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# Multiple choice question for engineering

## Set 1

1. The current amplification factor ϒdc is given by_________
a) IE/IB
b) IB/IE
c) IC/IE
d) IE/IC

Answer: a [Reason:] When no signal is applied, then the ratio of emitter current to base current is called as ϒdc of the transistor. As the collector is common to both input and output circuits, hence the name common collector configuration.

2. The relation between α and β is given by _________
a) 1/(1-α)=1- β
b) 1/(1+α)=1+ β
c) 1/(1-α)=1+ β
d) 1/(1+α)=1- β

Answer: c [Reason:] The current amplification factor (β) is given by IC//IB. When no signal is applied, then the ratio of collector current to the base current is called current amplification factor of a transistor. β is an ac base amplification factor. α is called as current amplification factor. The relation of IC and IB change as IC= βIB+ (1+ β) ICBO.

3. The CC configuration has an input resistance_________
a) 500kΩ
b) 750kΩ
c) 600kΩ
d) 400kΩ

Answer: b [Reason:] It has a high input resistance and very low output resistance so the voltage gain is always less one. It is used for driving a low impedance load from a high impedance source.

4. The application of a CC configured transistor is_________
a) voltage multiplier
b) level shifter
c) rectification
d) impedance matching

Answer: d [Reason:] The most important use of CC transistor is an impedance matching device. It is seldom used for amplification purposes. The current gain is same as that of CE configured transistor.

5. What is the output resistance of CC transistor?
a) 25 Ω
b) 50 Ω
c) 100 Ω
d) 150 Ω

Answer: a [Reason:] The CC transistor has a very low value of output resistance of 25 Ω. The voltage gain is always less one. It is used for driving a low impedance load from a high impedance source.

6. Increase in collector emitter voltage from 5V to 8V causes increase in collector current from 5mA to 5.3mA. Determine the dynamic output resistance.
a) 20kΩ
b) 10kΩ
c) 50kΩ
d) 60kΩ

Answer: b [Reason:] ro=∆VCE/∆IC =3/0.3m=10kΩ.

7. A change in 300mV in base emitter voltage causes a change of 100µA in the base current. Determine the dynamic input resistance.
a) 20kΩ
b) 10kΩ
c) 30kΩ
d) 60kΩ

Answer: c [Reason:] ro=∆VBE/∆IB =300m/100µ=30kΩ.

8. The point on the DC load line which is represented by ‘Q’ is called _________
a) cut off point
b) cut in point
c) breakdown point
d) operating point

Answer: d [Reason:] The point which represents the values of IC and VCE that exist in a transistor circuit when no signal is applied is called as operating point. This is also called as working point or quiescent point.

9. When is the transistor said to be saturated?
a) when VCE is very low
b) when VCE is very high
c) when VBE is very low
d) when VBE is very high

Answer: a [Reason:] When VCE is very low, the transistor said to be saturated and it operates in saturated region of characteristic. The change in base current IB does not produce a corresponding change in the collector voltage IC.

10. The input resistance is given by _________
a) ∆VCE/∆IB
b) ∆VBE/∆IB
c) ∆VBE/∆IC
d) ∆VBE/∆IE

Answer: b [Reason:] The ratio of change in base emitter voltage (∆VBE) to resulting change in base current (∆IB) at constant collector emitter voltage (VCE) is defined as input resistance. This is denoted by ri.

## Set 2

1. The base current amplification factor β is given by_________
a) IC/IB
b) IB/IC
c) IE/IB
d) IB/IE

Answer: a [Reason:] The current amplification factor (β) is given by IC//IB. When no signal is applied, then the ratio of collector current to the base current is called current amplification factor of a transistor.

2. In an NPN silicon transistor, α=0.995, IE=10mA and leakage current ICBO=0.5µA. Determine ICEO.
a) 10µA
b) 100µA
c) 90µA
d) 500µA

Answer: b [Reason:] IC=α IE +ICBO =0.995*10mA+0.5µA=9.9505mA. IB=IE-IC=10-9.9505=0.0495mA. β=α/(1-α)=0.995/(1-0.995)=199 ICEO=9.9505-199*0.0495=0.1mA==100µA.

3. A germanium transistor with α=0.98 gives a reverse saturation current ICBO=10µA in a CB configuration. When it is used in CE configuration with a base current of 0.22µA, calculate the collector current.
a) 0.9867mA
b) 0.7654mA
c) 0.51078mA
d) 0.23456mA

Answer: c [Reason:] Given, ICBO=10µA, α=0.98 and IB =0.22µA. IC=α/ (1-α) IB+ 1/(1-α) ICBO 0.01078+0.5=0.51078mA.

