Using the 860 as a modulator tube

[k2glo]

I desire to use a pair of 860 tubes as modulator output tubes in a special homebrew transmitter project.  I am aware that tetrodes are not the best choice for use in a modulator and especially a lowly inefficient 860, but  that is the tube that is most physically fitting for this particular project.  I am also aware that using the 860 will require an added screen and bias supply and a substitution of a triode or its sister 852 triode would be better choice but those are not options and the extra power supplies are no problem.  I cannot find any information on anyone using a pair of 860’s in a modulator and obviously no manufacturer’s published operating specifications as an AB modulator.  Have any of you used PP 860’s in a modulator either as tetrodes or  triode connected tetrodes?  Any thoughts and information on suggested operating perimeters would be most helpful.
Thanks,
Bob, K2GLO

[W7TFO]

There is no reason not to use a pair as mod amps, but there will be a god-awful high plate-to-plate impedance to deal with.

73DG

[Opcom]

Yes those like high voltage and not a huge current although they can deliver high peaks >2A! A whopping reserve. It makes me wonder if the gummint did something 15E-ish with them at VHF in WWII.

They are very linear. What will you modulate? What plate voltages being considered?

I have a spare pair here with the thick graphite-zircon-looking solid anodes, no plans to use them as yet. Also maybe an 852 Triode in another box somewhere. I have kept those nicely protected due their beauty. For some reason they show up in good cond. in the various auctions attended by antique radio collectors. (meaning collectors of old entertainment and SWL receivers for consumers). I guess they were collected due their appearance, and not used up.

DG is right about the p-p load, maybe 15K to 20K? going to have issues with a transformer's high end at such high Z, but probably OK for voice band. Multimatch units claim to go to 20K Ohms p-p loads but that's got to be pushing them. Maybe some transformer tricks can be used if you find a pair of 100W transformers with dual primaries so that each transformer does a share of the impedance. They probably do not even have to be identical because the windings would be put in series with this, although the first thought always tends to use identical parts.

This has my interest! Please elaborate on this design!

[k2glo]

Thanks for your input on using the 860.  I agree with you that the United Electronics 860 with the graphite anode are a beauty to behold.  Those stay on the shelf while the more common heat finned anode ones that I have will be the ones used. 

You asked what I intend to modulate with the 860's.  I have a Navy TAJ-19 sitting in the garage begging for some attention.  I really like the TAJ-19 with its classy tuning controls, ten panel meters, the beautiful green, blue and red indicator lights and most of all the front access panel door that allows you to view the 861 and three 860's even with the door shut. It is just a beautiful boat anchor! The original line up is 860 MO, 860 IPA, 860 audio osc. for MCW and 861 PA.  The transmitter is a CW and MCW unit covering 175kHz to 600kHz at about 500w.  It requires motor generators to power it.  In its present state, it is totally useless for amateur radio but I love the looks of it.  My thought was to make it a 160M AM rig using the 861 as the PA, one of the 860's as the IPA and the other two 860's as modulators.  Obviously, I would have to trim down the inductors quite a bit.  I would have never thought of your suggestion of placing a pair of modulation transformers in series to accommodate the high impedance.  That is a neat idea.  However, I have a NIB UTC Varimatch S-22 modulation transformer that is capable of 22K P-P on the primary. My initial request was to see if anyone has used these tubes as modulators either as tetrodes or triode connected tetrodes and get an idea of the triode connected tetrode plate to plate load impedance.

I could give up on the above idea and go back to using a modified original line up of 860 MO, 860 IPA, and 861 PA with the third 860 used as an audio osc. just sitting there as a spare.  If I did that, I would consider using 805's for the modulator.  I have a Techrad 350XM transmitter also in the garage that uses a pair of 813's modulated by 805's.  The 805 driver transformer has a 500-ohm primary that is intended to be driven by a separate (outboard) P-P 6L6 audio amplifier which Techrad also supplied. I do not have an original Techrad audio amplifier but it closely resembles a typical 6L6 amplifier with a 500-ohm output.  Some years ago, I visited Bud Bane in San Francisco.  He was one of the two gentleman who developed the 350XM.  Long story short, I came home with two HV plate transformers and an 805 driver transformer. I could dig through my transformer bone pile and find the driver transformer and build up an 805 modulator which would be more conventional but that is not as fun as trying to use the 860's as modulators.

With all said and done, I intend to only have it run at 375W AM to meet the legal 1500W limit so it is going to be a physically large transmitter for its output.

Suggestions are welcome.

