  I don't know what to think.  If I turn off auto range on the DVMs the CEs show hundreds of megs so I'm inclined to think the transistors are still OK and the phffft sound was the relay possibly beginning to chatter because of oscillation and the COR. Though the COR has never reacted to the oscillation.  Clues?

revisit.

May have a clue or 2.

tried it again but took it in steps. 

At 13.6V in there's about 155mA through the 82 ohm resistor.

cutting in half the value of the resistor to 41ohm with 155mA going through that value there should be about 6.3V across the 41ohms.  So advancing the PS so that the 41ohms is approaching a 6.0V drop the relay attempts to close.  Oscillation is very strong at this point and is activating the COR.

At 82 ohms the oscillation does not occur until the transistors are on.  And that is what is occuring with 41 ohms.

The next step is to try flipping the transformer and see what happens.

Bob,

  Since the AMP input and output is unloaded with the COR relay not energized, I think having those transistors all biased up in class AB mode like this is asking for trouble. Sure if you really want to do this, then only apply the bias when the COR is energized. This way both the input and output sees a load, hopefully in the order of 50 ohms.

  Maybe my way of thinking is flawed, but with the I/O unloaded (COR not on), isn't this similar to putting your car in neutral, and then putting a brick on the gas pedal?

BTW I just received my Retro75 + case kit. OH Boy!!

Jim
WD5JKO

Hi Jim,
I tried no bias on the transistors.   I ended up with a squarewave generator when RF was applied. No bias certainly squelches the oscillation but when driven there's clear indication of the oscillation on the output.  I need to build an attenuator.  It's possible I'm driving the piss out of the amp with the Retro.  The oscillation does show up when RF is applied and COR is activated by RF but does not appear when the COR is operated manually with no RF applied.  A 50 ohm load on the output does kill the oscillation but is evident when RF is applied.  Not sure what is occurring at this point.  There's never enough time to tinker.

How about NO bias only when the COR is not energized? That would mean interrupting the bias resistor 82 ohm, or 55, or whatever with an additional contact on the COR circuit, or some solid state means to accomplish the same thing.

Letting a high gain RF amp IDLE in AB mode when the I/O SWR is infinity (open circuit) is not something usually recommended. The other thought is to put two 510 ohm resistors, one across the input and and the other across the output (size for the power levels). Now the amp sees a 10:1 SWR instead of infinity and beyond..  :-\

Just a thought...

Jim
WD5JKO

yep, tried the no bias no RF drive and manually activated the relay.  No osciallation.

Also, no bias and no RF drive (stand by), no oscillation as expected since tranny's are cutoff.

It's when there's drive and bias that it oscillates. And of course it oscillates when there's bias and no RF drive as well.  I'll have to try the 510ohm trick.  Stand by.

So have you started your Retro yet?

Nope, just came in UPS. Getting the enclosure also gets you all the pots, connectors, etc. It was professionally silk-screened too.

Getting ready for a two week business trip to the Beverly, Mass. area. I could take the kit and do it at the factory I work at over the weekend, or instead go to Gloucester and take a "whale watch" three hour tour. I did that before, on a choppy day. After most of the passengers lost their lunch, and turned every shade of green, I ordered a greasy hot dog with all the fixings. Washed it down with a beer. A few folks after seeing me eat the dog had to go back out and repeat... ;D

Jim
WD5JKO

Sounds like the latest Hollywood diet.... FFFffwwarff  :-X

Hi

From the ARRL hand book
 LOW-Frequency Parasitic Oscillation ...bi-polar transistor exhibit a rising gain characteristic as  the operating frequency is lowered.
To preclude low frequency instabilities because of the high gain,
Shunt and degenerative feed-back are often used.Feed-Back (from collector to base) increases by nature of the feedback net work reducing the amplifier gain.( depends on the Frequency}
the feed back net work consist of R in series with C from collector to base of the transistor.
The value of C and R are usually chosen experimentally.C is usually between 220pf and 0.0015uf for HF band amplifiers,the net work has small effect at the desired frequency but has a pronounced effect on the lower frequency.
the R is between 52 to 5600 Ohm

Using R between emitter and ground develops a degeneration at low frequency ,and bypass it with a C for adequate RF bypassing for the intended operating frequency.The impedance of C rises progressively as the frequency is lowered   .
this R is seldom greater then 10 ohm and maybe as low as 1 ohm.

Using swamping resistor in the input amplifier usually from 3 to 27 ohm ,placing it as closed at the terminal base.


If You have the ARRL book 1997 edition, ,You can find this article in page13-27 and 13-28 more compelletly


Gito

got some limited success with negative feedback.  Got to jockey some values.  Still have a crappy looking output waveform but will mess with an attenuator for the Retro next.

Is it normal for bias to go to zero when RF is applied?  I don't remember it doing that and then again I can't remember if I ever watched the bias with RF applied.

Bob

Great news!

Some types of DC meters will not read correctly in the presence of RF. The base bias voltage should not go to zero... but remember: once you draw average collector current, you will also draw average base current... and, depending upon R4... the base voltage will drop somewhat.

Did you try reversing to input transformer leads to see if there is positive feedback coming in via the primary of the transformer? [Reversing the leads will convert that feedback, if present, from positive to negative feedback].

Stu

Stu,
What's your take on the meter readings I was seeing when checking the transistors CE junctions?   See reply #66: http://amfone.net/Amforum/index.php?topic=24494.msg183410#msg183410

It seems the input of the amp is not 50ohms.  There is a mismatch with at least a 3:1 SWR.  A tuner does not improve the output signal and I'm beginning to wonder if the transistors are toast.

Tried a 4dB attenuator.  that didn't do anything except lower the output.  Waveform is still poor.

Still poking.

Bob

With respect to your measurements on the transistors:

I think that you should assume, for now, based on the behavior you are observing (rather than the multimeter measurements) that one or both transistors are still working. If there was a collector-to-emitter short, there would be smoke rising due to the excessive collector-to-emitter current that would flow. If there was an emitter-to-base short, on either transitor... the bias voltage would be zero on both transistors. If one of the bases were open... you would see little or no effect from disconnecting the collector of that transistor (see below).

I suggest that you try the following:

Oscillation occurs because there is positive feedback somewhere. Adding a series RC between collector and base will introduce negative feedback (to cancel some or all of the positive feedback); and/or will introduce enough phase shift in the existing feedback to stop the oscillation. The value of C needs to be large enough to both: block the DC (which any capacitor will do), and to present an impedance that is less than the value of R. It is the R that provides the negative feedback. Since the RF output impedance of the amplfier is a few Ohms, and since the current gain is around 100: a relatively small value of R will be required... probably around a few hundred Ohms... maybe less. I.e. you want the fraction of the output current that flows into the feedback resistor (the rest flows into the RF load) to be comparable to the current gain of the transistor. However... before introducing negative feedback (which isn't in the existing design), it would be very helpful to know where the existing positive feedback is coming from.

a. Is the existing positive feedback coming from the output signal feeding back into the input transformer's input winding? To check, either reverse the input winding polarity, or disconnect the side of the input winding that is not grounded. [Shorting the input winding is inconclusive, because doing so also produces an RF short across the base of each transistor to ground]

If reversing the input winding (or disconnecting the side of the winding that is not grounded) has no significant effect on the oscillation... then that would indicate that the existing feedback, causing the oscillation, is not coming through the input winding of the input transformer.

b. You might want to see if the oscillation goes away (or changes significantly) when there is only one transistor being driven. Turning off one transistor might reduce the gain in the feedback path by 6dB... depending on what the feedback path is. This would also allow you to check whether one of the two transistors is "toast". You can turn off one transistor by removing the connection to it's collector.

Stu

Flipped winding of the primary on the input transformer.  Note: Secondary is not flippable.  Had a significant reduction in oscillation amplitude.   Reinstalled the RC networks for the Base Collector. (.01uf and 100ohm) and have a flat line again.  Also tried running with 0.0V bias by shunting the anode of the bias diode. Still can't seem to improve the output linearity.

Freq: 3880KHz
Upper waveform is output
Lower waveform is input

Scope settings:
x10 probes
Channel 1 (output) 5v/div
Channel 2 (input) 1v/div

The image doesn't reveal this but the input waveform isn't clean when the transmitter is keyed.

Bob

Ok... making progress!

In the picture you posted, where is the output probe? (I.e. is it across the dummy load?)

Also... please refresh my memory: what is the turns ratio of the output transformer (total turns on output v. total turns on input).

I.e. if the peak-to-peak output voltage across the dummy load is 100 Volts... and the turns ratio is X...  then the peak-to-peak voltage from center tap to either collector is 100/2X = 50/X (assuming both transistors are working and operating as a matched, push-pull pair). But the peak voltage from center tap to either collector cannot exceed the supply voltage (~ 13 volts)... so the square wave may be a result of "running out of headroom"

If (taking into account the 10x probe) you have a signal that is a 50 volt (peak) square wave across 50 Ohms... then you are running the amp at 50 watts of r.f. output (continuous). This amplifier may not be able to sustain that level of continuous output power without overheating the output transistors.

Try reducing the input level (use an attenuator... e.g. 100 Ohms from the center pin of the Retro75 to ground and 50 Ohms in series between the center pin of the Retro 75 and the center pin of the amp ) to see the impact on the output of the amplifier (shape and size)

It looks to me like the amplifier is saturating because it is running out of headroom (the collector-to-base voltage is approaching zero).

Stu

Amp looks saturated due to overdrive

Stu,
Agree on input power being to high and causing saturation. Got a 4dB attenuator in place plus tried the 50 and 100 ohm in addition to the 4dB attenuator then took a different approach to reducing the input to the amp.  Instead of using the Retro75 as an RF source I used another source, an RF signal generator.  2.0Vp-p is just enough to key the COR.  The sad part is the output waveform looks exactly the same on all accounts as above just reduced in size.  I'm stumped.

Scope probe at output jack of amp.
Transformers are 16:1 according to the instructions which I'm assuming is impedance ratio rather than turns ratio.  There are actually 4 turns in each transformer assembly.

Any concern about the input impedance not being ideal?

Bob

Assuming the output transformer has a ratio of total input turns to total output turns of 1:4; and if the peak-to-peak voltage across the dummy load at the output of the transformer is 100 volts...

Then the peak-to-peak voltage from collector to ground of either transistor should be 50 volts/4 = 12.5 volts. Therefore the peak voltage is 6.25 volts... and there should be no problem with saturation (assuming both transistors are working in a balanced way).

However... the picture you posted shows that there is a problem with saturation!

I am wondering if both transistors are, in fact, working.

Can you verify that both sides of the primary (the side facing the transistors) of the output transformer are properly connected to their respective collectors (i.e. the circuit board traces are ok, the solder joints are ok).

Likewise, can you verify that, in the absence of input rf, both transistors have approximately 0.6 volts between their base and their emitter (directly touching your probes to the transistor base and emitter).

Alternative: turn the power off, and check the continuity all the way from each transformer's center tap to the respective tabs on the transistors. Verify the each emitter has zero resistance to ground.

 Stu

Bob

Also:

Turn the COR on... (i.e. don't use the input rf to do it)

Now put a signal into the amp that is smaller than 2V peak-to-peak (e.g. 1 volt peak to peak)... and see if the saturation is significantly improved

Stu

Stu,
I'm not convinced the transistors are bad, at least not yet. Afterall, the amp does amplify just not cleanly.

