1. The collector current in a common base amplifier using n-p-n transistor is 24 mA. If 80% of the electrons released by the emitter is accepted by the collector, then the base current is numerically : ⇒ (NEET 2022 Phase 2)

A. 3 mA and entering the base

B. 6 mA and leaving the base

C. 3 mA and leaving the base

D. 6 mA and entering the base

The Correct Answer is ()

NEET 2022 Phase 2 Physics - Semiconductor Electronics Question 9 English Explanation

Given $I_{C} = 24$ mA

$α = 0.8 = \frac{I_{C}}{I_{E}}$

$I_{E} = \frac{24}{0.8}$ mA $= 30$ mA

$I_{B} = I_{E} - I_{C} = 30 - 24$

$= 6$ mA, entering the base

2. In the circuit shown in the figure, the input voltage V_i is 20 V, V_BE = 0 and V_CE = 0. The values of I_B , I_C and $β$ are given by ⇒ (NEET 2018)

NEET 2018 Physics - Semiconductor Electronics Question 22 English

A. I_B = 40 $μ$ A, I_C = 10 mA, $β$ = 250

B. $μ$ A, I_C = 5 mA, $β$ = 200

C. $μ$ A, I_C = 5 mA, $β$ = 250

D. I_B = 40 $μ$ A, I_C = 5 mA, $β$ = 125

The Correct Answer is (D)

From question, V_BE = 0, V_i = 20 V

V_CE = 0

V_b = 0 (earthed)

We need to find values of base and collector currents I_B and I_C and amplification factor $β$

3. In a common emitter transistor amplifier the audio sgnal voltage across the collector is 3 V. The resistance of collector is 3 k $Ω$ . If current gain is 100 and the base resistance is 2 k $Ω$ , the voltage and power gain of the amplifier is ⇒ ( NEET 2017)

A. 15 and 200

B. 150 and 15000

C. 20 and 2000

D. 200 and 1000

The Correct Answer is (B)

Given: V_i = 3 V, R_C = 3 k $Ω$ , R_B = 2 k $Ω$ , $β$ = 100

Voltage gain of the CE amplifier

A_V = $β \frac{R_{c}}{R_{b}}$ = 100 $(\frac{3}{2})$ = 150

Power gain = A_V $β$ = 150(100) = 15000

4. A npn 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 frop 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 ⇒ (NEET 2016 Phase 1)

A. 4, 4

B. 4, 3.69

C. 4, 3.84

D. 3.69, 3.84

The Correct Answer is (C)

Voltage gain = Current gain × Resistance gain

= 0.96 $\times$ $\frac{800}{192}$ = 4

Power gain = [Current gain] × [Voltage gain]

= 0.96 × 4 = 3.84

5. The input signal given to a CE amplifier having a voltage gain of 150 is V_i = 2cos(15t + $\frac{π}{3}$ ). The corresponding output signal will be ⇒ ( AIPMT 2015)

A. 2cos $(15 t + \frac{5 π}{6})$

B. 300cos $(15 t + \frac{4 π}{3})$

C. 300cos $(15 t + \frac{π}{3})$

D. 75cos $(15 t + \frac{2 π}{3})$

The Correct Answer is (B)

Given, V_i = 2cos(15t + $\frac{π}{3}$ )

and Voltage gain A_V = 150

For CE transistor phase difference between input and output signal is $π$ = 180^o

Using formula, A_V = $\frac{V_{0}}{V_{i}}$

$\Rightarrow$ V₀ = A_V $\times$ V_i

= 150 $\times$ 2cos(15t + $\frac{π}{3}$ + $π$ )

= 300cos $(15 t + \frac{4 π}{3})$

6. 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 be ⇒ ( NEET 2013)

A. $\frac{1}{3} G$

B. $\frac{5}{4} G$

C. $\frac{2}{3} G$

D. 1.5G

The Correct Answer is (C)

Formula for trans conductance $g$ _m is

$g$ _m = $\frac{β}{r_{i}}$

Formula for voltage gain A,

A = $β$ $\frac{R_{L}}{r_{i}}$

$\Rightarrow$ A = $g$ _mR_L

Given A = G

$∴$ G = $g$ _mR_L

$\Rightarrow$ G $\propto$ $g$ _m

$\Rightarrow$ $\frac{G_{2}}{G_{1}} = \frac{g_{m_{2}}}{g_{m_{1}}}$

$\Rightarrow$ G₂ = $\frac{0.02}{0.03}$ $\times$ G = $\frac{2}{3} G$

7. The input resistance of a silicon transistor is 100 $Ω$ . Base current is changed by 40 $μ$ A which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 k $Ω$ . The voltage gain of the amplifier is ⇒ (AIPMT 2012 Mains)

A. 2000

B. 3000

C. 4000

D. 1000

The Correct Answer is (A)

