# Transistors - Small signal and Large signal analysis

In analyzing transistor circuits, I know that you typically separate it into DC and AC analysis. In doing AC analysis, we often do small signal analysis, and from what I understand, it is the region where the transistor's V-I characteristic curve is linear. So, in AC small signal analysis, we assume that capacitors are shorted. My questions are:

1. What point would the capacitor value be if we assume that capacitors are shorted?
2. What is small signal analysis used for?
3. What is the difference between small signal and large signal analysis?
4. When do we use one or the other? Is there some sort of advantage?
5. In designing, which one is used?

From my questions, you can infer that I do not know much of the implications of doing so and so analysis.

• Who told you "we assume that capacitors are shorted"? To find the high-frequency limit of the behavior we might do that. But to find the low frequency limit, we'd assume capacitors are open. And to find the behavior at in-between frequencies, we'd use phasor analysis, which will depend on the capacitor value. Commented Nov 2, 2016 at 3:22
• apart from lectures, other sources told me capacitors act like short circuit in small signal analysis. Anyway, if one were to use phasor analysis or even laplace, what would the transistor model in either domain look like? Commented Nov 2, 2016 at 3:28
• @user128233 I think you are talking about externally placed coupling and bypass capacitors. We assume that they are shorted "if they are large enough". If you are talking about transistor's internal capacitances, they should be taken into account (with external caps as well) depending on the frequency range that the circuit is working at. Commented Nov 2, 2016 at 3:43
• @RohatKılıç oh yes, my bad. I thought it was implied that I was referring to coupling and bypass capacitors. What do you mean by large enough? And since it is large enough, what would it imply? Commented Nov 2, 2016 at 4:42
• @user128233 Generally you can see that the term "large enough" is mentioned in questions/exercises. Think about $X_C=1/2\pi f C$. If the cap is large enough then $X_C$ becomes small enough to assume that the cap is "shorted" even at low frequencies. Commented Nov 2, 2016 at 5:00