Electronic Devices - FET Amplifiers - Discussion

How did you get vgsq, when to substitute the value?

Correction to my previous comment,

K = Id(on)/(VGS(on) - VGS(th))^2.

Here's how I solved this:

1. Calculate k using the fomula:
K = Id(on)/(Vgs(on) - Vgs(th))
K = 240 x 10^-6

2. Solve for VGSq:
KVL on output side; Vdd - IdqRd - VDSq = 0
Since no DC current flows through 10 M ohms, VDSq = VGSq.
KVL becomes: Vdd - IdqRD - VGSq = 0
Substitute Shockley's equation for Id into Idq then solve for VGSq.
Vdd -k(VGSq - VGS(th))^2 - VGSq = 0
VGSq = 6.41 V

3. Solve for gm using the formula:
gm = k(VGSq - VGS(th))
gm = 1.64 mS

4. Solve for gain using approximate formula:
Av = -gm(Rd//rd)
Av = -2.96

@Sarah, how to get VDSq = 6.42 V?

@Gab Pesieto.

What is the Rf here?

Get VDSq first by plotting of the curves, VDSq = 6.42 V.

Get K, K = (Id(on))/(Vgs (on)-Vgs (th))^2, K = 240x10^-6.

Get gm, gm = 2K (VDSq-Vgs (th)), gm = 1.64 ms.

Get Av, Av = -gm (RD||rd), -2.96 approximately -2.95.

With rd,

Av = -gm*(Rf//rd//RD).

Without rd,

Av = -gmRD.

Please give me solution.

How to solve this question?