Transistors — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Transistors MCQs with step-by-step solutions (28 questions). Part of Semiconductor Electronics. Practise online on Prepizo — no login needed.
▶ Practise Transistors online (free)
Questions with solutions
Q1 — Transistors · medium · numerical
A n-p-n transistor is connected in common emitter configuration in which collector voltage drop across load resistance (800 Ω) connected to the collector circuit is 0.8 V. The collector current is
A. 2 mA
B. 0.1 mA
C. 1 mA ✓ Correct
D. 0.2 mA
Solution: $I_C = \frac{V_C}{R_C} = \frac{0.8}{800} = 10^{-3}$ A $= 1$ mA
Q2 — Transistors · easy · theory
For transistor action, which of the following statements is correct?
A. Base, emitter and collector regions should have same size
B. Both emitter junction as well as the collector junction are forward biased
C. The base region must be very thin and lightly doped ✓ Correct
D. Base, emitter and collector regions should have same doping concentrations
Solution: For transistor action the base must be very thin and lightly doped; the emitter is heavily doped and the collector moderately doped and slightly larger.
Q3 — Transistors · hard · numerical
In a common-emitter circuit with $R_B = 500$ kΩ (base), $R_C = 4$ kΩ (collector) and supply $V_{CC} = 20$ V, the input voltage $V_i$ is 20 V, $V_{BE} = 0$ and $V_{CE} = 0$. The values of $I_B$, $I_C$ and $\beta$ are given by
A. $I_B = 20$ µA, $I_C = 5$ mA, $\beta = 250$
B. $I_B = 25$ µA, $I_C = 5$ mA, $\beta = 200$
C. $I_B = 40$ µA, $I_C = 10$ mA, $\beta = 250$
D. $I_B = 40$ µA, $I_C = 5$ mA, $\beta = 125$ ✓ Correct
Solution: $I_B = \frac{V_i - V_{BE}}{R_B} = \frac{20}{500 \times 10^3} = 40$ µA
$I_C = \frac{V_{CC} - V_{CE}}{R_C} = \frac{20}{4 \times 10^3} = 5$ mA
$\beta = \frac{I_C}{I_B} = \frac{5 \times 10^{-3}}{40 \times 10^{-6}} = 125$
Q4 — Transistors · medium · numerical
In a common emitter transistor amplifier, the audio signal voltage across the collector is 3 V. The resistance of collector is 3 kΩ. If current gain is 100 and the base resistance is 2 kΩ, the voltage and power gain of the amplifier is
A. 200 and 1000
B. 15 and 200
C. 150 and 15000 ✓ Correct
D. 20 and 2000
Solution: $i_C = \frac{3}{3\text{ k}\Omega} = 10^{-3}$ A; $i_B = \frac{i_C}{100} = 10^{-5}$ A; $V_{in} = i_BR_B = 2 \times 10^{-2}$ V
Voltage gain $= \frac{3}{2 \times 10^{-2}} = 150$; power gain $= A_V \times \beta = 150 \times 100 = 15000$
Q5 — Transistors · hard · numerical
A n-p-n transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Ω is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 Ω, the voltage gain and the power gain of the amplifier will respectively be
A. 3.69, 3.84
B. 4, 4
C. 4, 3.69
D. 4, 3.84 ✓ Correct
Solution: $I_C = \frac{0.8}{800} = 1$ mA; $I_B = \frac{I_C}{0.96}$
$A_V = \frac{V_L}{I_BR_i} = \frac{0.8 \times 0.96}{10^{-3} \times 192} = 4$
$A_P = (0.96)^2 \times \frac{800}{192} = 3.84$
Q6 — Transistors · medium · numerical
For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is
A. 10 mV
B. 20 mV ✓ Correct
C. 30 mV
D. 15 mV
Solution: $A_V = \beta\frac{R_{out}}{R_{in}} = 100 \times \frac{2}{1} = 200$
$V_{in} = \frac{V_{out}}{A_V} = \frac{4}{200} = 20$ mV
Q7 — Transistors · medium · numerical
The input signal given to a CE amplifier having a voltage gain of 150 is $V_i = 2\cos\left(15t + \frac{\pi}{3}\right)$. The corresponding output signal will be
A. $300\cos\left(15t + \frac{\pi}{3}\right)$
B. $75\cos\left(15t + \frac{2\pi}{3}\right)$
C. $2\cos\left(15t + \frac{5\pi}{3}\right)$
D. $300\cos\left(15t + \frac{4\pi}{3}\right)$ ✓ Correct
Solution: A CE amplifier introduces a phase shift of $\pi$:
$V_0 = 150 \times 2\cos\left(15t + \frac{\pi}{3} + \pi\right) = 300\cos\left(15t + \frac{4\pi}{3}\right)$
Q8 — Transistors · hard · numerical
In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will
A. $\frac{2}{3}G$ ✓ Correct
B. $1.5G$
C. $\frac{1}{3}G$
D. $\frac{5}{4}G$
Solution: Voltage gain $G = g_mR_L$, so $G \propto g_m$:
$G_2 = \frac{0.02}{0.03}G = \frac{2}{3}G$
Q9 — Transistors · medium · numerical
In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. If the base resistance is 1 kΩ and the current amplification of the transistor is 100, the input signal voltage is
A. 0.1 V
B. 1.0 V
C. 1 mV
D. 10 mV ✓ Correct
Solution: $\Delta I_C = \frac{2}{2 \times 10^3} = 10^{-3}$ A; $\Delta I_B = \frac{10^{-3}}{100} = 10^{-5}$ A
$V_B = \Delta I_B \times R_B = 10^{-5} \times 10^3 = 10$ mV
Q10 — Transistors · medium · theory
The transfer characteristic [output voltage ($V_o$) vs input voltage ($V_i$)] for a base-biased transistor in CE configuration has three regions: I (cut-off, output high and flat), II (active, output falling linearly) and III (saturation, output low and flat). For using the transistor as a switch, it is used
A. in region III
B. both in region (I) and (III) ✓ Correct
C. in region II
D. in region I
Solution: As a switch the transistor is operated in the cut-off state (OFF) and the saturation state (ON) — regions I and III.
Q11 — Transistors · medium · numerical
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is
A. 500
B. 1000
C. 1250 ✓ Correct
D. 50
Solution: $A_V = \beta\frac{R_o}{R_i} \Rightarrow 50 = \beta \times 2 \Rightarrow \beta = 25$
Power gain $= \beta^2 \times \frac{R_o}{R_i} = 625 \times 2 = 1250$
Q12 — Transistors · easy · numerical
A transistor is operated in common-emitter configuration at $V_c = 2$ volt such that a change in the base current from 100 µA to 200 µA produces a change in the collector current from 5 mA to 10 mA. The current gain is
A. 75
B. 100
C. 150
D. 50 ✓ Correct
Solution: $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5 \times 10^{-3}}{100 \times 10^{-6}} = 50$
Q13 — Transistors · medium · numerical
The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain without feedback will be
A. 90
B. 10
C. 1.25
D. 100 ✓ Correct
Solution: $10 = \frac{A_V}{1 + 0.09A_V} \Rightarrow A_V = 100$
Q14 — Transistors · easy · numerical
A transistor is operated in common emitter configuration at constant collector voltage $V_c = 1.5$ V such that a change in the base current from 100 µA to 150 µA produces a change in the collector current from 5 mA to 10 mA. The current gain ($\beta$) is
A. 67
B. 75
C. 100 ✓ Correct
D. 50
Solution: $\beta = \frac{\Delta i_c}{\Delta i_b} = \frac{5\text{ mA}}{50\text{ µA}} = 100$
Q15 — Transistors · medium · numerical
A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produces oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be
A. $\frac{f}{4}$
B. $8f$
C. $\frac{f}{2\sqrt{2}}$ ✓ Correct
D. $\frac{f}{2}$
Solution: $f \propto \frac{1}{\sqrt{LC}}$
$\frac{f_2}{f} = \sqrt{\frac{LC}{2L \times 4C}} = \frac{1}{2\sqrt{2}} \Rightarrow f_2 = \frac{f}{2\sqrt{2}}$
Q16 — Transistors · medium · theory
A n-p-n transistor conducts when
A. collector is positive and emitter is at same potential as the base
B. both collector and emitter are negative with respect to the base
C. both collector and emitter are positive with respect to the base
D. collector is positive and emitter is negative with respect to the base ✓ Correct
Solution: For conduction the emitter-base junction must be forward biased (emitter negative w.r.t. base) and the collector-base junction reverse biased (collector positive).
