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# Multiple choice question for engineering

## Set 1

1. The current gain of BJT is_________
a) gmro
b) gm/ro
c) gmri
d) gm/ri

Answer: c [Reason:] We know, current gain AV=hfe. In π model, hfe is referred to β. We know, ri= β/gm. From this, β=rigm.

2. For the amplifier circuit of fig. The transistor has β of 800. The mid band voltage gain VO/VI of the circuit will be_________ a) 0
b) <1
c) =1
d) 800

Answer: c [Reason:] The circuit is PNP transistor, collector coupled amplifier. The voltage gain is unity for a CC amplifier. Hence on observation, the CC amplifier gives a unity gain.

3. In a bipolar transistor at room temperature, the emitter current is doubled the voltage across its base emitter junction_________
a) doubles
b) halves
c) increase by about 20mV
d) decreases by about 20mV

Answer: c [Reason:] The change in voltage with temperature can be found by, V(T) = 2.3m(∆T)VO . In a bipolar transistor at room temperature if the emitter current is doubled the voltage across its base emitter junction thereby doubles.

4. A common emitter transistor amplifier has a collector current of 10mA, when its base current is 25µA at the room, temperature. What is input resistance?
a) 3kΩ
b) 5kΩ
c) 1kΩ
d) 7kΩ

Answer: c [Reason:] We know, β/gm=ri = (IC/IB)/(IC/VT)=VT/IB=25m/25µ=1k.

5. For an NPN transistor connected as shown in below, VBE=0.7V. Give that reverse saturation current of junction at room temperature is 10-13A, the emitter current is_________ a) 30mA
b) 39mA
c) 29mA
d) 49mA

Answer: d [Reason:] When the collector and base are shorted, the transistor behaves as a normal diode. So, the diode equations imply. IE=IO(eV/V0-1). We get, IE=49mA.

6. The voltage gain of given circuit below is_________ a) 100
b) 20
c) 10
d) 30

Answer: c [Reason:] The gain for the given circuit can be found by, AV=RF/RS =100K/10K=10.

7. A small signal source V(t)=Acos20t+Bsin10000t, is applied to a transistor amplifier as shown. The transistor has β=150 and hie=3KΩ. What will be the VO? a) 1500(Acos20t+Bsin10000t)
b) -150(Acos20t+Bsin10000t)
c) -1500Bsin10000t
d) -150Bsin10000t

Answer: d [Reason:] AV=-hfe RLI/hie=3*150/3=-150. So, VO=-150V(t) But cos20t has low frequency so capacitors are open circuited. Only, the sine component is allowed. So, Vo =-150Bsin10000t.

8. Which of the following statements are correct for basic transistor configurations?
a) CB Amplifiers has low input impedance and low current gain
b) CC Amplifiers has low input impedance and high current gain
c) CE Amplifiers has very poor voltage gain but very high input impedance
d) The current gain of CB Amplifier is higher than the current gain of CC Amplifiers

Answer: a [Reason:] The CE amplifier has moderate input and output impedances. The CC amplifier has unity voltage gain. The common ba se amplifier has a unity current gain and high voltage gain.

9. The collector current is 2.945A and α=0.98. The leakage current is 2µA. What is the emitter current and base current?
a) 3mA and 55µA
b) 2.945mA and 55µA
c) 3.64mA and 33µA
d) 5.89mA and 65µA

Answer: a [Reason:] (IC – ICBO)/α=IE = (2.945-0.002)/0.98=3mA. IE=IC+IB . So, IB=3-2.495=0.055mA=55µA.

10. The change in collector emitter voltage from 6V to 9V causes increase in collector current from 6mA to 6.3mA. Determine the dynamic output resistance.
a) 20kΩ
b) 10kΩ
c) 50kΩ
d) 60kΩ

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

11. A transistor is connected in CB configuration. The emitter voltage is changed by 200mV, the emitter by 5mA. During this transition the collector base voltage is kept constant. What is the input dynamic resistance?
a) 30Ω
b) 60Ω
c) 40Ω
d) 50Ω

Answer: c [Reason:] The ratio of change in emitter base voltage (∆VEB) to resulting change in emitter current (∆IE) at constant collector base voltage (VCB) is defined as input resistance. This is denoted by ri. We know, ∆VEB/∆IE=ri =200/5=40Ω.

