# Transistors analysis

Discussion in 'Homework Help' started by ___, Apr 19, 2011.

1. ### ___ Thread Starter New Member

Apr 19, 2011
2
0
Hello all. I've been working on this question and it's killing me. I have solved the entire question (there are more parts ), 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 $\small V_{i}$?

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

2. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
300
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.

3. ### ___ Thread Starter New Member

Apr 19, 2011
2
0
Aha! So the operating point isn't calculated for a small signal input model?
Thanks a lot
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.