Bob, K2GLO

[Opcom]

That's a great looking transmitter (the definition of buzzardly might have a pic of it there..) and a good project to use it for 160M. Maybe you can do that without ruining the coils but switching in a substitute, as there are some experimental bands down low.. maybe some day more power will be allowed. I did look up the 861 in the RCA book but it is a good looking tube, that's a fine line up.

My method of guesstimating without going through the calculations is to find a tube with audio specs stated, that is similar in plate voltage and screen voltage for some use the two types have in common, and look at that other tube's audio ratings.

or example if a pair of 4-125's takes 2KV at 300mA and 350V on screen to make 350W into a 13600 ohm plate to plate load, then maybe the 860's with 2KV on plate and 300 on screen running at 200mA (2/3 the value) would make 233W and draw 200mA, and the load would be 20.4K (1.5x the value). The simple math seems to work because the plate swing and voltage stay the same, only the current changes. Anyone else use this short cut? Seems to work OK for me.

[k2glo]

Looks like a good short cut.  I did some searching and it looks like the 4-125 is about the only tube that as a good tube for comparison.  I have a couple questions re the short cut...  is the"1/3" a typo? Doing the math in the example you stated, it looks like "2/3" is the multiplier.  The other question is how does the 1.5 factor relate to the short cut?  Are there factors other than 1.5 that are used for other operating conditions or is it always 1.5?
Today, I brought the transmitter to the forefront in the garage to get seriously looking at the conversion.  I am getting excited about the project. Thanks for the replies and insight... It has been most helpful 

[AB2EZ]

Confirming the calculation that Opcom did:

From the specification sheet:

http://www.mif.pg.gda.pl/homepages/frank/sheets/049/8/860.pdf

Class B: B+ = 2000V, screen voltage = 300V, and average plate current = 60mA

If the average plate current for a single tube, operating in class B, with a sine wave driving the grid, is: 60mA, then the peak plate current is 60mA x pi = 60mA x 3.14 = 188mA

[Note: in his post (earlier in this thread) Opcom is assuming an average plate current of: 100mA per tube; which results in a calculated plate-to-plate load resistance value which is 0.6 x the value I will calculate below.]

If 188 mA (peak) flowing through the plate load resistance is going to cause a 1700V swing*... i.e. during the half cycle when the tube is conducting, the plate voltage goes from the 2000V B+ value (at 0 plate current) to the 300V screen voltage value (at 188mA of plate current)... then the plate-to-center tap (i.e. plate-to-audio ground) load resistance has to be:

1700V/188mA = 9.04k ohms

Therefore, the required effective plate-to-plate load resistance in push pull class B operation, to achieve full output = 4 x 9.04k ohms = 36.2k ohms.

[If I had assumed an average plate current per tube of 100mA, instead of 60mA, then my calculated value for the plate-to-plate load resistance would be: 0.6 x 36.2k ohms = 21.7k ohms... which is close the the value that Opcom obtained in his post.]

*From the curves in the specification sheet, the plate current will drop precipitously, and the screen current will rise precipitously if the instantaneous value of the plate voltage drops below the value of the screen voltage.

Reconciling the above with a pair of 4-125s operating in Class AB2:

From the Eimac data sheet:

Plate supply voltage = 2000V

Plate-to-plate load resistance = 13,600 ohms
[Therefore the plate-to-transformer center tap load resistance is 13,600/4 ohms = 3400 ohms]

DC (i.e. average) plate current (two tubes) at full signal (audio sine wave) = 300mA
DC (i.e. average) plate current (two tubes) at zero signal = 72mA

Implies:

The peak audio plate current swing in each tube is approximately: 150mA x pi= 471mA (approximate because the tubes are operating in Class AB)

The plate voltage during the half cycle that each tube conducts plate current swings by approximately: 471mA x 3400 ohms = 1600V (i.e. from 2000V down to 400V)

The audio power output (two tubes) is approximately: 1600V x 0.471A / 2 = 377 watts

The main difference between a pair of 4-125 tubes and a pair of 460 tubes is the significantly higher average plate current per tube (150mA v. 60mA)

Stu

[Opcom]

is the"1/3" a typo?

It is. I fixed it to 2/3.

1.5 is the inverse of 2/3.

2/3 the current, same volts = 1.5 times the R.


Stu's calculations get no objection here but I have some questions for understanding.

I am curious about them because the 60mA per tube is at carrier which is 1/4 the output level of the class B linear data.
So there is 30W @ carrier, and single tone or "RMS" output of 120W per tube at the top of the modulation cycle.

What is the average plate current at 120W output from the 860 in class B linear operation?
About 95mA? 
(@ 63% efficiency, not quite class B, maybe 20-30mA idling current)

How does the "average plate current @ carrier level * 3.14" come to be the same as the "peak current when the tube is fully driven in the given class B spec"? That is the relation I do not see so easily.

