NMOS small signal high frequency

Question

I'm trying to understand the transition frequency of a MOSFET.

The book defines it as the frequency at which the small-signal current gain of the device drops to unity while the source and drain terminals are held at AC ground. What does it mean by unity?

Given this circuit:

I understand that:

$$ I_o = g_m V_{gs}$$

But the expression for the input current is confusing me

(The book ignores Cgb since it is too small):

$$ I_i = s(C_{gs}+C_{gd})V_{gs}$$

Where does the s come from? Why are the capacitors added then multiplied by the gate to source voltage?

The final expression is: $$ f_t = \frac{g_m}{2\pi(C_{gs}+C_{gd})V_{gs}} $$

Answer 1

\$s\$ here is the Laplace variable. If you substitute \$j\omega\$ in its place that helps explain what they're doing. Substituting \$j\omega\$ for \$s\$ takes the expression from the Laplace domain into the frequency domain.

The step that might be missing for you is this one:

$$ I = V/Z $$ $$ i_i = \frac{v_{gs}}{1/(s(C_{gs}+C_{gb}))} = s(C_{gs}+C_{gb})v_{gs}$$

This comes from the Ohm's law impedance relationship \$I = V/Z\$ where in this case the impedance is of the parallel capacitors. \$Z\$ (impedance) for a capacitor is \$1/sC\$ (expressed in the Laplace domain).

Current gain is output current divided by input current:

$$ G_i = \frac{i_o}{i_i} $$

This is 1 (unity) when \$i_o = i_i\$

To get from angular frequency (\$\omega\$) to frequency in Hz, you divide by \$2\pi\$.

So working through the algebra, unity gain is where:

$$ i_o = i_i $$ $$ g_mv_{gs} = s(C_{gs}+C_{gb})v_{gs}$$ $$ s = j\omega = \frac{g_mv_{gs}}{(C_{gs}+C_{gb})v_{gs}} = \frac{g_m}{C_{gs}+C_{gb}} $$ $$ f_t = \frac{\omega_t}{2\pi} = \frac{g_m}{2\pi(C_{gs}+C_{gb})} $$

Answer 2

By unity gain, it means the output current is equal to the input current.

The input current expression comes from solving V=IZ for I and substituting the impedance of the capacitor (1/sC) for Z. Then you can factor out the common terms (s and Vgs) to get the expression your book has.

You get the final expression by equating input and output current, and solving for frequency (s= jw for the case of steady state AC input).