# Simple method to calculate output impedance of source follower

Output impedance of source follower without Rs would be 1/gm. However, if you add Rs the expression becomes a little complex and the result is as in the image below. I know how to get the result with brute-force method.
However, is the a simple way to get that expression taking advantage that we already know the output impedance of the circuit without Rs is 1/gm?

With R_s = 0, an M1 impedance contribution and an C_gs impedance are connected in parallel; the output impedance is $$Z_0 = {1\over{g_m}} || {1\over{s·C_{gs}}}$$ Neglecting C_ds, an R_s contribution is in series, only it should be scaled down. Common drain stage is a non-inverting current amplifier. Given a common drain stage's AC current gain $$A_{cd} = {1 + {g_m\over{s·C_{gs}}}}$$ we can calculate a current thru R_s and a scaled-down R_s contribution R_sScaled into the output impedance $$R_{sScaled} = {R_s\over{A_{cd}}} = {R_s\over{1 + g_m\over{s·C_{gs}}}} = {R_s·s·C_{gs}\over{s·C_{gs} + g_m}}$$ The total output impedance is $$Z = Z_0 + R_{sScaled}$$ $$Z = {1\over{s·C_{gs} + g_m}} + {R_s·s·C_{gs}\over{s·C_{gs} + g_m}} = {{1+R_s·s·C_{gs}}\over{s·C_{gs} + g_m}}$$ The only problem with this derivation is that it is only trustworthy for those who knows how to get the result with brute force method. For someone who has never analyzed an AC small-signal equivalent circuit of common drain configuration, it may be difficult to see a Thevenin behind the above equations.

• Thanks, I'll check it tomorrow as I'm not quite familiar current gain like voltage gain. Jul 17, 2020 at 17:00
• +1, this is easier in BJT as the current gain from base to emiter is constant (1 + β). Btw, you made a mistake in the current gain. In the numerator, it should be gm+s Cgs. Dec 10, 2023 at 17:00

anhnha - perhaps you like my approach?

From system theory (and from the classical feedback model) we know that - in case of feedback - the open-loop output impedance Zol without feedback must be divided by the expression (1-LG) with LG=loop gain. (We have a minus sign due to positive feedback)

This opens the way to a relatively simple and straight-forward calculation of the closed-loop output impedance Zout=Zol/(1 - LG).

1.) Disconnecting the gate from the rest of the circuit, the open-loop gain is

Aol=(Rs+1/sC)/(1/gm + 1/sC + Rs)

2.) If we open the loop at the gate node, the loop gain is

LG=Aol*Rs/(Rs+1/sC)=Rs/(1/gm + Rs + 1/sC)

3.) Without feeedback, the open-loop output impedance is

Zol=(1/gm)||(Rs+1/sC) = (Rs + 1/sC)/(1 + gmRs + gm/sC)

4.) Inserting these expressions into Zout=Zol/(1-LG) leads to

Zout=(1 + sRsC)/(gm + sC).

• This is also a good method to apply feedback theory. However, somehow I can't get the final expression in 4. Zout = ((gm + C s) (1 + C Rs s))/(gm + C s + C gm Rs s)^2 Jul 17, 2020 at 16:25
• In the denominator D(s) of the final expression we have D(s)=[1-Rs/(1/gm + 1/sC + Rs)]. Rewriting the "1" as a fraction (denominator like the second expression), Rs cancels out in the denominator. After some simple manipulations (simplify numerator and denominator) you arrive at the final expression.
– LvW
Jul 17, 2020 at 16:47
• I got it. You made a mistake in your final formula. The output impedance is reduced by LG so it should be divider instead. Btw, I got the same result with Extra Element Theory. Have you heard about that? Jul 17, 2020 at 16:59
• Where is a mistake? More than that, if you compare Zol and the closed-loop impedance Zout, you will see that Zout>Zol. This could be expected because of positive feedback. And - yes - I know the EE theorem.
– LvW
Jul 18, 2020 at 7:51
• You made a mistake in 4, just typo I believe. You wrote Zout=Zol(1-LG)  but it should be Zout=Zol/(1-LG) . Jul 18, 2020 at 11:03