Quote:
Originally Posted by kalee20
If Argus's concept of the upper valve acting as a perfect voltage follower were true, then with the grid at signal ground the cathode would indeed also be at signal ground and so would the anode voltage of the lower valve. But to get a perfect follower, the gm of the upper valve must be infinite (Rout = 1/gm). And as the valves are matched, the gm of the lower valve would also be infinite, so the minutest input signal would give infinite current into the upper cathode. And we know what happens with infinity times zero, we get anything we like - which in this case is 1.
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But it is just a conceptual thing, there is no point in heading to infinity and nothing is perfect. For the transistor case, Horowitz and Hill (pg 103) describe the upper transistor (they label as Q2) being interposed in the collector circuit of the lower transistor (Q1) to prevent Q1's collector from swinging (thereby eliminating the Miller effect) and in their words (which I believe): "passing the collector current through to the load resistor unchanged". So I remarked "pretty well rock solid"
They (H&H) didn't say how well the upper transistor prevented swinging, but clearly the voltage swing will be a fraction of what it will be across the load resistor.
You have pointed out, that this swing will be about equal to the signal input voltage. Also, that it appears then, the voltage gain of the lower transistor is -1.
However, my main point was that if the signal current in the load is generated by the lower transistor, and passes unchanged through to the load, then overall, the voltage gain of the cascode circuit, is attributable to the lower transistor, not the upper one, so I simply disagreed that the lower transistor contributed no voltage gain.
Also, this means if you wanted in a cascode transistor circuit, you could replace the upper transistor with a super beta transistor with an hfe of 1000 and the gain of the circuit would stay the same. So the gain (overall) is determined by the lower device.