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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Err... no, sorry.
Consider two decent NPN transistors (hfe = 100) in a cascode configuration, with the base of the upper one fixed, and 1mA collector current flowing through the lower one. All are at room temperature.
Then the lower one will have a gm at room temperature of 40mA/V; the internal emitter resistance will be 25Ω, and the input impedance will be 2.5kΩ.
A 1mV signal will give rise to a collector current signal of 40μA.
The upper transistor will be fed with 1mA quiescent emitter current, so its emitter resistance will be about 25Ω. The 40μA signal current will therefore give rise to an emitter voltage variation of 1mV - the same as the input voltage to the stage!
This will always be the case, whatever the current through the pair. It's hardly 'pretty rock solid' if it's the same magnitude as the input! In fact it's comparable with the valve version. Miller effect in the transistor will kick in, feeding back effectively 2 x Ccb.
Whether valve or transistor, if you have two similar devices in cascode, they'll have the same gm. So the voltage on the anode/collector of the lower device will be approximately equal to the input voltage. Why? Because the lower device will give a current variation equal to Vin x gm, and the upper device will have a 'looking-in' input impedance approximately 1/gm. So the voltage here is Vin x gm x 1/gm = Vin.