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Semiconductor Electronics Set-2 MCQ Online Practice Test | Class 12 | Quiz Tweek

Semiconductor Electronics Set-2

Practice Important MCQs for Semiconductor Electronics — Physics, Class 12.

Semiconductor electronics is central to both CBSE board exams and competitive tests (JEE/NEET) because it combines quantum-level material properties with circuit-level reasoning. Mastery of PN junction physics, carrier transport, diode/LED/Zener behaviour and BJT operation allows students to solve a wide variety of numerical, graph-interpretation and conceptual problems that frequently appear in higher-weight questions.

This chapter develops problem-solving skills used in device modelling (Shockley equation, depletion capacitance, small‑signal resistance), data interpretation (Arrhenius plots, semilog I–V), and design/analysis of bias networks and protection circuits. Practising multi-step numerical and assertion–reason problems from this topic sharpens the analytical reasoning required in board exams and competitive tests.

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Q1. An abrupt silicon PN junction at $T=300\,$ K has $N_A=10^{16}\,\text{cm}^{-3}$ and $N_D=10^{18}\,\text{cm}^{-3}$ . Using $n_i=1.5\times10^{10}\,\text{cm}^{-3}$ and $kT/q\approx0.02585\,$ V, compute the built‑in potential $V_{bi}$ from $V_{bi}=\dfrac{kT}{q}\ln\!\left(\dfrac{N_A N_D}{n_i^2}\right)$ . (Nearest value)

(A)

$0.58\ \text{V}$

(B)

$0.81\ \text{V}$

(C)

$1.06\ \text{V}$

(D)

$0.29\ \text{V}$

Q2. Statement‑I: Under steady uniform optical illumination of an intrinsic semiconductor the excess carrier concentrations satisfy $\Delta n=\Delta p$ , and the conductivity increment is $\Delta\sigma=q(\mu_n+\mu_p)\Delta n$ .
Statement‑II: Photons create equal electron–hole pairs but since $\mu_n>\mu_p$ in most semiconductors the numerical change in conductivity is dominated by electrons even though $\Delta n=\Delta p$ .

(A)

Both I and II are true and II correctly explains I.

(B)

Both I and II are true but II does not explain I.

(C)

I is true but II is false.

(D)

I is false but II is true.

Q3. A silicon diode at $300\,$ K has reverse saturation current $I_s=10^{-12}\,$ A. Using the Shockley equation $I=I_s\left(e^{\frac{qV}{kT}}-1\right)$ with $kT/q\approx0.02585\,$ V, estimate the forward current at $V=0.65\,$ V. (Closest)

(A)

$8.8\times10^{-3}\ \text{A}$

(B)

$1.0\times10^{-1}\ \text{A}$

(C)

$8.8\times10^{-2}\ \text{A}$

(D)

$1.6\times10^{-2}\ \text{A}$

Q4. On a semilog plot (natural log) of forward current $I$ vs forward voltage $V$ at $T=300\,$ K, two diodes P and Q show linear regions with slopes $s_P=19.35\ \text{V}^{-1}$ and $s_Q=38.70\ \text{V}^{-1}$ where $s=d(\ln I)/dV$ . Using $s=\dfrac{q}{n k T}$ , identify $n_P$ , $n_Q$ and the dominant transport mechanism in P and Q.

(A)

$n_P\approx1,\; n_Q\approx2;$ P diffusion, Q recombination.

(B)

$n_P\approx1,\; n_Q\approx1;$ both diffusion-dominated.

(C)

$n_P\approx2,\; n_Q\approx2;$ both recombination-dominated.

(D)

$n_P\approx2,\; n_Q\approx1;$ P dominated by recombination (depletion-region) and Q by diffusion.

Q5. Two reverse‑biased silicon diodes X and Y show sharp breakdowns at approximately $5.6\,$ V and $60\,$ V respectively. Experimentally, when temperature increases, $V_{BR,X}$ decreases while $V_{BR,Y}$ increases. Which statement best explains the observations?

(A)

X undergoes Zener (tunnelling) breakdown due to heavy doping (strong field) so $V_{BR}$ falls with $T$ ; Y undergoes avalanche breakdown in a lightly doped junction so $V_{BR}$ rises with $T$ .

(B)

X is avalanche and Y is Zener; temperature trends contradict known mechanisms.

