Alternate Half-Power for Quad Output

[martin manning]

This came up on another board; I'm curious as to what folks here think:

Suppose you have an amp with a quad of 6L6's working into a 2K a-a load, and you want to cut the power output by half. You could pull two tubes, but the remaining pair would need to see a ~4K impedance to maintain the same load lines as before.

Instead of removing two tubes, suppose a 10K resistor is placed between the cathode and ground of one tube on each side of the OT, causing those two to be biased into cut-off. Unlike the previous case, they are still in the circuit, but essentially not conducting any current.

The question is, what would happen to the load lines of the pair that are still connected normally?

MPM

[stokes]

If the two tubes are not conducting current,I would think it would have the same effect on the reflected impedance as removing them.This is just a guess,tho.If you leave the tubes plugged in and lift the cathodes you are stopping the tubes from conducting,so I assume it would be the same end to a different means.

[Chuck H]

Yeah, I thought about this and wasn't sure myself but that's what I was thinking too.

FWIW 1/2 power doesn't have nearly as much effect on volume as most designers and players would like. It might be an idea to simply lift the cathodes on two of the power tubes while simultaneously adding resistance to the OT secondary to bring the reflected primary impedance up to what a pair of tubes would like. This would produce 1/4 power and a more useful decrease in volume. Just a thought.

Chuck

[DIY_Guy]

We had talked about this a while back in a thread I started. I did what Chuck has mentioned- lifted the outer pair, and also added a big resistor. Works fine. Doesn't really cut volume per say, but it seems to break up the power tubes much easier than in 100W mode.

[martin manning]

Thinking this through some more I think it becomes clear: If half of the output tubes aren't trying to push current through the OT, the others will see it as a reduction in load impedance. With a 10k cathode resistor, I don't think more than a few mA will flow, even at high signal levels, and so the other tubes will see about half the load they see with all four cathodes connected to ground. Thanks for the replies!

MPM

[Wilder Amplification]

Thinking this through some more I think it becomes clear: If half of the output tubes aren't trying to push current through the OT, the others will see it as a reduction in load impedance.

First off...tubes don't PUSH current through an OT. The power supply PULLS current through the OT. The tubes just act like a voltage controlled "current valve" that uses an input signal to control the amount of current flowing through the OT primary at any given time. As the input signal swings positive/negative, it offsets the grid bias voltage by an amount equal to the input signal voltage, which "opens and closes the valve" so to speak.

With a 10k cathode resistor, I don't think more than a few mA will flow, even at high signal levels, and so the other tubes will see about half the load they see with all four cathodes connected to ground.

Not quite how things work.

The power supply voltage against the load (i.e. the OT primary) is what determines current flow. With 1/2 the required plate to plate load and only two tubes installed, the power supply will be pulling DOUBLE the current of a 50 watter, but now the only two tubes that are installed are being expected to pass the full amount of current that the power supply is pulling through the load rather than each one only seeing 1/2 the total load current like they would if all 4 were installed (this is why 100 watters have to have double the tubes in the first place...because the load is halved which causes the power supply to pull double the current). It is for this reason that the load must be doubled when only running two tubes...to cut the total load current in half. Reducing current to two of the tubes is only going to increase the amount of current that the other two have to pass by the amount that the current to the other two was reduced (current takes the easier path).

[martin manning]

Different thought process, same conclusion re pulling two of four tubes... It's my turbomachinery background showing through.

I was more interested in the difference between disconnecting two cathodes and inserting a 10k's to ground. I don't believe there will be much difference at all as regards how the load lines for the two active tubes will migrate.

MPM

[Wilder Amplification]

I don't believe there will be much difference at all as regards how the load lines for the two active tubes will migrate.

MPM

Um...the amount of tubes you have installed doesn't determine the load line. The value of the load itself against the power supply voltage determines the load line (Ohm's Law). As such, given the same load, pulling two tubes from the circuit has no effect on the load line itself.

