QB_PH_SEC8_023: Semiconductor Devices & Diodes

At room temperature ($T = 300\text{ K}$), the thermal voltage is $V_T = \frac{k_B T}{e} = 26\text{ mV}$. If an ideal silicon diode with ideality factor $\eta = 1$ carries a forward current of $I_1 = 1.0\text{ mA}$ at voltage $V_1$, by how many millivolts must the forward voltage increase to increase the current tenfold to $I_2 = 10.0\text{ mA}$? (Take $\ln 10 = 2.303$. Round off to one decimal place)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Shockley Forward Diode Equation** (for $V \gg V_T$): $I = I_s e^{V / (\eta V_T)}$ 2. **Ratio of Currents**: $\frac{I_2}{I_1} = e^{(V_2 - V_1) / (\eta V_T)} \implies \Delta V = V_2 - V_1 = \eta V_T \ln\left(\frac{I_2}{I_1}\right)$ 3. **Numerical Computation**: - $\eta = 1$ - $V_T = 26\text{ mV}$ - $I_2 / I_1 = 10.0\text{ mA} / 1.0\text{ mA} = 10$ $\Delta V = (1)(26\text{ mV}) \ln(10) = 26 \times 2.302585 = 59.867\text{ mV} \approx 59.9\text{ mV}$ 4. **Standard Range**: $[59.5, 60.5]$.