Why is the capacitance calculated with small-signal variables the same as the capacitance from the basic steady-state formula?
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. 531
Also looking at https://physics.stackexchange.com/questions/227063/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 https://physics.stackexchange.com/questions/227063/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 phasedifference 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! 628