Semiconductor Electronics-Materials, Devices and Simple Circuits MCQs with Answers – Part 6 (Class 12 Physics)

Explanation: \( \textbf{Given gain:} \) \(\beta=75\). \( \textbf{Base current:} \) \(I_B=0.60\,mA\). \( \textbf{Active-region estimate:} \) \[ I_C=\beta I_B \] \( \textbf{Substitution:} \) \[ I_C=75\times0.60\,mA=45\,mA \] \( \textbf{Load-limited current:} \) \[ I_{C(\text{sat})}\approx\frac{V_{CC}}{R_C} \] \( \textbf{Substitution:} \) \[ I_{C(\text{sat})}\approx\frac{12}{400}=0.030\,A=30\,mA \] \( \textbf{Comparison:} \) The active estimate \(45\,mA\) is greater than the load-limited value \(30\,mA\). \( \textbf{Operating implication:} \) The collector current cannot rise to \(45\,mA\) with this load and supply, so the transistor is driven into saturation. \( \textbf{Final answer:} \) Saturation is expected because the load allows only \(30\,mA\).

Explanation: \( \textbf{Current gain:} \) \(\beta=100\). \( \textbf{Collector load:} \) \(R_C=2.2\,k\Omega\). \( \textbf{Input resistance:} \) \(R_i=1.1\,k\Omega\). \( \textbf{Approximate voltage-gain relation:} \) \[ |A_v|\approx\beta\frac{R_C}{R_i} \] \( \textbf{Resistance ratio:} \) \[ \frac{R_C}{R_i}=\frac{2.2\,k\Omega}{1.1\,k\Omega}=2 \] \( \textbf{Substitution:} \) \[ |A_v|\approx100\times2 \] \( \textbf{Calculation:} \) \[ |A_v|\approx200 \] \( \textbf{Phase note:} \) A common-emitter amplifier normally has negative voltage gain when phase is included. \( \textbf{Final answer:} \) The magnitude of voltage gain is approximately \(200\).

Explanation: \( \textbf{Voltage gain:} \) \(A_v=-120\). \( \textbf{Input peak voltage:} \) \(V_i=15\,mV=0.015\,V\). \( \textbf{Output relation:} \) \[ V_o=A_vV_i \] \( \textbf{Substitution:} \) \[ V_o=(-120)(0.015\,V) \] \( \textbf{Calculation:} \) \[ V_o=-1.8\,V \] \( \textbf{Magnitude:} \) \[ |V_o|=1.8\,V \] \( \textbf{Sign meaning:} \) The negative sign represents phase reversal in the common-emitter amplifier. \( \textbf{Final answer:} \) The output is \(1.8\,V\) peak and \(180^\circ\) out of phase.

Explanation: \( \textbf{Collector load:} \) \(R_C=2.0\,k\Omega=2000\,\Omega\). \( \textbf{Collector-current increase:} \) \(\Delta I_C=4.0\,mA=4.0\times10^{-3}\,A\). \( \textbf{Collector-resistor voltage-change magnitude:} \) \[ |\Delta V_R|=\Delta I_CR_C \] \( \textbf{Substitution:} \) \[ |\Delta V_R|=(4.0\times10^{-3})(2000) \] \( \textbf{Calculation:} \) \[ |\Delta V_R|=8.0\,V \] \( \textbf{Direction of collector voltage:} \) If \(I_C\) increases, the voltage drop across \(R_C\) increases. \( \textbf{Collector effect:} \) A larger resistor drop makes the collector voltage fall. \( \textbf{Operating concern:} \) A large fall in collector voltage can push the transistor toward saturation. \( \textbf{Final answer:} \) The collector voltage tries to fall by \(8.0\,V\).

Explanation: \( \textbf{Given:} \) \(\alpha=0.975\) and \(I_E=8.0\,mA\). \( \textbf{Common-base gain relation:} \) \[ \alpha=\frac{I_C}{I_E} \] \( \textbf{Collector current:} \) \[ I_C=\alpha I_E \] \( \textbf{Substitution:} \) \[ I_C=(0.975)(8.0\,mA) \] \( \textbf{Calculation:} \) \[ I_C=7.8\,mA \] \( \textbf{Current relation:} \) \[ I_E=I_B+I_C \] \( \textbf{Base current:} \) \[ I_B=I_E-I_C=8.0\,mA-7.8\,mA=0.2\,mA \] \( \textbf{Final answer:} \) \(I_C=7.8\,mA\) and \(I_B=0.2\,mA\).

