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Title: Advice for final tank circuit Post by: W8ACR on January 05, 2010, 08:52:46 PM 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 Title: Re: Advice for final tank circuit Post by: DMOD on January 05, 2010, 10:12:31 PM 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 Title: Re: Advice for final tank circuit Post by: WA1GFZ on January 05, 2010, 10:58:41 PM 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. Title: Re: Advice for final tank circuit Post by: K9ACT on January 06, 2010, 12:03:37 AM 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 Title: Re: Advice for final tank circuit Post by: DMOD on January 06, 2010, 12:59:31 AM Quote 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. Quote 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 Title: Re: Advice for final tank circuit Post by: WA1GFZ on January 06, 2010, 10:36:06 AM 180 volts on the cathode will cut the tube off big time. you would make more power with a 6AQ5.
Title: Re: Advice for final tank circuit Post by: W3RSW on January 06, 2010, 11:07:42 AM 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. ;D Title: Re: Advice for final tank circuit Post by: W8ACR on January 06, 2010, 11:52:35 AM 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 Title: Re: Advice for final tank circuit Post by: KM1H on January 06, 2010, 11:55:12 AM 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 Title: Re: Advice for final tank circuit Post by: KM1H on January 06, 2010, 12:09:58 PM 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. Quote 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 Title: Re: Advice for final tank circuit Post by: Gito on January 07, 2010, 07:43:44 AM 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 Title: Re: Advice for final tank circuit Post by: K9ACT on January 07, 2010, 08:48:20 AM 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 Title: Re: Advice for final tank circuit Post by: K9ACT on January 07, 2010, 10:01:19 AM 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 Title: Re: Advice for final tank circuit Post by: KM1H on January 07, 2010, 11:36:59 AM Quote 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 Title: Re: Advice for final tank circuit Post by: Gito on January 07, 2010, 07:21:22 PM 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 Title: Re: Advice for final tank circuit Post by: K9ACT on January 07, 2010, 07:45:06 PM 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 Title: Re: Advice for final tank circuit Post by: KM1H on January 07, 2010, 09:50:21 PM 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 Title: Re: Advice for final tank circuit Post by: W8ACR on January 07, 2010, 10:07:24 PM 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 Title: Re: Advice for final tank circuit Post by: Gito on January 08, 2010, 01:54:33 AM 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 Title: Re: Advice for final tank circuit Post by: N2DTS on January 08, 2010, 09:05:17 AM 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 Title: Re: Advice for final tank circuit Post by: KM1H on January 08, 2010, 10:39:31 AM 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 Title: Re: Advice for final tank circuit Post by: K9ACT on January 08, 2010, 07:39:10 PM >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 Title: Re: Advice for final tank circuit Post by: DMOD on January 08, 2010, 11:19:21 PM Quote 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 Title: Re: Advice for final tank circuit Post by: Gito on January 09, 2010, 05:00:55 AM 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 Title: Re: Advice for final tank circuit Post by: Gito on January 09, 2010, 06:59:48 AM 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 Title: Re: Advice for final tank circuit Post by: K9ACT on January 09, 2010, 09:13:20 AM All I am suggesting is that one can not flatly accept everything we we read. Too frequently, someone makes a statement which may very well be true in his case and it gets repeated and passed on for generations like the Ten Commandments. This is surely such a case.
A 6146 is not a high power triode. Producing .6 W vs .2 W with a tube is trivial and costs nothing so perhaps not a bad idea. Producing 75W vs 25W is not trivial and is a major design issue. So, perhaps it is time to re-evaluate that rule of thumb. Let's find out what is really needed and not force people to over complicate a project because someone said so 50 years ago. My experience with power triodes is that the grid drive requirements are just about what the tube data suggests. This includes 811, 812, 8000 and 810. The only rule of thumb I could come up with would be to follow the tube data. js Title: Re: Advice for final tank circuit Post by: KM1H on January 09, 2010, 10:51:18 AM Yep, JS pretty much said it.
