Transistors analysis

Thread Starter

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Joined Apr 19, 2011
2
Hello all. I've been working on this question and it's killing me. I have solved the entire question (there are more parts :p), but I could only do it once I was told that \(\small V_{GS} = - V_S + V_G = - V_S\).
I was just wondering why \(\small V_{G}\) is at 0. I am having a hard time getting my head around that.
Shouldn't \(\small V_{G}\) simply be \(\small V_{i}\)?




\(dI = \frac{\partial I}{\partial V_{GS}} dV_{GS} + \frac{\partial I}{\partial V_{DS}}dV_{DS}\)
 

Adjuster

Joined Dec 26, 2010
2,148
You are asked to calculate the operating point. This is usually understood to mean the quiescent condition, where vi = 0.

Depending on whether the signal source is assumed to be able to pass DC, you might also want to consider whether the gate leakage current would develop a significant voltage in the gate resistor. Unless you can determine the gate leakage current, you might assume it to be negligible.

Edit: you are told the transistor is ideal, so definitely ignore gate current.
 

Thread Starter

___

Joined Apr 19, 2011
2
Aha! So the operating point isn't calculated for a small signal input model?
Thanks a lot :D
I have a couple of more questions regarding small signal model (Don't want to start a new thread just for that):

How do you decide whether a component is going to be a short or open circuit. For example, for midband frequencies a capacitor is a short. What about an inductor? How do you decide what is ground and what signals to keep?


And here:





My first response was that it behaves as a high pass filter, but could you please shed some light into this too?

Any help would be greatly appreciated.
 
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