Continuity looks good on both xformers.  Looks good going through the xformers to each transistor leg.

0.650 on each BE junction.

12.5V on the collectors.

Also tried  operating the COR manually with much lower input signal. In fact dropped the input level to the point of drawing a 1/2 amp from the PS. No change in wave form.  Looks like I have a pretty powerful square wave generator.   ;D

Bob

When you dropped the input level.. how did that affect the output level (putting aside the shape of the output waveform)?

Stu

it reduced the amplitude as one would expect.  No change in shape of waveform.  I'm wondering about the oscilloscope.

Bob

Swap the probes... so that the probe that is now attached to the input, and feeding scope channel 2... is connected to the output (and still feeding scope channel 2).

Separately: where are you connecting the input probe?

Stu

probes swapped. no change.  Input is at the input to the amp.  If the input probe is move to the Retro prior to the attenuator clean signal.  after the attenuator sinewave gets hacked up when amp is keyed.

Bob

Okay... looking again at the waveforms you posted above... both the input and the output waveforms are mirror image symmetrical around V= 0 volts... which strongly suggests that the amplifier is balanced (no even harmonics)... and therefore both transistors are working in a balanced, push-pull configuration.

Could you put the output (channel 2) probe on the base of one of the transistors (set to DC coupled mode)... while leaving input probe (channel 1) where it is now... and post the waveforms. Please do this with around 1 volt peak on the input side of the input transformer(i.e. where channel 1 is connected).

Also, please let me know what the DC level is on channel 2 (the base of the transistor) when the rf input is 0 (shorted)

The above should be done with the COR manually activated, in all cases.

Stu

Bob

P.S.

It's time to revisit the large electrolytic capacitor across the bias supply. Please reinstall a 0.1uF or larger capacitor in parallel with it.... as you did last week.

Stu

Bob

The reason I am concerned about the electrolytic capacitor is that the amplifier may be doing the following:

For some reason, there is no low impedance path from the center tap of the input transformer to ground (for RF).

Thus, the RF waveform being applied to the input winding of the input transformer cannot push current into one base unless it draws an equal and opposite current from the other base.


The current that can be drawn out of the base of either transistor is limited to the bias current.

Thus, as the input RF signal tries to push current into the top transistor's base, that current gets clipped at a level equal to the bias current in the bottom transistor.... and vice-versa.

This would result in the clipping (squaring) behavior you are seeing at the output of the amplifier.

In addition to adding the capacitor in parallel with the large electrolytic capacitor... make sure that the path from the center tap of the input transformer to the top of the electrolytic capacitor has a low DC resistance.  

Stu

Hi

In my opinion ,the square wave output ,it's not a problem ,maybe it must be like that,when I look at the schematic diagram it's a class D amplifier/broadband Amplifier.
Even if it's square ,it represent the amplitude of the carrier.
When we modulate ,the carrier amplitude goes up and down,now if we look at the square wave output,does the amplitude of  it follow as the input sine wave  up and downs,
like when You get a lower drive than You get a lower output,drive with a higher drive You got a higher output ,with the same square wave.

Remember,a square wave output is the most efficient  transmitter design,
It's the amplitude of it ,that's important does it follow the amplitude of the input amplitude after modulation as a whole.

In my opinion ,It's not "important" what the Carrier wave looks like,But the amplitude of the input carrier and the output carrier,does it follow liniearlly,when modulated.
Of course I may be wrong.

Gito.n

Hi Stu,
I never removed that capacitor.  It's still in place but it's not a 0.1uf but rather a 0.01uf.  You think I should up it by 10x?

I was looking at the app notes on CCI website.  They show this amp running an AM sig at about 50watts out with a pretty nice envelope with 1.5watts in.  What I also noticed in their app notes was the PCB layout.  The layout is much different than the current layout. I wonder if this has something to do with the waveform I'm seeing and even though there is a negative feedback network between the collector and base that this thing is still breaking out into oscillation during certain parts of the waveform cycle.  Maybe we haven't hit the sweet spot.  But then again maybe Gito has a point too.  Don't know.  Never worked on a broad band amp before.

I'm stumped.  This thing certainly looks like its running in saturation.  

Bob

You need an RF path to ground on the input transformer center tap whose impedance is 1 ohm of less at 3.885 MHz (corresponding the the r.f. input impedance from each transistor's base to ground)

.01uF at 3.885 MHz is 4 ohms.

Try 0.1 uF

Stu

P.S. You're getting there.

you might try a 1uf ceramic at the input ct. seems like the gain is very high.

Frank

With all due respect... I don't think its a gain problem. I think its a base current starvation problem.

One transistor's r.f. base current is being pulled from the other transistor (instead of from ground, via the center tap on the secondary of the input transformer) ... but that can only continue up to the level of the base bias. Beyond that, the transistor that is now starved for base current can only pull what it needs from the 10 ohm resistor from base-to-ground (10x the impedance of the base itself). This causes two effects.

1. The base current ... and thus the collector current abruptly levels off

2. The sudden change causes a glitch that stimulates oscillatory behavior

The 270uF electrolytic is, for some reason, not acting like a low enough impedance to ground. A 0.1uF is probably sufficient... but a 1.0 uF would be even better (if it behaves like 1uF at RF frequencies...i.e. 0.04 ohms to ground or so at 3.885 MHz)

Stu

Hi

Speaking of the bias suplly ,it used a MJE 243 conectet  as a replacement of 1N4497 diode ,and used as a "zener" diode to give a fix  + bias suplly ,to Make the Transmitter a class B or AB2 system.
It must have a stif bias supply.
So i'm curios  of the droping resistor is it R1 with a 100kohm value,if it so with a little current flowing in the base of the transmitter,the bias wiil go to Zero (droping resistor X current).

Maybe the right value is 100 ohm ,so there's about 100 ma flowing in the circuit of the bias supply that make it a Stif supply.

If i'm not mistaken 1 N4997 diode has a 50 watt power rating ,so it can disipated that current.

 Is the value of R1 and R4  wrong (interchanged)meaning the value of R1 is 33 ohm and R4 100 kohm

Gito.n

Hi

After looking at the schematic diagram R4 a 33 ohm 5 watt resistor .it is a shunt Resistor(fixed load) for the input transmitter ,to make the Driver transmitter a constand load.

So I  convinced that a 100kohm, value for R1(droping resistor) is wrong,since I think it is a droping resistor to make a stif bias for the base off the MRF 454 .
a Diode(1n4997) has a voltage of maybe 0.6 t0 ...0.8 volt (if I'm not wrong)between it's cathoda and anoda,so we can used it as Zener diode with high current
capability.
So R1 is the droping resistor to Diode (1n4997) conecting from the B+ ....R1 ---to diode ..to ground,and there we got a fixed + bias.

Gito

hi

I found a schematic diagram after googleling /searching in the Internet,it looks similliar
circuit .look at the min bias circuit and the clamping diode ,to make the transistor stable (to prevent run away current ) , the Diodes are in thermal contact with the transistor.

Gito.n

Gito
Bob [see the paragraph at the end of this long post]

Hi!

Keep in mind that these transistors are bipolar devices, not FETs. They are controlled by base current, not base voltage. Also, in this design, there is no emitter resistance to ground. The base current v. base-to-emitter voltage relationship is essentially exponential... so controlling the base-to-emitter voltage doesn't do a good job of controlling the base current.

As a result, it is a challenge to bias the transistors properly... and that is creating a lot of problems.

As you point out, there is a forward biased junction (actually another transistor, but shown as a diode in some versions of the schematic) that is acting in a role that is analogous to the way a Zener diode works (when it is beyond its breakdown voltage). This device is drawing current through R4 (82 Ohms in the present amplifiers... regardless of what some earlier versions of the schematic show as 33 Ohms)... and acting as a (roughly) 0.7 Volt voltage regulator. The thinking is that if this device is in thermal contract with the amplifier's RF power transistors, it will provide a bias vltage (again, around 0.7 Volts) that roughly tracks (with temperature changes) the base-to-emitter voltage of the RF power transistors.

This approach may work, to some extent... but it is still a crude method of achieving two goals:

a. Biasing the RF power transistors so that their resting (no RF applied) base current (and therefore their resting collector current) is some target value.

b. Allowing the total average base current (in the RF power transistors) to increase when RF input is applied to the amplifier. [Remember: the average collector current = H_fe x the average base current... where H_fe is approximately 100 for these RF power transistors]

Doing both a) and b) is hard. Doing just a) (using a current source) or just b) (providing a low DC impedance to ground) would be easy... but doing both is a challenge.

In this particular amplifier, the design was modified from its previous version [R4 is now 82 Ohms instead of its previous value of 33 Ohms] to fix sme shortcomings in that previous design. Unfortunately this change makes the amplifier unsuitable for AM operation.

In the new design, the large electrolytic capacitor (270uF) that is across the bias supply acts as a charge reservoir... to supply the required additional base current on modulation peaks (which is a separate issue from RF peaks). Thus this capacitor performs the role described above as b) during SSB modulation peaks (but it cannot perofrm that role on a continuous basis).

Between SSB modulation peaks, this reservoir of charge (contained in the electrolytic capacitor) refills by drawing current from the 13.8 volt supply ... via R4 (82 Ohms). R4 cannot deliver enough constant current to keep this capacitor charged up in AM operation (i.e. it cannot deliver the average base current needed to hold the amplifier at carrier output)... but it can deliver enough average current to support the positive peaks in base current associated with SSB operation.

For the above reason, one of the changes needed to operate this amplifer in AM mode is to decrease the vaue of R4 (not necessarily all the way back to the original 33 Ohm value).

Right now, Bob is dealing with a different problem. It is necessary that [in addition to the above issues that are related to AM v. SSB operation] the center tap of the input transformer (that provides the base current) have a low RF impedance to ground... i.e. less than 1 Ohm... at 3.885MHz. For some reason, the 270uF electrolytic charge resevoir capacitor is not providing a low enough impedance to ground at that point. It could be that this electrolytic capacitor does not normally have such a low impedance at RF. It could also be that the capacitor is damaged. In addition... I suspect that the biasing transistor (that acts like a forward biased diode) may be damaged. For AM operation, I would suggest that Bob replace that biasing transistor with a Schottky diode that is capable of handling 1 amp of average forward current. A Schottky diode will bias the RF transistors at a point which keeps their resting current essentially at zero... but which is close enough to the point where they would draw current. The application of RF at the input will cause the transistors to turn on... and if the applied RF is AM, the carrier level will keep the transistors (one or the other) turned on for most of each RF cycle. Thus, the amplifier will be operating in (approximately) push-pull Class BC (biased just below Class B)... but this is okay for AM operation.

Stu

Hi

Yes it's a bipolar transistor,the purpose of a fix/stif power suplly is used to set the operating Class of the transmitter,
 a class B transmitter needs a + bias suplly ,to make it a Class B it needs a standing current that is provided by biasing the MRf 454,

Yes  a  bipolar needs a current not voltage, that's why it must be a stiff power suplly /bias suplly,That does'n change it's voltage when current is drawn from it.(base current)
Say it used a Diode as a Zener with A 100 Ohm droping from the !2volt,than about 120ma is flowing in the R and trought the "Zener" , (0.8 V?)
when the grid is taking current,the voltage stay the same, because the grid is in parallel with the Zener, the current trought it is divided between the base current and the Zener.

