Worked example: MOSFET bias and switching

Semiconductor Devices · MOSFET · Example

Problem 1 — Find I_D in saturation

An enhancement NMOS has $k_n = 2\,\text{mA/V}^2$ and $V_T = 2\,\text{V}$ . The circuit sets $V_{GS} = 4\,\text{V}$ and $V_{DS} = 6\,\text{V}$ . Find $I_D$ and confirm the operating region.

Check that the transistor is ON.
Threshold $V_T = 2\,\text{V}$ , applied $V_{GS} = 4\,\text{V}$ . Since $V_{GS} > V_T$ , the channel exists.
Determine the operating region.
Overdrive voltage: $V_{GS} - V_T = 4 - 2 = 2\,\text{V}$ .
Check saturation condition: $V_{DS} \geq V_{GS} - V_T$ , i.e. $6 \geq 2\,\text{V}\;\checkmark$ . The MOSFET is in saturation.
Apply the saturation equation.
$I_D = \frac{k_n}{2}(V_{GS} - V_T)^2 = \frac{2}{2}(4 - 2)^2 = 1 \times 4 = 4\,\text{mA}$
Answer: $I_D = 4\,\text{mA}$ in the saturation region.

Problem 2 — Find V_GS for a target I_D

Using the same device ( $k_n = 2\,\text{mA/V}^2$ , $V_T = 2\,\text{V}$ ), find the gate-source voltage needed to set $I_D = 8\,\text{mA}$ in saturation.

Write the saturation equation and isolate the overdrive.
$I_D = \frac{k_n}{2}(V_{GS} - V_T)^2 \implies (V_{GS} - V_T)^2 = \frac{2\,I_D}{k_n}$
Take the positive square root (overdrive must be > 0).
$V_{GS} - V_T = \sqrt{\frac{2\,I_D}{k_n}} = \sqrt{\frac{2 \times 8}{2}} = \sqrt{8} = 2\sqrt{2} \approx 2.83\,\text{V}$
Add the threshold voltage.
$V_{GS} = V_T + 2.83\,\text{V} = 2 + 2.83 = 4.83\,\text{V}$
Verify by back-substituting.
$I_D = \frac{2}{2}(4.83 - 2)^2 = (2.83)^2 = 8.01\,\text{mA}\;\checkmark$
Answer: $V_{GS} \approx 4.83\,\text{V}$ .

Problem 3 — Triode region and drain resistance

The same device ( $k_n = 2\,\text{mA/V}^2$ , $V_T = 2\,\text{V}$ ) is biased with $V_{GS} = 4\,\text{V}$ and $V_{DS} = 0.5\,\text{V}$ . Find $I_D$ and the on-resistance $r_{DS}$ .

Confirm the operating region.
Overdrive: $V_{GS} - V_T = 2\,\text{V}$ .
Check: $V_{DS} = 0.5\,\text{V} < V_{GS} - V_T = 2\,\text{V}$ .
The MOSFET is in the triode (linear) region.
Apply the triode equation.
$I_D = k_n\!\left[(V_{GS}-V_T)V_{DS} - \frac{V_{DS}^2}{2}\right]$
$I_D = 2\left[(2)(0.5) - \frac{(0.5)^2}{2}\right] = 2\left[1 - 0.125\right] = 2 \times 0.875 = 1.75\,\text{mA}$
Find the on-resistance.
For small $V_{DS}$ , the MOSFET approximates a resistor:
$r_{DS} = \frac{V_{DS}}{I_D} = \frac{0.5\,\text{V}}{1.75\,\text{mA}} \approx 286\,\Omega$

Using the small-signal formula: $r_{DS} \approx \dfrac{1}{k_n(V_{GS}-V_T)} = \dfrac{1}{2 \times 2} = 250\,\Omega$ . The slight difference is because 0.5 V is not truly "small" compared to the overdrive.
Answer: $I_D = 1.75\,\text{mA}$ , $r_{DS} \approx 286\,\Omega$ .

By controlling V_GS you control r_DS — this is how MOSFETs work as analogue switches and pass-gates. A larger overdrive (V_GS − V_T) gives a smaller r_DS and faster switching.