Advice for final tank circuit

[W8ACR]

This is related to my other post on the Eimac 254W. I am building a single band 160 meter AM transmitter. The tube lineup is 6AG7 OSC, 6146 BUFFER, 254W final. It will be operated at 250 watts input, mainly because I only have a 125 watt mod transformer. It will run 1400V@180 mA. According to the tube data charts, I guess I'll need about -310V on the grid.

I'd like to hear opinions about how to couple the 6146 to the 254W and I'd especially like to hear opinions about how to design the final plate tank circuit. I'd appreciate comments on what value of Q to use, and I'd like to hear specific values of capacitance and inductance. I have a variety of single section and split stator capacitors to choose from. I also have a nice B&W 160TVL coil (inductance 90uH), but I'm not opposed to winding a homebrew coil.

Thanks, Ron W8ACR

[DMOD]

Okay, just speaking to the 6146 stage for now, I would come off the OSC stage plate with a 1000 pf coupling cap connected to a two resistor junction for the 6146 grid circuit. So its cap, resistor divider/grid. I.E., the 1000 pf cap goes to the junction of a 100 ohm resistor (1/2 watt) and a 68k resistor (1/2 Watt). The other end of the 68k goes to ground, while the 100 ohm resistor goes to the grid, pin 5.

The cathode of the 6146, pins 1,4,6, goes to a 1k ohm 5 Watt resistor and bypassed to ground with a 0.01 uFd cap at the cathode. Then connect to a 5 or 10 ohm 1/2 W resistor and then to ground. You can meter the current going through the tube via the voltage developed across the low value resistor (5 to 10 ohm) and ground. At the junction of the 1k and 5 to 10 ohm resistor, bypass with another 0.01 ufd cap.

Going from the B+ supply toward the plate, place 0.01 ufd at the B+ lead, and then go to a 2.5 mH RF choke (L1) bypassed to ground with 0.01 ufd cap. Then from the choke/cap junction, go to about a 125 uH tuning coil (L2) resonated by 50 pF or 10-125 pF variable cap to ground. Now I prefer to get the DC off the cap by having another coupling cap of 0.01 ufd between the plate/coil junction and the top of the variable cap.

A parasitic suppressor between top of L2 and the plate might be needed, say a 75 ohm two-watt carbon comp resistor shunted with 5 turns of #14 guage wire coiled around the resistor with the windings spread out.

The screen grid resistor from the B+ is determined by your B+; say your B+ for this stage   is 600 volts, then your SG resistor value would be about 50K to 68k at about 5 Watts. Place a 0.01 ufd cap between the SG pin 3 and ground.

Phil - AC0OB

[WA1GFZ]

125 uh and 450 pf what are you trying to do tune up on 600 meters?
1000 ohm cathode resistor? there will be no plate current.

[K9ACT]

I'd like to hear opinions about how to couple the 6146 to the 254W

First of all you need to think about how you are going to neutralize the triode.

You need a balanced plate or balanced grid and there are several options for both.

Balanced plate sort of eliminates the PI net output so you need to think about that a bit.

>and I'd especially like to hear opinions about how to design the final plate tank circuit. I'd appreciate comments on what value of Q to use, and I'd like to hear specific values of capacitance and inductance.

Having recently built two similar rigs and getting lots of help in the process, I think you would get a lot more satisfaction and fun if you worked those things out yourself and then played them by the group to see if it makes sense. Draw up a schematic to the best of your ability and post it for discussion.

The standard Q used for tank circuits is 12.  Every handbook printed has a simple chart where you consider plate voltage and current against the freq and come up with a cap value that produces a  Q of 12.  Plug this into another formula and get the inductance.

There are lots of on line caculators that take the grunting and error out of the calculations which is the biggest problem with the equations for me.

If you want a Pi out put, my favorite Pi calculator is at
http://www.qsl.net/wa2whv/radiocalcs.shtml

Take a look at my radio page for some ideas of how I did it.

Then study any pre- 60s handbook for the basics and the older it is, the more they will discuss triodes.
The more pictures you look at the more comfortable you will be.  All the schematics are pretty much the same.

Triodes are fun.  You are on the right track.

BTW, half the fun is winding your own coils.  In fact, the more you can make yourself, the greater the satisfaction.  I even made my own neut cap and knobs.

