Are 500 volts of signal going to be expected on the plate of a 6L6?

peteb

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It is common knowledge that the plate of a power tube is going to see hundreds of volts of signal, even on a smaller 6V6 amp.

now let’s consider a 50w 2X6L6 amp.

it is going to be higher.

I calculate 700 VAC is required to put 50W on the primary to get 50 W out of the secondary.

is that right? Is that possible?

if measured, is that what it is going to be?

700 V * 700 V / 10,000 ohm = 50 W

EDIT: 10,000 impedance is plate to plate. Each tube would only see 5,000 ohms of impedance.

revised numbers

50 W = V^2/ R = V^2 / 5,000

250,000 = V^2

V = 500
 
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Jowes_84

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I like a simple approach… so if I were you I would start with the secondary, you have a speaker impedance to work with. Then all you need to know is the turns ratio and there you have your AC on the primary.
 

peteb

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My first numbers were off. A pair of 6L6 works into a 10,000 ohm impedance, but each tube is only working into 5,000 ohms.

the full series:

a champ puts out 5 W and the 6V6 works into 5,000 ohms impedance.

5W = V^2 / R = V^2 / 5,000 ohms

25,000 = V^2

V = 158

a Princeton puts out 15 W and the pair of 6V6s work into 8,000 ohms of impedance, or 4,000 ohms per tube.

15W = V^2 / R = V^2 / 4,000

60,000 = V^2

V = 245

a Bassman puts out 50 W and the pair of 6L6s work into 10,000 ohms of impedance, or 5,000 ohms per tube.

50 W = V^2 / R = V^2 / 5,000

250,000 = V^2

V = 500
 

peteb

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Three points are to be made.

it is expected that the signal is at its highest level on the plate.

this signal voltage subtracts from the plate voltage, all three amps have plate voltages near 400 VDC, but only the Bassman sees a signal higher than the DC. The AC is subtracted from the DC, so at full power, and full signal current, the plate actually goes negative.


second point

i see an even progression, which is what I like to see.

this is what each tube plate sees at maximum plate current, or maximum power out:

champ
158 VAC and 32 mA of signal current

princeton
245 VAC and 61 mA of signal current

bassman
500 VAC and 100 mA of signal current

I like the way the voltage and the current evenly step up as power increases.

I tested a champ and found more than 158 VAC on the plate. This shows that you can put more power into the OT but you can only get the rated power out.

I am not going to bother measuring high plate signal on the Princeton or the bassman. I assume the math is correct.


the third point, let’s see if it all makes sense.

champ has a bias voltage of -23 and the gain of the 6V6 is 10. Working backwards.

max plate signal is 158 VAC

dividing by a gain of 10, max signal on the control grid is 16 VAC and bias is 23. A max signal voltage of 16 working against a bias voltage of 23 looks just about right.

my Princeton has a bias of -40 and the gain of the 6V6 is 10.
working backwards.

max plate signal is 245 VAC.

dividing by 10, the max signal on the control grid is 25. A max signal of 25 working against a bias voltage of 40 also looks about right.

my Bassman has a bias of -52 volts and the 6L6 has a gain of 8, I believe. Working backwards.

max plate signal is 500 VAC

dividing by 8, the max signal on the control grid is 62.5. I don’t like this result. The max signal on the control grid is more than the bias voltage, 62 versus 52. ???
 

2L man

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Your math does not work now because you do not understand how a push pull power stage work!!!
 

Bendyha

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1653086211190.png

…came across this ad in an old training manual, it had a lot of easy to learn lessons to offer, but this was the easiest lesson.
 
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peteb

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The math does work. My errors have been what impedance to use. There is still an error to fix.

has any one seen this done before, calculate plate signal voltage and current from the output power and the impedance? I have not.

the logic is rock solid.

for the OT to output 50 W, there must be at least 50 W of power put into the OT. We know the R or the impedance.

is there any reason that we cant use the output power and impedance and the relation,

power = V^2/R

to find the voltage?


I don’t think anyone can challenge the logic or the method.

please feel free to technically comment on the method if you want. Chitter chatter is not very helpful.


