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Q. Give a general account of common drain amplifier?
The output is taken over the source terminal and when the dc supply is replaced by its short circuit equivalent, the drain is grounded .That is why ,it is called common drain . The voltage source in the output circuit is expressed in terms of Vgd using Thevenin's theorem.
The output voltage
Vo = Rs /(Rs + rd/µ+1) * µVgd/(µ+1) = µRsVgd /(µ+1)Rs + rd
where Vgd =Vi ,the input voltage
Therefore,
Voltage gain
Av = Vo/Vi = µRs /(µ+1)Rs + rd
how over compensation by capacitive var rises voltage in a circuit?
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