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schematic

simulate this circuit – Schematic created using CircuitLab

I know that by using \$A_v=-G_mR_{out}\$, and \$G_m=\frac{I_{out}}{V_{in}}\$, we can get the voltage gain of this circuit.

After the small signal simplification, I got:

schematic

simulate this circuit

So, as node 1 is 0 volt, there is no current goes thru \$M_2\$, then \$I_{out}\approx g_{m1}V_{in}\$ ? So that \$G_m=g_{m1}\$ which is the \$g_m\$ of \$M_1\$? Also, the \$R_{out}\$ is \$R_{02}\$, and is the voltage gain of this circuit \$-g_{m1}R_{02}\$? Thank you!!!!

-------------------------------------------------------UPDATE

schematic

simulate this circuit

What about this circuit?

After the small signal, I got:

schematic

simulate this circuit

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  • 1
    \$\begingroup\$ This is not an amplifier. It's just a MOSFET incinerator. \$\endgroup\$ – Hearth Oct 24 '18 at 3:28
  • \$\begingroup\$ Are you comfortable thinking about loadlines, that involve a resistor. Multiplying the discrete resistor by the gm produces an estimated gain. \$\endgroup\$ – analogsystemsrf Oct 24 '18 at 3:57
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The voltage gain is 0, because \$V_{out}\$ is connected directly to a voltage source, so its voltage will never vary, no matter what signal is present at the input.

Furthermore, M2 will probably catch fire, because its gate and drain are connected directly to the Vcc supply, which will cause a large current to flow through the FET, while it also must drop the full Vcc voltage.

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Note:
This answer is for the updated schematic
Assumptions:
The quiescent voltages keeps the MOSFETS in saturation (pinch off) region

enter image description here

Here assuming that there is channel length modulation The gain at the first stage is \$-g_{m1}r_{o1}\$ . Since the voltage gain of the next stage \$M_{2}\$ depends on the load resistance \$R_{load}\$.
So the effective gain is \$g_{m1}g_{m2}r_{o1}(r_{o2}||R_{load})\$

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  • \$\begingroup\$ Derwin - are you aware that both drain nodes are connected to the power supply Vcc ? \$\endgroup\$ – LvW Oct 26 '18 at 12:18
  • \$\begingroup\$ Sorry I should have mentioned this in the answer . My answer assumed that the mosfets are biased properly (I know there is no feedback shown but i assumed that the quiescent voltages maintained them in saturation ) Also this answer is for the updated circuit where he used current sources \$\endgroup\$ – Derwin DV Oct 27 '18 at 11:34
  • \$\begingroup\$ OK - in case of current sources, you are right. But it is - as you know - a fundamental difference if I use voltage or current sources. \$\endgroup\$ – LvW Oct 27 '18 at 13:27
  • \$\begingroup\$ Yep I agree with that in case of voltage source the gain is zero and if Vcc is large the FETs are done \$\endgroup\$ – Derwin DV Oct 28 '18 at 6:10

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