Quote:
Originally Posted by Radio Wrangler
Quote:
Originally Posted by kalee20
The only way to double the gm of a transistor, that I know of, is to double the operating current (impossible in a series-connected cascode), or run the transistor at double the absolute temperature. You can't choose another type with a different gm (unlike valves). A cascode comprised of a BC107 and a 2N3055 will still display equal gm's and have the same signal voltage on the collector of the lower device as on the input. It's not a massive signal, but it's not rock-steady either!
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Quite true, but it seems counter-intuitive to people not familiar with the Ebers and Moll equation and onwards. You could also change the Gm by having one transistor in a different material with a different band-gap. Can't think why I'd want to do this though. |
I don't think this will matter! If you have two transistors in series, they'll have the same current passing through them (to within a couple of percent, if their hfe's are high). You could have one silicon, one germanium. The gm's will still match at the same current and temperature, so (gm lower) x (re upper) will still equal 1. And that means that (signal voltage on collector of lower) = (input voltage on base of lower).
Adding emitter degeneration by an added resistor will certainly change things, and the Miller pole will move up in frequency, as you say. That's moving the goalposts, of course!
Quote:
Originally Posted by Argus25
Secondly, in your analysis, you will note that your calculations did not have to refer to the gm of the upper transistor, just saying it was equal to the lower one.
You only needed its input impedance (25 Ohms) derived from its collector current, which ignoring base current, is its emitter current and depends on the lower device's collector current.
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Yup. But input impedance (25Ω) into the emitter of the upper transistor is 1/gm of that device. By knowing one, you know the other.
Quote:
Originally Posted by Argus25
If one regards the upper valve as a voltage follower, its cathode attempting to follow the grid voltage and the grid is at signal ground, if it were perfect in that application the plate voltage of the lower valve would not move at all. But since it has a source resistance, looking into the cathode of the upper device (valve), of around a few hundred Ohms typical, I agree it will move. In transistor cascode though, the collector voltage of the lower device is pretty well rock solid voltage wise, so I may have been thinking more of that than the valve case.
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That's the bit I'm querying! In a valve cascode (with identical valves), the anode voltage of the lower valve will move with the same amplitude as the input. It has to, as long as (gm lower) x (1/gm upper) = 1. And with transistors, the same applies (the transistors don't even have to be identical as long as you don't pair a bipolar with a JFET). There ain't no difference!
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.