Stu ,maybe I'm wrong but R4 is not the droping resistor it's a constant load for the the driver transmitter.

R1 is the droping resistor and it's a 100kohm resistor.
in the schematic that I atached before it used a 47 ufd  as a reservor and a 0.01 uf as a bypass capacitor in parallel,and I think the base current is in miliampere,maybe 10ma,So I think the 47ufd reservoar ,has enought ability ,to suplly it

Here I also attached the schematic diagram of EB 63.
unfortunatly it's small. picture


Gito.N

Gito

I think that you are not looking at an accurate schematic of the amplifier that Bob is using. You can see the schematic of Bob's actual amplifier by looking at his feedback reply post #47 in this thread that he submitted on June 27.

Separately... the Harris schematic that you posted above, is a good example of how to do the biasing. The adjustable voltage regulator in the Harris schematic [LM317] is delivering its current through an adustable resistance of between 10 Ohms and 110 Ohms... and the feedback from the output of this series resistor to the regulator's bottom (adj) input makes it an adjustable, regulated current source. In this case, it can deliver up to 1.25 Volts/10 ohms = 125mA of regulated current.

As we agree, the 1N4001 diodes in the Harris schematic act as a low voltage regulator to provide bias voltage (and therefore bias current) to the RF power transistors. The excess current that is flowing out of the regulated current source, but not flowing into the center tap of the input transformer, flows into these diodes.

About 50 mA flows from base to ground through each of the 12 Ohm base-to-ground resistors. So that means (in the Harris design) that, at most (if the 100 Ohm pot is set to 0 Ohms), 25 mA (total) is available for biasing the bases of the RF transistors.  25 mA of average base current will support 2.5A of average collector current (if H_FE is 100)... and that is sufficient to support around 10 watts of carrier in AM operation. [I suspect that the Harris amplifier is designed for SSB operation; and therefore, the average base current will be much lower... because there is no carrier]. I.e. 25mA of average base current => around 2.5A of average collector current. 2.5A x 13.8 volts (max) = 34 watts of DC power to the RF transistors. Assuming 33% efficiency (at carrier for a Class AB amplifier) => 11.4 Watts of RF output (at carrier).

Note that the 1N4001 diodes are serving a role that is similar to the role that I suggest could be served by a Schottky diode in Bob's design. The reason that I suggest a Schottky diode is to bias the RF transistors a little below the voltage at which base (and collector) current will begin to flow.

In the Harris design, there are two capacitors across the biasing bus (the center tap of the input transformer): 47uF and .01uF.

In Bob's amplifier: there was (before he added a capacitor) only a 470uF electrolytic (I said 270uF in some of my earlier posts... but it is shown as 470uF in the schematic that Bob posted).

A smaller capacitor in parallel with Bob's amplifier's 470uF capacitor... having a sufficiently low RF impedance (i.e. less than 1 Ohm at 3.885MHz) is needed. I am suggesting 0.1 uF (or something between 0.1uF and 1uF, that is rated for RF applications).

Stu

Hi

Stu of course a bigger reservoar capacitor and a higher RF bypass value is better,One thing that you and me are in common,is do we look at the same schematic diagram,the schematic I look at is R4 is not a dropping resistor,it's a load for the driver transmitter,
The dropping resistor is R1 that is a 100 kohm resistor,that's why I wrote in my reply it has a wrong value,it's more logical ,it's a 100 Ohm resistor

Gito.n

Gito

The schematic you are looking at is not the correct schematic (and you are also mis-reading the labels on the resistors because the schematic you are looking at has very poor resolution).

Please refer to post number 47 in this thread.

Once you and I are both looking at the schematic shown in post number 47 of this thread... I think we will be pretty much in agreement on most aspects of this.

Stu

The bias circuit as designed should work. Temperature tracking isn't great but it should work. I wonder if the output transformer has the wrong core material. I suggest you do some testing at 10 mhz. I have seen CCI amplifiers have problems below 3 mHz due to transformer reactance. Also you could try dropping the voltage to 6 or 8 volts on the collector supply. 

Frank,
that's a good idea.  the whole time I've been focusing on 75m.  I'll have to try a higher freq.  STand by.

Bob
Frank

Going to a higher frequency will also decrease the impedance of the (presently installed) 0.01uF capacitor that is in parallel with the 470uF electrolytic.

Before changing the frequency, I suggest that you increase the .01uF capcitor to 0.1uF.

That way, you can put aside the issue of whether or not the impedance from the center tap to ground on the input transformer is low enough (i.e. less than 1 Ohm).

Changing the frequency will change many things that are going on in the circuit... all at once.

Changing the frequency will also affect your measurments because (I believe) your scope is only good to 20MHz (therefore you will lose the harmonics of the waveforms that you observe in the measurment process... not in the waveforms themselves)

Changing the capacitor will only change one thing.

Stu

HI

Stu You are right,I'm looking at the wrong schematic.But after all we are talking about the min bias of the transmitter,so I'm not totally wrong .

It does not cross my mind that changing the 33 ohm to 82 ohm can improve the pateren  of the output of the transmitter,since the grid resistor is 12 ohm so in parallel it is 6 ohm (neglecting the internal resistance of the transistors) so the plus bias with 33 ohm is 6: (6 +33) x 12 volt is 1.8 volt with a "zener" in parallel  the voltage  is 0.8 volt
With a 82 Ohm  the plus bias is 6 : (6+82) x 12 volt is also 0.8 v0lt ,So the bias is the same.
the difference is,it lighten the burden of the power supply feeding the plus bias from around 400 ma to around 150ma

Speaking of power supply,something come into my mind,A 80 watt linear (class B} transmitter needs a 160 watt power input (50% efficiency) .

 with a 12 volt power supply it needs around 160 : 12 = 14 A.
Now a transformer has internal resistance in the primary and secondary winding, as the internal resistance(email cable resistance) of the primary winding reflected to the secondary winding + the resistance of the secondary winding it self  is .5 ohm for instance ,than there will be a 14 x 0.5 =  pullsating 8 volt drop in the power supply at the frequency of the transmitter.
With this drop it affect also the min bias of the transmitter ,that make the plus bias close to zero.
so at this point it' s not a class B transmitter ,but  close to  Class C operation.

I think You need a regulated power supply than can supply  maybe 20 A at 12 volt or a  12 volt car battery .That doesn't drop when taken high ampere currents.

again I may be wrong

Gito.n

Gito

I assume that the power supply that Bob is using can deliver the necessary current without a significant voltage drop. I.e., I assume that it can deliver something like 20 Amps without a significant voltage drop from its nominal output voltage.

Separately:

If the transmitter was operating "key down", and the amplifier was drawing a constant 14A of average collector current (using your example, for illustration only)... then the average base current would be 140mA (assuming the current gain, h_FE, =100)

As you pointed out, the 10 Ohm base to ground resistors (not 12 Ohms) will be drawing a total of at least 0.6 Volts/5 Ohms = 120mA.

So I ask you: how can the bias supply (that supplies both of these currents) deliver 260 mA (or more, if the base voltage is higher than 0.6 volts)? If the bias supply can't deliver this average current... where would it come from?

This is why the 82 Ohm resistor is a problem for any type of high duty cycle operation (e.g. AM).

Stu

I haven't read Stu's or Gito's recent responses yet because I was messing around with the amp since last post.  

Changed the .01 to a .1uf.  No change in linearity.

Upped the frequency to 14Mcps and the amp produces a nearly perfect sign wave.  It gets better as freq goes up.  Vout seems to increase as well.  

Bob

That's good ... but can your scope see any frequencies beyond 20 MHz? The output waveform has symmetry around the horizontal axis... so it only contans odd harmonics.

If you are operating at 14MHz... the next higher harmonic is at 42 MHz.

So, regardless of what the output waveform really is, it would look like a sine wave on your scope.

Also note that the fundamental of a square wave has a higher peak amplitude than the square wave itself.

Thus the peak output would look bigger at 14MHz as a result of this effect.

[If you start with a square wave of peak amplitude A, and if you filter out all but the harmonics... leaving just the fundamental (i.e. just a sine wave)... the amplitude of the resulting sine wave is A x (4/pi) ~ 1.27 x A]

Stu

Bob

If you go back to 3.885 MHz... can you post a dual trace scope picture of the base voltage (DC-coupled, either transistor) and the output voltage (across the dummy load)

Stu

unfortunately the scope I'm using has a bandwidth of 20MHz.  I do have a scope that is capable of 100MHz but the HV supply is toast. Another project. So I'm stuck for the time being.

Stand by for images.

Bob,
I wonder if you ever tried a low pass filter at the output? The harmonic energy could be the source of distortion.

Frank, Yes on the LPF. Not much of an improvement.

Chan2 is Base or bottom trace.

Chan1 is DL or upper trace.

F= 3880KHz
Chan1= .1V/div  probe set at x10 plus a x10 reduction at DL
Chan2= .1v/div probe set at x10

Both channels DC coupled.

Bob

The base voltage, and therefore the base current, is saturating, as I suspected. If you were to add the base voltage trace you are displaying to its inverted image, shifted by half a time slot (in order to account for both transistors)... the resulting waveform would look like the output waveform.

Note that on the trace that's shown... the base voltage drops to zero when the other transistor turns on... because the other transistor is drawing all of its base bias current away from it. Neither transistor has anywhere to draw base current from... except from the complementary transistor. This implies that the impedance from the center tap to ground is still too high.

Even 0.1 uF may not be enough across the 470uF electrolytic (even assuming that the electrolytic is still working*). Try 1uF or more**.

*I am also wondering if the electrolytic is damaged. I believe it is a low voltage electrolytic... and it could have been damaged by excess voltage at some time or other. [Excess voltage => shorted capacitor => excess current => toasted/open capacitor]

**If we want to draw (for example) 50mA of current from this capacitor for one half cycle (0.5 / 3885000 seconds)... then we would draw 6.4 nanoCoulombs during that one half cycle. If we wish to do this, without changing the voltage across the capacitor by more than 0.1 Volts (for example)... then the capacitance has to be 64 nanoFarads = .064uF. So... I would think that 0.1 uF is enough... but perhaps a 0.1 Volt change in the voltage across the capacitor in each half cycle isn't really good enough. Maybe we need to have enough stored charge to keep the change in voltage across the capacitor during each half cycle below 0.01 volts... in which case we need 0.64uF.

Stu

You should be getting about 300 volts peak to peak on the output. I see about 100 volts or so. What does each collector look like and what is the total input current?
You should be running 45 to 50% efficiency. I wonder if the output transformer is in saturation?  Are you sure the transformer is installed properly? How about the cap across the collectors, does it run hot?

