# I/O Resistance of common source MOSFET with source degeneration

Common Source MOSFET with source degenerations looks like this

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.

• What happens in a bipolar small-signal-model circuit, when emitter degeneration is installed? The small-signal-models are identical. – analogsystemsrf Mar 31 '17 at 2:46

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:

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.

$r_x = \frac{V_X}{I_X}$ 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 $r_{out} \approx R_D$

• 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 – Keno Mar 31 '17 at 21:02
• 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? – Keno Mar 31 '17 at 21:06
• Or not? Because it is already added to "rx"? – Keno Mar 31 '17 at 21:33
• @Keno When you are looking into the MOSFET drain terminal and the MOSFET work in saturation region. You will newer see 1/gm. – G36 Apr 1 '17 at 16:48