I'm not sure where I heard about this idea (certainly didn't come up with it myself). Before I rewire to test, how does the "wiper goes to grid" version compare in regard to tone and functionality?
Several years ago (maybe late 2006, early 2007 or thereabout) there was a thread talking about an amp with an unusual master volume circuit on a split-load inverter. I *think* it might have been a Peavey amp or something.
Anyway, I don't recall the exact arrangement, except that a 1M audio pot was used in place of the grid-reference resistor, and I *think* the wiper went to the grid.
Alright... you know the concertina essenitally provides no gain because of the large unbypassed resistor between cathode and ground (there's a way to get the gain back, but it's impractical in guitar amps and never used). The loss of gain is due to local feedback in the circuit, and there's an interesting effect as a result of the feedback called "bootstrapping". Aiken has some info and explanation on it when he talks about this type of inverter.
Anyway, the result of bootstrapping is that the input impedance of the split-load is much higher than you'd think from the 1M resistor, or even from the resistor and the cathode load. Imagine a tube that was being used as a split-load inverter, and that the gain of the tube with a 100k plate load is 50. However, you're using it as a split-load, so half the load is in the plate circuit and half is in the cathode circuit. If you drew a loadline for the inverter, you'd still draw a 100k loadline to estimate how the stage would operate. So if all 100k was in the plate of our imaginary tube and you applied a 1v signal to the grid, you'd expect a 50v change at the plate (since we said the gain of the stage was 50). Therefore, if you move half the load to the cathode circuit, you'd think that you'd get a 25v change at the plate and a 25v change at the cathode.
But we know from experience and being told that the gain of the split-load is less than 1 due to feedback, more like 0.98. So if we apply a 1v signal at the grid, 0.98v appears at the cathode. That leaves 0.02v across the 1M input resistor, and a current of 0.02uA through the 1M resistor. But it took a 1v signal to push 0.02uA through the grid reference resistor, so it therefore looks like R = e/I =1v/0.02uA = 50,000,000 or 50M! The feedback has raised the input impedance of the circuit.
One of the big reasons a tube stage has less gain than you'd think from either the amplification factor or (for pentodes) Gm*Rl is that the following stage's grid resistor loads the circuit down. Triodes have low internal plate resistance, so you can only benefit so much from the bootstrapped input impedance, but a pentode before this type of inverter can have massive gain due to the effect. There is a circuit that has to be shown in its schematic form rather than explained that takes advantage of this, but uses a 100k grid resistor rather than 1M.
The neat effect that PRR pointed out about this production amp with a 1M pot as the grid load for a split-load inverter preceeded by a triode was that withe the volume full up, the preceeding triode took advantage of bootstrapping to give higher than expected gain from the stage ahead of the split-load. But as the control was turned down, not only did less signal get fed to the inverter, but the gain of the preceeding stage also dropped because of a reduction in the bootstrapping. It was a 2-way gain reducer.
So try it a number of ways and see what works best for you. The complicated pentode-split-load circuit can be seen in High Fidelity Circuit Design by Norman Crowhurst on Pete Millet's website. I'm planning on playing with it some to see what I can accomplish.
Topic: Master Volume w/ split load / concertina PI??? (Read 10919 times)