# Designing a Common-source amplifier with a current source [closed]

I'm trying to design a common-source amp without using resistors in the Source\Drain.

I tried to design a common-source amp that will give ~35 gain (or less and I could just use two stages):

One circuit that gave me a very high gain (13k~) is:

My question is, How can I get a different gain from this amplifier?

Any tips\hints\help would be very much appreciated.

Edit ended up using a feedback loop to reduce gain. Thank you for you help everybody :)

• Why R28 is 7.5k if Ic20 is around 1mA?
– G36
Jun 1 at 15:19
• @G36 Actually, I don't know, I just tried using a resistor in the Source and saw that It added gain, I was hoping for an explanation how does that resistor come into play, Thank you for your comment :) Jun 1 at 15:41
• But notice that $7.5k\Omega \times 1mA = 7.5V$.
– G36
Jun 1 at 15:50
• @G36 Thank you. I derived that R28 should be ~0.5*R25 for the current mirror to function properly Jun 1 at 16:46
• I don't think this topology is going to work. You basically have two independent current sources series connected. Both the mirror and the MOS are trying to impose its own current and being equal is never going to happen unless you include some kind of closed loop control on either. Some unreliable operating point may be hit in simulation due to BJT and MOS output resistance but it's going to be so critical not to be useful in real world. Jun 1 at 17:17