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 Vout/Vin. Vin, in this case, is just Vg not Vgs since the source node is not connected to ground.

My analysis:

Since all of the current source's current is going thru r0 from Vout to GND,

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

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

I don't know what to do after this. My professor's solution says Vout/Vin=1. How?

  • 1
    \$\begingroup\$ The circuit as drawn has zero gain. That is, the output is completely independent of the input. Are you sure you drew it right? \$\endgroup\$ – Hearth Oct 20 '19 at 0:19

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.

enter image description here

Go back and review your common drain notes for details.


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