# What Plate Dissipation is - something doesn't seem right

Discussion in 'Amp Tech Center' started by peteb, Jul 28, 2017.

1. ### robrobPoster ExtraordinaireAd Free Member

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Somehow I missed jazzguitar's post at #120. Excellent post.

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2. ### petebFriend of Leo's

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If you look closer at my chart, you will see the DC voltage on the cath subtracted from the DC voltage of the grid, giving the DC voltage difference difference between them listed as the bias, this is the idle DC bias of the tube measured. The cathode to ground, as you say would be the cathode voltage minus zero, that would only tell you what the cathide volatge, I have no idea why you brought that up, it is irrelevant and irrelevant to my numbers?

BIAS on Fender Amps, note how they increase from left to right, that's because the signal is growing.

Amp------------preamp-------phase inverter----------power

Champ.............-2.......................NA......................-20

.......................................grid: 23
.......................................cath: 74
Princeton...........-2............bias: -51........................-35

.......................................grid: 68
.......................................Cath: 104
Bassman.............-2........... Bias: -36.......................-45

Twin....................-2....................?..........................-50

3. ### petebFriend of Leo's

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Danlad, you appear to understand this better than anyone else that has posted here, including myself.

you ARE saying the bias grows to fit its own need.

This means that a 10 volt Signal will work against the larger bias than a 1 volt signal. I would assume that the general relationship holds true, the bias is equal to or greator than the signal.

Bias is usually IDLE bias. Are you saying the idle bias is 1 volt, and it grows bigger as signal is applied?

I think that sounds reasonable, but what then about my data, above.

The voltages were taken at idle, and the potential DC difference between the cathode and the grid, appears to be no where close to one volt.

4. ### petebFriend of Leo's

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Excellent post old tele man, I like the bouy analogy!

So if a 10 volt signal is applied to the grid of the PI tube, then the bias is 10 + X.

Is that the right idea?

Thanks

5. ### Old Tele manFriend of Leo's

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Some things to clarify in that question:

1) The AC-signal "rides" on the +49Vdc (above ground) of the 56kΩ cathode-side resistor, which is passed to the control-grid thru the 1MΩ resistor (remember, no grid-current flows).

2) A 5V(pk) signal is half of a 10V(pk-to-pk) signal, and because we're "splitting" the signal in half, it's common practice to consider V(pk) rather than V(pk-pk).

3) Assuming that's a 5V(peak) signal, the PI control-grid will 'see' +54V(+5Vpk)/+44V(-5Vpk) AC-voltage riding above/below the +49Vdc point. However, the control-grid's time-averaged influence on tube conduction will vary *around* the -1.2Vdc(bias) voltage the tube experiences between its control-grid and its cathode.

4) The 5V(pk) input signal to the PI becomes 3.54V(rms) times tube gain (mu/(mu+1)) output across the plate-side(-half) and the cathode-side(+half) resistors (remember, output tube(s) respond to RMS levels #). When driven hard enough, the plate- and cathode-outputs can approach ½ of the B+ to PI.

# why? Because power output is commonly calculated using RMS-voltage and -current values.

Last edited: Aug 8, 2017
6. ### BendyhaFriend of Leo'sSilver Supporter

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Twisting facts again...

No...the signal goes up 5V...then down 5V, with such a small signal the 1v bias will be hardly changed, the difference between cathode and grid will remain close to 1V.

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Nah, am just making sense in my own terms of what others post.

The analogy of the waterline on a buoy rising and falling with waves is a good one. The buoy sits roughly as deep in the water regardless (our bias) even when the waves are high or low (our signal rising & falling) as it floats on top of the waves. It doesn't shoot 30 foot out the water or spontaneously sink, but the waves might rise & fall 30 foot. The buoy just pokes out the water just as much as it has been set up to regardless.

You seem to take bias as something that references ground, which it can do, but cathodyne PIs ride high on the tide.

Last edited: Aug 8, 2017
8. ### petebFriend of Leo's

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The question:

What is the bias on the cathodyne phase inverter?

1 volt?

Much higher than 1 volt?

Both?

9. ### petebFriend of Leo's

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Thank you old tele man, I think what you wrote above will make more sense to me when we all decide the answer to the question:

What is the bias on the cathodyne phase inverter?

1 volt?

Much more than 1 volt?

Both?

10. ### jazzguitarTele-Afflicted

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As has been stated before, the bias is about 1 milliampere which causes about 1 volt drop at the 1k bias resistor of the cathodyne phase inverter circuit of the Princeton.

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11. ### petebFriend of Leo's

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OK, I see where you are coming from.

I like the bouy analogy as it keeps the bias higher than the signal. One detail I would add, the water is the signal and the bias is NOT the bouy, but the TOP of the bouy, always out of the water.

I don't have a problem with the PI bias shifting up with the signal, but my idle bias measurements don't show the need for it. The idle bias is goodly high enough as it is, it doesn't need to go up.

That brings up the question of why doesn't it get measured? Why isn't it on the schematic like the other biases?

I have my own theory on that.

When I recently did a voltage survey on my amps (first complete one ever) I measured all the pins. The one pin I failed to get a meaningful voltage from was one of the PI grids, I kept getting a loud Nasty pop, and I am done measuring grids of PI tubes, but I did get my data as I will show in the next post

12. ### petebFriend of Leo's

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That's not what I'm finding at all.

13. ### petebFriend of Leo's

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In the first half of July I did these voltage surveys, the voltages supporting my claim that the PI bias is significantly more than one volt come from these surveys:

14. ### jazzguitarTele-Afflicted

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(Sigh)

The AA165 does not apply as the phase inverter is of a different type.

