# Trouble solving for voltage gain

I've trying to determine the voltage gain of this circuit (a small signal model of a MOSFET and a few resistors), but I'm having trouble getting enough equations to substitute. I see that ro and Rs are a voltage divider, so ... $$V_{in} = V_1 + (V_{out} + g_m*V_1*r_o)\frac{R_s}{R_s + r_o}$$

$$\longrightarrow V_{out} = \frac{V_{in} - V_1(1 + gm*r_o\frac{R_s}{R_s + r_o})}{\frac{R_s}{R_s + r_o}}$$

But from here, I need V1 in terms of vin or vice versa and i'm not sure how to go about that. I'm also interested if somebody has a more "clever" solution to determine vout/vin. simulate this circuit – Schematic created using CircuitLab

For the curious, this is how I got to this schematic: simulate this circuit

From the history of comments the original schematic was updated.

Rs and Rd are now in series because we're in small signal mode so that ground symbol doesn't matter. Then Rs+Rd is || to ro. Should be super simple from there.

$$Vout-V_{1-} = -gmV_1*(Rs+Rd||ro)$$ $$\frac{Vout-V_{1-}}{ro} = I_{ro}$$ $$gmV1 + I_{ro} = I_{Rs}$$ $$I_{Rs}*Rs = V_{1-}$$

Should be enough to solve it?

• I did a source conversion (that's why the equation for voltage is multiplied by ro). If I had used the original current source, ro would be in parallel with the source and the current would be gmV1. – Daniel B. Dec 13 '15 at 5:01
• @DanielB. Why's it being in parallel a problem though? – horta Dec 13 '15 at 5:05
• Because I thought it would be easier to solve this way; here I only have a single current loop rather than two. – Daniel B. Dec 13 '15 at 5:12
• @DanielB. Yeah, that's pretty common. (Bad pun I didn't mean to happen.) As long as there's not another ground symbol or Vdd or Vcc or some other related assumed infinite source or sink of current, then a ground symbol is nothing more than a reference voltage. Glad I could finally help. – horta Dec 13 '15 at 5:40
• It was. A current source in parallel with a resistor can be transformed to a voltage source in series with the resistor, and the voltage is the current times the resistance. – Daniel B. Dec 13 '15 at 5:52

This is how we were taught to solve these problems in my classes: 