4. In CE configuration, if the voltage drop across 5kΩ resistor connected in the collector circuit is 5V. Find the value of IB when β=50.
a) 0.01mA
b) 0.25mA
c) 0.03mA
d) 0.02mA

Answer: d [Reason:] IC=V across RL/RL=5V/5KΩ=1mA. IB=IC/β=1/50=0.02mA.

5. A transistor is connected in CE configuration. Collector supply voltage Vcc=10V, RL=800Ω, voltage drop across RL=0.8V, α=0.96. What is base current?
a) 41.97µA
b) 56.78µA
c) 67.67µA
d) 78.54µA

Answer: a [Reason:] Here, IC=0.8/800=1mA β= α/ (1-α)=0.96/1-0.96=24. Now, IB=IC/ β=1/24=41.67µA.

6. The collector supply voltage for a CE configured transistor is 10V. The resistance RL=800Ω. The voltage drop across RL is 0.8V. Find the value of collector emitter voltage.
a) 3.7V
b) 9.2V
c) 6.5V
d) 9.8V

Answer: b [Reason:] Here, IC=0.8/800=1mA. We know, VCE=VCC-ICRL =10-0.8=9.2V.

7. The relation between α and β is_________
a) β = α/ (1-α)
b) α = β/(1+β)
c) β = α/ (1+α)
d) α = β/(1- β)

Answer: b [Reason:] β is an ac base amplification factor. α is called as current amplification factor. The relation of IC and IB change as IC= βIB+ (1+ β) ICBO.

8. In ICEO, wt does the subscript ‘CEO’ mean?
a) collector to base emitter open
b) emitter to base collector open
c) collector to emitter base open
d) emitter to collector base open

Answer: c [Reason:] The subscript ‘CEO’ means that it is collector to emitter base open. It is called as the leakage current. It occurs in a reverse bias in PNP transistor. The total current can be calculated by IC=βIB+IC.

9. When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called_________
a) dc current gain
b) base current amplification factor
c) emitter current amplification factor
d) ac current gain

Answer: d [Reason:] The ac current gain is given by β=∆IC/∆IB. When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called ac current gain.

10. The range of β is _________
a) 20 to 500
b) 50 to 300
c) 30 to 400
d) 10 to 20

Answer: a [Reason:] Almost in all the transistors, the base current is less than 5% of the emitter current. Due to this fact, it is generally greater than 20. Usually it ranges from 20 to 500. Hence this configuration is frequently used when appreciable current gain as well as voltage gain is required.

## Set 3

1. Which of the following depicts the DC load line?
a)
b)
c)
d)

Answer: a [Reason:] In transistor circuit analysis, sometimes it is required to know the collector currents for various collector emitter voltages. The one way is to draw its load line. We require the cut off and saturation points.

2. For the circuit shown, find the quiescent point.

a) (10V, 4mA)
b) (4V, 10mA)
c) (10V, 3mA)
d) (3mA, 10V)

Answer: c [Reason:] We know, IE=VEE/RE=30/10kΩ=3mA IC=α IE =IE =3mA VCB=VCC-ICRL=25-15=10V. So, quiescent point is (10V, 3mA).

3. Which of the following depicts the load line for the circuit shown below?

a)
b)
c)
d)

Answer: d [Reason:] We know, IE=VEE/RE=15/5kΩ=3mA IC=α IE =IE =3mA VCB=VCC-ICRL=20-15=5V. So, quiescent point is (5V, 3mA).

4. For the circuit shown, find the quiescent point.

a) (6V, 1mA)
b) (4V, 10mA)
c) (10V, 3mA)
d) (3mA, 10V)

Answer: c [Reason:] We know, VCE=12V (IC)SAT =VCC/RL=12/6K=2mA. IB=10V/0.5M=20µA. IC= βIB=1mA. I VCE=VCC-ICRL=12-1*6=6V. So, quiescent point is (6V, 1mA).

5. Which of the following depicts the load line for the given circuit?

a)
b)
c)
d)

Answer: d [Reason:] We know, VCE=6V (IC)SAT =VCC/RL=10/2K=5mA. IB=10V/0.5M=20µA. IC= βIB=1mA. I VCE=VCC-ICRL=10-1*2=8V. So, quiescent point is (8V, 1mA).

6. The DC equivalent circuit for an NPN common base circuit is.

Answer: a [Reason:] In the common base circuit, the emitter diode acts like a forward biased ideal diode, while collector diode acts as a current source due to transistor action. Thus an ideal transistor may be regarded as a rectifier diode in the emitter and a current source at collector.