Be happy to help you.  I emailed a rough schematic to my elmer when I started on mine and sort of feel like I had something to do with designing it even though he made a lot of changes and filled in a lot of blanks.

Come back soon,

Jack    http://schmidling.com/radio.htm

[DMOD]

125 uh and 450 pf what are you trying to do tune up on 600 meters?
1000 ohm cathode resistor? there will be no plate current.

Thanks for the catch. I should have proof read better but I had to rush off.

The 1k cathode resistor produces a voltage drop of 180 volts for a plate current of 180 mils max. as a cathode (protection) bias in case osc RF exitation ceases. This is for  operation with 600 to 700 volts on the plate.

I have made the changes. One error was a typo and the other was a problem in MATLAB.

Then from the choke/cap junction, go to about a 125 uH tuning coil (L2) resonated by 50 pF or 10-125 pF variable cap to ground.

Just trying to help Ron get started. As always, he will have to experiment and tweak the circuits, which is half the fun of building.

Thanks

Phil - AC0OB

[WA1GFZ]

180 volts on the cathode will cut the tube off big time. you would make more power with a 6AQ5.

[W3RSW]

Don't let the 125 watt mod iron scare you off of running, say 300 watts in the RF amp.  Find a 20H choke or better, a 2 uf HV cap and run Heising modulation. 
That'll allow you to maximize value of your mod iron if you don't push your 'lows'.

Run audio at, say 100 cps gentle roll-off on low end (simple one pole, 6db/octave filter or equiv. coupling caps and don't worry about sounding too yellowie. This will be good for your limited mod. tranny.

If you fully modulated up to 100% you'll sound better than 90% of AM running out there anyway.

Watch the phasing, flip those mod. plate caps around, flip the D104 phasing, etc. (whatever) and you'll get over 100% with very little energy on the highs bothering the mod. tranny.  Few mikes seem to have the phasing differential scrot that the lowly D-104 has.  Fed into a decent 5 meg or better load makes all the difference in flattening the audio bandpass characteristics even though your rolling off the lows.

.. um, changing thoughts and wandering around so I better quit. 

[W8ACR]

Thanks for the replies.

Jack, I have most of the old ARRL handbooks from the 1940's and 50's, as well as a few of the old "Radio Handbooks".  I have read a lot about this subject. But there seems to be a lot of conflicting information. I had intended to use a conventional plate circuit with standard plate neutralizing. A plate voltage of 1400 @ 180mA gives an Rp of about 8000. I was planning a tank circuit Q of 12. According to the 1952 ARRL handbook, this would require a total tank capacitance of about 250pF. A coil of about 27uH would be required to resonate this capacitance at 1900kHz.

I understand that a push pull final would require a tank circuit Q of only 6 and would require only 1/4 of the capacitance of the single section capacitor, or in other words, each section of a split stator capacitor should be 1/2 of the single section value. In my example above, a push pull final would need a split stator capacitor 125pF per section and a coil of about 100uH.

The 1951 Radio handbook has a graph showing values for a Q of 15 instead of 12. This graph shows that a capacitance of 300 would be required for a Q of 15 which makes sense to me and seems to agree with the 1952 ARRL data.

However, the 1946 Radio Handbook has a graph on page 177 that indicates a single section capacitor of 70 pF is needed to achieve a Q of 12 at an Rp of 8000. I cannot reconcile these data. I must be missing something.

Whatever capacitance is correct, I assume that with a single ended final stage, a split stator needs TWICE the capacitance in each section. In my example above, a split stator would need 500pF per side to operate at the same conditions.

My tank circuit as currently designed, is using a National TMC 100D (split stator 100pF per side) with a padder of 50 pF and a B&W 160TVL coil (90 uH). It seems to resonate OK at 1915 kHz - I get a dip down to about 100 mA with the plates of the capacitor about half meshed, but I'm not sure what the Q of this circuit is. The B+ is on both stator and rotor in case that makes a difference (I don't think it does). 100mA is the most plate current I can get at dip which brings me to another problem.

