Semiconductor Electronics Set-1

Q1. A silicon sample at $T=300\ \mathrm{K}$ has intrinsic carrier concentration $n_i=1.5\times10^{10}\ \mathrm{cm^{-3}}$. It is doped n-type with donor density $N_D=1.0\times10^{16}\ \mathrm{cm^{-3}}$ and complete ionization is assumed. Using the mass‑action law $np=n_i^2$, estimate the hole concentration $p$ (in $\mathrm{cm^{-3}}$).

Q2. For a silicon diode at $T=300\ \mathrm{K}$ the reverse saturation current is $I_0=10^{-12}\ \mathrm{A}$. Using the diode law $I=I_0\left(e^{qV/kT}-1\right)$, compute the small‑signal (differential) resistance $r_d=\dfrac{kT}{qI}$ at forward bias $V=0.65\ \mathrm{V}$. (Use $kT/q\approx25.85\ \mathrm{mV}$.)

Q3. Assertion (A): In a p–n junction under moderate reverse bias the current is primarily due to majority carriers crossing the junction.
Reason (R): The reverse saturation current originates from thermally generated minority carriers (in neutral regions and depletion region) that are swept across the junction by the electric field, making the reverse current almost independent of reverse bias (until breakdown).

Q4. A diode shows forward currents $I_1=2.0\ \mu\mathrm{A}$ at $V_1=0.55\ \mathrm{V}$ and $I_2=2.0\ \mathrm{mA}$ at $V_2=0.75\ \mathrm{V}$ (measured at $T=300\ \mathrm{K}$). Assuming $I\propto\exp\!\left(\dfrac{qV}{nkT}\right)$, estimate the ideality factor $n$.

Q5. In a common‑emitter amplifier with the emitter bypassed by a large capacitor, the small‑signal emitter resistance is approximated by $r_e'\approx\dfrac{26\ \mathrm{mV}}{I_E}$. If $I_E\approx2.0\ \mathrm{mA}$ and $R_C=1.2\ \mathrm{k}\Omega$, estimate the magnitude of the voltage gain $|A_v|\approx\dfrac{R_C}{r_e'}$ (neglect transistor internal capacitances).

Q6. Assertion (A): For a BJT operating in the active region, increasing the reverse bias across the collector–base junction (keeping base current constant) increases the collector current.
Reason (R): This occurs because the widened collector–base depletion region increases the effective base width, so more minority carriers recombine in the base and the collector current rises.

Q7. A photodiode produces a short‑circuit photocurrent $I_{ph}=50\ \mu\mathrm{A}$. Its dark reverse saturation current is $I_0=10^{-12}\ \mathrm{A}$. Under open‑circuit the photogenerated current is balanced by diode forward current: $I_{ph}=I_0\left(e^{qV_{oc}/kT}-1\right)$. Estimate the open‑circuit voltage $V_{oc}$ at $T=300\ \mathrm{K}$.

Q8. From transistor output‑characteristic data in active region it is found that for base current $I_B=20\ \mu\mathrm{A}$ the collector current is $I_C=2.00\ \mathrm{mA}$. Using $\beta=\dfrac{I_C}{I_B}$, estimate the current gain $\beta$.

Q9. Assertion (A): As donor concentration in an n‑type semiconductor is increased substantially at $T=300\ \mathrm{K}$, the Fermi level moves closer to the conduction band and can enter it (degenerate doping).
Reason (R): Increasing donor density raises the equilibrium electron concentration; the chemical potential (Fermi level) shifts upward so that the occupancy of electronic states (as given by Fermi–Dirac statistics) corresponds to the higher electron density.

Q10. A diode displays forward currents $I_1=1.0\ \mathrm{mA}$ at $V_1=0.65\ \mathrm{V}$ and $I_2=50\ \mathrm{mA}$ at $V_2=0.80\ \mathrm{V}$. Assuming ideality factor $n\approx1$ and $kT/q\approx25.85\ \mathrm{mV}$, the ideal voltage change $\Delta V_{ideal}=(kT/q)\ln(I_2/I_1)$. The excess voltage $\Delta V_{series}=\Delta V_{measured}-\Delta V_{ideal}$ is due to series resistance $R_s=\dfrac{\Delta V_{series}}{I_2-I_1}$. Estimate $R_s$.