Hi

Stu
,I"m confused when You asked me where the current that supply the bias of the transmitter ,since I see the B+ running trough R1  trough the secondary winding of T1 is in series with R5 and R6 ( R5 in parallel with R6) ,
they form a voltage divider  with R1 33 Ohm in series with  5 Ohm ,so there's  about 300 ma flowing in this circuit ,so actually its a PLUS bias with 1.5 volt, but it was paralleled ,With A "zener" so the  bias voltage is 0.8.v Current is divided between R 5 Ohm and the "Zener"
So in static condition the PLUS bias supply  comes from the B+ trough R! ..through the secondary winding of T1 ... to the R5 and R6 and also to the base of the Transistor.
So of course the current is supplied from The B+ (12 volt)power supply) .
We are dealing with A plus bias not a min bias.

When the transmitter is in operation the more power needed that supply the base current comes from the power of the driver,which trough T1 (secondary winding) is flowing in the base of the Transistor.
So i still suspect the power supply of the the transmitter,with a 14 ampere needed.
it can drop when there's a resistance in Transformer itself ( email cable resistance)

It' just like supplying a min bias for Triode tubes audio amplifier/transmitter

The difference is A transistor if  the bias is is zero ,there's no current flow in the collector,so we must give a plus Bias to make the Transistor take current and depending the bias voltage to make the class  supposed to work.(class A,AB1,AB2,B,C)

In class B tube amplifier  not all class B is Zero bias,if we don't supply a min bias to it,the plate current will soars up,so we must give it a MIN bias,to control the standing current of the Tubes,

So I think there's no "difference" between this Transistor transmitter and tube transmitter ,when coming to Biasing them,what it needs is a Stiff power supply.

So actually this Class B Mrf454 is " The same" as a class B tube transmitter. Of course with different input and output coupling,but as a whole it is the"same"


Gito.N

Gito

Why are you referring to 33 Ohms. If you look at post 47, you will see that R4 is 82 Ohms in the verson that Bob is using. The bias source resistor is R4, not R1; it is 82 Ohms, not 33 Ohms. Lets talk about the same thing if we are going to talk about this.



Stu

Hi
when I read ,that the transmitter get a sine wave output,when the frequency goes higher,
than maybe the problem is in the material used in T1 and T2 is not the material that's needed for 3.7 Mc (the AL factor} or the output network is resonant at the higher frequency and not 'resonant at 3,7 mhz,

I believed even in broadband amplifier that use a C and a L network in parallel ,have a resonant frequency some where

Gito.N

Bob,
Download the EB63 app note from CCI. One of the scope pictures shows the output voltage swing at 100 watts am modulated. Input power is 2.2 watts.

Hi

Even if it used a 82 Ohm as R4 the current flowing in R5 and R6 is 12 volt divided by 87 ohm is around 150ma so it is enough to supply the base in static conditions.
As I wrote before the extra power that is needed when the transmitter is on is Supplied  by the driver transmitter.
Stu why  I used 33 0hm ,it' s from the original circuit  that you wrote in Your reply.
Or from Motorola schematic diagram.
And when I looked at the article ,the sine wave output is in 30 Mhz(the example)
Who knows the wave looks at 3.7 Mhz.

Gito.n

Sorry for attaching the picture twice

Bob

I was thinking about the scope trace you posted for the base voltage on one of the transistors.

If the center tap of the input transformer were at RF ground (at 3.885 MHz), the scope trace for the base voltage on either transistor would have mirror image symmetry with respect to the time axis.

But it doesn't

This implies that the center tap has an RF voltage on it (check it with your dual trace scope to compare the voltage on the center tap, using input 1, to the voltage on the base of either transistor, using input 2).

If there is RF voltage on the center tap (as I believe there is), that implies that you still don't have enough capacitance between the center tap and ground.

Increasing the value of the capacitor in parallel with the 470uF electrolytic capacitor to 1uF (or 10uF) should work to remove the RF on the center tap of the transformer.

Note: it is possible that the existing 470uF capacitor is not making a good contact to the ground plane of the board, or that it is not making a good contact to the center tap of the transformer. If so, tacking the extra capacitor on to the leads of the existing capacitor won't solve the problem.

Measure the resistance directly from the lead on the - side of the capacitor to the ground plane. Measure the resistance directly from the lead on the + side of the capacitor to the center tap of the transformer (or to the base of either transistor). 

Stu

Hi

I see there is a misunderstanding how really the bias supply works,

the purpose of it is to supply a fix and stiff voltage for a transmitter in static/standby mode and to the Class we intended to work with (class B).

Actually the bias supply needs only to supply a small current in the standby mode.,since the collector current is small at that time (depends on the amplification factor).
But to make it stiff ,it use a "bleeding "resistor in this case R4 and R5,R6 and the internal Base emitter resistance.IT use a bleeder that takes a high current in these case 300 ma,for what?not for feeding the MRF454 .

But To make it stiff supply so when more current flows trough  R5 and R6 and base of MRF 454 in case the transmitter is working/on,it has a little effect on the bias voltage .

Does the current that flows in the Base of the R5 and R6 when the transmitter is working is taken from the min bias?  not at all,the min bias duty is only to feed a small current and reference voltage for the MRF454.to work as class B transmitter.

So where does the current that supply the base of the MRF 454 came?
The MRF transmitter duty is To Amplify the driver transmitter,so the power from these driver is coupled trough T1 to the base of this RF 454 and this power/voltage that makes the current flow in the base of the MRF 454,
So the bias supply has nothing to do with the flow of the current in the base of this MRF454 when the transmitter is on.(no more current is taken from it)
But I see that the  capacitor is 500 uf(C22)  for the bias so I think we don't need to change it.

One question ,must the carrier of a transmitter be  a Sine Wave ,I think there are triangle ,square wave or in between these two wave and it's okay.
What you see in the example MRF 454 article is a product of AM modulation,that shows the envelope of it and because it is modulated with a sine wave ,naturally  the envelope is a sine wave.

But do You know the carrier wave is,Sine?,Square? it's still a question.

Gito.n

I'm wondering if there is a floating ground in the artwork or one of the mounting holes. This amp should be making a lot of power.
I have a couple 100 watt strips out of Raytheon marine rigs that I modified to extend to 30 MHz. They make a solid 100 watts to 10 meters.
Too bad I didn't know Bob needed one. I bet he could have bought both of them cheaper than the trouble he has been through with the CCI. 

Frank

I agree that there is most liely a bad connection between the - side of the 470uF capacitor and ground... or a bad connection between the + side of the 470uF capacitor and the CT of the input transformer. I suspect that (like most of us would have done) Bob is tacking the extra capacitor right across the easily assessible leads of the existing 470uF capacitor... and, of course, that would not lead to any improvement if one of those leads is not connected to the board.

As you suggest, probably a bad connection right where the capacitor leads go into the board... but it could also be a broken circuit board trace.


Gito

I respectfully disagree with your analysis (but that doesn't prove that I am right and you are wrong). On half of each cycle, the top transistor needs to be supplied with extra base current (i.e. in addition to its bias current). On the other half of each cycle the bottom transistor has to be supplied with extra base current. Both of those positive-going pulses of base current have to come from somewhere. I.e. current flows through a closed circuit. If an extra pulse of current is flowing into the base of the top transistor (and a half cycle later, into the bottom transistor)... it has to come from somewhere. The only places it can come from are as follows: [Remember the primary of the input transformer cannot provide current to the secondary of the input transformer... it can only produce a magnetic field that creates a voltage across the secondary of the input transformer]

a) It could be drawn from the base bias current of the bottom transistor.. i.e. when the base current in the top transistor goes up... and the base current in the bottom transistor goes down (and vice-versa). The problem with this (and I think this is what is happening now) is that the base bias current of the bottom transistor isn't big enough to provide the extra current needed by the top transistor (and vice-versa) when the top transistor is being driven above the bias current level. The extra base current that flows when the base of the top transistor is being driven positive is much larger than the bias current that can be "borrowed" from the base current in the bottom transistor.

b) It could be drawn from the 10 Ohm resistor from base to ground on the bottom transistor. There is plenty of current avaialble to borrow from the bias level current flowing through that resistor (around 60 mA is flowing through that resistor)... but in order to borrow that current, the voltage from base to ground of the bottom transistor would have to drop significantly below the 0.6 volt bias level. I think this is also what is happening now... and that is not desirable.

c) It could be provided by additional current (it would look like a rectified sine wave, and it would have a positive average value) provided via the center tap of the input transformer. This is how the circuit is supposed to work. Unfortunately... the center tap is (for some reason) unable to provide that current.

Why isn't the center tap able to provide that current (i.e. a positive-going  half-cycle pulse of current to drive the base of the top transistor, followed by another positive-going half cycle pulse of current to drive the base of the bottom transistor)?

I believe... based on the latest test/measurement results that Bob has provided... that the capacitor that is supposed to be connected from the input transformer's center tap to ground (i.e. the 470uF electrolytic) is not connected. It is probably not connected because there is a bad connection at the point where one of its leads it soldered into the board (e.g. a bad plated through hole or a cold solder joint)... or a broken circuit board trace.

Stu

Guys,
I tacked soldered the .1uf to the PCB at the bottom of R4..  Also tacked in another 470uf at the bottom of R4.  No change.  I did not remove the existing 470uf, not yet.

I will pull the PCB and check the feed thru's.  

Frank, transformers as far as I can tell are installed correctly.  In actuality, there is only 1 way to install them so it appears to be a no brainer, though please note, I have managed to find ways to F things up in the past but I don't believe I have at least not at this time.

I looked at the CCI notes on the 100w modulation.  I'm assuming that is based on the original design which has changed a tiny bit.  But note that envelope is at 30MHz.  Not 3.885Mhz.

A couple of scope measurements last night yielded the following:

I have equal and opposite (mirror image) waveforms at the Base of each transistor. So phase splitting appears to be working.  

The same goes for the Collector side or output transformer.  

So the transformers appear to be doing what they are supposed to be doing but maybe they are not the right 'value'. Not sure.

Please note that I have not read all the latest posting from last night to now.  I will catch up.

Bob

Since you have everything set up, including the scope... please measure (with the scope) the voltage from the input tranformer center tap to ground. It should be essentially 0.6 Volts with very little RF (because the RF is bypassed to ground). If not, there is a problem with the bypassing (as suspected).

Not only should the waveforms on either side of the transformer-to-ground (i.e. either base to ground) be equal and opposite...

The waveform on either side (i.e. just one side, taken by itself) should have mirror symmetry around the time axis. I.e., just like the output waveform across the dummy load, the waveform between either side of the transformer and ground should look like a positive-shaped pulse followed by an identical negative shaped pulse.

The fact that you are not seeing this symmetry of the base voltage waveform on either transistor (taken by itself) indicates that the center tap of the transformer is not at RF ground.

Stu

Bob,
Do you have a signal generator? It might be worth measuring the reactance of the output transformer secondary. I think it can be done from the output. Just put a 50 ohm resistor in series with the generator and feed RF into the transformer while you monitor both sides of the resistor with two scope channels. (POWER OFF)Then sweep from 1 MHz up to about 15 mhz. You can determine the reactance by the ratio of the voltages measured using the 50 ohms as a reference. Also you never posted the collector current. I would think the output transformer is saturating if the efficiency is below 45%.

Bob

One other thing:

Make sure that in wiring up the secondary of the input transformer to the board... that you did not accidently connect the center tap to one of the transistors, and the remaining winding end where the center tap should have gone.

