cathode voltage - shared/individual resistors

if i have a pushpull triode stage that is transformer coupled.
and a shared bypassed cathode resistor with a measured voltage of 4,5v.

does that suggest that each tube has a grid bias voltage of -4,5v or half?

and if i would separate the tubes with individual cathode resistors with double the resistance would i get the same bias voltage and bias current?

Yes, if you want separated cathode resistors value should double. Bias voltage stays the same this way. Biasing voltage looks like you have push pull line out, not power.
It would be interesting to know if there is any advantage by separating cathode resistors and caps with single ended.

yes its a pushpull line output for my eq.

im trying to decide if i want to go with the classic design of separate resistors, shared resistor or separate resistors with a smaller balance pot to dail in the best crossover distortions, especialy when inserting new tubes.

I don't have enough experiences with line level push pull stages to suggest which way you should go. But since we are mostly using new production tubes it is helpful having good control over parameters.
Local guitar amp repair guy who uses lots of tubes told me he has a lot of troubles with JJ's receiving tubes parameters. We both use EH which are much better in our experience.
You could search for schematics of new varimu compressors with such stages. One of them was redesigned by people here and has separated cathode resistors with trimmer, so why not try that, it is easy to change back to one resistor if needed.

Yea i have read about poorer balance in new tubes so i thought this could be neet.

I have channel set up with 100r balance pot - that can be mounted to the front panel - and 220r shared resistor and bypass 470uf at each cathode. It works of course and i could set for minimum 100hz hum.

But my main question is if my 4, 5v at the shared cathode resistor looks like 2, 25v or 4, 5v on each tube. Whats the headroom on each grid so to speak.

5v333 said:

But my main question is if my 4, 5v at the shared cathode resistor looks like 2, 25v or 4, 5v on each tube. Whats the headroom on each grid so to speak.

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You can generally reckon to swing the grid to about -0.5V before grid current starts to occur. and distortion rises quickly. So with -4.5V bias you can swing the input about 8V peak to peak and still remain in class A. This is about 2.8V rms or +11dBu. If your pp stage has 20dB gain, this corresponds to +31Bu at the anode. If you then have a 4:1 step down output transformer then the level at the output is 12dB lower at +19dBu.

There is not much point in using matched tubes. The manufacturer tests matching at one particular static bias point. It is anyone's guess whether the two section match at any other point on the curve. Using a common cathode resistor means only that both grid voltages will be the same. Each triode will have a slightly different current at that grid voltage. For an output stage you want to set the quiescent current because current swing is what determines output power. A common cathode resistor does not let you do this.  A trim pot between the cathodes will alow you to set both currents to the same value but you will need an additional small series sense resistor to allow you the measure the current in each cathode. It is debatable whether this gives any audible improvement.

Cheers

Ian

8v p-p at both grids? Or 8v accumilated across both grids?

5v333 said:

8v p-p at both grids? Or 8v accumilated across both grids?

Click to expand...

between each grid and 0V.

Cheers

Ian

so the signal applied to one of the grids can be as high as 8vp-p before clipping?

5v333 said:

so the signal applied to one of the grids can be as high as 8vp-p before clipping?

Click to expand...

Yes.

Note that using a single resistor (and a single cap) the signal current at the cap is much lower than with separate resistors and caps, as one pushes the other pulls so the current is balanced, so much lower AC current for the cap to handle. This means with 2 caps, the caps in parallel needs to be much bigger than the single cap, depending on how over valued the single cap is. Maybe the same value is enough, to be sure the RC should be lower than the dominant low frequency pole of your circuit so it doesn't affect the frequency response, but without going to far as the tube needs to charge it before blowing (if you put a 1 Farad supercap there the bias current will be way to high way to long till you get the 4.5V). You haven't mentioned the value of the resistor and caps used for me to help here, but 5 to 10 Hz pole on that RC circuit should be good, so after 1s the cap is charged over 99% of the final value.

JS

ok..!

so  how would a loadline look on the transfer sheet for a pushpull circuit?

my load is a 15k transformer. i guess that is 7.5k per plate! my B+ is 215v.

215/7500 = 28,7mA

i make a line between lowright 215V and highleft 28,7mA.

if i measure 4,5v across one common 220r cathode resistor i calculate 20,4mA cathode current and that is the sum of both triodes.

so my quesent point for one triode should be 4,5vg (allowing 8vp-p) and 10,2mA.

but this point is nowhere on my load line............?!?!?!

this is why i am repeating my self im so close to get the pushpull calculating and am excited to crack the code!!


