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Originally Posted by Argus25
I think you are right about this too. I have always been stuck on this notion, primarily because the conversion of anode current to a voltage (except for that lost in plate resistance) occurs outside the valve in the load.
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What if the load is infinite impedance i.e. not actually present? We are talking AC, so no need to worry about DC bias. Where does the 'conversion' of current to voltage take place then? The only impedance there is the anode impedance itself. Is that inside our outside?
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Originally Posted by Argus25
Doing that, the upper device has an impedance of (rp+R)/(u+1) looking into the cathode. Which also shows how u effectively lowers the input impedance of the upper device.
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But u is the raw voltage gain of the valve! How can a device which is not a voltage amplifier have a voltage gain defined?
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Once this is known it is all downhill easy, because in the cascode circuit you can then replace the upper device by an impedance of (rp +R)/(u+1) and replace the lower device by a voltage generator u(Vin) in series with an impedance rp.
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Same comment as above.
So to derive the equations you quoted you assume that the valves are voltage amplifiers with finite impedances.