Would it, for the 120W output, be: 95mA * pi = 290mA?
Or call it 94mA since 94 is 1/2 of Stu's 188mA.
and then 94mA * 3.14 is 295mA. So use that 295 mA as the "full signal average current"

and 1700V / 0.295A = 5795 Ohms
and two 5795 Ohm windings stacked on each other for a plate-to-plate load is 4* 5795 Ohms, or 23K Ohms plate to plate?

Still pretty close but what is the explanation about this? - where I would differ as to the plate to plate load?

How does it reconcile the difference in the amplifier between the "carrier average plate current" and the "(presumed) full signal average current", and the "plate to plate impedance"?

[AB2EZ]

Opcom

I think that you may be inadvertently comparing apples and oranges:

Although the math is similar in some respects, you cannot (well... at least, I cannot) analyze a linear RF amplifier, a plate modulated RF amplifier, and a class B push-pull audio amplifier at the same time.

For the case of a class B push-pull audio amplifier:

Each tube will have plate current flowing for one half of each audio cycle.

If the input signal driving the grid of each tube is a sine wave (one grid drive signal inverted with respect to the other) then the plate current (in each tube) will be the positive-going half of a sine wave.

The average value of the positive-going half of a sine wave current (of any frequency, f) is given by the relationship (using calculus to compute the area under the waveform of current v. time, from the beginning of a cycle to the end of a cycle... and then dividing the result by the time duration of one cycle):

The average value of the plate current = [2/(2 x pi x f)] / [(1/f)] x the peak value of the plate current = [1/pi] x the peak value of the plate current

Therefore the peak value of the plate current in each tube = pi x the average value (i.e. the DC value) of the plate current in each tube.

Note that the total average value of the plate current (both tubes) is 2x the average value of the plate current in each tube.

If the amplitude of plate voltage swing is 1700V, and the peak plate current is 188mA, then each tube will deliver an a average audio power to the load of:

1700V x 0.188A x 0.5 x 0.5 watts = 80 watts

In the above equation, the first factor of 0.5 takes into account the fact that the average value of the power dissipated in a resistive load, when a sinusoidal current flows through the load, is 0.5 x the instantaneous peak power dissipated in the load. The second factor of 0.5 takes into account the fact that each tube is delivering the power for only one half of each audio cycle.

The total average audio power delivered by the push-pull amplifier to the load is 2 x 80 watts.

The total average electrical plate input power  is 2000V x 0.120A = 240 watts

The plate efficiency of the amplifier = 160 watts / 240 watts = 67% (i.e. not taking into account electrical power dissipated as a result of screen current and filament power)

With respect to the "plate-to-plate load resistance (effective)":

I'm not sure what the official/standard definition is.

I tried Google, and I looked in my 1961 ARRL Handbook... but I couldn't find what I would consider to be an authoritative reference. Perhaps someone reading this thread can provide one... that can be traced back to something in (for example) a textbook on tube amplifier design.

I think the definition is something like:

The plate-to-plate load resistance (effective) = the recommended audio frequency load that the full primary of the output transformer must present to the push-pull amplifier (i.e. plate to plate <=> one side of the primary to the other side of the primary), for a specified value of DC plate voltage, in order to: achieve maximum output power from the push-pull pair... without exceeding a specified/moderate amount of distortion; and without exceeding the maximum allowable plate dissipation of each tube.

To my own "taste" I would ask the following in this situation:

"If the desired audio frequency plate load resistance seen by each half of a push-pull pair... i.e. between one side of the output transformer's primary and the center tap of the output transformer's primary... is:

1700V/0.188A = 9.04k ohms ...

...then what will be the corresponding audio frequency load resistance that should be presented by the full primary?"

The answer to that question is 4 x 9.04k ohms = 36.2k ohms

If the modulation resistance of the RF output stage is (for example) 5000 ohms, then the turns ratio (full primary : full secondary) required is: [the square root of (36.2/5)] : 1

I.e. 2.69 : 1

The peak output voltage from the secondary of the transformer, across the 5000 ohm secondary load resistance, will be 3400V/2.69 = 1264V, when the push-pull amplifier is producing its maximum output power.

Stu

[AB2EZ]

Note: I have added some comments to my earlier post (see the quote below) to reconcile my calculation with Opcom's estimate.

If I assume an average plate current, per tube, of 100mA... as Opcom did... then I obtain the same result for the plate-to-plate load resistance as Opcom did in his estimate.

