How to calculate the maximum signal voltage and current on the plate, using maximum output power and the impedance of of the OT and speaker

peteb

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in understanding how tube amps work, it is good to understand what is happening at maximum power and particularly in the high energy area where the power tube interacts with the amp.

knowing what to expect at the plate in terms of maximum signal voltage and maximum signal current is a big step in that direction. Many folks offer anecdotal information, but I prefer hard facts and data including numbers.


here is an underlying question.

we know the level of the DC current at idle. What happens to the plate current during full signal?
 
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peteb

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One underlying principle that must be understood is the conservation of energy principle.

The principle of energy conservation states that energy is neither created nor destroyed. It may transform from one type to another.


The conservation of energy principle applied to the output transformer implies that the output power cannot exceed the input power.

Losses. There are always losses to be encountered.


how efficient is a transformer? Let’s look it up.


The efficiencies of power transformers normally vary from 97 to 99 percent.

from



NOTE: you can trade in your voltage or your current for a higher or lower value, but not power.


lets assume 98% efficiency and round up to an even 100.
 

peteb

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CHAMP AMP
5W, single ended 6V6

Hammond transformers 1750C, 1760C and 1760CP all offer 5K and 8K impedance with a 4 ohm speaker.

5K

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

25,000 = V^2
maximum VAC voltage of the signal is 158
maximum AC current of the signal is 5W / 158 volts = 32 mA

8K

5W = V^2 / R = V^2 / 8,000 ohms

40,000 = V^2
maximum VAC voltage of the signal is 200
maximum AC current of the signal is 5W / 200 volts = 25mA
 

peteb

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PRINCETON AMP
15W, push pull 2X6V6

Hammond transformers 1750E, 1750EP and 1760E all offer 8.5K impedance with an 8 ohm speaker.

let’s consider one tube only so the impedance is 4,250 ohms and the power is 7.5 W.


7.5 W = V^2 / R = V^2 / 4,250 ohms

31,875 = V^2
maximum VAC voltage of the signal is 179
maximum AC current of the signal is 7.5W / 179 volts = 42 mA
 

peteb

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BASSMAN AMP
50W, push pull 2X6L6

Hammond transformers 1750L and 1760L both offer 4.2K impedance with a 4 ohm speaker load.

let’s consider one tube only so the impedance is 2,100 ohms and the power is 25 W. Then we will try both tubes to see if we get the same result.

single tube

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

52,500 = V^2
maximum VAC voltage of the signal is 229
maximum AC current of the signal is 25W / 229 volts = 109 mA

two tubes

50 W = V^2 / R = V^2 / 4,200 ohms

210,000 = V^2 (4 times as much, uh oh)
maximum VAC voltage of the signal is 458 (twice as much, but it is for two tubes)
maximum VAC voltage of the signal for a single tube is 229.
maximum AC current of the signal is 50 W / 458 volts = 109 mA
 
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peteb

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TWIN REVERB AMP
100W, push pull 4X6L6

Hammond transformers 1750W and 1760W both offer 2K impedance with a 4 ohm speaker load.

The TR is two tubes paralleled on each side of the OT primary. The calculation needs to change, but the individual tube should see very similar conditions to the tube in the BASSMAN. Let’s see how it works out.

let’s consider two tubes paralleled on one side of the OT. So the impedance is 1,000 ohms and the power is 50W.


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

50,000 = V^2
maximum VAC voltage of the signal is 224

if the voltage from two tubes in parallel put 224 VAC on one half of the primary winding, how much does each tube contribute?

voltages added in series are additive, voltages added in parallel are not additive, they are equal to the combined result.

Each tube puts out and sees 224 VAC, very close to the 229 of the BMAN.

maximum AC current of the signal is 50 W / 224 volts = 223 mA, but that is for two tubes, so each tube sees 223/2 = 112 mA maximum, very close to the 109 of the Bassman.
 

peteb

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Thank you for reading this far.