The load line along with the max dissipation rating of the tubes you're using will tell you what tubes you can safely run with that load, or how many of the same tube you would need to run into the load safely without overdissipating at 1/2 swing. If the dissipation at 1/2 swing = more than double the dissipation rating of the tube you either have to increase the load or add more tubes to share the extra current.

Tubes don't see impedance. They only see voltage and current. We use the voltage and current to CALCULATE the EFFECTIVE IMPEDANCE.

Basically the long and short of it is that the power supply voltage and the load govern everything, not the other way around. If you think about the tubes and OT like a simple adjustable voltage divider, with the top resistor in the divider representing the OT primary load, the grounded resistor in the divider being a potentiometer which represents the tube, and the rotation of the pot shaft representing the input signal, it should become crystal clear as to what's really going on in the circuit.

[Bruce / Mission Amps]

First off...tubes don't PUSH current through an OT. The power supply PULLS current through the OT. The tubes just act like a voltage controlled "current valve" that uses an input signal to control the amount of current flowing through the OT primary at any given time. ... (snipped out)

I'm not an electrical engineer but I'm so sure about that statement...
I was under the impression that current flows in the opposite direction of electron flow... if there even is such a thing as "electron flow".
If my assumption is correct (and many times it is not) then electrons are moving on an atomic level, atom to atom, from ground through the tube into the OT but "current" ( possibly the + space between the electrons?) is flowing from the OT through the tube to ground... in the other direction.

[Wilder Amplification]

I'm not an electrical engineer but I'm so sure about that statement...
I was under the impression that current flows in the opposite direction of electron flow... if there even is such a thing as "electron flow".
If my assumption is correct (and many times it is not) then electrons are moving on an atomic level, atom to atom, from ground through the tube into the OT but "current" ( possibly the + space between the electrons?) is flowing from the OT through the tube to ground... in the other direction.

Is the glass 1/2 full or 1/2 empty?

The positive charge pulls electrons through the tube. Electrons flow from the negative, boil off the cathode, and are pulled to the plate and return via the positive side of the supply. The PULLING FORCE that pulls the electrons eminates from the positive side and extends toward the negative.

At some point I'm sure the two meet in the middle if you wanna get right down to it.

But this has nothing to do with what I stated. A tube is just an electron current controller. The power supply pulls current through the load while the tube just controls how much current the power supply CAN pull through the load at any given point in the swing. The input signal controls the tube's effective plate resistance of the tube (which appears as a series resistance to the OT primary), by modulating the bias voltage at the control grid (signal appears as AC superimposed onto the negative DC bias voltage, which causes the grid bias to modulate more/less negative at input signal frequency).

The OP was talking as if the tubes PUSH the current through the OT primary, but in reality it's the pulling force of the positive side of the supply that PULLS electrons through the OT. The tubes just control this electron current through the OT primary.

[martin manning]

Um...the amount of tubes you have installed doesn't determine the load line....

So, why is the usual practice to run a pair of 6L6's with 4k primary impedance and a quad at 2k, if not to place the load lines appropriately on the plate curves?

MPM

[Wilder Amplification]

So, why is the usual practice to run a pair of 6L6's with 4k primary impedance and a quad at 2k, if not to place the load lines appropriately on the plate curves?

MPM

Because running 1/2 the load impedance allows the power supply to pull double the current. But you have to double up on output tubes to share that current because a single pair of them cannot handle double the current by themselves without over dissipating. With 1/2 the load pulling double the current and double the tubes controlling said current, each tube will see the same amount of current as it would see with double the load/1/2 the current and only a single pair of tubes.

The same reason why you run a single rectifier tube on a 50 watter, but if you want to use rec tubes on a 100 watter you double up on them in parallel so that you can safely pull double the TOTAL current, but each one would only have to pass 1/2 the total.

The amount of tubes don't control the load line. Rather, the load line controls what type of tube relative to max plate dissipation as well as the amount of said tubes required to safely run into said load without over dissipating.

[martin manning]

This is my understanding of running two tubes vs. four into the same Zpri- two different load lines.

MPM

[Wilder Amplification]

This is my understanding of running two tubes vs. four into the same Zpri- two different load lines.

MPM

There is no way that that load line can be the same load.