Stu

HI


Stu,

That's where sometimes we forget ,that because the power source is isolated from the  load because we use a transformer,that means there can be no current flow in the load( where does the current came from).

Example ,if we used this Transmitter Driver and its Frequency of  60 Hz, and the rms voltage output
 is 220VAC,than what do have,An AC line voltage.

No we connect it to power transformator ,for example  220 vac with an output of  12 volt AC rectified it ,than what we get? a 12 VDC power supply,we load it then there's Current flowing trough it.

You see there's a current flow in the load,even if the primary winding is isolated from the secondary winding.

So the 3.7 mhz driver transformer it,s also a power source,with a 3.7 mHz frequency,
Say the Rms voltage is 10 v couple it trough T1 than put a load in the secondary winding like a light bulb,the light bulb will will glow (if the turn ratio of T1 is righ).

Or just rectified it with a  diode that can be used with a high frequency , ground the CT ,than what we got a power supply that can deliver power to a load.

look at the picture that I attached,(a bad picture ) there you can see the current flowing in the base and  r5,r6 as a load even without connecting R 4 to any supply source.
A little modification, actually I must draw R5 and R6 parallel to each dioda (that represent the Base to Emitter of MRF 454)

Gito.n
  

Gito

Referring to your picture (the post above this one)

For every ma of average current that flows through the load you have shown, there is a milliamp of average current flowing through one of the two rectifier diodes. For every milliamp of average current that flows through one of the rectifier diodes, there is a miliamp of average current flowing through the center tap. If the center tap was not connected the ground (directly, or through the resistors you have shown) no average current would flow into the load.

The center tap in the transformer in your diagram in exactly analogous to the center tap of the secondary of the input transformer of the RF amplifier.

If the center tap of the input transformer were connected directly to ground (as one of the people posting in this thread has suggested), then the avarge base current would have some place to flow through. But that would result in zero bias on the transistors. So the center tap is connected to a bias supply. The bias supply must do two things:

a) provide the bias
b) provide a path through which the base current (the part of the base current that is in excess of the bias current) can flow from ground into the center tap

It is not necessary for the center tap to actually deliver power... obviously no power would be delivered via the center tap if the center tap is grounded. However, if the center tap is biased at 0.6 volts, then there will be power delivered by the bias supply (via the center tap) equal to 0.6 volts x the bias current + 0.6 volts x the additional average current (beyond the bias current) required to feed the base of each transistor.

Stu

Hi

 a diode needs at least 0.6 volt ac to conduct,but the rms voltage of the secondary of T1 is more than that,even when the CT is not directly grounded because the R ,it's a complete circuit  afterall,and R is small (combining R5 and R6 and the internal Bias resistant.)

And don't forget it's not the current that makes a diode to conduct,but the voltage as long as the voltage is higher than the conduction angle,more than 0.6 v or  0.8v ac ,and as long the RMS voltage between the secondary output is more than that voltage,between ground via R and the secondary output ,maybe 1 volt,the diode will conduct.

So it's the voltage first  and  ,high enough? Yes ,the diode will conduct and the current will flow,how much depended on the voltage and  if the voltage is 6 volt (just for example) and the R is 5 Ohm than the current flow is 1.2 A (only an example).
And the voltage across R  will vary depending of the varieing  RMS volatge and /current makes also the current flow in the Diode (base to emitter current ) to vary,and it makes also The Collector current to Vary.

Gito.N

Didn't have much time to troubleshoot tonight but I did take a couple of photos of the amp itself showing its current layout.  

You can see the .1uf cap next to R4 (black resistor).  The 470uf is below the input transformer.

You can also see the base to collector negative feedback RC networks sort of hanging in mid air.

Unfortunately, the resolution of my cheapo camera prevents closeups so detail shots are tough.

I did take a measurement of the input transformer center tap.  There is some thing on the tap but it is very low.  Looks like a rough sawtooth above and below the time axis.  About 140mVpp.

I'll have to revisit tomorrow or Friday when time permits.  

Side note: appreciate everyones input on this puzzle. 

Gito

In your previous post, you included a circuit with a center tapped transformer...and you made some assertions about how it behaved. I replied with the reasons why I believe it would not behave as you said it did.

Now you have posted a different circuit, and you are making some assertions about how it behaves.

Please note that:

All of the current has to flow through R.

In Bob's amplifier, the diode (of your circuit, above) would represent the base-to-emitter junction of one of the MRF transistors (except it would point in the other direction)... which (at proper bias), has a resistance of around 1 Ohm (according to the specification sheet).

In Bob's amplifier... the resistor, R, (of your circuit, above) would represent 10 Ohm resistor from base to ground of the other MRF transistor. Thus, current could flow out of the top of the input transformer's secondary... into the base of the top transistor... out of the emitter of the top transistor (which is connected to ground), up from ground through the 10 Ohm resistor associated with the bottom transistor... and back into the bottom of the input transformer's secondary. This is, a complete, closed circuit path. I agree with that.

The problem with this is that the RF voltage required to drive current through the series combination of: the top transistor's 1 Ohm base-to-emitter resistance and the 10 Ohm transistor going from ground to the base of the bottom transistor is 11x as large as the voltage required to drive the base of the top transistor using the center tap

To match into this, one would use a 2:1 (turns ratio) input transformer... not the input tranformer that is used in Bob's amplifier.



Stu  

Bob

140 mV peak-to-peak on the center tap of the input transformer is a lot. When the transistors are forward biased... base-to-emitter... the base voltage changes very little as you add the additional (RF) base current.

If one used 1 Ohm as the differenital base-to-emitter input resistance (as per the specification sheet) ... then 140 mV (RF peak to peak base voltage swing) corresponds to 140mA of peak-to-peak base current swing. 140mA of base current swing corresponds to 14A (if it could change that much without saturating) of collector current swing.

So... the question (in my mind) remains: why aren't the bypassing capacitors doing a better job of bypassing the center tap to ground?

Let's assume that (for some reason) the 470uF capacitor has lost its capacitance... or that it does not function as a 470uF capacitor at 3.885 MHz.

I think that we want to get the 3.885 MHz impedance of the bypass capacitor (whether a single capacitor, or two in parallel) down to 0.05 Ohms... so that the impedance from CT to ground is 1/20 of the base-to-emitter resistance.

Going through the formula: at 3.885 MHz you need a bypass capacitance of 0.82uF to get an impedance of 0.05 Ohms.

Therefore... assuming the connections are okay, and the bypass capacitor does what it is supposed to do at 3.885MHz... I suggest that you try 1uF.

Stu  

Hi

Stu just to make it simple to understand ,I draw  the picture of  a simplified diagram of a half of the schematic /one  driver of one Transistor.

 And maybe because the picture is bad and not easy to understand,

R is actually the representation of R5 and R6 that's is PARALLEL connected with the Base Emitter  resistant (1 ohm).

The Bias supply  flow from B+ ..R4...Ct ...trough winding of secondary of T2 to R5 and R6 to ground. The winding  coil of T1 resistor is "Zero",
And one of my reply /picture,I have wrote ....a little modification that the "R" is actually in parallel with the diodes ( representing Base to Emitter)

So does my schematic diagram represent this circuit,I always wrote that R is actually R5 and R6 and Base resistor in parallel .
So I draw a equivalent  circuit ,that maybe  is not easy to understand.


And sorry I'm looking the original schematic of Motrolola for my replies,because I find it's good and all the changes We made is based on this circuit.
I like to attached another picture,but since I don't have a scanner and I always use my Camera ,unfortunately I have a low Bat,maybe later I repost a modified picture later.


Gito.n

Bob are you sure you are driving the stage hard enough. You need about 10 volts RMS to drive the amp to full output. Again what is the collector current?
No going to bed  It is only 6:00PM

Hi

A picture I promised to Attached ,I should make the picture  looks like no 1,but because the Resistance of secondary winding is closed to Zerro,so I make a picture of number two,actually it's the "same circuit"  only for analysising  the Circuit.
For more complete schematic I make picture 3.



Bob maybe I'm wrong  ,my advice Changed the C  witch is parallel with the primary and secondary with a value that makes T2  resonant at the operating Frequency(tuning the output net work) since a resonant frequency can be find using a formula ,and I think it's a linear formula,by changing C you can move the resonant frequency,maybe using twice or more the value in the schematic the resonant frequency maybe lowered,But it is stiil a Broadband Transmitter,looking at T1 with it's primary and secondary winding closed coupled ,So we have a pass band output transformer but with a different center Frequency.

Gito.N  

Hi

After thinking ,make my head spin ,I came to a more simplified schematic diagram,
As I have   found out that actually there's no need  A CT of the Secondary winding Of T1 to make current flow i n the bases of MRF 454.

So what the use/purpose of the CT of T1
to deliver  the bias voltage of the MRF454.
And at the original schematic I wonder .I did not see a  RF bypass capacitor for RF from CT to ground (maybe 0.01 uf) in parallel with C 22 (500 UF).  to make equal voltage drive to each  MRF 454A,is it a mistake on the original schematic.? or  it is on purpose not to decouple the RF,as  I wrote  above no need to make  current flow  trough the Diode (base ...Emitter of MRF 454)  with A CT in the secondary winding of T1

Can someone answer me?

Gito.N

HI

Bob ,on second though I believed  that  the CT is purposely not by pass for RF.

The reason if CT is grounded by C ("short" for RF/ac) for the RF input than the current flows from Ground via half part of  T 1 going to Diode ( base to emitter of MRF 454) directly to ground so the diode is saturated.

When the whole secondary winding is used not decoupled for RF in the CT than the whole secondary RF voltage flows from upper connector of Secondary winding to diode  to R  back to the other bottom  point in the secondary winding .since there is an R the current can be limited. see at the picture.
upper diode is MRF 454 Base to emitter parallel with R5, the current trough the
 diode is limited by R 6 so there is a control to the current flowing trough it.
When the CT is grounded for RF (by a coupling C) than the curren/voltage that flows from grounded RF (CT)  via upper winding t  trough upper diode  directly to ground (no resistor to limit),so the diode (base to Emitter) MRF 454 maybe over loaded and destroyed.

I always wrote  that I maybe wrong.

Gito.N

Hi
Bob look at the schematic I attached.

difference current flow with a decoupling C for RF at CT and un decoupled RF at CT.
Am I right


Gito

Frank

I am wondering why it would take 10 Volts to drive the amp at 3.885 MHz.

If I use the "less than 2 Watts of input power" specification... then I, too, arrive at ~10 Volts. [I.e. 10 Volts (peak) of RF across a 50 Ohm resistor = 1 Watt]

But... if I consider that the input transformer is 8:1 between the primary input winding and half of the secondary winding (base to RF ground)... then 10 volts in would produce 1.2 volts out.

If I consider only the resistive component of the input impedance of the transistor, which is specified to be 1 Ohm (if biased correctly)... then 1.2 Volts of peak RF from base to emitter (ground) is much too large. To push an additional 50mA (peak) of current into the base (corresponding to an additional 5 Amps of peak collector current) would require only 50mV of peak RF base voltage. If I add some additional input voltage to take into account series inductance (looking into the base), and base-to-emitter capacitance (necessitating additional RF inout current from base-to-emitter) I still don't understand why one would need 1.2 Volts of peak RF to drive the transistor's base.

