How to get the gain for this MOSFET amplifier?

Question

This is a common drain MOSFET amplifier:

^{simulate this circuit – Schematic created using CircuitLab}

This is its small signal model at low frequency:

^{simulate this circuit}

I am at a loss as to how to find \$V_{\text{out}}/V_{\text{in}}\$. \$V_{\text{in}}\$, in this case, is just \$V_g\$ not \$V_{gs}\$ since the source node is not connected to ground.

My analysis:

Since all of the current source's current is going thru \$r_0\$ from \$V_{\text{out}}\$ to GND,

$$\frac{V_{\text{out}}}{r_0}=V_{gs} \times g_m$$ or $$\frac{V_{\text{out}}}{r_0}=(V_{in}-V_{out}) \times g_m$$. After some algebraic manipulation, you get $$\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{r_0 \times g_m}{1+r_0 \times g_m}$$

I know $$r_o \times g_m=\left(\frac{\delta I_d}{\delta V_{ds}}\right)^{-1} \times \frac{\delta I_d}{\delta V_{gs}}=\frac{\delta V_{ds}}{\delta V_{gs}}$$

I don't know what to do after this. My professor's solution says \$V_{\text{out}}/V_{\text{in}} = 1\$. How?

Answer 1

Vgs = 5V so RsOn is low and you have a SOURCE follower ( or common drain ) with some unknown load resistance say 100x RdsOn which is missing between Source and Vss.

So it's a drain follower, like a unity gain emitter follower with a gain=<1 depending on load regulation, Roin and a DC bias above Vgs(th).

But a rather sloppy logic diagram with errors... tisk.

Other _ - - Here is how I would make a 1GHz buffer Av =1 almost unity

I added Ciss and Rg to the model with very low RdsON Vt=1V and due to load regulation from Ron and insufficient Vgs bias from Vss the Vout DC is a bit more than -1V.

Go back and review your common drain notes for details.