Q17 — Transistors · easy · numerical
For a transistor $\frac{i_c}{i_e} = 0.96$, the current gain in common-emitter configuration is
A. 6
B. 12
C. 24 ✓ Correct
D. 48
Solution: $\beta = \frac{\alpha}{1 - \alpha} = \frac{0.96}{0.04} = 24$
Q18 — Transistors · easy · numerical
In a common-base configuration of a transistor $\frac{\Delta i_c}{\Delta i_e} = 0.98$, then current gain in common emitter configuration of transistor will be
A. 49 ✓ Correct
B. 98
C. 4.9
D. 24.5
Solution: $\beta = \frac{\alpha}{1 - \alpha} = \frac{0.98}{0.02} = 49$
Q19 — Transistors · easy · theory
If $\alpha$ and $\beta$ are current gains in common-base and common-emitter configurations of a transistor, then $\beta$ is equal to
A. $\frac{1}{\alpha}$
B. $\frac{\alpha}{1 + \alpha}$
C. $\frac{\alpha}{1 - \alpha}$ ✓ Correct
D. $\alpha - \frac{1}{\alpha}$
Solution: Since $i_e = i_b + i_c$: $\beta = \frac{\alpha}{1 - \alpha}$
Q20 — Transistors · medium · numerical
The transfer ratio $\beta$ of a transistor is 50. The input resistance of the transistor when used in the common emitter configuration is 1 kΩ. The peak value of the collector AC current for an AC input voltage of 0.01 V peak is
A. 100 µA
B. 0.01 mA
C. 0.25 mA
D. 500 µA ✓ Correct
Solution: $i_b = \frac{0.01}{1000} = 10^{-5}$ A
$i_c = \beta i_b = 50 \times 10^{-5} = 500$ µA
Q21 — Transistors · easy · theory
The correct relationship between the two current gains $\alpha$ and $\beta$ in a transistor is
A. $\beta = \frac{1 + \alpha}{\beta}$
B. $\alpha = \frac{\beta}{1 + \beta}$ ✓ Correct
C. $\alpha = \frac{\beta}{1 - \beta}$
D. $\beta = \frac{\alpha}{1 + \alpha}$
Solution: With $i_e = i_b + i_c$: $\alpha = \frac{i_c}{i_e} = \frac{\beta}{1 + \beta}$
Q22 — Transistors · easy · numerical
The current gain for a transistor working as common base amplifier is 0.96. If the emitter current is 7.2 mA, then the base current is
A. 0.29 mA ✓ Correct
B. 0.35 mA
C. 0.39 mA
D. 0.43 mA
Solution: $i_c = 0.96 \times 7.2 = 6.91$ mA
$i_b = i_e - i_c = 7.2 - 6.91 = 0.29$ mA
Q23 — Transistors · medium · theory
When a n-p-n transistor is used as an amplifier, then
A. the electrons flow from emitter to collector ✓ Correct
B. the holes flow from emitter to collector
C. the electrons flow from collector to emitter
D. the electrons flow from battery to emitter
Solution: In an n-p-n transistor, electrons injected by the emitter cross the thin base and most of them (≈95%) pass to the collector — electrons flow from emitter to collector.
Q24 — Transistors · easy · theory
An oscillator is nothing but an amplifier with
A. positive feedback ✓ Correct
B. negative feedback
C. large gain
D. no feedback
Solution: An amplifier with proper positive feedback (feedback voltage in phase with the input) generates oscillations without an external signal — that is an oscillator.
Q25 — Transistors · easy · theory
The part of the transistor which is heavily doped to produce large number of majority carriers is
A. emitter ✓ Correct
B. base
C. collector
D. Any of the above depending upon the nature of transistor
Solution: The emitter is heavily doped so it can inject a large number of charge carriers into the base; the base is lightly doped and thin, the collector moderately doped.
Q26 — Transistors · easy · theory
To use a transistor as an amplifier
A. the emitter base junction is forward biased and the base collector junction is reverse biased ✓ Correct
B. no bias voltage is required
C. both junctions are forward biased
D. both junctions are reverse biased
Solution: For faithful amplification, the emitter-base (input) junction is kept forward biased and the collector-base junction reverse biased.
Q27 — Transistors · easy · theory
In a common base amplifier the phase difference between the input signal voltage and the output voltage is
A. zero ✓ Correct
B. $\frac{\pi}{4}$
C. $\frac{\pi}{2}$
D. $\pi$
Solution: In the common-base configuration, the input and output voltages are in the same phase — zero phase difference.
Q28 — Transistors · easy · theory
Radiowaves of constant amplitude can be generated with
A. FET
B. filter
C. rectifier
D. oscillator ✓ Correct
Solution: An oscillator produces electrical oscillations of constant amplitude at a chosen frequency — from a few Hz to over 100 MHz.