(C)

Both are avalanche; difference in $V_{BR}$ trends is due to series resistance only.

(D)

The observed trends imply thermal runaway rather than Zener/avalanche mechanisms.

Q6. For the circuit: $V_{CC}=12\,$ V, $R_1=120\,$ k $\Omega$ (to base from $V_{CC}$ ), $R_2=30\,$ k $\Omega$ (base to ground), $R_C=4\,$ k $\Omega$ (collector to $V_{CC}$ ). Take $V_{BE}=0.7\,$ V and $\beta=50$ . Using Thevenin for the base divider, determine the operating condition and approximate collector current.

(A)

Transistor in active region with $I_C\approx3.54\,$ mA (no saturation).

(B)

Transistor remains in active region with $I_C\approx2.95\,$ mA and $V_{CE}\approx0.2\,$ V.

(C)

Transistor is in active region with $I_C\approx4.71\,$ mA, using $I_B=(12-0.7)/120\,\text{k}\Omega$ and ignoring the lower divider resistor.

(D)

Transistor saturates; available $I_B\approx70.8\ \mu$ A so $I_C$ is limited to about $(V_{CC}-V_{CE(sat)})/R_C\approx2.95\,$ mA and $V_{CE}\approx0.2\,$ V.

Q7. Statement‑I: For an $n$ ‑type semiconductor, increasing temperature moves the Fermi level $E_F$ toward the mid‑gap (away from the conduction band edge).
Statement‑II: As temperature rises the intrinsic concentration $n_i$ increases, making the dopant contribution relatively less important; thus $E_F$ approaches the intrinsic level $E_i$ , explaining the shift.

(A)

Both I and II are true and II correctly explains I.

(B)

Both I and II are true but II does not explain I.

(C)

I is true but II is false.

(D)

I is false but II is true.

Q8. An intrinsic semiconductor sample shows conductivities $\sigma_1=1.0\times10^{-6}\ \text{S m}^{-1}$ at $T_1=300\,$ K and $\sigma_2=2.5\times10^{-6}\ \text{S m}^{-1}$ at $T_2=330\,$ K. Assuming intrinsic conduction with $\sigma\propto e^{-E_g/(2kT)}$ , estimate the band gap $E_g$ (use $k=8.617\times10^{-5}\ \text{eV K}^{-1}$ ). (Nearest)

(A)

$1.12\ \text{eV}$

(B)

$0.52\ \text{eV}$

(C)

$0.26\ \text{eV}$

(D)

$0.75\ \text{eV}$

Q9. A diode biased at $V_D=0.7\,$ V has DC current $I_D=1.0\,$ mA and ideality factor $n=1$ . It is in series with $R=100\ \Omega$ . Using $V_T\approx0.02585\,$ V and small‑signal resistance $r_d=nV_T/I_D$ , a small increment $\Delta V_s=10\,$ mV is applied to the source. Using the small‑signal model, estimate the incremental diode current $\Delta I$ .

(A)

$7.9\ \mu\text{A}$

(B)

$794\ \mu\text{A}$

(C)

$79.4\ \mu\text{A}$

(D)

$7.94\ \text{mA}$

Q10. Two diodes D1 and D2 are in series and share the same forward current $I$ . For $T=300\,$ K the parameters are: D1: $n_1=1$ , $I_{01}=10^{-12}\,$ A; D2: $n_2=2$ , $I_{02}=10^{-9}\,$ A. They are forward biased together so that $V_{D1}+V_{D2}=0.80\,$ V. Neglecting the “−1” in the diode equation, solve approximately for $I$ and determine which diode drops the larger voltage.

(A)

$I\approx3.04\ \mu\text{A}$ ; $V_{D1}\approx0.415\,$ V, $V_{D2}\approx0.385\,$ V (D1 larger).

(B)

$I\approx0.10\ \text{mA}$ ; $V_{D1}\approx0.50\,$ V, $V_{D2}\approx0.30\,$ V.

(C)

$I\approx3.04\ \mu\text{A}$ ; $V_{D1}\approx0.385\,$ V, $V_{D2}\approx0.415\,$ V (D2 slightly larger).

(D)

$I\approx3.04\ \mu\text{A}$ ; $V_{D1}\approx0.385\,$ V, $V_{D2}\approx0.415\,$ V (D2 slightly larger).

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