Alternate Half-Power for Quad Output

[Wilder Amplification]

Thanks for the detailed explanation, John.

Isn't plotting the load line at half the slope (at 1/2 Zpri instead of 1/4 to represent four tubes vs. two) the same as scaling the vertical dimension of the curves by a factor of two? Each device is handling half the current at the same voltage, as if the load resistance were doubled.

To be clear, I'm plotting only one side of the output, and only the class B load line.

MPM

Whether you're running 2 or 4 tubes, if you're plotting only the B line, you need to use 1/4 the total Zplate-plate value as the load from plate to center tap in Class B mode = 1/4 Zp-p in Class B mode, NOT 1/2 the Zp-p. The load from plate to center tap only appears to be 1/2 the total plate-plate Z in Class A mode.

For 4 tubes, you leave the load line the same, but double the plate current scaling on the graph for a 4 tube amp.

The reason why the plate-center tap load is only 1/4 the plate-plate load is because the impedance ratio of an OT is the square root of the physical turns ratio.

Example...you have a transformer with a turns ratio of 15.8:1 from full primary to full secondary -

Turns Ratio = 15.8

15.8^2 = Plate-Plate Impedance Ratio = 250

Using a 16 ohm load, we reflect -

250 x 16 Ohm = 4K

Now let's cut the turns ratio in 1/2 to represent the "plate-center tap" turns ratio

15.8 / 2 = 7.9

7.9^2 = Plate-Center Tap Impedance Ratio = 62.41

Again, using a 16 ohm load, we reflect -

62.41 x 16 ohm = 1K (actually 998 Ohm but we'll round it off)

As you can see, the plate-center tap load impedance ends up being 1K, which is 1/4 the plate-plate load impedance.

Now when in Class A mode, you've got current being pulled through both halves of the primary at the same time, which causes the plate-center tap impedance to appear to be 1/2 the plate-plate load impedance, which causes the different slope in the Class A region of the load line.

[martin manning]

Whether you're running 2 or 4 tubes, if you're plotting only the B line, you need to use 1/4 the total Zplate-plate value as the load from plate to center tap in Class B mode = 1/4 Zp-p in Class B mode, NOT 1/2 the Zp-p. The load from plate to center tap only appears to be 1/2 the total plate-plate Z in Class A mode.

For 4 tubes, you leave the load line the same, but double the plate current scaling on the graph for a 4 tube amp.

I think I see where you're coming from now. Doubling the values on the current axis and plotting the current for two tubes will put the load line in the right place for a four-tube amp, but the result is no different than doubling the load and plotting the current for one tube as I did. It's like dividing the original load into two parallel loads, each of which must have twice the impedance of the original to be its equivalent when they are paralleled.

However, if you want to plot the load line representing the case where two of the four tubes have been removed, you have to revert back to the original single-device current scale, but you also have to move the left end of the load line up to the same current as before, which will rotate the load line clockwise about the (B+, 0) point on the Va-k axis. It will now run across the grid curves on a different path, with the result again being exactly the same as I plotted.

MPM

[Wilder Amplification]

I think I see where you're coming from now. Doubling the values on the current axis and plotting the current for two tubes will put the load line in the right place for a four-tube amp, but the result is no different than doubling the load and plotting the current for one tube as I did. It's like dividing the original load into two parallel loads, each of which must have twice the impedance of the original to be its equivalent when they are paralleled.

No...each of which would have the same impedance it would have in a 2 tube config. Paralleling them would cut the load impedance in 1/2, which is what you're doing in a 4 tube amp.

It's not the addition of more tubes that doubles the power. It's the fact that you're using 1/2 the load impedance across the same B+ voltage that doubles it. But you have to double the tube count because one pair of tubes cannot handle passing double the load current. By adding a tube in parallel with each tube, it makes it so that each tube only has to handle passing 1/2 the total load current. Since your load current has been doubled due to the load impedance being cut in 1/2, each tube will be passing the exact same amount of load current it would be passing in a 1/2 power config with double the load impedance because the total load current now has two paths (i.e. the two tubes) to take. You've basically turned a single lane one way street into a two lane one way street.

