RF Feedback

[AB2EZ]

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

[Gito]

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

[WA1GFZ]

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. 

[AB2EZ]

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

[W1RKW]

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.

[AB2EZ]

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

[WA1GFZ]

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%.

[AB2EZ]

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

[Gito]

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
  

[AB2EZ]

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

[Gito]

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

[W1RKW]

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. 

[AB2EZ]

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  

[AB2EZ]

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  

[Gito]

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

[WA1GFZ]

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

[Gito]

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  

[Gito]

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

[Gito]

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

[Gito]

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

[AB2EZ]

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

 

[AB2EZ]

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  

[Gito]

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

[WA1GFZ]

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.

[W1RKW]

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?