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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.

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    \$\begingroup\$ 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. \$\endgroup\$
    – The Photon
    Nov 2, 2016 at 3:22
  • \$\begingroup\$ 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? \$\endgroup\$
    – user128233
    Nov 2, 2016 at 3:28
  • \$\begingroup\$ @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. \$\endgroup\$
    – Rohat Kılıç
    Nov 2, 2016 at 3:43
  • \$\begingroup\$ @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? \$\endgroup\$
    – user128233
    Nov 2, 2016 at 4:42
  • \$\begingroup\$ @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. \$\endgroup\$
    – Rohat Kılıç
    Nov 2, 2016 at 5:00

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Here are my answers to your questions (1)...(5):

(1) Capacitors can be considered as short circuits if their impedance is much smaller than the resulting total (effective) ohmic resistance which appears in series or in parallel to the capacitor. The meaning of "much smaller" depends on the allowed calculation error (and the associated parts tolerances).

(2) Small-signal analyses are used for finding the gain and the input-/output impedances of a circuit. These 3 parameters are signal RATIOS (voltage/voltage current/current or voltage/current), which can be found only if these input/output quantities have the same sinusoidal waveform. Otherwise, we cannot calculate a ratio. And this is ensured for small-signals only (linearized V-I characteristics).

(3) Large signal analysis is required if the signals are so large that the preconditions for small-signal analyses (linearity) are not met anymore. However, in this case, no gain and no input/output impedances can be calculated. However, we can define and calculate input-/output power and efficiency values. (Examples: Transistorized power stages or switches).

(4) We have no choice. See (3).

(5) The design process depends on the particular circuit and its purpose (linear gain stage, power stage, switch,...)

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