7. The DC equivalent circuit for an NPN common emitter circuit is.

Answer: b [Reason:] In the common emitter circuit, the ideal transistor may be regarded as a rectifier diode in the base circuit and a current source in the collector circuit. In the current source, the direction of arrow points in direction of conventional current.

8. What is the other representation of the given PNP transistor connected in common emitter configuration?

Answer: d [Reason:] The emitter junction is forward biased with the help of battery VEE by which, negative of the battery is connected to the emitter while positive is connected to base. RE is the emitter resistance. The collector junction is reversed biased.

9. What is the DC characteristic used to prove that the transistor is indeed biased in saturation mode?
a) IC = βIB
b) IC > βIB
c) IC >> βIB
d) IC < βIB

Answer: d [Reason:] When in a transistor is driven into saturation, we use VCE(SAT) as another linear parameter. In, addition when a transistor is biased in saturation mode, we have IC < βIB. This characteristic used to prove that the transistor is indeed biased in saturation mode.

10. For the circuit shown, find the quiescent point.

a) (10V, 4mA)
b) (4V, 10mA)
c) (10V, 3mA)
d) (3mA, 10V)

Answer: c [Reason:] We know, IE=VEE/RE=10/5kΩ=2mA IC=α IE =IE =2mA VCB=VCC-ICRL=20-10=10V. So, quiescent point is (10V, 2mA).

## Set 4

1. The feature of an approximate model of a transistor is
a) it helps in quicker analysis
b) it provides individual analysis for different configurations
c) it helps in dc analysis
d) ac analysis is not possible

Answer: a [Reason:] The small signal model helps in quicker ac analysis of a transistor. The approximate model is applicable for all the configurations. The dc analysis is not obtained by using a small signal model of transistor.

2 A transistor has hfe=100, hie=2kΩ, hoe=0.005mmhos, hre=0. Find the output impedance if the lad resistance is 5kΩ.
a) 5kΩ
b) 4kΩ
c) 20kΩ
d) 15kΩ

Answer: b [Reason:] RO=I/hoe=1/0.005m =20kΩ.ROI= RO || RLI=20||5 =4kΩ.

3. A CE amplifier when bypassed with a capacitor at the emitter resistance has
a) increased input resistance and increased voltage gain
b) increased input resistance and decreased voltage gain
c) decreased input resistance and increased voltage gain
d) decreased input resistance and decreased voltage gain

Answer: c [Reason:] When a transistor is bypassed with a capacitor, it short circuits in the small signal analysis of transistor and the resistor too shorts. The input resistance becomes RI=hie. The value of the input resistance is decreased and the gain now will be increasing.

4. A transistor has hie =2kΩ, hoe=25µmhos and hfe=60 with an unbypassed emitter resistor Re=1kΩ. What will be the input resistance and output resistance?
a) 90kΩ and 50kΩ respectively
b) 33kΩ and 45kΩ respectively
c) 6kΩ and 40kΩ respectively
d) 63kΩ and 40kΩ respectively

Answer: d [Reason:] As the emitter is unbypassed, the input resistance Ri=hie+(1+hfe)Re =2+61=63kΩ. The output resistance RO=1/hoe=1/25MΩ=40kΩ.

5. A transistor has hie =1KΩ and hfe=60 with an bypassed emitter resistor Re=1kΩ. What will be the input resistance and output resistance?
a) 90kΩ and 50kΩ respectively
b) 33kΩ and 45kΩ respectively
c) 6kΩ and 40kΩ respectively
d) 63kΩ and 40kΩ respectively

Answer: d [Reason:] As the emitter is bypassed, the input resistance Ri=hie =1kΩ. The output resistance RO=1/hoe but the value is not given. So, hoe=0 and RO=1/0=∞.

6. In the given circuit, find the equivalent resistance between A and B nodes.

a) 100kΩ
b) 50kΩ
c) 40kΩ
d) 60kΩ

Answer: b [Reason:] RAB=RO||100Ω = (RSI+hie/1+hfe)||100 =9+1/100||100=100||100=50Ω.

7. Which of the following acts as a buffer?
a) CC amplifier
b) CE amplifier
c) CB amplifier

Answer: a [Reason:] The voltage gain of a common collector amplifier is unity. It is then used as a buffer. The CC amplifier is also called as an emitter follower. Though there is no amplification done, the output will be stabilised.

8. Which of the following is true?
a) CC amplifier has a large current gain
b) CE amplifier has a large current gain
c) CB amplifier has low voltage gain
d) CC amplifier has low current gain

Answer: b [Reason:] The CE amplifier has high current and voltage gains. The CC amplifier has unity voltage gain which cannot be regarded as high. The common base amplifier has a unity current gain and high voltage gain.