I assumed an average plate current per tube of 60mA... as given in the 860 tube specification for Class B operation, at 2000V plate supply voltage, as an RF amplifier (carrier conditions). At carrier (as a linear RF amplifier), the tubes would be operating at about 33% plate efficiency... so they would each be dissipating around 67% of 120watts = 80 watts.

In audio class B operation, with full sine wave output from a push pull pair... the tubes would be operating at about 67% efficiency... so (for the same 60mA average plate current per tube), each tube would be dissipating 33% of 120 watts = 40 watts.

Therefore, in a Class B push pull audio application, it would make sense to have the tubes operating at a higher average plate current per tube... and the 100mA value, per tube, that Opcom assumed probably makes more sense than the 60mA, per tube, value that I assumed.

Stu

Confirming the calculation that Opcom did:

From the specification sheet:

http://www.mif.pg.gda.pl/homepages/frank/sheets/049/8/860.pdf

Class B: B+ = 2000V, screen voltage = 300V, and average plate current = 60mA

If the average plate current for a single tube, operating in class B, with a sine wave driving the grid, is: 60mA, then the peak plate current is 60mA x pi = 60mA x 3.14 = 188mA

[Note: in his post (earlier in this thread) Opcom is assuming an average plate current of: 100mA per tube; which results in a calculated plate-to-plate load resistance value which is 0.6 x the value I will calculate below.]

If 188 mA (peak) flowing through the plate load resistance is going to cause a 1700V swing*... i.e. during the half cycle when the tube is conducting, the plate voltage goes from the 2000V B+ value (at 0 plate current) to the 300V screen voltage value (at 188mA of plate current)... then the plate-to-center tap (i.e. plate-to-audio ground) load resistance has to be:

1700V/188mA = 9.04k ohms

Therefore, the required effective plate-to-plate load resistance in push pull class B operation, to achieve full output = 4 x 9.04k ohms = 36.2k ohms.

[If I had assumed an average plate current per tube of 100mA, instead of 60mA, then my calculated value for the plate-to-plate load resistance would be: 0.6 x 36.2k ohms = 21.7k ohms... which is close the the value that Opcom obtained in his post.]

*From the curves in the specification sheet, the plate current will drop precipitously, and the screen current will rise precipitously if the instantaneous value of the plate voltage drops below the value of the screen voltage.

Reconciling the above with a pair of 4-125s operating in Class AB2:

From the Eimac data sheet:

Plate supply voltage = 2000V

Plate-to-plate load resistance = 13,600 ohms
[Therefore the plate-to-transformer center tap load resistance is 13,600/4 ohms = 3400 ohms]

DC (i.e. average) plate current (two tubes) at full signal (audio sine wave) = 300mA

Implies:

The peak audio plate current swing in each tube is approximately: 150mA x pi = 471ma (approximate because the tubes are operating in Class AB)

The plate voltage during the half cycle that each tube conducts plate current swings by approximately: 471ma x 3400 ohms = 1600V (i.e. from 2000V down to 400V)

The audio power output (two tubes) is approximately: 1600V x 0.471A / 2 = 377 watts

The main difference between a pair of 4-125 tubes and a pair of 460 tubes is the significantly higher average plate current per tube (150mA v. 60mA)

Stu

[Opcom]

I think that reconciles the apples and oranges into fruit, so maybe if not compared directly because one comes from single ended data and one regards extrapolation to push pull data, they make sense each itself.

I agree with the change of the plate load vs the multiplier for plate to plate as it applies to transformer windings. What I mean is, it is still two 9K windings, but when put in series it would be 38K, in the same way that a 16 Ohm winding's center tap in turns is 4 Ohms. The turns ratio vs Z ratio thing.

The difference I see is that only one 9K winding is used at a time, it is still a 9K winding or load as is suitable for the carrier condition of 60mA, if applied to be an audio amp as one half of the push pull circuit. I think with the reconciliation of the text in bold, the results agree well enough.

[Opcom]

Has this met the needs? Going to do it? Please post pictures as it goes, those are great looking tubes!

[k2glo]

I want to thank all who responded. It was a good discussion and I learned what I needed to pursue with the project using the 860's in audio P-P. If someone still has experience using the 860 in a triode connection mode, it would be nice to hear about it. Opcom requested that I post pictures as I go.  Well here is the first photo showing the 861 PA tube along with the other three 860's mounted on the unmodified TAJ-19 tube sub-chassis. I think it will look real impressive with all the filaments lit for combined, they consume a little over 200 watts of filament power.
Thanks again
Bob K2GLO

[Opcom]

That is very fine looking. Classic nice old tubes!