At higher frequencies... series inductance and base-to-emitter capacitance would come more strongly into play... and I can see how the input drive voltage requrement would be larger at (for example) 30MHz... but I don't understand why it would be so large at 3.885 MHz.

Stu

 

Frank

Expanding upon my question in my previous post:

Let's use the actual diode equation to describe the base current vs. the base-to-emitter voltage (excluding parasitic base and emitter series resistances for the moment):

I = I₀ [(exp eV/nkT) -1]

where
e= the electron charge
k = Boltmann's constant
T is the temperature (Kelvins)
I₀ is the reverse leakage current... whose value doesn't really matter for this purpose, as long as it is much less than the base current
n is a number between 1 and 2
I is the base current
and V= the base-to-emitter voltage

Then the question: how much do we have to increase V to increase the base current from 5 mA (500 mA of collector current) to 50mA (5A of collector current)?

A little arithmetic and a slide rule indicates that V has to increase by 2.3 x [kT/e] ~ 57 mV if n=1, and 114mV if n=2..

Thus, neglecting parasitic base series resistance and neglecting parasitic base emitter resistance, we would need a peak RF drive signal of somewhere between 57mV and 114mV from base to emitter to change the collector current from 0.5A to 5A.

If we add in the effects of parasitic base series resistance, parasitic emitter series resistance, parasitic base series inductance, and base-to-emitter capacitance... we get a bigger number for the required change in V... but those effects ought to be much larger at 30MHz than at 3.885 MHz (e.g. more current flowing through the base-to-emitter capacitance, more voltage drop across the parasitic series base inductance)

Stu  

Hi

Stu,you make me "dizzy" following ,a very good explaining,with all those mathematical formula.

I looked at another Way,  when You look at my schematic diagram ,that use no decoupling C from the CT to ground.
It is easier to find the Base current we need,say we need 120 ma,neglecting the voltage drop across the Diode(base to emitter) of the upper  Transistor.
The current flow is from one of the output of T1 secondary winding flowing to The "diode" .... trough ground .... trough R6  back to the other output of T1 secondary winding. since R6 is 10 ohm, We need a 1.2 volt across it.right,and since the current flows also in the "diode" that's Base to Emitter,so the Diode current is also 120 ma.
R6 is also parallel with the lower diode, but since these "diode" is back bias ,it does not  conduct so The R is still 10 ohm.
I believe You can get more acurate finding the value of the voltage needed to find the right voltage.

I believed that the transmitter design( motorola schematic diagram) ,purposely don't decouple the CT of secondary winding . For the reason I wrote above .
look at the schematic  at the  bottom picture

gito.n

Stu,
I was just looking at the app note that showed 2.2 watts of drive to make 100 watts out. I think it was at 30 MHz though. This simple amplifier may have poor input VSWR since there is no feedback. Also I bet the gain at the low frequency end is quite high without feedback. So there is a good chace the drive at the low end can be a lot lower. The trace inductance between the input transformer and transistor base has a big effect on performance. I found this out playing with ENI modules with MRF429s at 48 volts. Layout is very critical to get a flat input VSWR.
Much easier when you use FETs.

Frank,
I'll try to get a collector current for you. I have been hesitant only because it means trying to lift the Collector tab on the transistor or desolbering the transformer.   There's no easy way to get an individual measurement.  

Don't know if this is any help but the power supply indicates about 9amps current draw at 13.6V.  That is almost 50% less than what it was doing.  Note that the input is padded with a 4dB 50ohm resistive attenuator and the Base/Collectors have feedback networks attached.

I don't know what the input power is at this time.  I don't have a meter that is sensitive enough at that level and the input impedance is not 50ohms because there is indeed a mismatch between the Retro and amp when measured with a 3:1 SWR meter between the two.

For a 1uf cap, would an electrolytic suffice?  Probably not ideal in an RF app.  How about a tantalum.  Cap type matter at the center tap?

Thoughts on an RF choke in series with R4?

Spice simulation

I simulated the behavior of the amplifier for the case where the center tap is delivering bias from a current source (high impedance to ground at RF). I.e., the center tap does not have a bypass capacitor for RF.

Please refer to the schematic (attachment 1).

In this approximation, each transistor (one on the left and one on the right) is modelled as a diode in parallel with a 10 Ohm resistor. The fixed bias is set at 80mA... i.e. half the maximum current that the existing bias supply can deliver. I.e. (13.8 V - 0.7V) / 82 Ohms = 160mA total for both transistors. Without expalining why, I believe that the bias supply will deliver its maximum current when the amplifier is operated without a center tap RF bypass capacitor.

The diodes in the simulation are realistic, not ideal. So they represent an actual forward-biased emitter-to-base junction... at least to some approximation.

I assumed that the input transformer (4:1 turns ratio) transforms the 50 Ohm source (transmittor) output impedance to 50/16 Ohms ~ 3 Ohms.

Therefore, I am representing the secondary of the transformer as a voltage source with a 3 Ohm source resistance.

Since Bob's amplifier is currently delivering 50 volts peak to a 50 Ohm dummy load, the output current into the dummy load is 1A peak. Taking into account the 1:4 output transformer... the peak RF current (not including resting current) being produced by each transistor is 4A.

The RF portion of the base burrent is, therefore, 4A/100 = 40mA peak.

Still referring to attachment 1 (the schematic)... I adjusted the RF voltage source to produce a peak base current of 50mA... i.e. allowing for 10mA of DC base bias current. The required RF voltage was 0.55 Volts peak.

Attachment 2 shows the simulation of the voltage on the base of one of the transistors (i.e. positive side of the diode to ground): blue waveform. Note that this waveform is at least somewhat similar to what Bob is seeing on his scope.

Attachment 2 also shows the base current in one of the transistors: red waveform. Note that the peak base current is (as I targeted it to be) 50mA. The minimum base current (with the RF applied) is 0mA.

All of this is as I expected, for the case where the center tap does not have an RF bypass to ground. It's easier to simulate the waveforms rather than talk one's way through what they should be... but the result is the same.

Next, I will modify the schematic to simulate the case where there is an RF bypass to ground

[After I modified the schematic to add the RF bypass to ground (see my post #154)... I added the green curve in attachment 2 to show the voltage on the center tap in the absence of any bypass capacitance. Note that it is about 155mV peak to peak... just as Bob reported when he measured the voltage from center tap to ground. This strongly suggests that there is no RF bypass capacitance present in Bob's amp... even though they are physically attached to the board. Sounds like a broken trace.]

Stu

 

Bob

An electrolytic would suffice. I'm pretty sure that the problem is simply that the 470uF capacitor is not really there (for whatever reason)

An RF choke would serve little purpose... because R4 (82Ohms) is already a high impedance relative to the stuff on the other side of it... but it can't hurt.

Stu

Bob just measure the power supply current don't lift anything. I was just interested in the efficiency

Frank,
About 8.7amps according to the PS current meter.

Here's the simulation with a bypass capacitor from CT to ground. Note that I'm doing the simulation at 1000Hz (for convenience), instead of 3.885MHz. It doesn't make any difference... provided I scale the capacitor properly. I used a 1 Farad capacitor in the simulation... which scales to 257uF at 3.885 MHz.

The voltage on the primary side of the transformer is the same as it is in my earlier simulation (no center tap)... because the total voltage across the secondary (end-to-end) is the same.

Note the base voltage waveform. It is not symmetircal (as I thought it would be)... because of the effect of the series resistance of each of the pair the RF sources that represent the output of the input transformer. [Hmm... that sounds strange]

Note the increase in the peak base current (and thus the peak collector current) v. the previous case where there was no bypass capacitor from CT to ground. The peak base current is more than twice as large ... which means that the RF output power of the amplifier would be 4 times as large (e.g. 200 Watts instead of 50 Watts)

The brown trace is the voltage across the capacitor... essentially flat, as expected.

Stu

Yanked the 470uf out and replaced it with an 820uf @25v. Also installed a 2.2uf next to it.  Center tap V has decreased to about 50mVpp and to the point  where the scope  won't trigger properly.  Waveform is  dim and not pretty.  

Output no change.  Still got the near squarewave output.

Current draw from the PS  has increase slightly  ~9.1amps.

BTW, feed thru's between top and bottom are OK.

After I modified the simulation schematic to add the RF bypass to ground... I added the green curve in attachment 1, below, to show the voltage on the center tap in the absence of any bypass capacitance. I.e. I set the bypass capacitance to zero by removing it from the circuit.

Note that it is about 150mV peak to peak... just as Bob reported when he measured the voltage from center tap to ground. This strongly suggests that there is no RF bypass capacitance present in Bob's amp... even though they are physically attached to the board. Sounds like a broken trace.

Bob

If you are drawing 9A at 13.8 volts... then the electrical input power to your amp is 124 Watts.

Is the output across the dummy load still a 50 volts square wave?

This corresponds to 50 watts of RF output power.

That sounds about right if there is still headroom on the collector voltage.

I'm also wondering if the bypass capacitor at the output of the on-board power supply filter is doing its job.

With RF on... what is the waveform, including the DC  voltage, at the center tap of the output transformer or at the point where the on-board DC filter connects to the center tap of the output transformer. Measure the DC level with your scope, rather than a voltmeter.

Stu

Bob
Gito
et al.

Attached are the simulated currents on the base of each of the two transistors for

Case 1 (attachment 1) No bypass capacitor on the input transformer center tap

Case 2 (attachment 2) Bypass capacitor on the input transformer center tap

In either case, if you flip over one of the currents, and add them together... you get a nice sine wave. The current is larger with the input RF bypass capacitor in place, but the presence of the capacitor doesn't impact on the shape of the current (at least to the degree that the schematic in the simulation model is accurate...i.e.  in the absence of various parasitic inductances and capacitances... which probably aren't very important at 3.885 MHz if the amplifier is capable of operating at 30MHz).

Therefore... why does the output voltage across the dummy load look like a square wave?

That leads us from the input circuitry to the output circuitry. Again... is the center tap of the output transformer properly bypassed to ground... with respect to RF?

Stu

Hi

here I attached a class C Transmitter as You see at the equivalent schematic,
The current flows to the diode is uncontrolled  ,a  1 volt ac will makes the current flow of 1A in the base emitter of MRF  ,since the base emitter resistant is 1 ohm (like Stu wrote) 1 volt ac makes the base emitter to conduct( over 0.8 V),so 1; 1 = 1 A,How hard You tried ,the current that will flow ,even a 0.8 volt voltage (the turning on of the transistor) will give a 0.8 : 1 = 0.8 A and what happens when the gain is 100,the collector current will be 80 A?

What the difference of class C and class B ,the bias voltage,in class B the Ct is not directly grounded for DC voltage,but if You bypass it for RF ,than it's like grounding the CT  for RF and what happens, it work just like the Class C transmitter(for RF)

If you don't bypass for RF ,but bypass for DC only(500uf in the Schematic) ,than the circuit in the center ,look at the equivalent circuit ,at one half cycle ,in this case the upper Transistor is conducting, the current  trough the base emitter is limited by R6 (10 0hms) ..

the botom picture is a modified circuit,if the CT is grounded use a R1 and R2 limiting resistor.

If You used the the original circuit  DO NOt BYPASS THE CT FOR RF ,instate use an RFC or Ferrite RING  between bias supply to CT.to block the RF to ground ,it's my advised.

