Common Source MOSFET with source degenerations looks like this enter image description here

I am a bit confused about different input and output resistance statements (provided by different sources).

Some of them say that applying Rs to circuit DOES NOT change input and output resistances even a bit (which I hardly believe).

But the others say that Rs "boosts" AC output impedance which probably means that Rs increases output impedance.

But I can't find any formula which could explain what is happening with output resistance. (Such as for CS without Rs --> Rout = Rd || Rload || ro )

Can someone explains me what really happens with ouput resistance in CS source degeneration transistor circuit?

*I get the rest advantages as improved linearity, lower voltage gain, etc.

  • \$\begingroup\$ What happens in a bipolar small-signal-model circuit, when emitter degeneration is installed? The small-signal-models are identical. \$\endgroup\$ Mar 31, 2017 at 2:46

1 Answer 1


In general source degeneration resistor "adds" a negative feedback to the circuit (current-series feedback). In this case, we sample the output current (\$I_D\$) and return a proportional voltage in series with the input (\$V_{GS} = V_G - I_D*R_S\$). This type of a feedback increases \$Rin\$ and \$Rout\$. But notice that the MOSFET itself has a very large \$Rin =\infty\$, therefore \$Rin = R1||R2\$ remains unchanged.

The voltage gain also drops to \$Av = -\frac{R_D}{R_S + 1/gm} = -\frac{R_D||R_L}{\frac{1}{gm} +R_S||R_3} \$

This also improves linearity, because without \$R_S\$ voltage gain is \$gm*R_D\$ and as you should know \$gm\$ varies with drain current. Because \$gm\$ is a function of drain current (\$I_D\$), the voltage gain will vary with signal swing and the voltage gain also. But if we add external source resistance \$R_S\$ we notice that the \$R_S\$ does not change with the signal swing (\$I_D\$ swing)so, the overall voltage gain is stabilized and is more linear.

For \$R_S >> 1/gm\rightarrow A_V\approx \frac{R_D}{R_S}\$

Now let us look at \$rout\$. If we are looking from the load perspective we can see two paths for a AC current to flow:

enter image description here

First through \$R_D\$ resistor.

And the second one through MOSFET channel -->\$R_S\$ into GND.

As you can see now \$R_S\$ resistor is in series with the MOSFET channel.

So, to find resistance seen from the drain terminal into the MOSFET we need to use a small-signal-model.

enter image description here

\$r_x = \frac{V_X}{I_X}\$ and because \$V_G = 0V\$ we have:

$$V_{GS} = -I_X*R_S $$

And from KVL we have

$$V_X = I_{ro}*ro+I_X*R_S$$

$$I_{ro} = I_X - gm*V_{GS}$$

$$V_X=\left ( I_X - \left (gm\left ( -I_X \right )R_S \right ) \right )ro + I_XR_S $$

And solve for \$I_X\$ $$I_X = \frac{V_X}{ R_S + ro + gm*R_S*ro} $$ And finally we have $$r_x = R_S + ro + gm*R_S*ro = ro(1+gmR_S+\frac{R_S}{ro}) $$

$$r_x = ro*(1+gmR_S)+R_S $$

As you can see adding \$R_S\$ resistor increase the MOSFET resistance.

The \$ro\$ is boosted by a factor of \$(1+gm R_S)\$

So, the overall \$r_{out}\$ is equal to:

$$r_{out} = R_D||r_x $$

and because \$R_D<<r_x\$ we have \$r_{out} \approx R_D\$

  • \$\begingroup\$ I don't know what to say. I was really hoping for that kind of answer - you explained me everything! I hope this answer gets more up-votes :D \$\endgroup\$
    – lucenzo97
    Mar 31, 2017 at 21:02
  • \$\begingroup\$ One more thing: You said "rout = Rd || rx". What about "ro" - the channel resistance which equals "1/gm"? Can it be added in parallel with Rd and rx for more aproximate result/value? \$\endgroup\$
    – lucenzo97
    Mar 31, 2017 at 21:06
  • \$\begingroup\$ Or not? Because it is already added to "rx"? \$\endgroup\$
    – lucenzo97
    Mar 31, 2017 at 21:33
  • \$\begingroup\$ @Keno When you are looking into the MOSFET drain terminal and the MOSFET work in saturation region. You will newer see 1/gm. \$\endgroup\$
    – G36
    Apr 1, 2017 at 16:48

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