The AM Forum

I am bringing up my 2x813 ,I near as mentioned in other threads.

I can tune the amp on 80m and 40m but have not found a plate current dip on any other bands. Now what?

Is there a way to check the tank tuning with my grid dip meter?

The dips on the lower bands are very tiny with only a small amount of plate capacitor.

Thanks

Rich

Yes,  you can use a dipper.   If you have a vna / swr analyzer,  it's also doable.

I preset Ctune and Cloud,  making sure to get Ctune set taking Cstray into account.

Then set taps accordingly,  to get 50 ohms j0.   Use a 259b running into the antenna jack,  rf out relay in TX mode and a resistor from anode to ground.

This ensures Q is what you designed it a,  and any and all stray C and L is taken into account from strap to and from bandswitch,  etc.

Others have different methods,  I'm sure.

--Shane
KD6VXI

Rich

Assuming that you are using a pi network at the output, the problems you describe are generally caused (at each band where the problem occurs) by too much inductance.

When you reduce the inductance, you can compensate for the associated reduced Q (i.e. you can bring the effective Q back up) by increasing the loading capacitance.

When you measure the behavior of the pi network, using an MFJ 259, as Shane described, you should chose the value of the resistor that will be temporarily placed from the parallel plates-to-ground to be equal to the RF load impedance you are trying to place across the output of the tubes.

I.e, the value of this resistor should be approximately: 0.5 x the DC plate voltage you are using / the average plate current that will flow when the transmitter is producing its maximum output.

Naturally, all of this is done with the transmitter completely powered off.


Stu

I'll vouch for this!  all the junk and built-but-never-worked-right amps I end up with, it's gold!

Went out and bought myself an MFJ259.  I was able to find good coil tap points for 80m and 40m but trouble on the higher bands.  I found a tap for 20m but it was right at the end of of L1, the tubular coil.  I added a 500pF padder on the output side and that moved the tap back a few turns.  I know you can calculate preset positions for C1 and C2 but I can't find any combination what would work.

Not sure what to do now.  I have a 1500 ohm resistor from the plates to ground (tubes in place).  (2500V / 0.4A * 2) / 2   (two tubes)

Rich

Rich

Are you sure that your numbers are correct?

Maybe I need to look at your earlier posts regarding this amplifier

2500V x .4A = 1000W. This seems very high for the peak electrical input power to a single 813.

If this is a linear amplifier running AM, then the electrical power input, at carrier, to each 813, will be around 2500V x 0.2A = 500W. If the efficiency (at carrier) is 33%, then the plate dissipation, at carrier, for each tube, will be 335W. But the rated maximum plate dissipation (per the RCA specification sheet) is 100W CCS and 125W ICAS.

If this is a plate modulated class C amplifier, and if 2500V is the peak of the modulated voltage (at 100% peak modulation), and 400mA is the peak of the modulated plate current (per tube)... then the electrical power input, at carrier, for each tube (for each tube) will be 250W. At 75% efficiency, the plate dissipation at carrier (for each tube) will be 62.5W... which is okay.

If this is a plate modulated class C amplifier, and if the peak of the modulated voltage (at 100% peak modulation) is 5000V, and 400mA is the peak of the modulated plate current (per tube)... then the electrical power input, at carrier, for each tube, will be 500W. At 75% efficiency, the plate dissipation at carrier (for each tube) will be 125W... which is probably okay. If this is the case, the temporary resistor you are using should have a value of: 0.5 x the unmodulated B+ / the average plate current (for two tubes) at carrier = 0.5 x 2500 / 0.4A = 3125 ohms


Stu

Stu,
Of course I am not sure this is correct, in fact it is very likely to be wrong!  This is a linear for SSB service.  Calculating the plate load resistance is as clear as mud to me.  I see lots of different ideas on how to do it out there...

The 400mA is for two tubes at 200mA each.  Not sure this is correct but seemed like the maximum current, at least during tuning.  Right now my transformer is inadequate, with a 400V sag from 2900V to 2500V under load.

Thank you for the help!