Analysis of the plate voltages.

amp———Max plate signal (VAC)——-Plate voltage (VDC)

champ(5k)———158—————————350-400
champ(8K)———200—————————350-400

princeton——-—179—————————350-400

bassman———--229—————————410-460

Twin rev——-—-224—————————-410-460


conclusion:

the maximum AC signal voltage on the plate is approximately half of the the DC voltage on the plate.

The AC voltage subtracts from the DC voltage, so overall peak voltage at the plate is not a concern.
 
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peteb

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Analysis of the plate currents

amp—idle current(mA, DC)—max signal current (ma,AC)

champ(5K)———-45-50————-32
champ(8K)———-45-50————-25

princeton————-15-20————-42


bassman—————28-33————109

twin———same as bassman

conclusion:

CHAMP
the maximum AC signal current is a little more than half of the idle DC current.

PRINCETON
the maximum AC signal current is double or more than double the idle DC current.

BASSMAN and TWIN REVERB
the maximum AC signal current is around three time the idle DC current.

The AC current adds to the DC current, so overall peak current, combined AC and DC, at the plate could be a concern. Sometimes a plate will red plate at idle. Imagine the heat when the AC plate current is added to the DC plate current. One plus three times is four times.

the screen does not have AC voltage on it but the screen pulls AC current. In this respect, we can only assume that the current on the screen is going to behave like the current on the plate. We can assume the AC screen current could be three times more than the DC screen current. The signal could cause the screen current to increase by four times.

this could be a hazard for the screen and the screen resistor.
the Power handling of the screen resistor depends on the rating. They don’t dissipate much power be because the current is low and the resistance is relatively low.

screen dissipation could be a serious concern.
 
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peteb

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Screen max dissipation power.

I don’t know about the 6L6, but I saw a spec for the 6V6 that the max plate dissipation is 12 W and the max screen dissipation is 2 W.

let’s look at the Bassman.

Idle plate current is 30 mA

6% of 30 is 1.8, ,let’s go with 2 mA of DC screen current at idle.

my Bassman screens are both at 447 VDC at idle. i will use 450.

screen idle dissipation calculation

0.002 amps times 450 VDC is 0.9 watts


lets make the .9 watts one watt.

the screen current could increase four times.

I see the possibility that the screen dissipation could peak at around 4W and that may be close to the max screen power dissipation for a 6L6 tube.


conclusion about screen failure.

it is probably the screen and not the screen resistor that will fail from too much power dissipation.

NEXT STEPS

I have three defunct 6V6 tubes.

I believe it is time to crack them open and inspect for screen failure.
 

peteb

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Thank you for reading.

Please correct me if you read anything that is wrong including all assumptions.
 

peteb

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Thoughts on gain/AC/DC

gain of tube type = idle VDC on plate/screen / idle VDC on grid

and

gain of tube = VAC signal on plate / VAC signal on grid(180 out of phase)

In a real general sense, both the AC on the plate and the AC on the grid are going to peak out at max clean power of the amp, at a level that is approximately half of the VDC in that location, whether it is the plate or the grid. It could be more than half but less than the DC.

Once you know the DC level, it is fairly easy to understand the AC level anywhere in the amp.
 

bebopbrain

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What happens to the plate current during full signal?

One can put a current probe on an amp and see the current on an oscilloscope.

An amp just before clipping puts out a sine wave (roughly) at rated power.
An amp at full power puts out a square wave (roughly) at twice rated power.
I think this is a mistake on your part.
A 5W amp produces a 10W square wave output at the speaker.

Consider a push pull amp at full power.
One power tube is off (cut off) and the voltage on its plate is B+.
The other power tube is saturated (fully on) and the voltage on the plate is near 0V.
The transformer sees B+ between the saturated tube and the center tap.

I = P/V = 2 * rated power / B+

Let's say rated power = 40W and B+ = 400VDC
Instantaneous plate current (for the conducting side) is = 200mA

In reality current is slightly higher to make up for losses and because the voltage will be less than shown, since saturation isn't a perfect short circuit.
 




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