9. In an NPN silicon transistor, α=0.995, IE=10mA and leakage current ICBO=0.5µA. Determine ICEO.
a) 10µA
b) 100µA
c) 90µA
d) 500µA

Answer: b [Reason:] IC=α IE +ICBO =0.995*10mA+0.5µA=9.9505mA. IB=IE-IC=10-9.9505=0.0495mA. β=α/(1-α)=0.995/(1-0.995)=199 ICEO=9.9505-199*0.0495=0.1mA==100µA.

10. In CB configuration, the value of α=0.98A. A voltage drop of 4.9V is obtained across the resistor of 5KΩ when connected in collector circuit. Find the base current.
a) 0.01mA
b) 0.07mA
c) 0.02mA
d) 0.05mA

Answer: c [Reason:] Here, IC=4.9/5K=0.98mA α = IC/IE .So, IE=IC/α=0.98/0.98=1mA. IB=IE-IC=1-0.98=0.02mA.

## Set 5

1. The collector current (IC) that is obtained in a self biased transistor is_________
a) (VTH – VBE)/RE
b) (VTH + VBE)/RE
c) (VTH – VBE)/RE
d) (VTH + VBE)/RE

Answer: a [Reason:] The collector current is analysed by the DC analysis of a transistor. It involves the DC equivalent circuit of a transistor. The base current is first found and the collector current is obtained from the relation, IC=IBβ.

2. The collector to emitter voltage (VCE) is obtained by_________
a) VCC – RCIC+RBIB
b) VCC – RCIC-REIE
c) VCC + RCIC
d) VCC + RCIB

Answer: b [Reason:] The collector to emitter voltage is obtained in order to find the operating point of a transistor. It is taken when there is no signal applied to the transistor. The point thus obtained lies in the cut off region when the transistor is used as a switch.

3. The thermal runway is avoided in a self bias because_________
a) of its independence of β
b) of the positive feedback produced by the emitter resistor
c) of the negative feedback produced by the emitter resistor
d) of its dependence of β

Answer: c [Reason:] The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the emitter resistor in a self bias. The IC which is responsible for the damage is reduced by decreased output signal.

4. When the temperature is increased, what happens to the collector current after a feedback is given?
a) it remains same
b) it increases
c) it cannot be predicted
d) it decreases

Answer: d [Reason:] Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the drop across the emitter resistor increases with decreasing collector current and the thermal runway too.

5. What is the Thevenin’s voltage (VTH) in a self bias shown below?

a) VCCR2/R1+R2
b) VCCR1/R1+R2
c) VCCR2/R1-R2
d) VCCR2/R1-R2

Answer: a [Reason:] The base current cannot be obtained directly from the KVL or KCL applications. The VCC and VBE cannot come under a single equation. So, the circuit is changed with a Thevenin’s voltage (VTH) and Thevenin’s resistance.

6. What is the Thevenin’s resistance (RTH) in a self bias shown below?

a) R1R2/R1+R2
b) R2/R1+R2
c) R1R2/R1-R2
d) R1/R1-R2

Answer: a [Reason:] The base current cannot be obtained directly from the KVL or KCL applications. A potential divider network is formed by R1 and R2.The VCC and VBE cannot come under a single equation. So, the circuit is changed with a Thevenin’s resistance.

7. The stability factor for a self biased transistor is_________
a) 1 – RTH/RE
b) 1 + RTH/RE
c) 1 + RE/RTH
d) 1 – RE/RTH

Answer: b [Reason:] The stability of the circuit is inversely proportional to the stability factor. The emitter resistor is very large when compared to the Thevenin’s resistance. When β is not that large, then S=(1+ β)( RTH+ RE)/ (1+ β)RE+ RTH.

8. In the circuit, the transistor has a large β value (VBE=0.7V). Find the current through RC.

a) 0.5mA
b) 2mA
c) 1mA
d) 1.6mA

Answer: c [Reason:] We know, IC=VTH-VBE/RE =9*3/9=3V. IC=3-0.7/2.3=1mA.

9. A silicon NPN transistor is used and it has a large value of β. Find the required value of R2 when IC=1mA.

a) 10kΩ
b) 20kΩ
c) 30kΩ
d) 40kΩ

Answer: d [Reason:] For silicon, VBE=0.8V, VCE=0.2V. IC=VTH-VBE/RE. By pitting the values, we have VTH=1.3V. R2 can be found from, VCCR2/R1+R2. We get R2=40KΩ.

10. The value of αac for all practical purposes, for commercial transistors range from_________
a) 0.5 to 0.6
b) 0.7 to 0.77
c) 0.8 to 0.88
d) 0.9 to 0.99