Why are capacitors treated as shorts in small signal models?

If I understand correctly, a small signal model (e.g. hybrid pi modal of MOSFETs) is just a linear approximation of a nonlinear circuit centered at the DC bias point. At DC, ideal capacitors act like open circuits and linear approximations are generally only accurate for small deviations from the linearization point, which is the DC point in this case. Hence, it seems like it would make more sense to treat capacitors as open circuits, not shorts. So why do we do the opposite? How does it make sense for a linear approximation centered at the DC point to not use the DC behavior of capacitors? If it makes a difference, I'm specifically thinking of the case of MOSFET amplifier circuits, since that's what we're currently covering in my class on analog and digital circuits and is what prompted this question.

\$\begingroup\$ @Mikayla, Your first sentence is extremely well-written! Nice. The small signal model is exactly that: a linearized (first order term of the partial differential) version of the large-scale non-linear behavior, with its 1st order slope centered on the DC quiescent bias point. As far as the capacitors go, everything important happens within an order of magnitude (between X/10 and 10X) of the region where the capacitor impedance is the same as some other nearby impedance. DC is usually far away, so open there. Daylight is usually far away, so closed/short there. \$\endgroup\$

\$\begingroup\$ @periblepsis, what means "daylight" here? does it refer to the range of frequencies of interest? the ones that you want to keep in a filter. \$\endgroup\$

\$\begingroup\$ @dandavis That's not really a sensible description: a capacitor never is an open circuit. Instead it is close to an ideal voltage source (resistance close to zero), with the imprinted voltage changing over time in proportion to the charge stored. An open circuit will not admit any current even when a voltage is applied whereas a capacitor will accept arbitrarily high currents with the voltage only changing over time in response to currents. \$\endgroup\$

7 Answers 7

The stable DC operating point of a capacitor is characterized by zero current: in that respect it resembles an open circuit. But a momentary current does not change the voltage across the capacitor: you have to apply a current over time to get a voltage change. So momentarily, the capacitor acts as a short once you subtract its current DC value, just like an ideal voltage source would. Just how momentarily, depends on the capacitance and the current we are talking about. A DC current will not stop changing the voltage, so for DC currents we have no stable operating point. An AC current will have the voltage fluctuating around the operating point, with less fluctuation for higher frequencies (and, naturally, for lower currents).

\$\begingroup\$ Very accurate observations accompanied by very good intuitive explanations here and especially in your comments above. I enjoy reading them. \$\endgroup\$

When you treat them as short circuits you are making the assumption the have negligible reactance at the frequencies you are interested in. This is usually true for the coupling capacitors in an amplifier circuit.

There are also capacitors you treat as open circuits because they have very large reactance at the frequencies of interest.

When you start calculating the bandwidth of the circuit, you start adding them back in.

There's a bit of confusion regarding the role of capacitors in small-signal models. So, let's clarify.

1. DC Bias Point and Linearization: Indeed, the small signal model is a linearized model about the DC bias point. This means that any component behavior is linearized around its DC condition, i.e., the condition when a steady-state DC voltage is applied.
2. Capacitors at DC: At DC steady state, capacitors behave like open circuits. This is because once a capacitor is fully charged, no current flows through it. When you're analyzing a circuit to find the DC operating point (sometimes called the Q-point), you indeed treat capacitors as open circuits.
3. Small-Signal Analysis and AC: Once the DC bias point is determined, small-signal analysis looks at how the circuit behaves for small deviations around this point due to small AC signals. Even though the model is linearized around the DC point, this doesn't mean we are still in DC conditions. Instead, we are in conditions where small AC signals are superimposed on the DC bias.
4. Capacitors in Small-Signal Analysis: Since we're now analyzing the behavior under AC conditions (albeit small signals), capacitors no longer behave as open circuits. They have a reactance given by \$X_C = \frac\$ , where \$\omega\$ is the angular frequency of the signal. For many small-signal analyses in circuits, it's common to deal with mid-band frequencies where the capacitive reactance is significant. This is why capacitors are typically included in small-signal models.
5. Why Not Open Circuit?: If you treated capacitors as open circuits in small-signal models, you would ignore crucial coupling and bypass roles that capacitors play in amplifier circuits. For instance, coupling capacitors allow AC signals to pass while blocking DC. Meanwhile, bypass capacitors provide AC ground for certain nodes, improving amplifier performance.

So, when setting the DC operating point, capacitors are open circuits. But when studying the circuit's response to AC signals (small-signal analysis), capacitors play a vital role in the circuit's behavior. Treating them as open circuits during this analysis would result in an incorrect representation of how the circuit behaves in the presence of AC signals.

You are completely correct that the DC point should use the DC behavior of capacitors. In fact, that's exactly what you do. Capacitors are only short circuits when you consider the "small signal" component after you found the DC linearized point. So capacitors are open when considering the DC component, then shorts (or at least small negative imaginary impedance) when solving for the non-DC small signal response.

In the small signal model, your signal is some DC component plus a time varying component whose magnitude is small relative to the DC component. The signal is therefore something like: \$V(t) = V_+v(t)\$

First you find the Q (or bias) point by considering capacitors as open circuits, exactly like you say. That gives you the values of hybrid-pi model at that DC bias. That handles the \$V_\$ part.

Then you use the hybrid-pi model to see what happens to \$v(t)\$ . Here you are using the linearized model, like you said. You linearized the transistor, now you can just solve the circuit using \$v(t)\$ as the signal input. That means capacitors act like shorts (or at least \$Z=1/(i\omega C)\$ ).