Those of us that have spend years or decades actually building things tend to look warily at old published info. The ARRL repeated 30's designs almost into the 60's and never bothered to revaluate them with actual test equipment. These days it is so easy for even the individual to have test equipment that was affordable only by commercial labs in 1960. Ive been assuming that Rons project is all on one chassis, compact, and with minimial lead lengths. If not then my earlier statement of 2 sets of coils with links and up to 4 variable caps will hold. It never hurts to design the driver for its maximum efficiency as its easier to reduce drive than create it out of the air. The comment about the 833A is interesting also. If a Viking II with a pair of 6146's can drive a pair of 833A's to full power than one MUST assume that a single 6146 can drive a tiny little 254 almost out of its socket. Just dont expect as a high efficiency at 1400V compared to 3000V. Carl KM1H Title: Re: Advice for final tank circuit Post by: W8ACR on January 09, 2010, 02:23:29 PM Hello all,
I guess it's time to describe my project in more detail. Years ago, I acquired a GK500 that was in very poor shape. The PS and MOD decks were shot (they had actually been under water!!). The RF deck was mostly intact but a few key parts were missing. I was able to build usable PS and MOD decks with parts on hand, but they were significantly different from the original design. I used a UTC S22 mod transformer which did not fit on the original MOD deck. I therefore had to add a fourth deck - one just for the S22. It was all a quick and dirty hack job, and quite ugly, but the rig actually worked half decent, and It was used on the air for the past ten years or so without any major problems. I have always wanted to do a better restoration, and in particular, I wanted to get the mod transformer back on the MOD deck. I recently acquired a 125W mod transformer that will fit on the MOD deck, so it was time to do the project. I tore down the RF deck to nothing but sockets and transformers and rebuilt the OSC and BUF stages according to the original GK500B schematic. I decided I wanted to redo the final stage with a tube more suited to the new lower power level. Obviously, a 4-125, 4-65 or 813 would have been more suited to the original circuit, but I didn't have any of these on hand, so I chose the 254W since it was about the right size for the power level and it presented the challenge of designing a new circuit. The RF deck has been rebuilt, and the OSC and BUF stages seem to be working properly. For simplicity, I left out the bandswitching ciruitry and decided to make it a single band 160 meter rig. Plate and grid voltages on the 6146 are correct, but it dips down to 10 mA at resonance and I can't get it to load up any more than this. If I tune it off resonance, say up to about 25 mA, then the 254W grid current will rise to about 20 mA, but that's as far as it will go. I was hoping to use the simple capacitance coupling of the original design, but no go. It seems to be acting like an impedance mismatch, and that is why I thought adding a tuned circuit at the 254W grid might solve the problem. Once I solve the 6146-254W coupling problem, then I'll tackle the 254W plate tank issue which should not be difficult. I have at my disposal a military voltmeter capable of reading up to 5000VDC, an O-scope, a frequecy counter, a signal generator, and an HP VTVM. I could draw and post a schematic, but think of it this way, ahead of the 254W it is a GK500B in the OSC and BUF stages, and from there on it is a plate neutralized triode amplifier with conventional plate tank and variable link output to the antenna. Hope this helps you to understand what I am trying to do. Thanks for you comments and suggestions. 73, Ron W8ACR Title: Re: Advice for final tank circuit Post by: N3DRB The Derb on January 09, 2010, 06:20:42 PM Ron,
try some link coupling. I always had more success at doing that. It's inherently more flexible. Title: Re: Advice for final tank circuit Post by: K9ACT on January 09, 2010, 08:07:05 PM I could draw and post a schematic, but think of it this way, ahead of the 254W it is a GK500B in the OSC and BUF stages, and from there on it is a plate neutralized triode amplifier with conventional plate tank and variable link output to the antenna. First of all, a tuned grid on your PA is not an option. Not even ARRL handbooks build triode amps that way. Secondly, words are not a substitute for a schematic. I for one, am getting tired of words. How can you, (let alone us at the other end of your keyboard), possibly understand what you are doing without making/seeing a schematic? js Title: Re: Advice for final tank circuit Post by: W8ACR on January 09, 2010, 08:42:01 PM Jack,
Sorry for being lazy, I'll post a schematic later tonight. Thank you for your input. I do have several schematics that I am working from, it's not just in my head. I think I used the wrong term when I spoke of a tuned grid. I simply meant that I needed to have a grid tank, just as you have in your 8000 rig. The Globe King 500 has no grid tank, it feeds the grid of the 4-400 directly from the plate tank of the 6146 via a coupling capacitor. I was hoping to do the same with the triode, but I guess it won't work. I will post the schematic of the GK500 RF deck as well as the circuit that I was trying to use. If you want to look up the GK500 schematic it is on the BAMA site at edebris.com. 73, Ron W8ACR Title: Re: Advice for final tank circuit Post by: Gito on January 09, 2010, 09:38:15 PM Hi
I think the RSGB and ARRL hand bpook is a "profetional" hand book and accepted world wide and the article it wrote must come from theory and practise,they must have experiments and building transmitter to find what they wrote.It's not the made of one man,but of many man experience. So I believed it. Just one thing that we must miss, The power dissipated in the grid leak resistor. If we used an automatic bias ,than the power turn as heat is R X I (grid current) and if i'm not wrong the data book does'not include it on the driving power needed. For me designing a transmitter first find the tube with the power output We need,find the input impedance of the tube,if it possibe find a driving tube that's capable with 3 times yhe power needed to drive the Final tube,find a tube with enough HV so when loaded gets a output impedance like the input impedance of the final tube,so we need only one resonance tune circuit at the Grid circuit of the final Amp. I've made several transmitter,one is a 833 final drive with two 807 at 500 v,using a pi section to match the 833,I must load it (807) to it's limit to drive the 833, It' all home made maybe a bad design? But it's running for 3 years without any complain. Gito Title: Re: Advice for final tank circuit Post by: Gito on January 09, 2010, 10:02:22 PM Hi
another transmitter I build three 833 .3000 V HV loaded to 900 ma that's an input of 2700 watt,with a 813 driver 2000 v on plate. All homemade and it.s been running for three and a half year. As it said before maybe not a good design. Gito Title: Re: Advice for final tank circuit Post by: K9ACT on January 10, 2010, 12:03:46 AM Kudos to Gito.
