# Input capacitance of common source amplifier

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?

• The Miller effect only applies during the amplifier's linear range where Vout = Vin*some negative gain. So the first point to establish is what Vout is doing.
– user16324
Dec 15, 2021 at 16:24
• @user_1818839 But if you built this circuit in real life, and wanted to measure the input capacitance, wouldn't you try to do like I have done, and measure around 0.15pF? And does it even make sense to talk about an input capacitance, if it changes over time?
– Carl
Dec 15, 2021 at 20:17
• Sure. But first I'd verify the output was about Vdd/2, to see the Miller capacitance.
– user16324
Dec 15, 2021 at 21:50
• There are a few answers to something very similar, here. Dec 16, 2021 at 11:35

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