QB_PH_SEC8_010: Semiconductor Devices & Diodes

An n-channel JFET has a saturation drain current of $I_{\text{DSS}} = 8\text{ mA}$ and a pinch-off voltage of $V_P = -4.0\text{ V}$. What is the transconductance $g_m$ (in $\text{mS}$) at a gate-to-source voltage of $V_{\text{GS}} = -2.0\text{ V}$?

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Shockley's Drain Current Equation** (in saturation/pinch-off regime): $I_D = I_{\text{DSS}} \left(1 - \frac{V_{\text{GS}}}{V_P}\right)^2$ 2. **Transconductance Definition**: $g_m = \left.\frac{\partial I_D}{\partial V_{\text{GS}}}\right|_{V_{\text{DS}}} = -\frac{2 I_{\text{DSS}}}{V_P} \left(1 - \frac{V_{\text{GS}}}{V_P}\right) = \frac{2 I_{\text{DSS}}}{|V_P|} \left(1 - \frac{V_{\text{GS}}}{V_P}\right)$ 3. **Evaluation**: - Maximum transconductance at $V_{\text{GS}} = 0$: $g_{m0} = \frac{2 \times 8\text{ mA}}{4.0\text{ V}} = 4.0\text{ mS}$ - At $V_{\text{GS}} = -2.0\text{ V}$: $g_m = 4.0 \times \left(1 - \frac{-2.0}{-4.0}\right) = 4.0 \times (1 - 0.5) = 2.0\text{ mS}$ 4. **Official Integer Answer**: 2.