Great looking stuff. If it works it's a good design in my book. Thanks for sharing these pics, Jack Title: Re: Advice for final tank circuit Post by: K9ACT on January 10, 2010, 12:19:04 AM One other "detail".... I never made the connection between GK500 and a Globe King 500 and assumed it was some exotic old rig with some weird tube in it.
My Elmer K9WEK/Dan replaced the RF deck in his GK500 with a pair of 8000's pretty much like mine and can be of a lot more help on your project. I have asked him to join in this discussion. Jack Title: Re: Advice for final tank circuit Post by: W8ACR on January 10, 2010, 02:10:08 AM OK Jack,
Thanks, that would be great if Dan could join in on the discussion. I am attaching three schematics with this post. First is the Globe King 500 schematic. Next is a schematic from the 1952 ARRL handbook showing a triode final amplifier fed with simple capacitative coupling to the driver. Next is a hand drawn schematic of the circuit I built and which did not work. Ron Title: Re: Advice for final tank circuit Post by: W8ACR on January 10, 2010, 02:15:33 AM Here are three more hand drawn schematics of proposed fixes to my drive problem.
Ron Title: Re: Advice for final tank circuit Post by: KX5JT on January 10, 2010, 02:23:26 AM It was easy to turn this monitor on it's side here at work, it's a flat panel LCD but working the mouse when the cursor moves in different directions is another story!
Title: Re: Advice for final tank circuit Post by: K9ACT on January 10, 2010, 10:14:50 AM OK Jack, Thanks, that would be great if Dan could join in on the discussion. In the mean time, famine to feast. My knee jerk reaction was to suggest that it could be a triode/tetrode thing but after finally getting that schematic right side up, all I can do is again wonder if the Handbooks are invincible. I looked through a number of old ones and could not find a triode without a tuned grid. I am curious about RFC5. Unless it's a parasitic choke, there would be no RF on the grid. The GK500 is a tetrode and plays by different rules as it obviously works. At this point I will defer to Dan. Jack Title: Re: Advice for final tank circuit Post by: k9wek on January 10, 2010, 12:28:27 PM Hi Ron,
Hi Ron, I think you have gotten some pretty good advice from the guys on this forum. Start with the tube data sheet as has been suggested. I'm not familiar with the 254W tube, but I'll make some general suggestions that should apply to any tube amplifier. Since you are running the 254W at reduced plate voltage, if you reduce the plate current proportionally you know it will work because this is what happens during modulation. However, the power input (and output) may be reduced more than you'd like. You will not need as much grid current, or power, to drive the tube under these conditions. Decide on what level you want to operate the tube and assume a grid current and driving power to begin your calculations. In general, the grid current should be kept proportional to the plate current given in the tube data sheet. If several operating conditions are given you can usually interpolate between them. Now you need to design the grid circuit. Calculate the input impedance (resistance) from the formula in all the ARRL Handbooks. Input impedance (ohms) = driving power (watts) x 622,000/(grid current)^2. Now, for a Q of 12 calculate (or use the chart) for the tuned circuit values. For a simple tuned circuit (single ended, not balanced) the reactance of L and C will be 1/12 of this value. There is some confusion on how to go from the single-ended values from the chart to values for a push-pull (balanced) circuit. The reason you can use half of the C value in each section is because the grid load is across only half of the tuned circuit, and therefore an impedance of 4 times that value is reflected across the whole tuned circuit. This makes L be 4 times as big and C be 1/4 as big. This accounts for why C of each half can be half the value for the single ended circuit. This factor of 4 also applies to the L and C values for the plate circuit when you go from single ended to balanced. Remember that the load impedance (resistance) that you should use to calculate the plate circuit values should be 1/2 the DC plate resistance. The charts already have taken this into account. I would throw out option 1 and option 2 as they don't have any advantage over the original circuit. The original circuit (balanced plate circuit) is the one recommended by all the old timers, and should be the simplest to neutralize. In order to use a pi network output circuit you need to use a balance grid circuit in order to get out of phase voltage for neutralization. Option 3 is what Jack is running. You can also bring the bias to the center tap of the inductor with a bypass capacitor to ground there instead of the coupling capacitor and RF choke. I prefer grounding the center tap of the capacitor (the degenerative circuit that takes more drive, but the drive apparently shows up in the output) because it has proven to work and be more stable for me. In any case, remember to use an RF choke if there is RF at that point in the circuit. This is not shown on your schematics. Also, only ground (for RF) one center tap of either the coil or capacitor on either the plate or grid tuned circuit. I won't take sides on the debate about how much extra power your driver should be capable of. Once you have your amplifier working, you can determine what voltage and current you will have to operate the driver at. Link coupling between the driver and final has advantages and disadvantages. With 2 tuned circuits there is likely to be more loss and more adjustments to make to change frequency. In some cases you can't get practical L and C values for a Q of 12 in the grid circuit, so you will use more capacitance and higher Q which will waste driver power. You have to get the amplifier working and then design the driver. It would be nice to have lots of extra driving power so that you can experiment. You may want to link couple a separate transmitter (one or two 6146's?) to the grid of the final during development. I hope all this is relavent and gives you some ideas. Dan K9WEK Title: Re: Advice for final tank circuit Post by: W8ACR on January 10, 2010, 07:54:03 PM Hello Dan,
