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This is the problem:

And this is my attempt to solve it:

For (1): I don't know how to get vgs or how to use the hint to get rid of it.

For (2): I assume that my solution is right but should I add details on how did I get Rout ?

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    \$\begingroup\$ You're on the right track, you need to substitute in for your unknown (V_gs) using the hing you're given, and solve for V_out/V_in While solving feedback circuits (this is a voltage feedback, also called an source degeneration circuit) you'll generally need to use this kind of a trick. \$\endgroup\$ – deadude Dec 2 '14 at 20:48
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You have an equation for \$v_o\$ of the form \$v_o = kv_{gs}\$ but you need \$v_{o}/v_i\$. Use the hint (which is just KVL) to substitute for \$v_{gs}\$: $$v_o = kv_{gs} = k(v_i - v_o)$$ and solve for \$v_o/v_i\$.

For \$R_o\$ you have forgotten the contribution of the dependent current source. To calculate \$R_o\$ set \$v_i = 0\$, apply a test voltage \$v_t\$ across the output, and calculate the test current \$i_t\$ flowing from that test voltage. Then \$R_o = v_t/i_t\$. Since \$v_i = 0\$, \$v_g = 0\$ and therefore \$v_{gs} = -v_s = -v_o = -v_t\$. The dependent current source therefore supplies a current \$-g_m v_t\$. Now use KCL at the output node to calculate \$v_t\$ in terms of \$i_t\$ and re-arrange to find \$R_o = v_t/i_t\$.

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  • \$\begingroup\$ I followed what you said. Is Rout correct now? i.imgur.com/ouriQFw.jpg \$\endgroup\$ – ammar Dec 2 '14 at 21:15
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    \$\begingroup\$ @ammarx Yes, that looks correct. You notice you get a much lower answer than what you originally calculated -- this is good because the circuit is a source follower (buffer) and you want a low \$R_o\$ for a buffer. \$\endgroup\$ – Null Dec 2 '14 at 21:19

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