I'm having trouble getting enough drive to the 254W. The driver is a 6146 with 500V on the plate and a screen dropping resistor of 25000 ohms. I get a very nice dip in plate current down to 20mA, which gives me 10 watts input at the driver stage. I need 25 watts output to fully drive the 254W. Even if I remove the screen resistor and run the screen at 500V, I can only get the 6146 plate current up to 25mA at dip. I assume that my problem is in the plate tank of the 6146, but I guess that it may also be a drive problem at the grid of the 6146. The buffer circuit that I am using is based on the circuit used in the Globe King 500B. It has occurred to me that that circuit is designed to drive a 4-400 which only requires about 5 watts of drive power. Perhaps I need a different buffer plate circuit. The tube data sheet calls for a grid current of 40-45 mA. At my current drive level, I am getting 20mA of grid current.

Lastly, I am confused about the value of grid bias needed on the 254W. I have read that for plate modulated AM, one should have bias set at 2X cutoff plus 35%. If I am interpreting the graph correctly, at Eb of 1500V, cutoff for a PL254W (I assume this is the same as an Eimac 254W) is about -50V.    2X50+35%=135. Yet the tube data sheet recommends a grid voltage of -330. This is 6.5X cutoff!!!! I assume that a very negative grid voltage would affect the driving requirements.

There is some good news as well. I have been able to verify that each stage is operating at 1.9 mHz, that is, I'm not tuning any of the stages on a harmonic frequency. Also, the 254W did neutralize easily.

I guess to boil it down to basics, I need more drive to the 254W and I need some confidence that my final tank Q is in the ballpark.

Thanks again, Ron W8ACR

[KM1H]

Most important is what type of antenna will you be using?  The various ARRL and E&E Handbooks have tons of output circuits with values for any type of plate network.  With a Pi net  or shunt fed link coupled you can use much smaller plate spacing in the Tune cap. Id suggest a fairly high output Q for harmonic reduction, you dont want to be bothering 75M and wont be doing big QSY's on 160 AM.

Its much better to have as much drive as possible from the 6146 and control it with a pot in the screen as needed and as the tube ages.

Interstage coupling can take any of the popular forms and with 3 stages you will have plenty of driver stages harmonic suppression. On 160 neutralization may not even be required if you swamp the drive a bit with about 5-10K across the coil and use a series grid resistor of around 20-30 Ohms to also stabilize. Dont use a grid RF choke as they have a bad habit of setting off LF parasitics especially with a low band amp. Use about a 1-2K carbon resistor bypassed at the cold end with a .1uF.

If it does need to be neutralized do it cold with a mica trimmer or small variable. Apply some low drive and resonate the final to get some power indication.  Then kill the HV and put a scope or RF voltmeter at the antenna connector. Apply drive, and adjust the trimmer for a null; dont touch any other controls.  Measure the trimmer value and use a length of solid dielectric RG 58 and trim as needed to get the same C. The center conductor goes to the plate and what remains of the shield to the input. No absolute perfection required

[KM1H]

Thanks for the replies.

Jack, I have most of the old ARRL handbooks from the 1940's and 50's, as well as a few of the old "Radio Handbooks".  I have read a lot about this subject. But there seems to be a lot of conflicting information. I had intended to use a conventional plate circuit with standard plate neutralizing. A plate voltage of 1400 @ 180mA gives an Rp of about 8000. I was planning a tank circuit Q of 12. According to the 1952 ARRL handbook, this would require a total tank capacitance of about 250pF. A coil of about 27uH would be required to resonate this capacitance at 1900kHz.

I understand that a push pull final would require a tank circuit Q of only 6 and would require only 1/4 of the capacitance of the single section capacitor, or in other words, each section of a split stator capacitor should be 1/2 of the single section value. In my example above, a push pull final would need a split stator capacitor 125pF per section and a coil of about 100uH.

The 1951 Radio handbook has a graph showing values for a Q of 15 instead of 12. This graph shows that a capacitance of 300 would be required for a Q of 15 which makes sense to me and seems to agree with the 1952 ARRL data.

That is a good value to use

[/b] Ignore the 1946 info.



However, the 1946 Radio Handbook has a graph on page 177 that indicates a single section capacitor of 70 pF is needed to achieve a Q of 12 at an Rp of 8000. I cannot reconcile these data. I must be missing something.

Whatever capacitance is correct, I assume that with a single ended final stage, a split stator needs TWICE the capacitance in each section. In my example above, a split stator would need 500pF per side to operate at the same conditions.

