Why are capacitors treated as shorts in small signal models?

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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{1}{j\omega C}\$, 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.

Answer 4

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_{DC}+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_{DC}\$ 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)\$).

Answer 5

In the case of e.g. input capacitors, the assumption is that the capacitor is large enough to not (edit: charge significantly enough to present a significant opposing voltage to) the applied signal. As long as electrons are flowing into one side, electrons are flowing out of the other side.

Answer 6

Basic idea

A capacitor is neither an open circuit nor a short connection; it is a "duplicating voltage source" (a "voltage clone"). Imagine the simplest capacitive circuit - a capacitor connected to a DC voltage source. The capacitor is charged to the source voltage and no current flows in the circuit because two sources of equal but opposite voltage are connected in a loop.

Operation

Emulated capacitor: We can actually replace the charged capacitor with a voltage source Vc and thus conveniently examine it through an DC ammeter and the CircuitLab DC Live simulation.

DC input voltage

If you start to slowly change the source voltage, the "capacitor" will also change its voltage; their difference will remain equal to zero and no current will flow. We can mimic this by setting the same voltages to both Vin and Vc:

Vin = 1 V; Vc = 1 V -> Ic = 0

^{simulate this circuit – Schematic created using CircuitLab}

Vin = 2 V; Vc = 2 V -> Ic = 0

^{simulate this circuit}

Vin = 3 V; Vc = 3 V -> Ic = 0

^{simulate this circuit}

So the source has the illusion that there is nothing connected to it, but it is, and it is a voltage source. There is also an ammeter connected with almost zero resistance, but the source "sees" an infinitely high resistance as if by some magical means the ammeter resistance is increased many times over. This trick is known in circuitry as bootstrapping.

Slowly changing DC input voltage

Conceptual circuit: We can automate this experiment by replacing the input voltage source and the emulated capacitor by two equal low-frequency AC voltage sources.

^{simulate this circuit}

As above, their difference remains equal to zero and again no current flows.

Practical circuit: Let's finally explore a true capacitive circuit. Since the capacitance is small enough, the capacitor quickly charges...

^{simulate this circuit}

... and its voltage follows the input voltage; the current is low.

AC input voltage

If you start rapidly changing the input voltage, the "capacitor" will not be able to change its voltage Vc, and the current will change significantly. To limit the current to a reasonable value and to "cheat" the simulator, I have set a low resistance (only 1 Ω) to the ammeter.

Vin = 2 V, Vc = 1 V, Ic = 1 A

^{simulate this circuit}

Vin = 3 V, Vc = 1 V, Ic = 2 A

^{simulate this circuit}

Vin = 4 V, Vc = 1 V, Ic = 3 A

^{simulate this circuit}

Thus the input source will have the illusion of being shorted, but in fact it is not.

Swept input voltage: To automate the experiment, let's sweep the input voltage around Vc...

^{simulate this circuit}

AC practical circuit: ... and finally explore the real capacitive circuit with an AC input voltage.

^{simulate this circuit}

As you can see, Vin varies only with +100 mV above and below 2 V...

... but the current significantly changes from -500 mA to 500 mA.

Generalization

Now we can imagine what happens in the more complex circuits with capacitors.

Capacitor acting as an open circuit

When the power supply is turned on, the capacitors are charged to the (quiescent) voltages between the respective nodes that would be without capacitors. There are no voltage differences, currents do not flow, as if the capacitors are disconnected; figuratively speaking, the circuit "softens". This open-circuit capacitor property is used when the operating (quiescent) point is set.

Capacitor acting as a short circuit

As the regulating element begins to vary its current, the voltages between the nodes begin to change. Currents begin to flow and the capacitors are "connected" to the circuit; figuratively speaking, the circuit "hardens". This short-circuit capacitor property is used when an input AC voltage (no matter with small or large amplitude) is applied.

Applications

Let's consider (the output part of) a typical circuit of an AC common-emitter amplifier stage with blocking capacitor in the emitter and coupling capacitor between the collector and load. In this conceptual circuit, I have emulated the collector-emitter (drain-source) part of the transistor with variable resistor Rce. Thus a voltage divider is formed by the three resistors Rc, Rce and Re.

Initial state

By adjusting Rce we set the desired quiescent collector voltage (10 V) and emitter voltage (5V).

^{simulate this circuit}

Connecting voltmeters

Then let's measure the voltages between the node pairs by connecting voltmeters in the place of the future "capacitors" (voltage sources). Remember their readings.

^{simulate this circuit}

Connecting "capacitors"

Then replace the voltmeters by voltage sources that emulate the capacitors. Adjust the source voltages equal to the voltmeter readings above ("charge the capacitors"). As you can see, nothing changes (hover over the circuit devices to see the voltages and currents by the help of the CircuitLab DC Live-Simulation feature).

^{simulate this circuit}

Decreasing Rce

Now decrease the resistance Rce of the regulating element ("transistor"). The divider's current increases so the emitter voltage across Re increases and becomes higher than the capacitor voltage Ve. The "capacitor" is "connected" and the divider passes a current through it.

^{simulate this circuit}

At the same time, the voltage drop across Rc increases. The voltage between the "transistor collector" and load RL decreases and becomes lower than the capacitor voltage Vc. The "capacitor" is "connected" and it passes a current through the divider and load RL, but since the "capacitance" is significant, the voltage across the "capacitor" almost does not change. So it transfers the "shifted" voltage decrease to the load, and the load voltage goes below zero.

Increasing Rce

Then increase Rce. The divider's current decreases so the emitter voltage Re decreases and becomes lower than the capacitor voltage Ve. The "capacitor" passes a current through Re but its voltage almost does not change.

^{simulate this circuit}

The voltage drop across Rc also decreases. The voltage between the "collector" and load RL increases and becomes higher than the capacitor voltage Vc. The divider passes a current through the capacitor and load RL but the voltage across the "capacitor" almost does not change. So it transfers the voltage increase and the load voltage goes above zero.

Thus the "collector voltage" variations around the 10 V quiescent voltage are "moved" to around ground.

Answer 7

May I try to give an answer in only four sentences?

• It is correct that the linear small-signal model of a transistor is valid only for a fixed DC operating point, since the relevant transistor parameters depend on DC conditions.
• The capacitors contained in an amplifier circuit are usually selected in size so that they represent only a negligible small resistance (1/wC) for the signals to be amplified (signal coupling, separation between DC and AC signals).
• For a linear small signal analysis (no DC sources/voltages; valid for sinusoidal signals only) it is, therefore, reasonable to consider these capacitive branches as short circuits.
• The situation is different in electrical filters, where these capacitors - together with resistors - are supposed to cause certain time constants and frequency-dependent phase shifts.