# Step-by-Step Solution with Rendered Equations ## Given Data - **Forward bias voltage:** \[ V_A = .7\ \text{V} \] - **High temperature:** \[ T_H = 300\ \text{K} \] - **Low temperature:** \[ T_L = 250\ \text{K} \] - **Measured current ratio:** \[ \frac{J(T_H)}{J(T_L)} = 375 \] - **Assumption:** Diffusion constant \( D \) and lifetime \( \tau \) are **independent of temperature**. --- ## Step 1: Diode Current Expression For a forward-biased pn junction, the current density is proportional to the square of the intrinsic carrier concentration and the exponential voltage dependence: \[ J \propto n_i^2\, e^{\frac{qV_A}{kT}} \] The temperature dependence of current comes from: 1. \( n_i \) (intrinsic carrier concentration) 2. Exponential term \( e^{qV_A/kT} \) --- ## Step 2: Temperature Dependence of \( n_i^2 \) The intrinsic carrier concentration varies as: \[ n_i^2 \propto T^3 e^{-\frac{E_g}{kT}} \] Thus, diode current can be written as: \[ J(T) \propto T^3\, e^{-\frac{E_g}{kT}}\, e^{\frac{qV_A}{kT}} \] Combine the exponents: \[ J(T) \propto T^3\, e^{\frac{qV_A - E_g}{kT}} \] --- ## Step 3: Form the Current Ratio Take the ratio at two temperatures: \[ \frac{J(T_H)}{J(T_L)} = \left( \frac{T_H}{T_L} \right)^3 \exp\left[ \frac{qV_A - E_g}{k} \left( \frac{1}{T_H} - \frac{1}{T_L} \right) \right] \] --- ## Step 4: Substitute Known Values ### Temperature Ratio Term \[ \left( \frac{300}{250} \right)^3 = (1.2)^3 = 1.728 \] ### Exponential Term Setup \[ \frac{1}{T_H} - \frac{1}{T_L} = \frac{1}{300} - \frac{1}{250} = \frac{250 - 300}{300 \times 250} = \frac{-50}{75,\!000} = -.0006667 \] Boltzmann constant: \[ k = 8.617 \times 10^{-5}\ \text{eV/K} \] --- ## Step 5: Insert into Equation \[ 375 = 1.728 \cdot \exp\left[ \frac{.7 - E_g}{8.617\times10^{-5}} \times (-.0006667) \right] \] Divide both sides: \[ \frac{375}{1.728} = 217 \] Take the natural logarithm: \[ \ln(217) = 5.38 \] --- ## Step 6: Solve for \( E_g \) \[ 5.38 = \frac{(.7 - E_g) \times (-.0006667)}{8.617\times10^{-5}} \] Multiply both sides by \( 8.617 \times 10^{-5} \): \[ 5.38 \times 8.617 \times 10^{-5} = (.7 - E_g) \times (-.0006667) \] \[ .000463 = (.7 - E_g) \times (-.0006667) \] Divide both sides by \( -.0006667 \): \[ \frac{.000463}{-.0006667} = .7 - E_g \] \[ -.695 = .7 - E_g \] So, \[ E_g = .7 + .695 = 1.395\ \text{eV} \] --- ## Final Answer \[ \boxed{E_g \approx 1.4\ \text{eV}} \] --- ## Interpretation - The calculated bandgap is **close to silicon** (\( \approx 1.12\ \text{eV} \) at room temperature). - The slight difference is due to: - Approximations in the calculation - Assumed temperature independence of \( D \) and \( \tau \)