As you can see the voltage you read from grid (pin 7) to ground in the 6G2 Princeton is about 23 volts which is caused by the low internal resistance of your meter. Reading this voltage accurately requires either a meter with much higher internal resistance than the typical 10M, or some other method reading it correctly (with a VTVM a direct reading may be possible).

This I have stated before, and I think it is also mentioned in Bendyhas explanation of the cathodyne phase inverter.

This cathode follower circuit is extremely sensitive at its input (the grid, pin 7 in this case) so any resistance to ground from here lower than say 100 megaohms or so will pull the voltage down. If you read the cathode voltage at the same time (with another meter) you will find it close, it "followed" the input voltage.

If you look at the 6G2 schematic you will see it correctly lists the cathode voltage at 56.5 volts with the grid voltage (via the 1M resistor, the grid current is negligible) correctly at 55 volts which means by design -1.5 volts bias voltage, same as the other preamp tubes in the amp. This is probably a nominal voltage implied by the engineer, not read.
Actual amps have close to 1V, depending on the actual tube used.

a) do not listen to advice given you here freely by many
b) do not know how to read the bias voltage of a cathode follower.

Q.E.D.

Hint: do that reading again but only the voltage across that 1k resistor. I.e. red probe to pin 8, black probe to pin 7.

Last edited: Aug 9, 2017
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15. ### Old Tele manFriend of Leo's

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The FIRST thing I immediately see are the two DIFFERENT PI circuits, 6G2 Princeton (cathodyne) vs. AA165 Bassman (Long-tail); are you still trying to compare APPLES and ORANGES?

This is what happens and why inside the cathodyne (split-load) PI tube circuit (in physical chronology):
1) the filament heats up developing a "space charge" cloud of electrons around the cathode.
2) until the plate voltage (B+) is established, the tube does not conduct current.
3) plate voltage ramps up and reaches it full voltage (B+).
4) because there is NO voltage across the control-grid-to-cathode (0=Vgk) yet, the tube tries to conduct at almost maximum current.
5) the quickly rising plate current thru the plate-load resistor(56kΩ) and the cathode (1kΩ+56kΩ) resistors develop their respective voltage "drops."
6) the control-grid "sees" its grid-to-cathode voltage drop from 0.0VDC to -1.0VDC(BIAS) as the tube current reaches 0.877 milliamps, because...
7) the cathode voltage "drops" are: (a) Vk(top): 50VDC = (56kΩ+1kΩ)×0.877mA; (b) Vk(bottom): 49VDC = 56kΩ×0.877mA; (c) Vg(bias): -1.0VDC = 49VDC-50VDC.
8) in the plate circuit, the same 0.877mA current "drops" a similar 49VDC, ie: Vp = (B+)-56kΩ(plate)×0.877mA = (B+)-49VDC.
9) with no ac-input signal, the tube is now in its quiescent (idle) condition.
10) As the components (resistors) heat up, their values can drift, resulting in slight variations in voltage drops...and, thus, BIAS voltage.

NOTA BENE:
• Measurements make between component(s) and chassis are 'ground-referenced.'
• Measurements make across a component are 'device-referenced' and can be above, at, or even below ground potentials.

Last edited: Aug 9, 2017
16. ### BendyhaFriend of Leo'sSilver Supporter

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23V volts on the grid seems about right. A useless measurement, almost as useless as this thread.

It is a measurement of the cathode resistors voltage drop being split by the voltage divider you create with the 1M grid bleed and the (probably) 1M input imp. of your meter, after the cathode voltage is shifted by the insertion of this added load changing the current draw across said resistor and the shift of tube bias at the grid changing the Q point and current draw over said resistor. A very usefull number to have, try comparing it to the amount your lights dim when you place two fingers on to voltage carrying leads running to the bulb, you will find it has the same ratio, and is just as usefull to know, at just as instructive.

17. ### petebFriend of Leo's

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Thank you, all three of you, for your well written replies.

I agree, the measurements may be meaningless.

My side: the AC signal voltage is supposed to have a lower magnitude than the DC bias voltage, as the AC is subtracted from the DC to open up the tube.

Is anyone willing on the otherside to claim this statement:

A one volt DC bias is large enough bias for a tube to amplify a signal as large as 10-20 VAC.

I would like any of you three to either claim that statement is true, or say it can't be true.

I'm ready for the thread to end, all people can do now is state their position, and then we can let it rest.

18. ### BendyhaFriend of Leo'sSilver Supporter

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Do you mean....you want a dead person to somehow "cross over" from the otherside and give you the sign ?

PETE.........PETE.........IT'S TRUE........A TUBE CAN DO THAT IF YOU DESIGN THE CIRCUIT AROUND IT TO DO THAT....

Yes, it happens all the time. If the circuit is set up like a standard cathode follower or Cathodyne PI, or many of the other ways we could easily confuse you with.

Whats more, the measurment you have taken from you amps prove this fact.

19. ### petebFriend of Leo's

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Thank you Bendyha,

Anyone else?

20. ### petebFriend of Leo's

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Just another thought that I forgot to post.

People above, have pointed out that a long tail phase inverter is different than a cathodyne follower phase inverter.

But I want to point out that they both have the same curious phenomena.

Instead of the current attacking the cathode with a full frontal assault, the current is divided, and a flanking manuver allows the current to reach the grid via an alternate route. Here the grid and the cathode are connected thru resistance on both types of PI, this is why it is hard to measure the voltage on the grid. When the grid and cathode are connected, it is a form of feedback.

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