- 8,032
- 870
GreenPrint said:So then [itex]V_{GS}[/itex] is about -.974 V?
The equation looks like a pain in the but to solve algebraically so i graphed it.
Why is it a pain? It's a simple quadratic in V_S.
GreenPrint said:So then [itex]V_{GS}[/itex] is about -.974 V?
The equation looks like a pain in the but to solve algebraically so i graphed it.
GreenPrint said:Oh you I think that was for the previous problem. Apparently he uses
[itex]\frac{R_{D}}{1 + g_{m}R_{s} + \frac{R_{D} + R_{S}}{r_{d}}}[/itex] is this wrong?
This is the reason why I think it's wrong.
Oh I mean it's just a lot of math so I just plugged it into my calculator and graphed it.
However in my book for a Self-Bias Configuration with Rs bypassed they get this
[itex]Z_{O} = \frac{1 + g_{m}R_{S} + \frac{R_{S}}{r+{d}}}{1 + g_{m}R_{S} + \frac{R_{S} + R_{D}}{r_{d}}}R_{D}[/itex]
and a self bias configuration with Rs bypassed has the same exact small signal equivalent circuit as the one in this problem. The only difference is that RG is RTh in this problem
but the electrician already said how it was wrong because when you take the limit as rd goes to infinity you don't get RD
weird I'm learning a lot from this lol
GreenPrint said:He doesn't show how he got the original expression but he does this see attached.
when i solved for VGS i got -0.974 V and gm = 5.403 mS
but this
[itex]Z_{O} = \frac{1 + g_{m}R_{S} + \frac{R_{S}}{r+{d}}}{1 + g_{m}R_{S} + \frac{R_{S} + R_{D}}{r_{d}}}R_{D}[/itex]
gives me 1730.826