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I'm not sure did I solve this problem right.

I made \$ r_{o2} \$ , which is resistance looking at Q2 drain. So I can get $$ g_{m,Q3} = 1.8m(A/V) $$ from $$A_v = \frac{r_{o2} || r_{o3}}{1/g_m + (r_{o2} || r_{o3})}$$

and I could get W/L by

$$ g_{m,Q3} = \sqrt{2I_D \mu C_{ox}(W/L)} $$ then $$ (W/L)_{Q3} = 16.2 $$

And I can find out \$ V_{GS} = 0.8V , V_{GS} = V_{DS}\$ at Q1

so

$$I_{REF} = 1/2 \mu C_{ox}(W/L)_{Q1}(V_{GS}-V_T)^2(1+ \lambda V_{DS})$$

then I get $$ (W/L)_{Q1} = 20.576 $$ and I can guess (W/L){Q2} = (W/L){Q1} because their \$V_{GS}, V_{DS}\$ is same.

I know in Q2 $$g_m = \frac{2I_D}{V_{GS}-V_T} = \mu C_{ox}(W/L)(V_{GS}-V_T) $$

But this equation is not true when I substitute my values. What's the problem?

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  • \$\begingroup\$ While I think this could be a "good" homework question because it shows your work and allows someone to help point out where you went wrong, I think the title desperately needs to be revised. \$\endgroup\$ – JYelton Jun 15 at 18:14
  • \$\begingroup\$ @JYelton Do you have some idea about that?? \$\endgroup\$ – Pare Kanes Jun 15 at 18:43
  • \$\begingroup\$ You need to treat the current mirror Q1, Q2 separately from Q3. Determine W/L of Q1 and Q2 such that \$I_{REF}\$ is flowing. Now you can determine the output resistance \$r_{o}\$ of Q2. Then we know that small signal equivalent of the current mirror: an ideal current source with value \$I_{REF}\$ and a resistor \$r_{o2}\$ in parallel with it. Now draw the small signal equivalent circuit of Q3 and use the model of the mirror connected to its source. \$\endgroup\$ – Bimpelrekkie Jun 15 at 18:56
  • \$\begingroup\$ if the gain is 0.9 you need to have ro2||ro3 = 9*gm3 \$\endgroup\$ – G36 Jun 15 at 18:57
  • \$\begingroup\$ The gain of 0.9 V/V is the result of the output impedance of the source follower (roughly 1/gm) and \$r_{o2}\$, these make a voltage divider. From \$r_{o2}\$ determine what the output impedance of the source follower needs to be. Then you know what the gm of Q3 needs to be, from that determine W/L (you know the biasing current already). \$\endgroup\$ – Bimpelrekkie Jun 15 at 18:58

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