Common Source amplifier

Thread Starter


Joined Jun 6, 2009
Am I correct about the exact (not approximated) output resistance of this amplifier?

Rout = (RD || ro) * (1 + gm * RE)

Thank you.



The Electrician

Joined Oct 9, 2007
alphacat has some misplaced parentheses:

Rout = (RD || ro) * (1 + gm * RE)

should be:

Rout = RD || (ro * (1 + gm * RE))

Jony130 appears to intend this same expression, but has missing parentheses.

This expression, Rout = RD || (ro * (1 + gm * RE)), is a good approximation provided that ro >> 1/gm.

tnk has it exactly right.
This circuit is used as a phase splitter in audio practice:

and was first used in vacuum tube circuitry:

If RD and RE are identical, then the voltage gain from gate to drain is identical to the voltage gain from gate to source, thus providing a pair of outputs identical in amplitude but 180 out of phase.

This circuit has been the subject of much controversy in the audio world because the output impedance (driving point impedance) at the drain is different than at the source. It would seem that if the two outputs (drain and source, or plate and cathode if tubes are used) are loaded with identical impedances, the balance would be upset, but this is not the case.


Joined Feb 17, 2009
Yes, my mistake.
This equation Rout=ro * (1 + gm * RE)) I took from the book and it was not a single word about simplification.
So I decided to find Rout by myself.

Rout = V2/I2 for V1 = 0V Vgs = (-I2)*Re

V2 = ( I2 - (gm*(-I2)*Re) )*ro + I2*Re

I2 = V2 / ( Re + ro + gm*Re*ro )

Rout = Re + ro + gm*Re*ro

And that means that tnk equations are OK.


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