Input capacitance of common source amplifier

Question

I am trying to find the input capacitance for the common source amplifier stage below: -

I have tried to find \$C_{in} \$ by simulation. I apply a linearly increasing voltage source \$v_{in}(t) = 1 \text{V} \cdot t \$. By doing this, I should be able to find \$C_{in} \$ with the formula $$i_{in} = C \frac{dv_{in}(t)}{dt} \Leftrightarrow i_{in} = C \cdot 1\text{V}$$

From the simulation, the current \$i_{in} \$ settles at \$-150\text{fA} \$ which suggests that \$C_{in} = 150\text{fF} = 0.15\text{pF} \$ which corresponds to adding the two capacitors \$C_1 \$ and \$C_2\$ in parallel, \$C_1+C_2 = 0.05\text{pF} + 0.1\text{pF} = 0.15\text{pF} \$.

However, the solution states that \$C_{in} = 2.45\text{pF} \$ because you have to take the Miller effect into account. But why don't I see this when I simulate the circuit? Is the solution incorrect (probably not) or am I not doing something correct with my simulation?

Answer 1

If you use an AC analysis, it's pretty easy to perform an input capacitance test. Using LTspice syntax:

Parallel input capacitance: -Im(I(vin)/v(in))/omega
Parallel input resistance: -1/Re(I(vin)/v(in))

In LTspice, the above equations are entered in the graphical portion of the program.