Moreover, You cannot graph two tubes onto the same graph like that. They don't work that way. Two tubes in push pull are graphed via a composite load line. This is where the load line for one tube is graphed using 1/2 the Zp-p load for the Class A load line, and 1/4 the Zp-p load for the Class B load line. Once that load line is graphed, it is then inverse mirrored to make one big load line (the 2nd graph gets inverse mirrored since the tubes work out of phase). This "rolls both the tubes in a push pull output section into one big tube" as Merlin likes to state.

Here is an example of a Class AB load line with a 600V B+ and a 5K Zp-p using 2 x KT88s. The dotted "arc" in the graph is the maximum rated dissipation curve of each tube while the dot in the center of the graph represents the idle bias point, which I've set on the graph at 50mA per tube, which would be right at 70% of the plate dissipation at a 600V plate voltage. You can also see that at full power out, the total peak-peak swing of the voltage across the full primary = double the B+. Since a push pull output section is essentially a difference amplifier and the two work out of phase, this provides a summing effect since 600V - (-)600V = 1200V.

In this graph, you can see that in the region where the graphs overlap, the load line has a slightly different slope than the portion of the load line that is above the overlap region. This is because when the amp is operating in the Class A region, both tubes are on and makes both halves of the primary = 1/2 the total primary Z. The overlapping region of the graphs is the "Class A" region where the two tubes are both conducting at the same time (coincidentally, this is what we call the "overlap" region) in Class A mode. The portions of the load line above the overlap are the Class B load line, with the transition from one to the other happening just above/below the overlap region. The borders of the graph overlap region are the area where one tube goes into cutoff, which begins the transition to Class B mode. Once the tube transitions to Class B mode, the other goes into cutoff, which takes that side of the OT out of the circuit, and the impedance of the conducting half of the primary = 1/4 total primary Z, which causes the steeper slope in Class B mode -



From this load line you'll also notice that contrary to popular belief, maximum dissipation DOES NOT occur at the peaks of the swing, but rather at 1/2 the swing of the signal. This is evident due to the load line being the furthest above the max dissipation curve represented by the dotted arc when the plate voltage drops to 1/2 the total B+ at 1/2 swing.

In order to graph a 4 tube amp, each graph would have to be adjusted to represent a pair of paralleled tubes (instead of each graph representing 1 tube each in a 2 tube push pull section), which involves doubling the plate current scaling of the graph. Everything would look the same and your load line would appear to have the same gradient, but everything is effectively doubled since the graph's plate current scaling has been doubled (plate current gets doubled since the effective plate resistance of two tubes in parallel = 1/2 the plate resistance of one).

On your graph, there's no way that the two load lines could be the same load. This is because once the tube gets driven to peak swing (i.e. just before saturation), it's effective plate resistance is at or near zero (actually in the hundreds of ohms, but not enough to consider relative to the load line), plate voltage drops to zero, the power supply sees just the load of 1/2 the OT primary and as such will draw the same current whether you have two or four tubes installed given the same load and the same B+ voltage.

The long and short of all of this is that in order to double the output power given the same B+ voltage, you have to 1/2 the load impedance to draw double the current. One pair of tubes alone cannot handle controlling double the load current by themselves, so you parallel two tubes across the other two tubes, which allows them both to pass double the current, with each tube passing 1/2 the total load current (current within parallel circuit branches gets summed while voltage across the two branches remains the same). If you want to pull 1/2 the total load current to cut the output power in half, at the same B+ voltage you must DOUBLE the load, which negates the need for the two additional paralleled tubes.

[martin manning]

Thanks for the detailed explanation, John.

In order to graph a 4 tube amp, each graph would have to be adjusted to represent a pair of paralleled tubes (instead of each graph representing 1 tube each in a 2 tube push pull section), which involves doubling the plate current scaling of the graph. Everything would look the same and your load line would appear to have the same gradient, but everything is effectively doubled since the graph's plate current scaling has been doubled (plate current gets doubled since the effective plate resistance of two tubes in parallel = 1/2 the plate resistance of one).

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