Solved Examples (Dedicated Problem Set)

How to Use This Problem Set

Your full workout for Semiconductor Electronics, grouped by theme: band gaps & photons, carrier densities & mass action, conductivity, junction & bias reasoning, diode circuits & dynamic resistance, rectifiers, and the Zener/LED/logic-gate extras.

Keep these handy:

• Photon-gap bridge: $E_g$(eV) = $\dfrac{1240}{\lambda(\text{nm})}$; gaps: C 5.4, Si 1.1, Ge 0.7 eV
• Intrinsic: $n_e = n_h = n_i$; doped: majority ≈ dopant density; minority = $\dfrac{n_i^2}{n_{maj}}$
• Conductivity: $\sigma = e(n_e\mu_e + n_h\mu_h)$; donor ionisation ~0.01 eV (Ge), 0.05 eV (Si)
• Bias: forward barrier $V_0 - V$ (mA); reverse barrier $V_0 + V$ ($\mu$A); cut-in 0.2 V Ge / 0.7 V Si
• Dynamic resistance: $r_d = \Delta V/\Delta I$; rectifier ripple: half-wave $f$, full-wave $2f$
• Zener: $I_s = \dfrac{V_{in} - V_Z}{R_s}$, $I_Z = I_s - I_L$; gates: check corner rows (0,0) and (1,1)

State the condition, then substitute.

Solved Examples - Band Gaps & Photons

Example 1. Threshold wavelength for pair creation in Si ($E_g$ = 1.1 eV)?

Solution: $\lambda = \dfrac{1240}{1.1} \approx 1127$ nm (near IR).

Example 2. In Ge (0.7 eV)?

Solution: $\lambda = \dfrac{1240}{0.7} \approx 1771$ nm.

Example 3. Can 500 nm light create pairs in diamond (5.4 eV)?

Solution: photon = $\dfrac{1240}{500} = 2.48$ eV < 5.4 eV → no.

Example 4. An LED emits at 620 nm; its band gap?

Solution: $E_g \approx \dfrac{1240}{620} = 2.0$ eV — red-orange.

Example 5. Order C, Si, Ge, Sn by band gap and classify each.

Solution: C 5.4 (insulator) > Si 1.1 (semiconductor) > Ge 0.7 (semiconductor) > Sn 0 eV (metal).

Solved Examples - Carrier Densities & Mass Action

Example 6. Si doped 1 ppm with As ($5 \times 10^{28}$ atoms/m³, $n_i = 1.5 \times 10^{16}$): find $n_e$, $n_h$. (NCERT Example 14.2)

Solution: $n_e \approx N_D = 5 \times 10^{22}$ m⁻³; $n_h = \dfrac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = 4.5 \times 10^9$ m⁻³.

Example 7. Ge with $n_h = 5 \times 10^{21}$ m⁻³ ($n_i = 2.4 \times 10^{19}$): find $n_e$ and the type.

Solution: $n_e = \dfrac{(2.4 \times 10^{19})^2}{5 \times 10^{21}} = 1.15 \times 10^{17}$ m⁻³; $n_h \gg n_e$ → p-type.

Example 8. What donor density triples $n_e$ over $n_i$ in Si?

Solution: need $n_e = 4.5 \times 10^{16}$; with $n_e \approx N_D +$ (intrinsic balance), for $N_D \gg n_i$ regime the approximation $N_D \approx 4.5 \times 10^{16} -$ correction; to first order $N_D \approx 3n_i - n_i^2/(3n_i) = 2.67n_i \approx 4 \times 10^{16}$ m⁻³. For exam purposes: $N_D \approx 3n_i$ when doping dominates.

Example 9. By what factor does 1 ppm doping (Example 6) raise $n_e$ above intrinsic?

Solution: $\dfrac{5 \times 10^{22}}{1.5 \times 10^{16}} \approx 3.3 \times 10^6$ — a million-fold from one ppm.

Example 10. In Example 6, the ratio $n_e : n_h$?

Solution: $\dfrac{5 \times 10^{22}}{4.5 \times 10^9} \approx 10^{13}$.

Solved Examples - Conductivity

Example 11. Intrinsic Si with $n_i = 1.5 \times 10^{16}$ m⁻³, $\mu_e$ = 0.135, $\mu_h$ = 0.048 m²/Vs: find $\sigma$.

Solution: $\sigma = en_i(\mu_e + \mu_h) = 1.6 \times 10^{-19} \times 1.5 \times 10^{16} \times 0.183 \approx 4.4 \times 10^{-4}$ S/m.

Example 12. The doped sample of Example 6 ($n_e = 5 \times 10^{22}$): find $\sigma$ (ignore holes).

Solution: $\sigma = en_e\mu_e = 1.6 \times 10^{-19} \times 5 \times 10^{22} \times 0.135 \approx 1.1 \times 10^3$ S/m — over a million times the intrinsic value.

Example 13. Resistivity of the doped sample?

Solution: $\rho = 1/\sigma \approx 9 \times 10^{-4}$ ohm m — still within the semiconductor range.

Example 14. Why does Si's conductivity rise with temperature while copper's falls?

Solution: Si: carrier number grows exponentially (bond breaking) — dominates. Cu: carrier number fixed; phonon scattering grows — resistance rises.