Thank you for your input. Here are the parameters for the grid driving circuit of the 254W tube as taken from the tube data sheet. I am extrapolating somewhat for a plate voltage of 1400V. Plate Voltage 1400V Plate current 180mA Grid Voltage -330V Grid current 45mA Driving Power 24Watt Platew power input 252Watt Grid input resistance is therefore 24/2025X622000=7371 Ohm If I use a single section capacitor, the proper values for a Q of 12 would be 140pF and 50uH. If I use a split stator capacitor of 70 pF each side, that is a total capacitance of 35pF. This would require an inductance of 200uH for resonance. Is that correct? That seems like a pretty large value of inductance. I do want to use the split stator capacitor. Do my calculations sound right to you? Thanks, Ron W8ACR Title: Re: Advice for final tank circuit Post by: k9wek on January 11, 2010, 08:34:22 AM Ron, Yes, your calculations are right. You can use a smaller coil (meaning lower L) and bigger cap, but that would mean a higher circuit Q than 12. This especially happens on the higher bands where you can't get the cap small enough because of tube capacitance. Shooting for those calculated values gives a Q high enough for good grid voltage swing flywheel action) into the grid current region. This is necessary for best efficiency as you can see from the tube curves. If the grid waveform flattens out on peaks, the tube won't conduct as well. A higher Q wastes more driver power heating up the coil. Its a compromise. The Q of the coil (maybe 200) needs to be much greater than the Q of the circuit (12 or more) or its losses become more significant. Therefore use a physically large coil for best Q. Dan K9WEK Title: Re: Advice for final tank circuit Post by: N2DTS on January 11, 2010, 09:27:00 AM I think I would try option one, that looks like it would work well to me.
I don't think the Q is critical in the grid stage, you can likely fudge around with that if the coil gets too big. Triodes are a LOT different in the drive catagory from tetrodes! My 2X813 rig takes very little drive for full power out, and I run it way into class C by increasing the grid leak resistor, and it still takes much less drive power then the pp812a rig did. The 812a rig took twice the drive I think, for half the power out, and the 812a is a wimpy triode. The 833 is a monster triode, so maybe using an 813 to drive it is not unreasonable if you want long life. Still, the 813 is good for 300 watts of carrier, which seems excessive to me....300 watts to drive an 833 or even two seems excessive, even for ccs broadcast ratings.... Brett Title: Re: Advice for final tank circuit Post by: K9ACT on January 11, 2010, 10:13:11 AM I am sort of running Option 3 but with a Pi output tank.
Perhaps it's just familiarity but it certainly looks simpler and cleaner than the other two AND as mentioned, it only requires about half the drive specified in the tube date because of the regeneration effects. Having said all that, it has always puzzled me why one needs two tank circuits on top of each other so to speak. Either one will tune grid and following plate. Discussing this with Dan yesterday, it seems to have more to do with physical location than any absolute technical need. If the driver is on the same chassis, inches away, it presents an entirely different scenario than if they are far apart connected with coax. The Globe King in one example that works with a single tank circuit. The problem seems to be impedance matching and transmitting high voltages over a coax. I am just paraphrasing Dan and perhaps oversimplifying things but it made sense when he suggested this. Perhaps he has given more thought to this or someone else has some ideas on the subject. Jack Title: Re: Advice for final tank circuit Post by: k4kyv on January 11, 2010, 11:45:25 AM ...it has always puzzled me why one needs two tank circuits on top of each other so to speak. Either one will tune grid and following plate. Discussing this with Dan yesterday, it seems to have more to do with physical location than any absolute technical need. If the driver is on the same chassis, inches away, it presents an entirely different scenario than if they are far apart connected with coax. The Globe King in one example that works with a single tank circuit. The problem seems to be impedance matching and transmitting high voltages over a coax. One of the advantages of using two separate tank circuits with link coupling between them is that each stage can have a common grounding point for all the components in that stage, allowing each stage to "float" relative to other stages in the transmitter. There can be differences in rf potential from one point on the chassis to another just a few inches apart, not to mention what can exist between two separate chassis bases in a rack mounted rig. These rf potential differences can induce regeneration if every component in each stage is not grounded to a single point. Link coupling