Or use a limiting resistor between T1 and base  of the MRF 454 if the CT is "grounded" for RF look at the bottom picture.


Gito.n

8 amps sounds like there may be the wrong core material in the output transformer.
MRI amplifiers don't work below 40 meters because they use type 61 material in the transformers. The firat thing you observe is high power supply current and crappy waveform. Again it may be worth checking it at a higher frequency. at 8 amps you should be making a clean 50 watts output

Frank
Bob

Core saturation of the output transformer sounds like a plausible explanation for the shape of the output waveform... but why doesn't the saturation go away (or become less) when Bob reduces the peak-to-peak input drive signal?

If the transformer is saturating at the current that leads to 50 volts peak output... why doesn't Bob see something that looks more like a sine wave when he reduces the input dirve voltage by (for example) 6dB?

Stu

HI

The input wave may looks good a sine wave,since a representation of the output of the driver, but what if the base to emitter current 0f the MRF 454 saturated ?

look at the pictures ,and analysis of the circuit I attached above
I wonder If I'm wrong tell me ,that I'm Wrong.
And if I'm Right ,tell me that I'm right.

Gito.n

Frank
Bob

Here are some even more interesting simulations for the entire amplifier:

Attachment 1 is the schematic used in the simulation. The transistors are similar to the MRF's, except that they have a current gain of 20 (instead of 100)

Attachment 2 is the collector current in each transistor (you need to flip one, and add them to get the total current in the primary of the output transformer) for the case where the center tap of the output transformer is at RF ground (i.e. the center tap bypass capacitor is sufficiently large)

Attachment 3 is the collector current in each transistor when the center tap RF bypass capacitor on the output transformer centre tap is small  (scaled to .003885 uF).

Look familiar?

Check the bypassing on the output transformer CT. I'm beginning to wonder if there is a systemic problem on the circuit board that is causing an open circuit on these bypass capacitors.

Stu

Hi

Stu , I looked at your stimulation picture You used a 3 ohm (R1) in series with the power generator source , so when the peak voltage is 0.8 volt the Diode is conducting trough 3 ohm trough internal resistance of diode depending on the internal resistance of the diode (in this case 1 ohm) total 4 ohm what happen? a 0.8v divided by 4 Ohm = 200ma,that's why the wave is not good.Can the signal supply it?
If You used a signal generator with a signal that' below 0.8 volt ,the diode won't conduct,so what you see is a sine wave across the 10 ohm resistor,naturally  a sine wave.

 The  base emitter current is to high ,as the transmitter have a gain of 100( ?) than the collector current is 20 A.(saturated?)
As the signal generator needs at least 0.8 volt for the diode to conduct.,How hard You try if the internal resistance of the diode is 1 ohm ,You will end with a 200 ma diode current .
if R1 is 10 ohm You got a 80ma diode current ,changed it to 100 Ohm than You got a 8 ma  diode current.

So this diode current is actually the current of base to Emitter of MRF 454 that wiil be amplified by 100 (collector current).


Gito.n

Gito

My suggestion is that: as you think about these things, you should separate (in your thinking) the average values of the voltages and currents... such as: the average base current, the average base voltage, etc.... from the RF waveforms (with their average values subtracted out)... such as: the RF voltage on the base of a transistor (with its average value subtracted out)

For example: there is a certain average base current, flowing in each transistor that must be provided by the bias supply and/or the 10 Ohm resistors.

In the simulation results that I have posted, I have set the total average current flowing from the bias supply to be 160mA. That's what I set it to be, for the purpose of those simulations. In the simulation results I posted, I used a pair of 80mA current sources (one from base to emitter of each transistor) to represent the total current from the bias supply. I have other simulations that I have done, where I have used a center tapped transformer to supply the bias... but, again, for the simulations I have posted so far: I placed an 80mA current source from base to ground (emitter) on each transistor. What the simulator calculated is based on those two 80mA current sources. The average base current in each transistor must be 80mA minus the average current flowing in its corresponding 10 ohm resistor. That is Kirchoff's current law.

According to the simulation, when RF input is applied, the base current on each transistor is varying from 0mA to 88mA (in my latest simulations)... with an average value of around 30.5 mA.  If I turn off the RF, the average base current in each transistor is around 28.5 mA. That's what the simulator predicts for the particular transistors I am using in the simulation.

Likewise, there is a certain average base voltage on each transistor.

When no RF is applied, this average base voltage, for the transistors I used in the simulation, happens to be around 515 mV. That is, of course, consistent with the average base current (with no RF applied) being around 28.5mA. I.e. 80mA of average bias supply current (per transistor) - 28.5mA of average base current (per transistor)= 51.5 mA of current flowing through each 10 Ohm resistor. 51.5 mA x 10 Ohms = 515 milliVolts.

When RF is applied, the base voltage varies from 382mV to 578mV. It is not a perfect sine wave, and the average value is not exactly in the middle. The average value is 495 mV. i.e.when RF is applied, the average value of the base voltage drops from 515mV to 495mV. Thus the average voltage across the corresponding 10 Ohm resistor drops by 20mV... corresponding to a 2mA reduction in the average current through that 10 Ohm resistor... and a 2mA increase in the average base current (from 28.5 mA to 30.5 mA).

That's the way those particular transistors behave... as per the simulation results... with a fixed 80mA current source from base to ground (emitter) of each transistor.

Note that, with RF applied, the peak-to-peak voltage (i.e. not including the average value) on the base of each transistor is 578mV -382mV = 196 mV peak-to-peak (not a perfect sine wave). That is the peak-to-peak RF drive voltage that the secondary of the input transformer must deliver.

Since the transformer has a 4:1 turns ratio... the peak-to-peak voltage on the primary of the input transformer must be (in this simulation, with these components): 784mV.

Stu


 

Bob
Frank
et al.

I'm learning a heck of a lot from the simulations.

Playing around with a few parameter values... and observing the results of the simulation:

1. It is very important that the center tap of the output transformer have a good RF ground. I'm using a value that corresponds (when scaled to 3.885MHz) of 10uF for the output transformer center tap bypass to ground. If it's missing (bad circuit board trace, bad capacitor, or whatever) one sees the output waveform that Bob has been observing.

2. If you want to get high power out... you need to have enough average current from the base bias supply. When I used 82 Ohms... the output power was limited because the transistors end up with too low of an average base voltage*... and they operate deep into class C. When I switched to 33 Ohms in the simulation... everything was much better.

*The reason that the average base voltage drops is because: as you increase the RF input drive to get up to the target output power level, more and more average base current has to be provided by the associated 10 Ohm resistor. In order for the 10 Ohm resistor to provide that average current, the average voltage across the 10 Ohm resistor has to decrease. E.g. to increase the average base current by 30mA, the average voltage across the 10 Ohm resistor (which is also the average base voltage) has to drop by 300mV.
 
Stu.

Stu,
I think the transformer is just saturating so the turnoff voltage is distorted into a square wave. Also Bob has blown a set of devices so seems to me the load is collector load is wrong causing excessive current when the drive is increased.
Increasing the DC current through the promary will drive the reactance even lower  
Easy to verify or prove me wrong by increasing the frequency.

Frank

In my latest simulation (schematic and results attached below), there are no transformers to saturate. The inductors in the simulation are generic Spice models, and they behave as ideal components.

However, when I make output center tap bypass capacitor (C1 in the schematic) too small... look at what happens to the collector current waveforms: compare attachment 2 (C1=0.001uF) to attachment 3 (C1=10uF bypass)

[All capacitor values are quoted by scaling the frequency in the simulaton to 3.885MHz]

I agree that Bob will have to be cautious about increasing the average base current. Too small a value of R4 will enable the desired level of peak RF output power during transmit...but will mean that the transistors are drawing too much current when the RF input is removed. I.e. they will idle at too high a current. Too large a value of R4 keeps the idling current down... but starves the transistors for base current during transmit. The simulations were done with R4=82 Ohms.

This is very much like trying to build a transformer-coupled push-pull linear amplifier with tubes that draw no plate current unless their grid is positive with respect to the cathode.

I think we agree that properly biasing this type of amplifier is a PIA.
 
Stu

Hi

Stu,I admit, You are good in Explaining,and make a big input to me.But I think there's a  big difference
Between by passing the CT and un bypasing the CT,

When You Bypass the CT (grounded for RF) . There's two individual circuit,the upper circuit and the bottom circuit ,that can be  separated from each other ,
Since there's no Connection to Ground to make the current flow,it must take it current from the bias supply. when the bases of the transistors takes current(to make a complete circuit)

But if You un bypass the CT .The RF voltage and current  is flowing around a complete  circuit so there's no need to take the current from the bias supply
 the current is flowing through the Base of the MRF 454 from the bias supply .(transformer .to upper transistor trough R6 ...back to transformer)

So since it does not take it's bias current from the Bias supply ,The bias voltage more stable.

I have proven that without CT "grounded" the current can be make to flow in the bases of the transistors,if the CT is not bypass for RF.("grounded")
It needs no other source to make the current flow in the bases of the transistors.

So after all we are back to the biasing topic.
enough bias to make the transistor work as class B,it is stable?
In case 1 where the CT is "grounded" like Stu wrote it takes current from the bias supply so with a "big" current flowing in the bases of MRF 454 like 100 ma, than the bias voltage  with 33 ohm will drop to 33 X (0.1 + 0.3)= 13.2 volt since the power supply is 12 volt,than What is the class of this transmitter working?
The 0.3 is the current Flowing from B+ through R4(33 ohm) trough R5 and R 6 in parallel to ground.

And in the second case  if we un ground the CT for RF ,the Bias voltage is more stable ,since it does not  take the current from the bias supply.

Than the only thing that we must do is Variable bias supply that's stiff enough and well regulated to feed the CT with it, omitting the  Zener diode and R4.

and if we used it ,since it's stable and can supply enough Current without changing it 's output voltage.
Just set the power supply to it's intended voltage to make The transistor work as class B,or AB2.

With these bias supply both circuit ,grounded and ungrounded CT.I believed will work


Gito.N

Output is what appears in the attachment. Sorry for the blurred image.  Some improvement from doing the following:

Dropped R4 to 50ohms.  At 13.6V, bias is sitting at .701V.  Current draw from PS is 940mA (standby mode).  Transmit current has increased from about 8.5A to 11.3A.  Bias does drop to .647V.  Previously, bias was around .660V with a 300mA draw from the PS at 13.6V (standby).    I don't believe I can go any lower with R4 unless the bias method is changed.

Checked bypassing.  Tried different values on the output side. Made no difference.  Resoldered all feed thru's and compared grounding to schematic.  Don't see any abnormalities.

Bob

Can you post the following two pictures?

1. Dual trace: output RF at dummy load (same as last post) and the voltage waveform on the output transformer's center tap.

2. Dual trace: output RF at dummy load (same as last post) and the voltage on either transistor's base

Stu

Revised
Here you go Stu.  At first I had them mixed up but they are correct now.

Image 1. Dual trace: output RF at dummy load (same as last post) and the voltage waveform on the output transformer's center tap.