My tank circuit as currently designed, is using a National TMC 100D (split stator 100pF per side) with a padder of 50 pF and a B&W 160TVL coil (90 uH). It seems to resonate OK at 1915 kHz - I get a dip down to about 100 mA with the plates of the capacitor about half meshed, but I'm not sure what the Q of this circuit is. The B+ is on both stator and rotor in case that makes a difference (I don't think it does). 100mA is the most plate current I can get at dip which brings me to another problem.

I'm having trouble getting enough drive to the 254W. The driver is a 6146 with 500V on the plate and a screen dropping resistor of 25000 ohms. I get a very nice dip in plate current down to 20mA, which gives me 10 watts input at the driver stage. I need 25 watts output to fully drive the 254W. Even if I remove the screen resistor and run the screen at 500V, I can only get the 6146 plate current up to 25mA at dip. I assume that my problem is in the plate tank of the 6146, but I guess that it may also be a drive problem at the grid of the 6146. The buffer circuit that I am using is based on the circuit used in the Globe King 500B. It has occurred to me that that circuit is designed to drive a 4-400 which only requires about 5 watts of drive power. Perhaps I need a different buffer plate circuit. The tube data sheet calls for a grid current of 40-45 mA. At my current drive level, I am getting 20mA of grid current.

Lastly, I am confused about the value of grid bias needed on the 254W. I have read that for plate modulated AM, one should have bias set at 2X cutoff plus 35%. If I am interpreting the graph correctly, at Eb of 1500V, cutoff for a PL254W (I assume this is the same as an Eimac 254W) is about -50V.    2X50+35%=135. Yet the tube data sheet recommends a grid voltage of -330. This is 6.5X cutoff!!!! I assume that a very negative grid voltage would affect the driving requirements.


At that plate voltage -75V will just cutoff the tube. I use 2.5X as the bias for AM so that means ~ -190V.


There is some good news as well. I have been able to verify that each stage is operating at 1.9 mHz, that is, I'm not tuning any of the stages on a harmonic frequency. Also, the 254W did neutralize easily.

I guess to boil it down to basics, I need more drive to the 254W and I need some confidence that my final tank Q is in the ballpark.

Thanks again, Ron W8ACR

[Gito]

hi

the Q factor on an parallel circuit in resonance is depended on the reactance of the C and L,
The higher value of C and lower value of inductance in a parallel  resonance circuit have a higher Q factor,a small C and high value of inductance(L) has a lower Q value.
In a low Q  circuit sometime it's difficult to couple the power output of a buffer to the Final RF tube,and the harmonics is not suppressed /filtered  enough.
In high Q it's more "easy" to couple  the Buffers power to it's final.
To high a Q of a parallel Circuit,there's power loose in the circuit because it's heating the  parallel resonance in the circuit .
We can not choose at will, it depends on the load it works into,the input impedance of the Final tube,the output impedance of the driver tubes

To drive a final RF tube ,we must know the input impedance of it,  in this case 254W final has  about 13 kohm load ,theformula in ARRL handbook driving power watt(25) divided by (d.c grid current x grid d.c current) X 622 X 1000  = about 13 kohm.
So the 6146 must drive a load of 13Kohm ,a full loaded 6146 ( 500 v X .1 A = 50 watt input,an output of 35 watt) has a 5 kohm impedance,so to drive this 254W we can't directly couple it to final tube with C coupling ,but must use a step up coupling ,by using link coupling to the grid circuit of the final tube.
And considering the looses in the circuit coupling ,we must need  a 1,5 to 2  drive to get the needed grid current /grid drive.

So 6146 is not a good choice ,use a higher voltage tube  that can deliver  50 0r 60 watt output,like 813 with a 1000 VDC  if loaded to 80ma has an output impedance of 12.5 kohm that "match" the input impedance of th 254W final ,we can used capacity coupling from the 813 to grid resonant circuit of 254W.

Grid bias,the cut off bias is different with the operating bias.We used cutoff bias to protect our tube ,so when there's no drive the tube does not take excessive current.
The grid current of 254 is about 55ma ,so to get to the operating value (-330 v) we need a grid leak resistance of  (330 - cutoff bias voltage) : grid current.
330 v = (R X 55ma) + cutoff/protective bias.