isolates each stage from the others. With capacitive coupling, the rf potential differences between points on the chassis cannot be isolated, since the rf ground return from the capacitive coupling is via the chassis or ground bus. So with one tank circuit and capacitive coupling, the stages are more likely to be flaky. I have had that exact problem with the BC1-T on 160m. Gates just grounded everything to the nearest convenient point on the chassis or cabinet frame, and I noticed some flakiness in tuning the original configuration on 1230 kHz into a dummy load. There was even more after the 160m conversion. I re-routed the grounds in the rf driver and final stages to use single grounding points as much as possible, but with components mounted on three levels, at the driver stages, 833A's, and the upper deck with PA tank components, it was not practical to find one single grounding point for the entire driver-PA circuitry without a spider web of ground wires. Re-routing the grounds helped, but there is still some flakiness in the driver tuning. For example, if I don't dip the 807 driver plate current exactly, the grid drive to the 833s may increase according to the grid current meter, but the 833As run orange hot and rf output drops off. This tells me that I still have a parasitic oscillation somewhere, and that the extra grid drive is at some frequency other than the operating frequency. I need to check the waveform of the rf drive to the grids and see if I can find exactly what is going on. I suspect it has something to do with rf potential differences between the grounding points in the driver stage and the final. Another advantage of two separate link-coupled stages is that one of the links can be made variable, which makes it very easy to adjust the grid drive to the following stage. Of course, link coupling leaves you with the disadvantage of having to adjust two separate tuned circuits between the stages. One way around that problem is to use "unity coupling" between the stages. The plate coil is a tuned circuit in the conventional sense, but the grid coil is wound on the same form as the plate coil, preferably interleaved between turns of the latter. It is not tuned, but the tight inductive coupling between coils allows it to function just like a single tapped coil, and yet still provide isolation between stages that would not be possible with capacitive coupling. Title: Re: Advice for final tank circuit Post by: KM1H on January 11, 2010, 12:06:32 PM The 1952 HB amp is OK but Id either shunt RFC-4 with about 3K or replace it entirely with a resistor. C-11 should be at least a .02. RFC-5 is fine, it is simply a parasitic choke resonant way above 160M. A low value carbon resistor does the same as well as may eliminate the need for neutralizing; Ive mentioned these things before.
Lets skip Circuit 1, there are several problems with it. As for Options 1, etc you certainly like to eliminate bypass caps! Since it has been established that its all on one chassis I see only one need for multiple circuits and link coupling. That would be for stagger tuned broadbanding since you have power to burn in a properly functioning 6146 driver. The same can be done with two coupled coils as mentioned a few times earlier. In either case a Q lower than 12 is desirable. What has not been answered is the condition of your tubes as well as the amount of drive to the 6146 grid. Until that has been answered we are laboring with too many unknowns. Carl KM1H Title: Re: Advice for final tank circuit Post by: W8ACR on January 11, 2010, 03:06:13 PM Thanks again for all the comments and suggestions. It is painfully obvious that I have a lot to learn here.
The 6146 is a good tube, and the 6AG7 seems to be working just fine with a nice dip to about 10 mA at resonance. Plate, grid and screen voltages are correct on all tubes as well. If I use inductive coupling to drive the 254W, should I use twin lead, coax, or a single wire with the other side to chassis for the coupling lead? Also, should I also use a Q of 12 to design the plate tank of the 6146? The tuning cap of the 6146 is grounded on the rotor in the original design, so I'll have to redo the plate tank on the 6146 if I want to use inductive coupling. Thanks, Ron Title: Re: Advice for final tank circuit Post by: K9ACT on January 11, 2010, 06:48:34 PM >The tuning cap of the 6146 is grounded on the rotor in the original design, so I'll have to redo the plate tank on the 6146 if I want to use inductive coupling. Another interesting situation here. We think of inductive coupling as a ground isolation trick but in most cases, the link goes to a coax connector mounted on the chassis. Then a coax to the other chassis and another connector and piece of coax to the next link. There are all sorts of theories about which end of which coax to ground, adding even more confusion to the subject. If you use a link to twisted pair and a receiving link, there is no ground connection at all and no need to even think about how the driver is tuned. To confuse things further, I am using a Pi tank in the driver to a link in the power amp grid with coax between. Again, where to ground the coax? Does it really matter? js Title: Re: Advice for final tank circuit Post by: WU2D on January 11, 2010, 07:57:36 PM Have you decided on Rice (grid) as shown in the original schematic or Hazeltine (plate) neutralization? You can use the Rice on either a link or Pi output and it has many advantages in implementation and power rating of components, but the Hazeltine is a little better system according to many handbooks and most BIG RIGS use that.