Current gain ( $β$ ) :

$β$ = $\frac{Δ I_{C}}{Δ I_{B}}$

= $\frac{2 \times 10^{- 3}}{40 \times 10^{- 6}}$ = 50

Voltage gain of the amplifier is

A_V = $β \frac{R_{L}}{R_{i}}$

= 50 $\times$ $\frac{4 \times 10^{3}}{100}$ = 2000

8. In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k $Ω$ is 2 V. If the base resistance is 1 k $Ω$ and the current amplification of the transistor is 100, the input signal voltage is ⇒ (AIPMT 2012 Prelims)

A. 0.1 V

B. 1.0 V

C. 1 mV

D. 10 mV

The Correct Answer is (D)

Here, R_C = 2 k $Ω$ = 2 × 10³ $Ω$

V₀ = 2 V, R_B = 1 k $Ω$ = 1 × 10³ $Ω$ , $β$ = 100

The output voltage, across the load R_C

V₀ = I_CR_C = 2

The collector current (I_C)

I_C = $\frac{2}{2 \times 10^{3}}$ = 10^-3 A = 1 mA

Current gain( $β$ ) = $\frac{I_{C}}{I_{B}}$ = 100

$\Rightarrow$ I_B = $\frac{I_{C}}{100}$ = $\frac{10^{- 3}}{100}$ = 10^-5 A

Input voltage, V_i

= I_BR_B = (10^–5 A) (1 × 10³ $Ω$ ) = 10^–2 V = 10 mV

9. A transistor is operated in common emitter configuration at V_C = 2 V such that a change in the base current from 100 $μ$ A to 300 $μ$ A produces a change in the collector current from 10 mA to 20 mA. The current gain is ⇒ (AIPMT 2011 Prelims)

A. 50

B. 75

C. 100

D. 25

The Correct Answer is ()

Current gain, $β$ = $\frac{Δ I_{C}}{Δ I_{B}}$

= $\frac{(20 - 10) \times 10^{- 3}}{(300 - 100) \times 10^{- 6}}$ = 50

10. 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 ⇒ (AIPMT 2010 Prelims)

A. 500

B. 1000

C. 1250

D. 50

The Correct Answer is (C)

Voltage gain= $β$ × Impedance gain

$\Rightarrow$ 50 = β × (200/100)

$\Rightarrow$ $β$ = 25

Power gain = $β$ × Voltage gain

= 25 × 50 = 1250

11. The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain without feedback will be ⇒ (AIPMT 2008)

A. 1.25

B. 100

C. 90

D. 10

The Correct Answer is (B)

Negative feedback is applied to reduce the output voltage of an amplifier. If there is no negative feedback, the value of output voltage could be very high. In the options given, the maximum value of voltage gain is 100. Hence it is the correct option.

12. 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 $μ$ A to 150 $μ$ A profuces a change in the collector current from 5 mA to 10 mA. The current gain $β$ is ⇒ (AIPMT 2006)

A. 50

B. 67

C. 75

D. 100

The Correct Answer is (D)

Current gain, $β$ = $\frac{Δ I_{C}}{Δ I_{B}}$

= $\frac{(10 - 5) \times 10^{- 3}}{(150 - 100) \times 10^{- 6}}$ = 100

13. For a transistor $\frac{I_{C}}{I_{E}}$ = 0.96, then current gain for common emitter is ⇒ (AIPMT 2002)

A. 12

B. 6

C. 48

D. 24

The Correct Answer is Option (D)

$\frac{I_{C}}{I_{E}}$ = 0.96 = $α$

$∴$ Current gain, $β$ = $\frac{I_{C}}{I_{B}}$ = $\frac{α}{1 - α}$ = 24

14. For a common base circuit if $\frac{I_{C}}{I_{E}}$ = 0.98 then current gain for common emitter circuit will be ⇒ (AIPMT 2001)

A. 49

B. 98

C. 4.9

D. 25.5

The Correct Answer is Option (A)

$\frac{I_{C}}{I_{E}}$ = 0.98 = $α$

$∴$ Current gain, $β$ = $\frac{I_{C}}{I_{B}}$ = $\frac{α}{1 - α}$

15. The correct relation for $α$ , $β$ for a transistor ⇒ (AIPMT 2000)

A. $β = \frac{1 - α}{α}$

B. $β = \frac{α}{1 - α}$

C. $α = \frac{β - 1}{β}$

D. $α β = 1$

The Correct Answer is Option (B)

$β$ $\frac{I_{C}}{I_{B}}$ = $\frac{I_{C}}{I_{E} - I_{C}}$ = $\frac{\frac{I_{C}}{I_{E}}}{1 - \frac{I_{C}}{I_{E}}}$ = $\frac{α}{1 - α}$

⇒ Semiconductor

Topic 7 : Application of Transistor