I will now fix the bassman calculation and add the twin.
 

peteb

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My first error with the bassman was using the plate to plate impedance instead of just one side. My second error is that a Bassman uses 4,200 ohm plate to plate, where I mistakenly used 10,000. Now I understand why the bigger the amp, there is less impedance. The twin reverb uses 2,000 ohm.

BASSMAN
50 W output, 4,200 ohm plate to plate impedance,

50 W = V^2 / R = V^2 / 2,100 ohms

105,000 = V^2

Signal Voltage = 324 volts
Signal current = 50 / 324 = 154 mA

Twin Reverb
100 W output, 2,000 ohm plate to plate impedance,

100 W = V^2 / R = V^2 / 1,000 ohms

100,000 = V^2

signal voltage = 316
signal current = 100 / 316 = 316 mA. (Not a typo, V = mA)

analysis

i like how the Twin reverb and the Bassman have nearly the same signal voltage as the TR is two Bmans in parallel.
 

peteb

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Further analysis of the Bassman

measured on my AA165 Bassman:
plate voltages equal screen voltages all around 447 VDC
screen grids are at -52 VDC

maximum signal voltage and current at the plate, calculated for 50 W output

Max signal voltage is 324 VAC
Max signal current is 154 mA

I like how the max signal on the plate is smaller but not significantly smaller than the VDC on the plate.

I believe the DC Plate voltage sets a practical upper limit for the plate signal. I don’t think we want the plate going negative. This is a reason to not have too high of an impedance.

I am not sure why this is working out so nicely, but this next part also works out nicely.

the gain of the 6L6 is 8.

max signal on the plate 324 / 8 = 40.5, max signal on the grid.

I like how 40.5 is lower than the bias of 52, but not significantly lower than 52.

this helps confirm that the bias sets a practical upper limit for the signal on the grid.


the last check, screen voltage 447 / 8 = 56 volts required for bias, on the cold side. My bias is 52. Sounds just right.


why bother?


my goal for a long time is to be able to look at any Fender type schematic and be able to just know what the expected signal level is at any place in the amp.

I believe we are there.

I could not have done it without the help of the TDPRI!

it takes a group effort.

thank you TDPRI
 

peteb

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I see a problem.

i used the impedance from one half of the primary, because I wanted to know the signal for one tube only.

if only one out of two tubes is being considered, then only half of the output power should be used.

right?


we would then see significantly less signal on the plate.

in calculating the gain, Vout / Vin, I was surprised that the specified gain of the tube worked. In the pre amp, the 12AX7 has a theoretical gain of 100, but once the parts are connected the gain drops to about half of 100.

maybe the signal voltage is less than I calculated, and the gain is less, leaving the maximum signal voltage on the grid about the same.


I am confident that the max signal on the grid is more than half the bias but less than the bias.
 

2L man

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Pete, you use strange way to approach when you first decide the power of 50W and try to "proof" how it comes. The power is second last what comes out of audio amplifier!

Last is when loudspeaker turns power to sound pressure but on workshop a non inductive resistor should be used because it behave the same thru whole frequency range and it is how we make power the last and huge impedance variable, what instrument loudspeaker represent, gets out of system.

Study "the loadline principle" and it would be more interesting to read your work and participate when you "speak same language all tube quys speak"

Also define Voltages (Currents and Powers too) better using Vp (peak) Vpp (peak to peak) and Vrms (root mean square) always there is a possibility to confusion, not to forger VAC and VDC. Just V when you write of grid/g1 and anode is not enough when signal require Vp or Vpp.
 

peteb

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Thanks 2L man, that is a nice reply.

I made too many mistakes, using the wrong impedance, using the impedance incorrectly, and interpreting the result wrong.

I am going to start with a clean sheet:



Also define Voltages (Currents and Powers too) better using Vp (peak) Vpp (peak to peak) and Vrms (root mean square) always there is a possibility to confusion, not to forger VAC and VDC. Just V when you write of grid/g1 and anode is not enough when signal require Vp or Vpp.

unless other wise stated, VAC is always assumed VAC.

it is true that specifying VAC versus VDC is clearer, but often the context is enough.
 




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