The small-signal model (SSM) of a resistor is obviously the same resistor. Here I got a problem with understanding what SSM of inductor and capacitor would look like. Here's my analysis and please correct my if you see something wrong in my analysis:

To find SSM of the inductor we develop our model and write small-signal voltage as follows: \begin{eqnarray*} v_L=L\ \frac{d i}{dt} \\ v_{SSM}= \frac{\partial v_L}{\partial i_L}.i_L\\ v_{SSM}=\frac {\partial (L\ \frac{di_L}{dt})}{\partial i_L}.i_L\\ v_{SSM}=\frac {\partial }{\partial i_L}(L\ \frac{di_L}{dt}).i_L=L\frac{d}{dt}(\frac {\partial i_L}{\partial i_L}).i_L\\ v_{SSM}=L \ \frac{d}{dt}(1).i_L=0 \end{eqnarray*} So I infer that the SSM of the inductor is much the same as that of a voltage source, which is shorted out. Similarly for the SSM of the capacitor one can show, through the same kind of analysis, that \begin{eqnarray*} i_{SSM}=0,\ \end{eqnarray*} which says that the SSM of a capacitor behaves like an open to our circuit model. Do you think that my analysis is wrong? Where is the bug in my reasoning?


  • \$\begingroup\$ The steady state model of an inductor is a short, the small signal model is the same as the large signal model. Passive linear circuits tend to behave the same regardless of whether you are in the small signal or large signal region. (I say linear because non-linear devices will do different things depending on the signal - transformer saturation for example) \$\endgroup\$
    – Sam
    May 5, 2016 at 9:34
  • \$\begingroup\$ Alright. I want to show it mathematically like we did for a resistor. We know a resistor is a linear device so its small signal model is the same as it's large, and we can prove it. So at least we would expect to get the same result for inductor and capacitor as they are linear devices. \$\endgroup\$
    – dirac16
    May 5, 2016 at 9:43
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    \$\begingroup\$ VL = L+di/dt, it doesn't matter what size the signal is, the equation still holds, there is no other equation... in much the same way as V=IR, there is no other equation for small signals \$\endgroup\$
    – Sam
    May 5, 2016 at 9:49
  • \$\begingroup\$ True. now I can understand. As inductor is linear we could care less what the size of the signal is because we would get a linear response out of a linear device, irrespective of the input signal. However, for non-linear devices things get complicated and we would have to consider linearization methods to get a linear response. Alright? \$\endgroup\$
    – dirac16
    May 5, 2016 at 10:01
  • \$\begingroup\$ Yep, you got it, that's exactly right, being linear means it will respond the same way under all conditions, being non-linear means that it will behave differently under different conditions hence the need to use linearization methods to make things simple to solve (you could use non-linear maths, but that's a good way to give yourself an aneurysm) \$\endgroup\$
    – Sam
    May 5, 2016 at 10:04

1 Answer 1


Your analysis is wrong.

Small signal models consist in a linearization of the non-linear equations describing a certain device (as a transistor or a diode). Of course, it makes sense only if such device is non-linear, i.e. described by a set of non-linear equations.

It is called small-signal analysis (and the models are called small-signal models) because they are only accurate if the signals you're considering have "small" amplitude (as the linear approximation of a non-linear equation can be used to make accurate calculations only if you consider quantities nearby the point around which your analysis is centered). Read about Taylor series for further info.

The reason for the linearization of a non-linear equation: the second is a lot harder to handle and/or to solve.

The equation you wrote for the capacitor and the inductor are already linear, i.e. there's no need for linearization.


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