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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:

enter image description here

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 :)

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    \$\begingroup\$ Why R28 is 7.5k if Ic20 is around 1mA? \$\endgroup\$
    – G36
    Jun 1 at 15:19
  • \$\begingroup\$ @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 :) \$\endgroup\$
    – Edward Josef
    Jun 1 at 15:41
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    \$\begingroup\$ But notice that \$7.5k\Omega \times 1mA = 7.5V\$. \$\endgroup\$
    – G36
    Jun 1 at 15:50
  • \$\begingroup\$ @G36 Thank you. I derived that R28 should be ~0.5*R25 for the current mirror to function properly \$\endgroup\$
    – Edward Josef
    Jun 1 at 16:46
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    \$\begingroup\$ 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. \$\endgroup\$
    – carloc
    Jun 1 at 17:17

1 Answer 1

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With the high source resistance of 7.5k the DC setup is wrong, the current mirror is not operating in the desired range, the current is just not reached. A value of 2.2k seems to create an acceptable dynamic range for U_OUT, 5VDC in my simulation. The gain is very high and has distortions, so I inserted a resistor R5 in series with C2 to control the gain and improve linearity.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you for the answer :), unfortunately I'm not allowed to use resistors that "touch" the Source in small signal, So I have to leave the capacitor by itself (R5=0Ohm) so that in small signal, the resistor R28 is shorted \$\endgroup\$
    – Edward Josef
    Jun 1 at 16:07

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