So We can used different bias as protective bias.
 the operating bias is actually the Grid current X grid leak resistance of the 254W.( if using automatic bias)


Gito

[K9ACT]

So 6146 is not a good choice ,use a higher voltage tube  that can deliver  50 0r 60 watt output,

There is a very easy way around that problem that you won't find in the handbooks but we discussed it here in detail.

The drive requirements of triodes can be varied by as much as 100% depending on how the grid tank is biased and grounded.

If you ground the tuning cap frame you will need about the drive power suggested.  However, if you float the cap, you can get buy with less than half of that.  I am driving a pair of 8000's with about 12 watts.  K9WEK built up the same RF deck but with the cap grounded and requires about 30 watts.

The reason for this is that in one mode, the circuit is re-generative and steals a bit of output power to drive the grid.  The other is de-generative and  adds the grid drive to the output.

If you look at the schematic for my 811 rig,

http://schmidling.com/811_am.png

you see that it is in the degen mode and worked just fine.  This same rig was upgraded to 2 811's and then a single 810.  I found that I did not have enough drive from my 807 exciter and switched to a 6146 and higher plate voltage.  This solved the problem for the 811's but still not enough for the 810.

I then rewired it to the re-gen configuration and now have more than enough drive even for the two 8000's.

I don't think I have a schematic on line and I will post one but you simply ground the center tap of the coil, feed the bias through an RF choke to the grid and float the cap.

This assumes of course, that you go with a balanced grid circuit which also allows a Pi output if desired or you can go with a simple link coupling and eliminate a variable.

js

[K9ACT]

Here is the schematic of the regenerative grid drive arrangement for 8000's rig.

12 Watts in 400 Watts out.  Can't do much better with a tetrode.

You can ignore the core.  I am using a toroid but an air coil works just as well.  It was an experiment that stuck.  I thought isolation between grid and plate would be improved but this is pretty much of a non-issue with triodes.

js

p.s. Would someone please teach the spell checker how to spell triode and tetrode

[KM1H]

So 6146 is not a good choice ,use a higher voltage tube  that can deliver  50 0r 60 watt output,like 813 with a 1000 VDC  if loaded to 80ma has an output impedance of 12.5 kohm that "match" the input impedance of th 254W final ,we can used capacity coupling from the 813 to grid resonant circuit of 254W.

There is absolutely no reason that a 6146 cant be used. As I stated above it pays to look at the H&K 254 tube curves using the link I provided in another thread about the 254. Suggesting an 813 is ludicrous, the 254 has a gain of 25 and we dont need brute force to make it work!

http://scottbecker.net/tube/sheets/114/h/HK254.pdf

Using the curves it shows that approximately -75V is needed for cutoff and applying the accepted 2.5X cutoff rule we wind up with -190V bias for CW. This is not a large value and should be the starting point. Since you know the voltage then design the input circuit with a sufficient ratio to obtain it and with sufficient overhead for losses. Since the actual drive power is the same on CW and AM the only difference is voltage. If insufficient modulation cant be reached at -190V (this is probable) then up the ratio and go for the level needed. You also have plenty of additional range left in the 6146 to help, at the 25W level it is loafing in the 2E26 range.

The old plug in grid coils seldom were sufficient and construction articles often required modifications. A scope will be a big help in monitoring the peaks and ones that cover 160M are about free. It also takes out the guesswork.

Carl
KM1H

[Gito]

Hi

Carl You are miss understanding what I wrote a 254W needs a 25 watt input,
A 6146 at 500vdc can give an 35 watt output,and as we know there's loses in coupling the power to the 254 We must at least have maybe 50 watt power from the driver tube.Even if the gain of the tube is 25,we still ned a 25 watt input.

So in my opinion ,better used a tube that can give You more power about 60 watt power. output.

An 813 is just an example ,look at the voltage I wrote and the plate current of it 1000v X 60ma= 60 watt input or about 42 watt output the point is to get An output impedance that matches  the input impedance of the 234W using this input circuit (using a double section variable capacitor so you can neutralize  the transmitter)
Of course We can use a smaller tube ,that when loaded has the output impedance that "matches" the input impedance of the 254W.
So don't look at the maximal ratings of the tube( a 250 watt output power tube) it's just an example.

A 6146  loaded fully has an output impedance of 5 kohm to give an output of 35 watt.