Don't forget to think about the driver - it must be neutralized especially if you go with a bottle rocket like the 6146. Mike WU2D Title: Re: Advice for final tank circuit Post by: Gito on January 11, 2010, 08:15:14 PM Hi Bret.
why using a 813,because after the 807, we got 6146 and to jump to 813, so as I used the 3 to 1 formula driver wattage ,if i used a 807 or 6146 ,I need 2 807 aor 2 6146, when using two tubes in parallel ,sometimes one of the tubes goes weaker,so I must replaced /find another tube to have the same power,it happens to the two 807 in my 833 transmitter ,which I regret later. So to over come it I used the only option I have a single 813,of course not with it full rating,you can used a 1500 v supply load it to 120 ma giving a 180 watt input or 126 watt output, that's about 3 times the drive power needed for two 833.(40watt) . so follow the leader (ARRL) of course any body can have different opinion ,but I only follow the Leader. Sometimes,we over looked the parameter data of a tube. Like the 254W a "small" tube and small output power comparing a "monster" tube like 833 tube. The input drive of 254W tube needs is 25 watt drive power . The input drive of 833 tube is 20 watt drive power. So, You can clearly see which tube needs a higher drive ,even 833 gives a higher output power compare to 254W ,25 watt is 25 watt and 20 watt is 20 watt right. Ron attached a schematic diagram of a 175 watt transmitter on 160 meter band from the ARRL hand book it used a 6L6 as a driver that is capable of 20 t0 30 watt output (Class C) an a 812 final that needs only 6 watt drive power.So the driver is more than 3 times the 812 needs Ron You can used this circuit but with a pi-section coupling between the driver and the final,but of course with different values that;s needed Because with pi-section ,you can match and load easily the output impedance of the driver and the input impedance of the final and the power needed. I re attached the circuit of the 175 watt transmitter in upright position. Gito Title: Re: Advice for final tank circuit Post by: k4kyv on January 11, 2010, 08:16:36 PM We think of inductive coupling as a ground isolation trick but in most cases, the link goes to a coax connector mounted on the chassis. Then a coax to the other chassis and another connector and piece of coax to the next link. There are all sorts of theories about which end of which coax to ground, adding even more confusion to the subject. But with link coupling, there still is isolation between tank circuit and coupling coil. It is not the grounded coax braid that causes the problems, but the rf return path via the chassis that is required for capacity-coupling between stages. But there is no need to ground the coax braid, or even use coax for that matter. I don't use coax for any of my links. I have had good result with zip cord, sometimes with the conductors separated. The pair doesn't have to be twisted. One trick I have used is to strip off the outer jacket and braid from scrap coax and use the inner conductor as the insulated wires for the link. I just taped them together or used tie wraps, on the assumption that the insulation used to make coax might be a better dielectric than plain old hookup wire. But in any case, if the stages are only a few inches to a couple of feet apart, almost any kind of insulated wire will produce negligible losses in the link. I use stranded wire in the case of variable links because it is more flexible. Quote If you use a link to twisted pair and a receiving link, there is no ground connection at all and no need to even think about how the driver is tuned. That's the best way to do it. Title: Re: Advice for final tank circuit Post by: Gito on January 14, 2010, 03:27:43 AM Hi
Because I thing it is important to know ,the drive needed(drive power) for RF tube, I found it on The RCA transmitting Tubes hand book(TT4) page 32 and 33 ,the bottom line is..." They can be estimated with reasonable accuracy for straight -through amplifiers..At frequency up to about 30 megacycles per second,total tube and circuit LOSSES are approximately TWICE the DRIVING POWER..." (data book) That means the driving power needed = Losses( twice the driving power/data book) + driving power (data book)= 3 times driving power in the technical data book. Since it is written in The RCA Transmitting tubes(the maker/producer of transmitting tubes.It must be true. Gito Title: Re: Advice for final tank circuit Post by: K9ACT on January 14, 2010, 10:48:37 AM That means the driving power needed = Losses( twice the driving power/data book) + driving power (data book)= 3 times driving power in the technical data book. Perhaps a language problem here? I do not see the jump from 2 to 3 in the text. The 2X includes all the losses. >Since it is written in The RCA Transmitting tubes(the maker/producer of transmitting tubes.It must be true. I make no such assumption but I am a born-again cynic. Keep in mind that RCA is/was selling tubes and wants customers to be happy with them and not be plagued with returns and complaints. It costs THEM nothing to suggest more overhead than is typically needed. We, on the other hand have to pay for that extra power so we (I) tend to want to know how things work in the real work. As mentioned, my experience with a number of different power tubes points to the fact that the overhead is excessive. I would be more interested in knowing what your experience is with all the transmitters you have built. Put a power meter between your driver and power amp and tell us how this compares with the tube data. Jack Gito Title: Re: Advice for final tank circuit Post by: KM1H on January 14, 2010, 01:14:25 PM I think its more than a language problem. Sine the 254 is not even in TT-4 or any other RCA manual why should a generalization even apply? RCA had their own way of writing specs while others specifically listed that the driving and output power includes expected losses in a properly engineered circuit.