However, if you want to plot the load line representing the case where two of the four tubes have been removed, you have to revert back to the original single-device current scale, but you also have to move the left end of the load line up to the same current as before, which will rotate the load line clockwise about the (B+, 0) point on the Va-k axis. It will now run across the grid curves on a different path, with the result again being exactly the same as I plotted.

MPM

This is exactly what happens if you remove two tubes WITHOUT doubling the load. If you pull two tubes but DON'T double the load, then yes the load line shifts clockwise about the (B+/0) axis, which is what you DON'T want to happen. Doubling the load once the two tubes have been removed puts the load line back in the right spot where it crosses the grid curves right at the Vg1=0 line and doesn't exceed the maximum dissipation curve.

But...as you can see, it's NOT the amount of tubes you have in the amp that shifts the load line. It's the doubling/halving the load impedance that shifts it. The amount of tubes you have in the amp determines whether the load impedance needs to be doubled or halved to place the load line in the right spot.

[martin manning]

This is exactly what happens if you remove two tubes WITHOUT doubling the load....

Which is exactly what I've been talking about since post #1: I asked what is the difference between simply pulling two tubes and inserting a 10k between two of the cathodes and ground, as far as the load line is concerned. Note the two load lines I plotted are labeled Two Tubes, 2k and Four Tubes, 2k.

MPM

[Wilder Amplification]

Which is exactly what I've been talking about since post #1: I asked what is the difference between simply pulling two tubes and inserting a 10k between two of the cathodes and ground, as far as the load line is concerned. Note the two load lines I plotted are labeled Two Tubes, 2k and Four Tubes, 2k.

MPM

And what I stated was that with two tubes on a 2K load, assuming a typical B+ of 450 volts, you would hit a dissipation of 100 Watts...well over double what any typical guitar amp power tube (i.e. 6L6/EL34/6550/"KTXX"), which would cause the average dissipation to exceed the maximum rated dissipation. As such, you would be forced to double your load to keep the tubes from going overcurrent.

So yes, your steeper load line illustrates what would happen if you removed two tubes.

Now my point is, whenever you remove two tubes from the circuit you must double the load. I don't care how you do it...be it through 10K cathode resistors pushing the tubes into cutoff, lifting the cathode grounds, killing the screen voltage, or physically removing them from the amp...the load MUST be doubled to cut the power output in 1/2...plain and simple. You can see from the steeper load line on the two tubes w/2k load that it's not correct...as such the load must be doubled to put the load line in the same spot. Power output is a function of B+ voltage across a load, not a function of how many tubes you have in the amp. The fact that you have to add two tubes to get double the power is a byproduct of the fact that cutting the load impedance in 1/2 allows double the current to flow, and two tubes alone cannot handle passing double the current by themselves.

Kinda like when you put bigger tires/wheels on a lifted truck you have to gear down the rear end. It's a two part action...you have to change one when changing the other to keep everything matched.

About the only way to drop power output headroom without having to double the load would be to cut the B+ in 1/2.

[martin manning]

Ok, glad that's straightened out. If we ever meet I'll buy the beverages!

MPM

[ingrast]

double plate load

Ok, glad that's straightened out. If we ever meet I'll buy the beverages!

MPM

Sorry,

Does your output transformer have taps for 4 / 8 / 16 ohms ?

The simplest way to step up load impedance to match 2 instead of 4 output tubes is to connect 8 ohm load to 4 ohm tap and so on. Same RMS voltage squared divided by twice load impedance = half power and half current on half number of tubes so same current by tube.

Rodolfo

[Dave H]

Alternate Half-Power

I have a 4 x 6V6 amplifier which has a half power switch to disconnect two tubes at the cathodes but there was hardly any sound difference between full and half power so for an alternative half power switch I changed it to run 2 x EL34 at 4k for full power or 2 x 6V6 at 8k for half power. I can hear the difference between full and half power now. I just connect the 8 ohm speaker to the 4 ohm jack as ingrast said to make the load 8k for the 6V6s.