Rich

Rich

If this is a linear amplifier, biased for class B (or slightly into class AB) operation, with a pair of 813's, each running at 200mA of peak plate current (i.e. on an SSB modulation peak)... with 2500VDC plate voltage... then the temporary resistor should have a value of approximately (not exactly):

H x 2500V / (2 x 0.2A) = (for H=0.64) 3979 ohms = (approximately) 3900 ohms

Where the number H is 2/pi = 0.64

You may ask (actually you probably won't) where the number 0.64 comes from. It comes from the fact that, in a class B RF amplifier, the peak value of the fundamental frequency component of the RF plate current, is pi/2 x the average plate current (averaged over each RF cycle). The RF impedance, looking from the tube into the pi network (at the fundamental RF frequency), has to equal the B+/(the amplitude of the fundamental frequency component of the RF plate current)... so that the tube's plate voltage just touches (goes down to) 0 volts in each RF cycle. Therefore the RF impedance, looking from the tube into the pi network (at the fundamental frequency) has to be 2/pi x the B+ / the average plate current (averaged over each RF cycle at peak power output).

Stu

Pete

You might also try disconnecting the plate choke from the plates of the tubes... during this test with the power off.

The plate choke should look like a very small value capacitor at higher RF frequencies... but if it doesn't, it might be messing up the behavior of the pi network.

Stu

Thank you Stu, I will give that a try.

I have tried using the spreadsheet here:

http://www.ifwtech.co.uk/g3sek/pi-l-net.zip

When I enter the parameters for 20m I get a very small value for C1...

I have attached it here.

Rich

Rich

In the spreadsheet, you appear to have entered a "maximum DC anode current of 0.8A"... but, I think you said (in your prior post) that it is 0.2A per tube.

If you make that change (see the attached, revised spreadsheet), the target load impedance (for good output power and reasonable efficiency) will be close to 4000 ohms.

This leads to a reasonable (but small) value of C1 (at 14.2MHz), given all of the tube and stray capacitances you have entered.

Also, all of the above is based on the "perfect" plate choke values you have entered... particularly the value of 0 for parallel capacitance.

Note that the "k" factor in the spreadsheet is: 1/(what I called H in my earlier post)

For ideal class B operation, k= pi/2 = 1.57. Using k=1.7 is fine for class AB linear operation.

Stu

Stu,
  That all makes sense. This is turning into a good learning experience.  I will try to preset C1 and C2 with these values and try again.

Rich

Keep in mind, the self resonant frequency (SRF) of ALL the elements in the PI plate matching network system come into play. So if you start out with real ideal values from the PI or PI-L networks, don't be surprised if they are QUITE off. If you really want to get a handle on how off this can be, put the SRF of the choke, plate blocking cap (the doorknob), the SRF of the tune and load C's into a program like LT Spice. Then run the analysis exactly like the process the others have posted. Fortunately, the required values for the C-L-C for example in the PI net sorta work in your favor as you move from the 80 M band towards 10 M. That is the L and C's  tend to go down in value, moving the SRF of the elements UP a bit. 

Alan

Stu,
When I increase the frequency to say 21.3Mhz the C1 value again goes negative.  Why are the parameters such that C1 is always so small?

Rich

Hi Rich,

I am not familiar or tried the spreadsheet program. Looks like he takes into account some additional stray elements in the design, which is good. I suspect if you assign inappropriate value(s) for these strays, you can wind up with an impossible solution, i.e. minus C.

I did a hand calculation just for the PI match, yes ideal. I assigned a LOADED Q for the PI of 12. This is a typical value and its value affects a number of performance parameters. At 21.3 MHz, use Stu value of 4 K, I get 22pF for the plate tune cap, 2.75 uH for the L, and the output cap, so called, C2 at 200 pF. In the sheet, you should get similar values if you make all the strays ZERO. Start off say, with the ideal case, then add in the strays. Hope this helps.

2.75uh coil 2" dia. 2" long @1/4" spacing 9 turns #10 solid wire.

Rich, Stu,

I had a chance to look at the spread sheet program. Very nice routine and addresses many of the elements we overlook and assume to be zero. Indeed, the guess that the strays are creating a negative C value, say in C1 if I recall, is true. Example, the default value in Cp for the design is 30 or so pF. After calculating the required C for the pi or pi-L and subtracting off Cp, YES, you will get a negative value. Moral, get a grip on your parasitic strays!! I did the case last posted and had to drop Cp down to 10 pF. Then reasonable values result.

Again, a nice SS routine, TNX for the heads up.

Alan 

Rich
Alan

Yes... that is exactly the issue. If the total required tuning capacitance, at 21.3MHz is 25pF*...and if the total parasitic capacitance from plate to ground is 26pF (including the plate-to-ground capacitance of the two tubes)... then you can't make that pi network design work.

