# Design for a common source FET Amplifier Circuit

Here's a design question our prof gave us to study for the exam. I'm having trouble putting it together.

A common source amplifier is shown. The FET is working in saturation and has gm = 0.5mS, Kn = 0.5 mA/V^2, Vtn = 1v and lambda = 0.02V^-1. The resistance of the signal source is 2kohms. There is a maximum gain of 112, and Id = 0.24mA. The circuit has output resistance of 20.9kohm, and input resistance of 892kohms. Determine Vds. Alright so my thoughts when trying to come up with a solution were as follows:

1) It's given that the MOSFET is in saturation therefore: 0.24 = 0.25(Vgs - 1)^2(1+0.02Vds) 2) We know Vd = 12 - (Rd)(0.24) 3) We know Vs = 0.24(R5 + R4)

Also, this is DC analysis, so the capacitors act like open circuits; technically only the middle part of the circuit is off concern.

I don't know how to take it from here; we're not given any of the resistor values and it seems like there's simply not enough equations to cover the unknowns. I presume it's got something to do with the provided input and output impedences, but I just don't know how to apply them. Any help would be appreciated

• Something sounds wrong here: I don't think that the FET can be in saturation if you want any gain to happen. Given that the gain is stated as 112 (which I also don't believe is correct either), the FET can't be in saturation. – Dwayne Reid Nov 9 '15 at 0:59
• Best advice I can offer is: don't believe either the saturation comment OR the gain of 112. Find Vgs and work from that. – Dwayne Reid Nov 9 '15 at 1:01
• How would you go about finding Vgs given you know none of the resistor values ? – user3479118 Nov 9 '15 at 1:26
• It has been a while since I had to do these types of problems. But Rin is R1 in parallel with R2, right? And Id is given, and the trans-conductance is given. So isn't that enough to figure out R1 and R2 and Vg and Vgs? Once you have Vg and Vgs, you can figure out Vs, which will allow you to solve for the sum of R4 and R5. Once you have that sum, I think you might have enough information to solve for everything required. – mkeith Nov 9 '15 at 3:43