Explanation: \( \textbf{Given:} \) \(\beta=60\), \(I_E=6.10\,mA\). \( \textbf{Current relation using \(\beta\):} \) \[ I_C=\beta I_B \] \( \textbf{Emitter current relation:} \) \[ I_E=I_B+I_C \] \( \textbf{Substitute \(I_C=\beta I_B\):} \) \[ I_E=I_B+\beta I_B=(\beta+1)I_B \] \( \textbf{Base current:} \) \[ I_B=\frac{I_E}{\beta+1} \] \( \textbf{Substitution:} \) \[ I_B=\frac{6.10\,mA}{61} \] \( \textbf{Calculation:} \) \[ I_B=0.100\,mA \] \( \textbf{Unit conversion:} \) \[ 0.100\,mA=100\,\mu A \] \( \textbf{Final answer:} \) The base current is closest to \(100\,\mu A\).

Explanation: \( \textbf{Supply voltage:} \) \(V_S=9.0\,V\). \( \textbf{\(LED\) forward drop:} \) \(V_{LED}=2.0\,V\). \( \textbf{Desired current:} \) \(I=14\,mA=0.014\,A\). \( \textbf{Voltage across resistor:} \) \[ V_R=V_S-V_{LED} \] \( \textbf{Substitution:} \) \[ V_R=9.0\,V-2.0\,V=7.0\,V \] \( \textbf{Series resistance:} \) \[ R=\frac{V_R}{I} \] \( \textbf{Calculation:} \) \[ R=\frac{7.0}{0.014}=500\,\Omega \] \( \textbf{Protection meaning:} \) The resistor drops the extra voltage and limits the \(LED\) current. \( \textbf{Final answer:} \) The nearest suitable series resistance is \(500\,\Omega\).

Explanation: \( \textbf{Photon energy:} \) \(E=2.4\,eV\). \( \textbf{Useful constant:} \) \(hc=1240\,eV\,nm\). \( \textbf{Photon relation:} \) \[ E=\frac{hc}{\lambda} \] \( \textbf{Rearrangement:} \) \[ \lambda=\frac{hc}{E} \] \( \textbf{Substitution:} \) \[ \lambda=\frac{1240\,eV\,nm}{2.4\,eV} \] \( \textbf{Unit cancellation:} \) The unit \(eV\) cancels. \( \textbf{Calculation:} \) \[ \lambda\approx516.7\,nm \] \( \textbf{Final answer:} \) The wavelength is approximately \(517\,nm\).

Explanation: \( \textbf{Dark current:} \) \(I_d=3\,\mu A\). \( \textbf{Measured current:} \) \(I_{\text{total}}=11\,\mu A\). \( \textbf{Photocurrent per intensity level:} \) \(0.40\,\mu A\). \( \textbf{Photocurrent due to light:} \) \[ I_p=I_{\text{total}}-I_d \] \( \textbf{Substitution:} \) \[ I_p=11\,\mu A-3\,\mu A=8\,\mu A \] \( \textbf{Intensity level relation:} \) \[ \text{Intensity level}=\frac{I_p}{0.40\,\mu A} \] \( \textbf{Calculation:} \) \[ \text{Intensity level}=\frac{8}{0.40}=20 \] \( \textbf{Final answer:} \) The light-intensity level is \(20\).

Explanation: \( \textbf{Terminal voltage:} \) \(V=0.48\,V\). \( \textbf{Load current:} \) \(I=0.30\,A\). \( \textbf{Incident power:} \) \(P_{\text{in}}=0.90\,W\). \( \textbf{Electrical output power:} \) \[ P_{\text{out}}=VI \] \( \textbf{Substitution:} \) \[ P_{\text{out}}=(0.48)(0.30)=0.144\,W \] \( \textbf{Efficiency relation:} \) \[ \eta=\frac{P_{\text{out}}}{P_{\text{in}}}\times100\% \] \( \textbf{Calculation:} \) \[ \eta=\frac{0.144}{0.90}\times100\%=16\% \] \( \textbf{Final answer:} \) The efficiency is \(16\%\).

