Understanding "grid drive" to drive, 2, 4, 6 power tubes

[nsubulysses]

This doesn't add up.

3 x 110mA = 330mA. A 1.6k OPT is 400 ohms each side so power out =0.33^2*400/2 = 22W under these conditions.

What is your supply voltage?
How did you measure 110mA?
How did you measure power out?
Do you have the OPT specs?
What is the screen voltage?

4 power tube amp, 1K screen grid, 1K in place of choke, 585V B+, 560V screen

6 power tube amp, 1K screen, 650R in place of choke, 585V B+, 560V screen

current draw measured over 1 ohm 1% cathode resistor

OT for 6 power tubes is 200W at 1.6K primary. That's all I know

The 150W OT for 4 power tubes is 1.9K primary

[Dave H]

This doesn't add up.

3 x 110mA = 330mA. A 1.6k OPT is 400 ohms each side so power out =0.33^2*400/2 = 22W under these conditions.

What is your supply voltage?
How did you measure 110mA?

Yes, how was that 110mA measured? The equation is for peak current isn't it? Wouldn't it need a peak current of 1A for 200W (400W pk) and 400V peak voltage with a 1.6k OT? 330mA is a voltage swing of only 132V.

I think reducing the OT impedance would only yield more power if the PT is up for it. Power comes from the PT not the OT.

Edit: I should have read the last post.

"current draw measured over 1 ohm 1% cathode resistor"

[Helmholtz]

current draw measured over 1 ohm 1% cathode resistor

With meter in DCV mode, this gives averaged DC cathode current, containing 20..30mA of screen current. DC cathode current cannot be used for power calculation.

The plate/screen current values I mentioned above (125mA/25mA) were taken from the 6CA7 datasheet, using the "class B, 500V" example.
http://www.tubezone.net/pdf/6ca7.pdf
I am quite sure these are also averaged DC values (reducing to idle numbers for zero input voltage) and their sum (150mA) should be comparable to the measurements of nevetslab.
150mA is actually the max. allowable average cathode current.

[Helmholtz]

Data sheet numbers are usually for BOTH tubes at idle, but PEAK for one tube, in Class AB operation.

Numbers in the 6CA7 data sheet would be way too low for peak current.

[Dave H]

It should read (0.33^2)*400/2 =22W.

I'm taking the 330mA figure as the peak current but I'm wondering if it was supposed to mean something else?

We now know 330mA is the average DC cathode current each side so perhaps the equation should be -

(0.30*1.11*2)^2 * 400 = 177W ?

I've subtracted 30mA screen current and multiplied by 1.11 to convert average to RMS as it's a sine wave input.

[Helmholtz]

We now know 330mA is the average DC cathode current each side so perhaps the equation should be -

(0.30*1.11*2)^2 * 400 = 177W ?

I've subtracted 30mA screen current and multiplied by 1.11 to convert average to RMS as it's a sine wave input.

I don't think this can work. Cathode current will not be a perfect half-sine in a class AB amp. Also average screen current would be more like 75mA for 3 tubes.

Why not simply scope plate voltage swing and measure peak?

[Dave H]

Why not simply scope plate voltage swing and measure peak?

Because I don't have the amp. Average cathode current was all I had to work with

[Old Tele man]

OOPS...typo on my part...the diving control-grid signal is PEAK, the output plate current is RMS:

"a) Peak AF grid-to-grid voltage is a voltage for two tubes (Vgg); the grid-voltage seen by each
single tube is one-half of this value (Vg). The product of Vg and gm is RMS plate current[2].
b) Zero-signal numbers are for two tubes; the value for a single tube is one-half of those
values.
c) Maximum-signal numbers are RMS values for both tubes (one conducting, one in cutoff),
the average value for a single tube is one-fourth of those values.
d) Average value of combined zero- and maximum-signals is approximated as:

X(avg) = (Xq/2 + ∆X/4).

NOTE: ∆ represents AC-change in voltage (VAC) or current (IAC) and q denotes quiescent or
idle values (VDC or IDC). "

[Dave H]

Cathode current will not be a perfect half-sine in a class AB amp.

True, I assumed it was class B for the estimate because at full power there's only a small overlap. It almost looks like it's full wave rectified.

[nickb]

I have learned a lot about phase inverters and screen grid dissipation and many other things along the way. Thanks!

Could anyone care to speculate on this question?

SO I have an amp with 6x6CA7 and it does 180W at clipping. Power tubes clip at 110mA or so. OT primary impedance is 1.6K.
A 4 power tube amp with same power supply voltages and 1.9K OT primary does 144W and power tubes clip at 130mA or so.

If I order a 200W OT with 1.1 or 1.2K primary impedance will my power tubes clip at the expected-ish 130mA and will I achieve 200-210W at clipping?

After some thought, the best answer I can give you is, "I really don't know."

But you can find out easily enough and at minimal cost. Stick with the transformer you have and just tweak the value of the dummy load. So, if you have 1.6k transformer running into 8 ohms, then for a 1.2k transformer make the load 8 x 1.2/1.6= 6 ohms by whatever means you have to hand. Running at 1Khz means there won't be an issue with the transformer saturating at the higher power. There may be a difference in transformer losses but I'd expect it to be insignificant.

[Helmholtz]

^^^Great idea!

[Helmholtz]

True, I assumed it was class B for the estimate because at full power there's only a small overlap. It almost looks like it's full wave rectified.

[ATTACH=CONFIG]51984[/ATTACH]

Does the picture show cathode currents?

[Helmholtz]

"a) Peak AF grid-to-grid voltage is a voltage for two tubes (Vgg); the grid-voltage seen by each
single tube is one-half of this value (Vg). The product of Vg and gm is RMS plate current[2].
b) Zero-signal numbers are for two tubes; the value for a single tube is one-half of those
values.
c) Maximum-signal numbers are RMS values for both tubes (one conducting, one in cutoff),
the average value for a single tube is one-fourth of those values.
d) Average value of combined zero- and maximum-signals is approximated as:

X(avg) = (Xq/2 + ∆X/4).

NOTE: ∆ represents AC-change in voltage (VAC) or current (IAC) and q denotes quiescent or
idle values (VDC or IDC). "

What is the origin/source of this text?

[Old Tele man]

This paper: http://thermionic.info/mccaul/McCaul...heets_2008.pdf

Hosted here: http://thermionic.info/

and posted earlier in this thread: https://music-electronics-forum.com/...l=1#post518965

[Helmholtz]

Thanks a lot!