Example 15. A Ge rod's resistance halves when warmed. What does this indicate about carrier generation?

Solution: carrier density roughly doubled — thermal generation across the 0.7 eV gap is steeply temperature-sensitive; the semiconducting signature.

Solved Examples - Junction & Bias Reasoning

Example 16. Barrier 0.7 V; forward bias 0.4 V. Effective barrier and depletion trend?

Solution: 0.3 V; depletion narrows.

Example 17. Same junction, reverse bias 4.3 V?

Solution: barrier 5.0 V; depletion widens; current = $\mu$A saturation.

Example 18. Field in a 0.2 $\mu$m depletion layer holding 0.6 V?

Solution: $E = \dfrac{0.6}{2 \times 10^{-7}} = 3 \times 10^6$ V/m.

Example 19. Why does reverse current double when temperature rises ~10 °C but barely change with voltage?

Solution: it is minority-carrier limited; minority generation (∝ $n_i^2$) is exponentially temperature-sensitive, while even small V already sweeps all available carriers.

Example 20. In which bias does 'minority carrier injection' occur and what does it produce?

Solution: forward bias; electrons injected into p and holes into n diffuse toward the contacts — the mA forward current.

Solved Examples - Diode Circuits & Dynamic Resistance

Example 21. From a Si characteristic: V 0.7 → 0.8 V takes I 10 → 20 mA. $r_{fb}$? (NCERT Example 14.4a)

Solution: $r = \dfrac{0.1}{10 \times 10^{-3}} = 10$ ohm.

Example 22. At -10 V, I = -1 $\mu$A: $r_{rb}$? (NCERT Example 14.4b)

Solution: $10/10^{-6} = 10^7$ ohm.

Example 23. Ideal diode + 300 ohm + 6 V battery (forward): current?

Solution: 6/300 = 20 mA.

Example 24. Same circuit with a real Si diode (0.7 V drop)?

Solution: $(6 - 0.7)/300 \approx 17.7$ mA.

Example 25. Two ideal diodes: D1 forward, D2 reverse, each in series with 100 ohm, both pairs in parallel across 2 V. Total current?

Solution: D2's branch open; only D1's branch conducts: I = 2/100 = 20 mA.

Example 26. A diode (0.7 V) in series with 230 ohm across 5 V: power dissipated in the resistor?

Solution: I = 4.3/230 ≈ 18.7 mA; $P = I^2R \approx 0.080$ W.

Solved Examples - Rectifiers

Example 27. Input 50 Hz: output ripple of half-wave and full-wave rectifiers? (NCERT Exercise 14.6)

Solution: 50 Hz and 100 Hz.

Example 28. Input 60 Hz to a bridge rectifier: ripple frequency and inter-hump gap?

Solution: 120 Hz; gap = 1/120 ≈ 8.3 ms.

Example 29. Secondary 15 V rms, ideal-diode half-wave rectifier: peak output?

Solution: $\sqrt{2} \times 15 \approx 21.2$ V.

Example 30. Full-wave (centre-tap) with total secondary 40 V rms: rms input per diode?

Solution: each diode works on half the winding: 20 V rms (peak ≈ 28.3 V).

Example 31. Filter design: full-wave, 50 Hz, $R_L$ = 500 ohm. Minimum C for time constant 10× the hump gap?

Solution: gap = 10 ms; need $R_LC$ = 0.1 s → $C = \dfrac{0.1}{500} = 200\ \mu$F.

Example 32. Why must the diode's $V_{br}$ exceed the peak secondary voltage?

Solution: in its blocking half-cycle the diode bears the full peak in reverse; reaching $V_{br}$ un-limited would destroy it by overheating.

Solved Examples - Zener, LED & Logic Gates [JEE/NEET Extras]

Example 33. Zener $V_Z$ = 9 V, $R_s$ = 500 ohm, input 15 V, no load: $I_Z$?

Solution: $(15 - 9)/500 = 12$ mA.

Example 34. Add a 1.5 kohm load: find $I_L$, $I_Z$.

Solution: $I_L = 9/1500 = 6$ mA; $I_Z = 12 - 6 = 6$ mA.

Example 35. Input range 12-18 V, $V_Z$ = 9 V, $R_s$ = 500 ohm, load 6 mA: $I_Z$ range?

Solution: $I_s$: 6-18 mA → $I_Z$: 0-12 mA (regulation just barely holds at 12 V input).

Example 36. LED band gap 2.5 eV: emission wavelength?

Solution: $\lambda = \dfrac{1240}{2.5} = 496$ nm — blue-green.

Example 37. Inputs A = 1, B = 0 into OR, AND, NAND, NOR: outputs?

Solution: 1, 0, 1, 0.

Example 38. Simplify $Y = \overline{\overline{A} \cdot \overline{B}}$.

Solution: De Morgan: $Y = A + B$ — an OR gate built from three NANDs/inverters.

Example 39. A NOR gate's inputs are tied together: resulting gate?

Solution: $Y = \overline{A + A} = \bar{A}$ — NOT.

Example 40. Output of an AND gate feeds a NOT gate. Inputs (1,1): trace the final output, and name the composite gate.

Solution: AND → 1; NOT → 0. The composite is a NAND gate.