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enter image description here What will be the gain of circuit 1) We dont include channel length modulation? 2) we include channel length modulation? If we include channel length modulation will the gain be 1? as vds should remain constant as current source is ideal so increase in vg should have increase in vs and same increase in vd? Moreover why we bias at constant current then if vgs cannot be changed? How will gmvgs current increment gets generated?

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    \$\begingroup\$ Maybe this is what you are having trouble with, but the circuit as presented won't do anything useful. If the current source is truly ideal, it will prevent the input signal from having any effect on the output voltage. \$\endgroup\$ – The Photon Jul 21 '18 at 4:27
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The ideal current source in the source path has an infinite internal source resistance Ri. Hence, you have a circuit with a an infinite large feedback resistor (because every resistor in the source path provides negative feedback).

This can be seen for evaluating the gain formula with feedback (gm: transconductance):

A=Rc/[(1/gm) + Ri)]

In case you want to consider channel length modulation we have Rc||ro instead of Rc only.

This means: Gain A=0 for Ri approachung infinity.

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The input signal doesn't have any effect on the output voltage. This can be seen from the small signal model of the circuit:enter image description here

The current of dependent source all goes through Ro, and by KCL at point A, the small signal current through RL will be zero, hence vout=0*RL=0V, so no amplification at all and gain will be 0.

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  • \$\begingroup\$ Dirac...do you really believe that - in reality - there will be current as shown in your diagram? I doubt. \$\endgroup\$ – LvW Jul 22 '18 at 8:34
  • \$\begingroup\$ @LvW Of course no. Actually I drew an ideal model of the circuit. But, right, there will be some very small current through RL due to the finite impedance of the current source in real world. So we have gain but that's waay negligible. \$\endgroup\$ – dirac16 Jul 22 '18 at 9:32

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