*For an antenna/output load impedance of 50 ohms; an input impedance of 4000 (looking into the pi-network from the tube side); and an overall network Q of 12.

There are alternatives.

For example, you can increase the target Q from 12 to 14. Although perhaps non intuitive**... this results in a decrease in the value of the required inductor from 2.41uH to 2.11uH. It also results in an increase in the required loading capacitance from 140pF to 181pF. However, the value of the total required tuning capacitance goes up from 25pF to 29pF... which is now compatible with the 26pF of total parasitic capacitance from plate to ground.

**I.e. the impedance of the inductor goes down, but the effective load resistance (produced by the 50 ohm antenna/output load in parallel with the loading capacitor) goes down even more... resulting in an increase in the Q.

Stu

Thanks guys for your thoughts on this.  I only had 10 minutes to look at it this morning.  I was able to find a tap for 20m but it was at the last 4 turns of the copper tubing coil L1.  Trying the next band, there was no coil left!  If I tapped right at C1 then I could just start to find an SWR dip.

So parasitic capacitances are where I am going to look next.  My layout is pretty funky, necessary as the tubes are in the front. 

The MFJ259 is really becoming handy, I can use it to find pF capacitances at various frequencies.

Rich

Sounds like your fighting Cstray....   And a lot of it.   The 813 has a bit of Cout,  and that also comes into play.

You can wind a small choke to go between Cblock and Ctune.   Using the Cout and Cstray,  this will drop your effective Zout of the tube,  allowing less Ctune for the same Q.

Value of the small choke can be determined experimentally.....   Or you can use the first turn or so of the ten meter coil as the step down as well.

--Shane
KD6VXI


edit:  Fixed muh typos

Thank you Shane, I will try that. 

Rich

Looking at the 63 Handbook, there is the one-band KW amplifiers using a pair of 813's in parallel, grid driven. Nevertheless, a couple of things worthy of note. His PI nets are dedicated to each band. No band switch. On 15 meters his plate tank cap is 35pF and his PI inductor is 1/4 inch copper tube, 2 t.p.i. and 3 inches long. 2.5 inch diameter. Finally, his Cload is just 325 pF. So he has just 6 T of 1/4 copper tubing. One last item to note, is his plate caps. No use of the Millen ceramic plate caps. Instead just the standard clip on plate caps void of ceramic. I have read, the use of the ceramic plate caps adds enough shunt C to be problematic. Since the Dk of ceramic is ~ 10... Maybe so. Not sure I buy that, but might be worthy of note, dispense with the fancy ceramic plate cap.

Shane,
I added a small coil between the blocking cap and the tuning cap. It had the opposite of the intended effect, requiring me to move the 20m even farther towards the end of L1. I put the coil between the blocking cap and both the tune cap and the rest of the tank, perhaps it should go only to Ctune.

I had never heard the ceramic plate caps could cause such additional capacitance, seems counter intuitive.

Rich 

Rich

Can you (re)post a copy of the schematic... or a link to the schematic

Stu

http://www.w7ekb.com/glowbugs/projects/JunkerAmp.pdf

Rich

As a comment on the design shown in the article... I would suggest that L1 be oriented perpendicular to L2. If L1 is oriented parallel to L2, then... when all of the turns of L2 are shorted (end-to-end) by the band switch, a portion of the magnetic flux traveling along the axis of L1 (produced by the RF current passing through L1) will also pass through the axis of L2. This will result in an extra load on the tank circuit that will disturb the behavior of the tank circuit.

Stu

The motivation for the statement is due to surrounding the bare metal of the plate cap with ceramic whose dielectric constant is about 10. So it is not counter intuitive, however it does beg the question, where are the set of metal plates creating such a shunt C on the plate. One is the plate itself, but where is the 2nd plate. Again, I suspect this is not the issue in your case, however if you are trying to chase stray C you begin to question all possibilities. The original article looks straight forward. Rich, you may want to reduce your troubleshooting to the most basic circuit elements. Hence, just consider the PI network only terminated in the total Rp of the tubes alone, 4K I believe. And see if that is straight forward to match to 50 ohms. For a Q of 12-15, it is easy to determine the presetting of the two C's at 21.3MHz. Then attempt to find a tap point on the PI net inductor... if possible, that would be L1 I think.

Along Stu's suggestion, remove L1 all together. Is L2 just to large to begin. The mutual coupling present even after "shorting" all of L1 is effectively still increasing L2. 

It's nice when there is a way to get all this sorted out without stressing the electronic circuits.