Explanation: \(\textbf{Given open-circuit voltage:}\) \(V_{oc}=0.62\,V\). \(\textbf{Given short-circuit current:}\) \(I_{sc}=0.50\,A\). \(\textbf{Loaded terminal voltage:}\) \(V=0.50\,V\). \(\textbf{Loaded current:}\) \(I=0.40\,A\). \(\textbf{Required quantity:}\) Useful electrical output power at the loaded operating point. \(\textbf{Power relation:}\) \[ P=VI \] \(\textbf{Open-circuit check:}\) In open circuit, current is zero. \[ P_{open}=V_{oc}\times 0=0 \] \(\textbf{Short-circuit check:}\) In short circuit, terminal voltage is zero. \[ P_{short}=0\times I_{sc}=0 \] \(\textbf{Loaded condition:}\) Both voltage and current are non-zero, so useful output power is obtained across the load. \(\textbf{Substitution:}\) \[ P=(0.50)(0.40) \] \(\textbf{Calculation:}\) \[ P=0.20\,W \] \(\textbf{Final answer:}\) The useful electrical output power is \(0.20\,W\).

Explanation: \( \textbf{Dynamic resistance relation:} \) \[ r_d=\frac{\Delta V}{\Delta I} \] \( \textbf{First interval voltage change:} \) \[ \Delta V_1=0.68\,V-0.64\,V=0.04\,V \] \( \textbf{First interval current change:} \) \[ \Delta I_1=10\,mA-4\,mA=6\,mA=0.006\,A \] \( \textbf{First interval resistance:} \) \[ r_{d1}=\frac{0.04}{0.006}\approx6.7\,\Omega \] \( \textbf{Second interval voltage change:} \) \[ \Delta V_2=0.72\,V-0.68\,V=0.04\,V \] \( \textbf{Second interval current change:} \) \[ \Delta I_2=22\,mA-10\,mA=12\,mA=0.012\,A \] \( \textbf{Second interval resistance:} \) \[ r_{d2}=\frac{0.04}{0.012}\approx3.3\,\Omega \] \( \textbf{Comparison:} \) The second interval has the smaller dynamic resistance because the current rises more for the same voltage change. \( \textbf{Final answer:} \) The second interval has \(r_d\approx3.3\,\Omega\).

Explanation: \( \textbf{Input resistance:} \) \(R_i=1.5\,k\Omega\). \( \textbf{Collector load:} \) \(R_C=4.5\,k\Omega\). \( \textbf{Current gain:} \) \(\beta=80\). \( \textbf{Input signal:} \) \(V_i=10\,mV=0.010\,V\). \( \textbf{Approximate gain relation:} \) \[ |A_v|\approx\beta\frac{R_C}{R_i} \] \( \textbf{Resistance ratio:} \) \[ \frac{R_C}{R_i}=\frac{4.5}{1.5}=3 \] \( \textbf{Voltage gain magnitude:} \) \[ |A_v|\approx80\times3=240 \] \( \textbf{Output magnitude:} \) \[ |V_o|=|A_v|V_i=240(0.010)=2.4\,V \] \( \textbf{Phase note:} \) The common-emitter output is normally phase reversed relative to the input. \( \textbf{Final answer:} \) The approximate output voltage magnitude is \(2.4\,V\).

Explanation: \( \textbf{Given expression:} \) \(Y=\overline{A\cdot B}+A\cdot C\). \( \textbf{Input values:} \) \(A=1\), \(B=1\), and \(C=1\). \( \textbf{First product:} \) \[ A\cdot B=1\cdot1=1 \] \( \textbf{Complement of first product:} \) \[ \overline{A\cdot B}=\overline{1}=0 \] \( \textbf{Second product:} \) \[ A\cdot C=1\cdot1=1 \] \( \textbf{Final output:} \) \[ Y=0+1=1 \] \( \textbf{Final answer:} \) The output is \(1\).

