MOSFET DC analysis

MOSFET DC analysis finds a MOSFET circuit’s Operating point (the quiescent $I_D$, $V_{GS}$, and $V_{DS}$) using the MOSFET large-signal model. Every small-signal/amplifier calculation assumes a known bias point, so this comes first.

The recipe

You have three relations to work with:

1. The overdrive definition: $V_{GS}=V_t+V_{OV}$.
2. The square-law: $I_D=\frac{1}{2}{k}_n{V}_{OV}^2=\frac{1}{2}{k}_n(V_{GS}-V_t{)}^2$.
3. KVL/KCL around the circuit.

Steps:

1. Assume saturation. It’s the usual amplifier region and gives the simple square-law.
2. Write KVL through the gate–source loop. The gate draws no current (insulated by the Gate oxide), so nothing flows in the gate path, which cuts the loop equation way down.
3. Combine KVL with the square-law. This gives a quadratic in $V_{GS}$ (or in $V_{OV}$). Solve it.
4. Pick the physical root: the one with $V_{GS}>V_t$ (equivalently $V_{OV}>0$). The other root is spurious.
5. Compute $I_D$, then $V_{DS}$ from the drain-loop KVL.
6. Check the assumption: is $V_{DS}\ge V_{OV}$? If yes, saturation holds and you’re done. If $V_{DS}<V_{OV}$, the device is actually in the MOSFET triode region — discard and redo using the triode equation $i_D=k_n[V_{OV}{V}_{DS}-V_{DS}^2/2]$.

Choose $R_D,R_S$ so $I_D$ is at a target with the device saturated.

Worked example

A MOSFET with $V_t=0.7\text{V}$ and $k_n=1{\text{mA/V}}^2$ has its gate held at $V_G=5\text{V}$ through a large resistor, a drain resistor $R_D$ to a $V_{DD}=10\text{V}$ supply, and a source resistor $R_S$ to ground. Take $R_S=3\text{k}\Omega$. Find $I_D$, $V_{GS}$, $V_{DS}$ and verify saturation.

The gate current is zero, so no drop occurs in the gate path and $V_G$ appears directly at the gate. KVL gate → source → ground:

$V_G=V_{GS}+I_D{R}_S$

Substitute the square-law $I_D=\frac{1}{2}{k}_n(V_{GS}-V_t{)}^2$:

$5=V_{GS}+(3000)\cdot \frac{1}{2}(10^{-3})(V_{GS}-0.7{)}^2$

$5=V_{GS}+1.5(V_{GS}-0.7{)}^2$

Let $x=V_{GS}-0.7$ (so $V_{GS}=x+0.7$):

$5=x+0.7+1.5x^2\Rightarrow 1.5x^2+x-4.3=0$

$x=\frac{-1+\sqrt{1+4(1.5)(4.3)}}{2(1.5)}=\frac{-1+\sqrt{26.8}}{3}=\frac{-1+5.177}{3}\approx 1.39\text{V}$

(taking the positive root for $V_{GS}>V_t$). So $V_{OV}\approx 1.39\text{V}$, $V_{GS}\approx 2.09\text{V}$, and

$I_D=\frac{1}{2}(10^{-3})(1.39{)}^2\approx 0.97\text{mA}$

With $R_D=2\text{k}\Omega$:

$V_{DS}=V_{DD}-I_D(R_D+R_S)=10-(0.97\times 10^{-3})(2000+3000)\approx 10-4.83=5.17\text{V}$

Check: $V_{DS}=5.17\text{V}\ge V_{OV}=1.39\text{V}$ ✓ saturation holds, so the answer stands.

Solve $I_D,V_{GS},V_{DS}$ via KVL + square-law + saturation assumption; verify it holds.

The general drain-loop result used above is $V_{DS}=V_{DD}-I_D(R_D+R_S)$, and the gate-loop result $V_G=V_{GS}+I_D{R}_S$ is the basis of source-resistor MOSFET biasing: if $V_G\gg V_{GS}$ then $I_D\approx V_G/R_S$, almost independent of the wobbly transistor parameters.