Shortcut Methods
Numerical Problems on PN Junction Basics:
1. Forward Bias:
 Shortcut: For silicon diodes, the forward bias voltage is approximately 0.7 V. Thus, the voltage drop across the diode can be assumed to be 0.8 V  0.7 V = 0.1 V.
 Calculation: Using Ohm’s law, I = V/R, the current through the diode is I = 0.1 V / 10 Ω = 10 mA.
2. Reverse Bias:
 Shortcut: For germanium diodes, the reverse saturation current is typically in the range of 1 nA to 10 nA.
 Calculation: Using the equation I_R = I_S (e^(V_R/V_T)  1), where I_R is the reverse current, I_S is the saturation current, V_R is the reverse bias voltage, and V_T is the thermal voltage (approximately 26 mV at room temperature), the reverse current is: $$I_R = 10 nA * (e^(10 V/26 mV)  1) \approx 2.4 \times 10^{36} A$$
3. Depletion Region:
 Shortcut: The width of the depletion region can be approximated using the formula: $$W = \sqrt{\frac{2\epsilon (V_{bi} + V_R)}{qN_{d}N_{a}}}+\frac{\epsilon}{qN_{a}W}+\frac{\epsilon}{qN_{d}W}$$ where ϵ is the permittivity of silicon, V_bi is the builtin potential, V_r is the reverse bias, q is the electronic charge, and $N_d$ and $N_a$ are the doping concentrations on the nside and pside, respectively.
Calculation: $$W = \sqrt{\frac{2 \times 1.04 \times 10^{12} F/m (0.7 V + 0 V)}{1.602 \times 10^{19} C \times 10^{15} cm^{3} \times 10^{17} cm^{3}}} = 1.14 \times 10^{6} m$$
4. Diffusion Current:

Shortcut: The diffusion current can be approximated using the formula (I_d = qAn_id_p/\tau_p + qAD_n n_p/ \tau_n), where q is the electronic charge, A is the area of the junction, n_i is the intrinsic carrier concentration ((\approx)1.5 × 10^10 cm^3 for silicon at room temperature), (D_p) and (D_n) are the hole and electron diffusion coefficients (generally taken as 104 cm2/s for silicon at room temperature), (\tau_p) and (\tau_n) are the hole and electron lifetimes, respectively (generally taken as 10^6 s for silicon at room temperature), and n_p is the electron concentration on the pside.

Calculation: $$I_d = (1.602 \times 10^{19} C)(10^{3} cm^2)(10^{17} cm^{3})(10 cm/s)/(10^{6} s)+ (1.602 \times 10^{19} C)(10^{3} cm^2)(1.5 \times 10^{10} cm^{3})(10 cm/s)/(10^{6} s) \approx 2.46 \times 10^{9} A$$
5. Capacitance:

Shortcut: The capacitance of a pn junction can be approximated using the formula (C = \frac{\epsilon A}{W}), where ε is the permittivity of silicon, A is the area of the junction, and W is the width of the depletion region.

Calculation: $$C = \frac{1.04 \times 10^{12} F/m \times 10 mm^2}{0.5 \times 10^{6} m} = 2.08 pF$$
6. Breakdown Voltage:

Shortcut: For abrupt junctions, the breakdown voltage is given by (V_{BR} = \frac{\epsilon E^2}{2qN_a}), where ϵ is the permittivity of silicon, E is the critical electric field (around 3 × 10^5 V/cm for silicon), and N_a is the doping concentration on the pside.

Calculation: $$V_{BR} = \frac{1.04 \times 10^{12} F/m \times (3 \times 10^5 V/m)^2}{2 \times 1.602 \times 10^{19} C \times 10^{16} cm^{3}} \approx 9.2 V$$
Note: These numerical problems are just a few examples to give you an idea of the types of questions you may encounter. The actual numerical values can vary depending on the specific context and parameters involved in the problem.