0
\$\begingroup\$

In most of the books it looks like small signal is used as a steady state analysis, for example a capacitor is modelled as: $$Xc=\frac{1}{j \omega C}$$ Which clearly indicates a steady state analysis.

Can we use small signal model approach for transient analysis of a non-linear circuit with some energy storing device or there are some other techniques that are used for such type of input (superposition of DC bias and a small signal)?

\$\endgroup\$
5
  • \$\begingroup\$ This depends on the nature of the non-linearity, however, the describing function is sometimes applicable to frequency response analysis. \$\endgroup\$
    – Chu
    Aug 17, 2021 at 10:00
  • \$\begingroup\$ Just to clarify, there is no "small-signal model" for a capacitor. Linear devices have just one model, and it works for both small and large signals. \$\endgroup\$ Aug 17, 2021 at 10:38
  • \$\begingroup\$ Can you provide a source / example for the statement "In most of the books it looks like small signal is used as a steady state analysis". Perhaps an excerpt from a book highlighting portions which you think prevents one from using small signal analysis for transient behavior of a circuit (where signals are small). \$\endgroup\$
    – AJN
    Aug 17, 2021 at 12:25
  • \$\begingroup\$ @AJN "It is not explicitly mentioned but analysis related to steady state like frequency response are done by this method but I don't find transient transient using this method", \$\endgroup\$
    – user215805
    Aug 18, 2021 at 4:44
  • \$\begingroup\$ transient analysis \$\endgroup\$
    – user215805
    Aug 18, 2021 at 5:06

1 Answer 1

3
\$\begingroup\$

As you would expect, the small signal model is valid as long as small signal conditions apply, which means changes in state variables are small enough that the linear approximation used to establish the small signal model is good enough.

Under those conditions, small signal frequency response will predict transient response, settling time, etc.

Small signal is a linear model, therefore it cannot give any information with regard to nonlinear effects like distortion.

However if the transient you want to study is large enough to cause a significant change in component parameters like gm, hFe, junction capacitance, etc... or even worse some active devices change state, turn off, become saturated... then the small signal model, which assumes all this stuff is constant, is no longer valid. For example, slew rate and clipping are large signal phenomena. The slew rate spec of an opamp assumes an input signal step large enough to overwhelm the input stage and turn off one transistor in the input pair, and clipping will most likely involve saturation or a state change in a component designed to avoid saturation, like a Baker clamp diode turning on, along with some transistors turning off.

superposition of DC bias and a small signal

This is always the case, since the small signal model comes from linear approximation of circuit behavior at its operating point. If you did not specify a DC operating point by setting the DC value of your input source, you will get the default which is usually 0V.

The capacitor equation you quote is always valid... as long as capacitance is constant. In an opamp, for example, when output voltage gets close to the rails, output transistors (and some others in the circuit) get low Vce or Vds, which means their capacitance increases, transconductance decreases, etc. So, frequency response, phase margin, and all other small signal parameters do depend on output voltage and output current.

If you use an opamp model which includes "real" transistor models, or if you build an amplifier circuit in the simulator, to observe this effect, you can run AC analysis while stepping the output voltage from 0 to just before clipping. If the load is reactive, also pay attention to output current. This will give information on the stability of the circuit close to clipping.

For example if the output of your opamp is loaded by a resistor to GND, and it is a single supply rail to rail opamp, when the output gets near 0V, the bottom transistor of the output stage is almost saturated, so it will be very slow... but it has almost nothing to do and very little current to pass, so that somewhat compensates. However, when the output gets close to the positive supply, the top transistor gets close to saturation, and it has all the load current to handle. On top of that the bottom transistor, which has all the Vce, is off. This means you'll get quite a different frequency response when the output gets close to one rail or the other, and that depends on the type of load and how it is connected.

\$\endgroup\$
3
  • \$\begingroup\$ Thanks for answer!"which means changes in state variables are small enough " Is there any method to determine it mathematically? "The capacitor equation you quote is always valid" But problem is this is valid for steady state analysis, but for transient isn't we have to use differentiation or integration model of capacitor? \$\endgroup\$
    – user215805
    Aug 22, 2021 at 6:48
  • \$\begingroup\$ "small enough" is a subjective: if device characteristics like gm and capacitance change, say, a few % between the top of the waveform and the bottom, there will be some distortion, but that will only change the frequency response by a few % which is a tiny fraction of a dB. So you could calculate how "small" from how much error you allow on the frequency response. \$\endgroup\$
    – bobflux
    Aug 22, 2021 at 7:02
  • \$\begingroup\$ The capacitor equations written as Zc=1/jCw and Ic=jCw V are just another way to write i=Cdv/dt ; this works if C is constant. Small signal impulse response and small signal steady state frequency+phase response are two ways to look at the same thing, you can calculate one from the other by Fourier transform. You'd say the capacitor is under large signal conditions if its capacitance varies significantly with voltage, like a X7R ceramic cap. In this case, i=dq/dt, q=cv, so i=cdv/dt + vdc/dt, the extra term which models capacitance change means it is no longer equivalent to Ic=jCwV. \$\endgroup\$
    – bobflux
    Aug 22, 2021 at 7:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.