So in this circuit the input impedance of 254W is about 12 to 13 kohm( there is a formula in the ARRL handbook to find the input impedance of the tube).

If an 6146 has to load into 12 kohm load, it can't be load to it's full capacity,at 12 kohm load 6146 can only give about 15 watt output and the 254w needs a 25 watt input drive.At least in this circuit.
Unless using a matching circuit like pi-section coupling.or step up link coupling,that make a complicated circuit feeding the parallel resonance input circuit of 254W

In my opinion the bias needed in a RF tube is the bias that is produced in the operating condition of the tube
So if the data tube stated We need a 330V min bias  and 4o ma grid current we simply can find the Grid leak resistor of that tube ---330 V:40ma= 8.25 Kohm

So as long we used a 8.25 kOhm grid leak resistor and a 40ma current flowing in it we Have the correct bias and the correct operating point.

If we used fixed bias,the main purpose is to protect the tube from excessive plate Current when there's no drive.

There are 3 different type of bias automatic bias,protective bias and kathode bias if I"m not mistaken.
A tube Using a protective bias ,is a combination of fix bias and automatic bias.
In my opinion ,it's not important how high the fixed bias is as long as it can protect the final tube from excessive plate Current when there' no drive,and gets the operating bias needed as there is a Drive.

Why must we calculate the  min bias needed in operating condition ,since it's clearly stated  in the Data books.(class C telephony or class C telegrafy, Class Ab....)
The point is how must we achieved that conditions .

Gito



 

 

[K9ACT]

Hi

Carl You are miss understanding what I wrote a 254W needs a 25 watt input,
A 6146 at 500vdc can give an 35 watt output,and as we know there's loses in coupling the power to the 254 We must at least have maybe 50 watt power from the driver tube.

I think those fudge factors are nice if you can afford it but in the real world, if the tube data says 25W, that's what you need.

How can you possibly lose 50%  between a 6146 and the grid?

It's inches away.  Do you lose 50% between your final and the antenna? I hope not.

If 10% feels good then 30W will certainly do the job and probably, so will 25W.


js

[KM1H]

And I fail to understand where 500V on the 6146 came from, it is rated up to 750V and an easy 50-60W out. The step up required to drive the 254 grid is simple math and poor efficiency is simply poor design or construction. Its the basic conservation of energy law. Convert power to voltage and not heat.

From reading the full H&K spec sheet it is obvious that the 25W drive is actual power into a real circuit; not theory plus losses TBD later.

There is nothing complicated about a step-up double tuned transformer circuit which Ive been suggesting all along, 2 coils and 2 variables. No links necessary, its a simple transformer. This could even be handled by a toroid for maximum efficiency.

Carl
KM1H

[W8ACR]

Thanks again for all the comments. Carl, here is my plan to remedy the situation.

1. Reread the section of the ARRL handbook dealing with driving power amps.

2. Raise the plate voltage on the 6146.

3. Couple the 6146 to the 254W inductively with tuned circuits at both the plate and grid.

I'll update this site with results when available

73 Ron W8ACR

[Gito]

Hi

Why 500 volt DC, because Ron said so in his article(used 500v at the plate)
With a different plate voltage ,it's a different story.

But still You have to used double tuned link coupling circuit that means needing two variable capacitor two coil. Tuning this two coupling circuit,which interact each other,but of course it can be done

Compare it to using a single coupling capacitor and single tuned Grid circuit .
which one is easier to adapt and to tune.  

Please read the ARRL handbook" feeding excitation to the Grid.....".The tube tables give approximate figures for the grid driving power required for each various operation conditions.these figures,HOWEVER,do not include  CIRCUIT LOSES.In general,the driver for class C AMPLIFIER should be CAPABLE  of supplying AT LEAST THREE TIMES the driving power  shown on typical operating conditions.....

You can find this article in the ARRL handbook


Gito

[N2DTS]

Three times the driving power seems crazy to me.

While grids do take some power, I thought the voltage swing was important also, and if the grid tuning and coupling  circuit is not right, you need a lot more power for no reason.

I would NOT want to have to tune two controls to peak grid drive, and think you could come up with something to give the voltage swing, even if that is increasing the 6146 plate voltage (and run less current).

