Can someone help me with getting the AC form of this MOSFET circuit? I am having trouble understanding why in class they took Q1 and Q2 out of the AC form:
The AC form they showed us in class (the FET you see here is Q3):
Electronic – AC form for a FET circuit
analogcircuit analysismosfet
Related Solutions
After step 6, your circuit will be as shown below.
simulate this circuit – Schematic created using CircuitLab
Which can be further reduced as a 16V battery in series with 3k and 5k resistors.
Now, you will have 10V across this 5k resistance, ie, 10V across each parallel 10k. Splitting 10k into 6k+4k, you will have 4V across 4k resistance. From the polarity of v1 marked, V1= -4V.
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?
Best Answer
Given the above, all you are left with is Q3 being part of the small signal AC analysis. The two capacitors are shorted because, their dominant use is to prevent DC levels being affected by external inputs and outputs hence, in small signal AC analysis, we can ignore and short them out.