Why is the capacitance calculated with small-signal variables the same as the capacitance from the basic steady-state formula? [closed]

Question

As an electrical engineering student I am currently working at a capacitance model for transistors. To verify my model I use Sentaurus TCAD to simulate my device and create capacitance data. I struggle to interpret the generated data, because I feel like I don't understand the definition of capacitances in general. Also looking at What is capacitance, in general? did not help. A big part of my confusion lies in the different capacitance formula.
TCAD calculates the capacitance values, like explained in What is capacitance, in general?, by setting a DC operating point and then superimposing small signal excitation voltages on top, so that the capacitance values can be extracted from the admittance like this: $$ \frac{\delta i_{i}}{\delta v_{j}} = y_{ij} = a_{ij} + i\omega C_{ij}\quad \Rightarrow \quad C_{ij}=\frac{Im(y_{ij})}{\omega}=\frac{Im(y_{ij})}{2\pi\nu} $$
$\nu$ being the excitation frequency*
In general (steady-state, DC) capacitances are defined like this: $ C=\frac QV $
using indices for the electrodes I can rewrite it to: $ C_{ij} = \frac{Q_i}{V_{j}} $ assuming that electrode i is at ground.

Now comparing this definition to the one of the small signal (or with phasors) they don't seem to match, because the steady state one describes the relation of charge to voltage, while the other describes the impact of an admittance on the phase difference between a (small signal) current and voltage. $$ C_{ij}=\frac {Q_i}{V_j} =\frac{dQ_i}{dV_j}\cdot \frac{dt}{dt}= I_i\frac{dt}{dV_j} = I_i\left(\frac{dV_j}{dt}\right)^{-1} \overset ?{\neq} \frac 1{2\pi\nu}\cdot Im\left(\frac{d\underline I_i}{dt}\cdot\left(\frac{d\underline V_j}{dt}\right)^{-1}\right) \\= \frac 1{\omega}\cdot Im\left(\frac{d\underline I_i}{d\underline V_j}\right) = \frac {Im\left(\underline Y_{ij}\right)}{\omega} = C_{ij} $$ the underlined letters depict phasors

But still the capacitance is notationalwise the same and should be interchangable. Why is this possible? Why is not differentiated between small-signal and steady-state capacitance?

I am very grateful for every answer and input you guys can give me!

Answer 1

The definition of capacitance in a linear dielectric is calculated indeed by $\frac{Q}{V}$ but it is a simplification of a situation where the relationship is nonlinear that is abstracted from the more general case of a constitutive material relationship between the locally induced polarization and the local electric field $\mathbf P = \mathbf {P(\mathbf{E})}$. (EEs prefer the $D$ over $P$ and write $\mathbf D = \epsilon_0 \mathbf E + \mathbf P$, $\mathbf D = \mathbf {D(\mathbf{E})}$.

In the nonlinear case the linear "definition" is not useful except for being as an operating point around which one linearizes the same way as you have surely learned of the "gain" of a non-linear amplifier makes no general sense unless it is meant to be at a fixed operating point around which you linearize locally.

Answer 2

In things like transistor junctions, the geometry changes with applied voltage. Dielectric constants in things like ceramic capacitors are functions of the electric field. In those cases, $Q/V$ is not a constant as a function of voltage. The small-signal capacitance gives you the change in $V$ for a small change in $Q$ (or vice-versa) when they aren't proportional for large changes.