Brett

[KM1H]

We are dealing with a triode Brett which may require neutralization. Since the usual configuration between a transmitter an amplifier is via 2 tuned circuits and 2 links with 2 to 4 variables Im suggesting less complexity since it will all be on one chassis. Since this amp will only be used on 160M AM it will be likely that once set the caps wont need adjustment again.

Monitor the 6146 screen voltage and current and use a WW pot to adjust the voltage. At 750V on the plate it may be a little much dissipation for long winded transmissions so find a happy balance between Ep, Ip, Esg, and Isg.

You also mentioned possible low drive to the 6146. Id solve that first. Will the 6AG7 be crystal controlled or driven by a VFO?  If you have big crystals a 6L6 could be used altho a 6146 certainly doesnt need much drive. I use a 5763 driving one right thru 10M and have to back it down at times. Is your 6146 a known good tube? Do you have a scope to check levels all the way thru?

Carl
KM1H

[K9ACT]

>But still You have to used double tuned link coupling circuit that means needing two variable capacitor two coil. Tuning this two coupling circuit,which interact each other,but of course it can be done

Not sure what Ron is doing as he has another thread going where he is discussing an existing transmitter.

Aside from that, one would assume that the driver is on a separate chassis and these sort of exciters are typically Pi output or simple link tank with a coax leading to a grid tank circuit.  What scares you about two tank circuits?

>Compare it to using a single coupling capacitor and single tuned Grid circuit .
which one is easier to adapt and to tune.

A schematic would help here as that sentence does not produce circuits in my brain.  How does a coupling capacitor tune the plate of the 6146?

>Please read the ARRL handbook" feeding excitation to the Grid.....".The tube tables give approximate figures for the grid driving power required for each various operation conditions.these figures,HOWEVER,do not include  CIRCUIT LOSES.In general,the driver for class C AMPLIFIER should be CAPABLE  of supplying AT LEAST THREE TIMES the driving power  shown on typical operating conditions.....

I hate to state the obvious, but one need not believe everything one reads even in an ARRL publication.  That particular statement is utter rubbish.

First it was 2X, now it's 3X.  I am still waiting for someone to tell me where all this power is going.  How can it possibly take 75W to get 25W a few inches to a grid in a transmitter?

js

[DMOD]

3. Couple the 6146 to the 254W inductively with tuned circuits at both the plate and grid.

I don't see why this is needed. The 6146 circuit description I gave you earlier, with only one tuned circuit at the 6146 plate, was used to drive two 4-400's in parallel. Yes, I know the 4-400's are tetrodes but still it drove those grids in parallel (through parasitic supressors) without complaining.

The OSC plate had a 2.5 mH choke to B+ with a 1000 pf cap going from the OSC plate to the grid circuit of the 6146.

It would be helpful if you showed us your propsed or current driver circuit.

Phil - AC0OB

[Gito]

Hi

Don't blame me.It's in the ARRL books the"holy book" of radio amateurs it's not me who wrote "needs three times .......),if You doubt it please complain to the ARRL devision.
But for me I believed it .

The 254 W needs 25 watt drive  for a 330 watt output right ,
The 833 tube needs 20 watt drive  to get a 800 watt output,logically if 6146 can drive a 245W,than it can drive 833 easily  ,that needs a 20 watt input,smaller than 25 watt ,right.
But I never find a 833 transmitter with a 6146 driver.
So don't look at the output, look at the input needed to the drive the RF tube.

A 4400 needs only 3,5 watt drive per tube ,and a 12ma grid current,that means two  4400 needs a 7 watt and 24 ma grid current using the formula in the ARRL  books the input impedance  is about 7500 Ohm.(CCS operations)

So how much power can a 6146  deliver   to a 7500 ohm-----750 : 100 ma= 7500hm,
So a 750 plate volt loaded to 100 ma gives exactly  7500 ohm.

How much power can we get from 750 loaded to 100 ma....75 watt input or about 50 watt output that means 7 times the power needed to drive two 4400

That's why a 6146 can easily drive a pair of 4400.and needs only one tuning circuit.

using one coupling  capacitor means the coupling from the driver tube to the parallel resonance circuit (one tuning circuit)at the control  grid of the 254W.


Gito

[Gito]

Hi

another article taken from RSGB(RADIO SOCIETY OF GREAT BRITAIN).The radio communication handbook.
It's only apart of an article.That told the driver power needed.

Gito