If you look at the actual H&K spec sheet (the developer of the 254) it includes a complete PP 254 circuit with values for ham bands. It states that a driving power of less than 50W is needed for 750W out. I think we are thru wasting time on the 2X or 3X issue. Carl KM1H Title: Re: Advice for final tank circuit Post by: w1vtp on January 14, 2010, 01:19:59 PM Here are three more hand drawn schematics of proposed fixes to my drive problem. Ron I like the option with the link coupling. Make it swinging and you can adjust the RF drive right where you want it Al Title: Re: Advice for final tank circuit Post by: Gito on January 15, 2010, 03:21:55 AM hi.
there's no need to put a power meter between the driver and the final,since I put a plate current meter in each of my drivers and You can clearly see the watt input of the driver(input power) and how much must I load the driver to drive the final to it's rating.and it confirms what the book told us. Also I have read and analyze find most of the schematic diagram of transmitters confirm with the book. 254W as Ron wrote with low voltage on the plate .needs a 25 watt input . 254 W with 400 watt output(high output) with 4000 V (data sheet) on plate it,only needs 12 watt input so 50 watt driver is more than 3 X 12 watt. Yes the losses are twice the driving power ,if we read it carefully ...that may be lost in tube socket or in component and wiring circuit of the driving circuits.or tube losses due to electron -transit time phenomena,internal lead impedance,or other factors . so the power wasted is twice the power driver (needed). so if we drive with 20 watt ,can the losses be 40 watt?of course it can't So we must drive with 60 watt power....... 40 watt losses and 20 watt that drives the final tube. It's my conclusion and confirmed by the ARRL & RSGB hanbook ---3 to 1 drive power. Maybe it a waste of time for this Issue. But it's the basic that at least I must know before building a transmitter. I don't force anyone to follow what I wrote. To build a Transmitter we must have something to learn from,and I believed most of us ,learn from the ARRL handbook. Gito Title: Re: Advice for final tank circuit Post by: K9ACT on January 15, 2010, 08:53:58 AM there's no need to put a power meter between the driver and the final,since I put a plate current meter in each of my drivers and You can clearly see the watt input of the driver(input power) and how much must I load the driver to drive the final to it's rating.and it confirms what the book told us. The tube data lists RF power required, not DC input. For the purpose of this discussion you must know the RF drive. You have just thrown a possible 50% clunker into the discussion. >254W as Ron wrote with low voltage on the plate .needs a 25 watt input . 254 W with 400 watt output(high output) with 4000 V (data sheet) on plate it,only needs 12 watt input so 50 watt driver is more than 3 X 12 watt. I don't follow this at all. >so the power wasted is twice the power driver (needed). The power wasted is the drive power (RF) produced minus the power (RF) needed to drive the grid to proper grid current/voltage. >Maybe it a waste of time for this Issue. It's an important issue. The usual applies to those who think it's a waste of time: read something else. >To build a Transmitter we must have something to learn from,and I believed most of us ,learn from the ARRL handbook. I don't wish to stir the pot on this but I prefer the Radio Handbooks or the so-called West Coast Handbooks. However, none of them are perfect so we have to use our own brains to sort things out. Jack Title: Re: Advice for final tank circuit Post by: K9ACT on January 15, 2010, 10:10:40 AM As a point of interest, I browsed my 1941 edition of the Radio Handbook and found the attached pertinent paragraph.
I presume "tube data" means data sheets and not extraneous "X factors". We could argue about the definition of "slightly" but clearly, 2X or 3X would not fit. Jack Title: Re: Advice for final tank circuit Post by: KM1H on January 15, 2010, 04:41:01 PM I think its safe to say that Gito is completely confused ;D I just hope others arent.