Explanation: \( \textbf{Given expression:} \) \[ Y=(A+\overline{A}B)C \] \( \textbf{Useful Boolean identity:} \) \[ X+\overline{X}Y=X+Y \] \( \textbf{Apply identity with \(X=A\) and \(Y=B\):} \) \[ A+\overline{A}B=A+B \] \( \textbf{Substitute back:} \) \[ Y=(A+B)C \] \( \textbf{Meaning:} \) The expression is not simply \(AC\) or \(BC\), because either \(A\) or \(B\) can satisfy the bracketed condition before \(C\) is applied. \( \textbf{Final answer:} \) The simplified expression is \((A+B)C\).

Explanation: A \(NAND\) gate gives \(Y=0\) only when both inputs are \(1\). For \(A=0,B=0\), the product \(A\cdot B=0\), so \(Y=1\). For \(A=0,B=1\) and \(A=1,B=0\), the product is also \(0\), so \(Y=1\) again. For \(A=1,B=1\), the product is \(1\), and the \(NAND\) output becomes \(0\). Since the transistor switch turns on when the gate output is \(1\), it turns on for all input pairs except \(A=1,B=1\).

Explanation: In a capacitor-input filter, ripple decreases when the capacitor can supply the load for longer between charging peaks. For the same load current and rectifier frequency, a larger capacitance means slower voltage fall between peaks. A common proportional idea is that ripple voltage is inversely related to capacitance under similar conditions. Doubling the capacitance therefore roughly halves the ripple voltage. The ripple frequency remains \(100\,Hz\) because the rectifier type and supply frequency have not changed.

Explanation: \( \textbf{Load current:} \) \(I_L=40\,mA=0.040\,A\). \( \textbf{Ripple frequency:} \) \(f=100\,Hz\). \( \textbf{Capacitance:} \) \(C=1000\,\mu F=1000\times10^{-6}\,F=1.0\times10^{-3}\,F\). \( \textbf{Approximate ripple relation:} \) \[ V_r\approx\frac{I_L}{fC} \] \( \textbf{Substitution:} \) \[ V_r\approx\frac{0.040}{(100)(1.0\times10^{-3})} \] \( \textbf{Denominator:} \) \[ (100)(1.0\times10^{-3})=0.10 \] \( \textbf{Calculation:} \) \[ V_r\approx\frac{0.040}{0.10}=0.40\,V \] \( \textbf{Final answer:} \) The approximate ripple voltage is \(0.40\,V\).

Explanation: \( \textbf{Average filtered voltage:} \) \(15\,V\). \( \textbf{Ripple:} \) \(1.0\,V\) peak-to-peak. \( \textbf{Minimum input to regulator:} \) \[ V_{in,\min}=15\,V-\frac{1.0\,V}{2}=14.5\,V \] \( \textbf{Zener voltage:} \) \(V_Z=12\,V\). \( \textbf{Series resistance:} \) \(R_S=150\,\Omega\). \( \textbf{Minimum series current:} \) \[ I_{S,\min}=\frac{V_{in,\min}-V_Z}{R_S} \] \( \textbf{Substitution:} \) \[ I_{S,\min}=\frac{14.5-12}{150}=\frac{2.5}{150}=0.0167\,A \] \( \textbf{Load current:} \) \[ I_L=10\,mA \] \( \textbf{Minimum Zener current:} \) \[ I_{Z,\min}=I_{S,\min}-I_L=16.7\,mA-10\,mA=6.7\,mA \] \( \textbf{Final answer:} \) The minimum Zener current is approximately \(6.7\,mA\).

Explanation: \( \textbf{Secondary peak voltage:} \) \(18\,V\). \( \textbf{Bridge rectifier conducting diodes:} \) \(2\). \( \textbf{Drop per diode:} \) \(0.7\,V\). \( \textbf{Total diode drop:} \) \[ V_D=2\times0.7\,V=1.4\,V \] \( \textbf{Approximate capacitor peak voltage:} \) \[ V_{\text{peak,out}}\approx18\,V-1.4\,V=16.6\,V \] \( \textbf{Ripple peak-to-peak:} \) \(2.0\,V\). \( \textbf{Minimum filtered voltage:} \) \[ V_{\min}=16.6\,V-1.0\,V=15.6\,V \] \( \textbf{Regulator condition:} \) This minimum value is the lowest input available to the Zener stage during the ripple cycle. \( \textbf{Final answer:} \) The minimum input voltage to the Zener regulator is approximately \(15.6\,V\).