RC is 220r cathode resistor with a 470uF.

although im trying out this circuit with an extra 100r balancing pot before the 220r resistor and one 470uF at each cathode.

the lowest cutoff in the amp is around 2 - 2,5hz from the input transformer.
output transformer is 10hz.

thanks ian and joaquins for the knowledge!!! keep it coming!

I think you are mixing AC signal stuff with DC bias stuff.

At each grid, 0VDC, 8VAC(pp), no considerable current at the grids

At each cathode, 4.5VDC, there should be 0VAC if the decoupling caps are big enough. You said 220Ω so 20mADC, 10mADC each tube. As long as you have some current you are still in class A, so let's say 15mAAC(pp) (*correct to 11mAAC(pp))

At the plates, 215VDC, same 10mADC, same 15mAAC(pp) 112VAC(pp). That's a bit much, Ian estimated 20dB over the 8Vpp, so 80Vpp, so more like 11mAAC(pp)

  Then, If you want to split the resistor you should use 220Ω*2=440Ω each, you can find 470Ω more easily than 440Ω so you could go with that. Use 470µF caps and you can call it a day. If you want to balance the current you could use a pot, wiper to ground, each end to one of the cathode resistors. 100Ω pot is usually good enough, in such case you should use 390Ω resistors. If the tubes are so different you need to go out that 100Ω is probably good to change the tubes but you could use a bigger pot and smaller resistors. 500Ω pot and  220Ω resistors would be an extreme case, I guess.

  One of the benefits of matching the bias current is the transformer sees no DC current and then LF fits more confortable on the transformer linear range. If you have excessive DC at the transformer LF response and distortion would be compromised.

JS

The dc load line and ac load line are different. With a transformer, its dc resistance is very low so the plate voltage (dc) is constant and equal to the HT voltage. The dc load line is therefore a vertical line at the HT voltage. As you are biased at -4.5V, where this vertical line intersects the -4.5V plate volts vs plate current graph is the dc operating point.

The ac load line must pass through this point but its angle is determined by the ac load seen by the plate which is the load reflected back through the output transformer.

There are good explanations of all the basic stuff in on line tube books.

Cheers

Ian

ok!

so that explains the mixing up  of ac and dc loadlines.

dc resistance of my transformer is 655r + 655r.

215V / 655r = 328ma (and even if this is too high for a 6sn7 it cancels out when equal current is put on both windings..!)

each grid-cathode is biased at 4,5VDC 10mADC allowing each grid to welcome signal swings of about 8Vp-p.

and if an individual cathode resistor circuit would be used, the resistance of each would be the double:

440r x 10ma = 4,4VDC


if anybody has more to say about advantages for shared/individual setups would be interesting.

thanks again!!!

5v333 said:

ok!

so that explains the mixing up  of ac and dc loadlines.

dc resistance of my transformer is 655r + 655r.

215V / 655r = 328ma (and even if this is too high for a 6sn7 it cancels out when equal current is put on both windings..!)

Click to expand...

At the dc bias point, the effective plate resistance is in series with the dc resistance of the transformer secondary.  The effective plate resistance depends on the bias point. Your current bias point leads to a current of about 10mA. The effective plate resistance is therefore about 215V/10mA which is 21K5. This is very large compared to the dc resistance of the transformer so for setting the bias point you can ignore the transformer dc resistance.

You can think of a tube as a variable resistor with the wiper controlled by the grid to cathode voltage.

Cheers

Ian

something is wondering around my head instead of inside my head... it hearts to learn!

how do we make a dc load line for one tube tube in a pushpull transformer coupled amp?

does the DC loadline change depending on how much the plate is drawing?


i thought it would be like with a resistive coupled SE amp. B+ / Rload

no... i think im on the right track using the DCR of the transformer winding as a DC loadline it matches with the bias point of 4,5VDC and 10mADC.

im searching through RDH4 and googling the net about all this but info is scattered and quite often in an english thats to advanced for me.

once again, thanks for your responds! its good to combine chatting with people and read about it!

5v333 said:

no... i think im on the right track using the DCR of the transformer winding as a DC loadline it matches with the bias point of 4,5VDC and 10mADC.

Click to expand...

That is right, but the transformer dc resistance is so small you can assume it is zero which is why the dc load line is a vertical line

im searching through RDH4 and googling the net about all this but info is scattered and quite often in an english thats to advanced for me.

once again, thanks for your responds! its good to combine chatting with people and read about it!

Click to expand...

I know what you mean. Right now I am trying to learn technical German and it is  not easy.

Cheers

Ian

That is right, but the transformer dc resistance is so small you can assume it is zero which is why the dc load line is a vertical line

Click to expand...

thanks for confirming!!!!

Right now I am trying to learn technical German and it is  not easy.

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good for you! its fun with languish. opens up new things in life!