So lets get back to the original problem....how to get the 6146 driver to transfer sufficient power plus reserve to the 254 grid. The steps were already covered I believe. 1. Ascertain that the 6AG7 and 6146 can produce sufficient power into a dummy load using a pi-network or other network. 2. Provide the proper LC circuitry to drive the 254. Adding links wastes power on that small chassis. A 2 coil transformer with tuned primary and secondary shouldnt be any harder than an IF transformer....the principle is the same. 3. Something I forgot to mention earlier is to use just the single tuned circuit that started all this and see if tapping down or up doesnt work. The Up means the variable C is tapped down on the coil and the sweet spot on the coil is found for the 254. 4. This isnt rocket science. Carl KM1H Title: Re: Advice for final tank circuit Post by: Opcom on January 15, 2010, 08:38:14 PM As a point of interest, I browsed my 1941 edition of the Radio Handbook and found the attached pertinent paragraph. I presume "tube data" means data sheets and not extraneous "X factors". We could argue about the definition of "slightly" but clearly, 2X or 3X would not fit. Jack Does the quote mean that the available driving power is to be considered to be what is present at the drive terminals of the driven stage, that is, after the coupling? I tend to think it means that. Also, for all of this, there is a difference in coupling methods used. parallel circuit, link, pi, etc. and these are not specified within all those statements, nor are their efficiencies. Do the statements refer to AM or FM operation? It does not say but the drive requirement changes with amplitude modulation and not with FM. Could that be the root of the confusing statements? Just my 2 cents. Title: Re: Advice for final tank circuit Post by: K9ACT on January 15, 2010, 10:17:13 PM Does the quote mean that the available driving power is to be considered to be what is present at the drive terminals of the driven stage, that is, after the coupling? I tend to think it means that. I tend to think the difference is in the noise. What is so complicated about measuring the output of a driver/exciter and then adjusting this to the tube data value for drive requirement and then connecting this to the PA grid and see what happens? It is even easier if you just connect a power meter between driver and grid and determine what power is required for specified output. Forget how you do this... just DO IT and let's talk about the results, not speculation or parroting books. I have done it many times and I know what happens or we wouldn't be having this discussion. js Title: Re: Advice for final tank circuit Post by: Gito on January 16, 2010, 12:50:29 AM Hi.
Jack when We know the input power of the driver,of course we know the output of the driver it.s about 70% right ,I think I don't have to explain it . Okay I give up,but for me I don't dare to say that I 'have more knowledge than all of You,Especially to the author of the ARRL and RSGB handbook,the RCA transmitting tube handbook. I got the article from ARRL 1978 hand book ,and the fourth edition RSGB hand book.1969 handbook. So it's all my opinion bases on the references come from books that is accepted in most of the world.Not my personal opinion And with pracktise ,I found it's confirm what it wrote. So I don't force anyone to follow/believe what I've wrote Thanks Gito, Title: Re: Advice for final tank circuit Post by: Gito on January 16, 2010, 06:46:38 AM Hi
Ron a suggestion to make a transmitter you want, I changed the schematic from the ARRL handbook you attached ,with a pi section coupling between driver and final Amp. it's only basic ,you can changed the 6L6 with 6146 ,807 with one or two tube in parallel ,I think 6AG7 have enough power to drive two 807 or 6146. Also changing the HV voltage for the driver tube,Change the final tube HV voltage ,the bias voltage to suit the need of the triode final tube. You can find the final tube input impedance according the formula that Dan has wrote . R= (W : I : I) X 620000 W=drive input power ,I=grid current. And the driver power is Plate voltage X Plate current =input power. in class C the efficiency is about 70 to 75 % . The output power of the driver is input power X 70%. R impedance of the driver is Plate :plate current. Knowing these two impedance and the frequency of the transmitter, You can design the Pi-Section values.There's also a formula to calculate in the handbook. I hoped I'm not wrong Gito Title: Re: Advice for final tank circuit Post by: W8ACR on January 16, 2010, 04:21:12 PM Thank you Gito, Jack, and all others who weighed in on this topic. My plan at the moment is to use a series fed plate tank circuit on the 6146 coupled inductively to the grid of the 254W. There will be a balanced grid tank at the grid also, and I will use grid neutralization. I don't know yet what type of plate circuit I will use on the 254W. With grid neutralization, I have the option of either a pi network or a conventional circuit with swinging link output.
A 6146 can run 750V at 150 mA, which should be plenty of power to easily drive the 254W. If and when this project has a successful conclusion, I will post pictures (hopefully right side up). Can somebody give me a starting point to try on how many wraps to use on the inductive link both at the plate tank of the 6146 and at the grid tank of the 254W? Thanks again, Ron W8ACR Title: Re: Advice for final tank circuit Post by: Gito on January 16, 2010, 05:52:57 PM Hi Ron, AMfone - Dedicated to Amplitude Modulation on the Amateur Radio Bands
my advice ,download the TT4 RCA transmitting tube, It's not only a Tube sheet transmitting tube, in it You can find all the theory of how to treat a Tube,what is class C,how to find the operating parameter,how to modulate. and in my opinion it's a very good reference book. On link coupling page 40 it wrote ......in this method of coupling in fig41,substantially IDENTICAL "LINK" windings of A FEW TURNS each are inductively coupled to the plate- tank coil and to the grid tank coil of the following stage........ So there's no "rule" of how many turns You must have....it can be three ,four tuns it doesn't matter,It uses few tutns to get A "very" low impedance connected together through.a suitable transmission line..... with little danger of excessive radiation( because very low impedance?) .......... . And You can read it at the page that I attachced Gito modification Sorry as I read the ARRL handbook there are 2 different link coupling With flat lines like Co-Ax or parallel feed cable ,we must have the reactance of the coupling coil as high/the same as the impedance of feed cable at the transmitter frequency. with unmatched lines,it can be used without much trouble if the line is kept very short comparing the Frequency of the transmitter (300000 : F =... meter) |