X10 Probe
Channel 1: Center tap of Output Transformer  (lower trace)
0.2V/div
Channel 2: Dummy Load  (upper trace)
2.0V/div


Image 2. Dual trace: output RF at dummy load (same as last post) and the voltage on either transistor's base.

X10 Probe
Channel 1: Base of output transistor   (lower trace)
50mV/div
Channel 2: Dummy Load   (upper trace)
2.0V/div

One additional tidbit.  The input power is approximately 1watt.

And found a source for the original bias diode (1n4997).

Bob

That's a big signal on the center tap of the output transistor (about 1.7 volts peak). Remember, the B+ supply is 13.8 Volts.

What causes this noise?...

It could be a number of things... but, when each transistor tries to draw collector current, it is helpful if that transistor's current can all flow up to the collector from the center tap ... rather than flowing out of the other transistor's collector (or not having anywhere to flow from).

In my latest simulaton (just making little tweaks to the model)... if you really have 10uF from CT to ground (as per the schematic)... the voltage waveform on the CT should be essentially invisible.

Here's a posibility. Could it be that the physical distance from the center tap to the 10uF bypass capacitor is too long... and therefore the inductance of that path is coming into play? This is a relatively low impedance point (compared to what we're used to with tube designs).

At 3.885 MHz, and around 10 Amps of amplitude... it only takes 4 nanoHenries (0.1 Ohm of inductive reactance) to produce a 1 volt signal on the center tap (the waveform will be at twice 3.885 MHz)

Can you add a 10uF capacitor that is physically closer to the center tap ... and going to ground close to where the emitters go to ground? [10uF presents a reactance of 0.004 Ohms at 3.885 MHz]

Stu
 

it could also be the way the scope probe is grounded.

Hi

Back to bias, if we used Stu schematic (grounded CT for RF),And if You looked it carefully The "Zener" diode is in Parallel with the base Emitter of MRF 454.
Three diode in Parallel

So omitting this "Zener" ,the transmitter will still work,but why using this "Zener"
since if the transistor gets warm ,the bias voltage is changing ,and the colector current start to rise, So there must be a second active devise to stop it ,its the "Zener" that have a negative cofiecient. and it must have a  thermal contact with these transistor.
 
So the Bias current is totally depended of R4,and the internal impedance of The Transistor.
And since the bias current is taken from B+ ,variying the base current is also variying th base voltage to some extend.
 So in actullay the zener voltage and the MRF transistor,must have the same treshold voltage (because manufactoring. proses) which the diodes start to cunduct.,a slight diference, will make different current flowing through each diodes.

remember to used equalizing resistor in parallel diode conection.

Just a though.

Gito.n

Guys, if Bob can't get a clean output with 900 ma of bias, then bias is not the problem. The shunt diode type is not all that critical because this is a sloppy circuit anyway. Based on amps I have played with, diode temperature compensation will have bias change a good amount as the heat sink heats up. A good thermal contact is pretty critical. The diode will do its job if the stage doesn't run away. Also I saw similar waveforms in the ERB amps before I added more primary turns to the output transformer to get it down to 160 meters from 40 meters. Most MRI amps work down to 40 and I have modified a number of different ones. I have seen a poor collector waveform effect the base wave shape.
Running the DC through the transformer makes the primary reactance even lower.

Hi

Not saying the schematic diagram is wrong,I don't have the right to say so,but these diagram is bothering me ,when the CT is grounded for RF ,direcy connect1ng the output of T1 to the bases and the Emitter is directly grounded to ground,and if the internal resistance of MRF 454  is 1 Ohm,
Then a 0.1 volt apearing  at T1 output will give .... 0.1 volt : 1= 100 ma.base current
          0.2 volt apearing  at T1 output will give ..... 0.2 volt : 1 = 200ma base current
In these case the MRF454 is in conducting state   becuase the biasing of it.

So the voltage of secondary winding must supply is  around 0 to 0.2 volt. to make the base current flow around 100ma and 200 ma So the turn ratio of T1 is very impotant,when using 16 t0 1 ratio ,it  ,make  a 3.2 reflecting volt in the primary .if the primary has a 50 ohm input than the current flowing trough it is 3.2 : 50 = 0.06 A,so the watt needed is 0.06 X 3.2 = 0.19 watt?
This does not make sense to me.

If turn ratio 50 to 1,than You get a 10 volt in the primary of T1,than with a 50 ohm  input load  the current is around 10 :50 = 0.2 A so the watt needed 0.2 X 10 volt is 2 watt ,more likely,
It 's more make sense to drive the amplifier with a 2 watt power output.

I must be wrong. But where ..,can You tell me?

Gito.N

 Hi

Now I realised having an Idea that the base Emitter resistance is 1 ohm is wrong,(where did I get it ?) ,all what I wrote is based on this asumption.

first I begin to think if the resistance is 1 ohm ,having a  0.1 voltage on it gives a 100ma current.
Second  it used a 12 ohm in parralel ,what the use of it,since the base emitter is "one ohm".It still give a resistance close to 1 ohm.

So the base emitter resistance must have a enogh resistance,at least at rf .
With it than the transmitter circuits make sense.
The purpose of the 12 ohm resistor in my opinion is a constant load for the RF driver.

the parralel input impedance is 1 ohm when the Transsistor delivered 80 watt output,
that means at that time the base emitter is taking current,as we know Impedance ,say for instant  of a transmitter  is B+ divide  current = impedance(the formula is depending on the operation class,)
so is the input impedance of the transitor when taking the max current (base current} to delivered a 80 watt input power of the transistor.
So when there's no current flowing trough the base,it's impedance must be higher.

 

Gito.n

Guys,
Ok put a 10uf right on the C-tap.  The noise dropped about 25 to 30%.  Probed the grounds and the noise is the same all over the PCB however, when I probe the heatsink which the PCB is screwed to by 9 screws the heatsink exhibits very little noise. It's nearly flat line.  Figuring the anodizing had something to do with it I scraped the anodizing off of a spot and probed again no change.  I wonder why the heat sink would be nearly clean but the traces that the screws go threw are not.

BTW, absolutely no change in the output with the extra 10uf.

Bob

I think it is time to test the modulation linearity of the amp.

[The output is not a sine wave... but the fraction of the total output power in the fundamental component of the output is not too bad... and one will always need a low pass filter of some kind on the output of this kind of amplifier. I.e. one would not expect the output to be a clean sine wave, unless you pass it through a low pass filter]

If you decrease the amplitude of the sine wave at the input (before the input attenuator) by a factor of 2, how much does the amplitude of the output (across the dummy load) change.

Note: it would be best to make this measurement with some kind of low pass filter between the output of the amp and the dummy load. You wil need one eventually... but if you don't have one now, you can put a low pass filter between the dummy load and the scope by making a simple 5MHz RC filter. For example: If the scope probe capacitance is specified at 15pF to ground, put 2200 Ohms in series between the dummy load and the probe. If the scope probe capacitance is specified at 22pF to ground, then put 1500 Ohms in series between the dummy load and the probe.

Stu

Let me ask you this.  I've constructed a 50ohm 4dB attenuator at the output of the Retro75.  Can I assume that a 4dB decrease at the input of the amp will result in a 4dB decrease at the output?

I do have LPFs. No problem there.

"Let me ask you this.  I've constructed a 50ohm 4dB attenuator at the output of the Retro75.  Can I assume that a 4dB decrease at the input of the amp will result in a 4dB decrease at the output?"

Yes...

As you make this test... it is important that the impedance that the Retro is looking into doesn't change... and it is important that the source impedance that the amplifier sees... i.e. looking back toward the Retro... doesn't change.

If you are using the Retro in this test, it would be even better to just modulate the Retro at
1kHz... and to compare the RF output of the Retro (trace 1) to the RF output of the amp (trace 2), while sweeping the scope at around 0.2 ms/div.

If all is well, the envelope at the output of the amp will look like the modulated envelope at the output of the Retro.

If you see flattening on positive peaks at the output of the amp... then reduce the power level at the input of the amp with the attenuator.

If you can find an input power level at which the envelope of the RF at the output of the amp tracks reasonably well with the envelope of the RF at the output of the Retro... then you're "good to go"... except, perhaps, for the power output of the amp being a little lower than you were hoping for.

Stu

OK fine business.  I will make a go of it and report back shortly.  

No problem on the lower power output.  It will still be better than 2 watts.

Now go an enjoy the rest of the weekend.  I've taken a boat load of your time with this.

Stu,
Despite the linearity of the sinewave (carrier) the input and output appear linear to each other.  That's a good sign.

Bob

Okay...

Then whatever the Retro would sound like on the air... the amplified signal should sound the same.

Maybe it's time for an on-the-air comparison.

Stu

I'm wondering if I should knock the gain down some more so output is about 45w.  Output is about 70watts carrier right now with the attenuator.  Having the attenuator in place effects the receive a tad on the Retro which isn't spectacular to begin with. Or should the attenuator be used  when in transmit only rather than mess with the gain.

Bob

I would suggest that you put the attenuator between the relay and T1... so that it only is in the path on transmit.

Use a 50 Ohm pi attenuator made from 3 non-inductive resistors:

100 Ohms to ground on each side (2 watt or larger resistor on the input side of the pi, 1/2 watt or larger on the output side of the pi)

~67 Ohms (1 watt or larger) on the horizontal part of the pi.

The attenuation will be around 10dB from the output of the Retro to the input of the amp.

On transmit, the load on the Retro will be close to 50 Ohms, even if the impedance looking into the amp (T1) is higher or lower than 50 Ohms. The attenuator will do a good job of isolating the Retro from the amp.

Stu

Hi Stu,
This begs a question and I'm sure this has been discussed here.  In this day and age is it possible to get non-inductive resistors?  2 watts or higher, carbon types, seem like a thing of the past.  Are metal film resistors considered non-inductive? 

Anything wrong with putting the attenuator at the Retro instead.  Space is a premium on the amp.  It would be easy to put a relay in place on the Retro and switch the attenuator in and out during rcv and xmit modes.

Bob

Ok to put the attenuator any place convenient... and to use a TR switch

Metal film resistors are also non-inductive. I've been using 5W and 2W Xicon metal oxide resistors for things like terminations and attenuators.

I used 3 of them (150 Ohms, in parallel, on the input and the output; and 37.5 Ohms in series) to make a 50 Ohm 6dB attenuator. With a 50 Ohm termination on the output, the attenuator looks like 50 Ohms at the input (measured with my MFJ antenna analyser) at frequencies up to 30MHz. It was still pretty good at 148MHz (56 + j23 Ohms)

Stu

OK on the resistors.  I don't have what I need on hand but Mouser seems to have a very good selection of metal oxide types at various power levels.  An order will be on the way. 

Hi

A beter circuit using MRF 454 ,with UNGROUNDED CT/WITHOUT CT at all in T!
using LIMITING RESISTOR R1.R2

using negative feed-back from T2(L3 at T2 to Q1 and Q2.

Using biasing circuit ,that proofs the importance of the biasing point
See the complete article inAN-762 sharp 300.

It's also wrote
The biasing source is ajustable from 0.5 t0 0.9 volt,which is sufficient from Class B to Class A operation
In class B the bias voltage is equal to the Transistor Vbe and there's no  COLLECTOR IDLING CURRENT PRESENT (except small collector -emitter. leakage Ices)

Gito.n