## Set 2

1. The input resistance in a CB transistor is given by _________
a) ∆VCE/∆IB
b) ∆VBE/∆IB
c) ∆VBE/∆IC
d) ∆VEB/∆IE

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

2. The output resistance of CB transistor is given by _________
a) ∆VCB/∆IC
b) ∆VBE/∆IB
c) ∆VBE/∆IC
d) ∆VEB/∆IE

Answer: a [Reason:] The ratio of change in collector base voltage (∆VCB) to resulting change in collector current (∆IC) at constant emitter current (IE)¬ is defined as output resistance. This is denoted by ro.

3. Which one of the following depicts the output characteristics for a CB transistor?
a) b) c) d) Answer: b [Reason:] A graph of IC against VCB is drawn. The curve so obtained is known as output characteristics. The emitter current (IE) is kept constant.

4. The input characteristics of a CE transistor is_________
a) b) c) d) Answer: c [Reason:] A graph of IE against VEB is drawn. The curve so obtained is known as input characteristics. The collector base voltage (VBC) is kept constant.

5. A transistor is connected in CB configuration. The emitter voltage is changed by 200mV, the emitter by 5mA. During this transition the collector base voltage is kept constant. What is the input dynamic resistance?
a) 30Ω
b) 60Ω
c) 40Ω
d) 50Ω

Answer: c [Reason:] The ratio of change in emitter base voltage (∆VEB) to resulting change in emitter current (∆IE) at constant collector base voltage (VCB) is defined as input resistance. This is denoted by ri. We know, ∆VEB/∆IE=ri =200/5=40Ω.

6. When the collector junction is reverse biased and emitter junction is forward biased, the operating region of the transistor is called_________
a) inverted region
b) active region
c) cut off region
d) cut in region

Answer: b [Reason:] In the active region, for small values of base current, the effect of collector voltage over collector current is small while for large base currents this effect increases. The shape of characteristic here is same as that of CB transistors.

7. Which of the following corresponds to the output circuit of a CB transistor?
a) VBE
b) IB
c) VCB
d) VCE

Answer: c [Reason:] Here, the quantity collector to base voltage corresponds to the output circuit of a CB transistor. The complete electrical behaviour of a transistor can be described by stating the relation between these quantities.

8. The input of a CB transistor is given between_________
a) collector and emitter terminals
b) base ad collector terminals
c) ground and emitter terminals
d) emitter and base terminals

Answer: d [Reason:] The name of the CB transistor says that it’s a common based one. The input is given between the emitter and base terminals and the output is taken between collector and base terminals.

9. The current gain of the CB transistor is_________
a) less than or equal to unity
b) equal to unity
c) greater than unity
d) remains same

Answer: a [Reason:] The input current flowing into the emitter terminal must be higher than the base current and collector current to operate the transistor. Therefore the output collector current is less than the input emitter current.

10. The input characteristics of a CB transistor resembles_________
a) Forward biased diode
b) Illuminated photo diode
c) LED
d) Zener diode

Answer: b [Reason:] The input characteristics resemble the illuminated photo diode and the output characteristics resemble the forward biased diode. This transistor has low input impedance and high output impedance.

## Set 3

1. The input characteristics of a CE transistor is_________
a) b) c) d) Answer: b [Reason:] A graph of IB against VBE is drawn. The curve so obtained is known as input characteristics. The collector emitter voltage (VCE) is kept constant.

2. 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.

3. Which of the following depicts the output characteristics of a CE transistor?
a) b) c) d) Answer: d [Reason:] A graph of IC against VCE is drawn. The curve so obtained is known as output characteristics. The base current (IB) is kept constant.

4. The output resistance is given by _________
a) ∆VCE/∆IB
b) ∆VBE/∆IB
c) ∆VBE/∆IC
d) ∆VCE/∆IC

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

5. Which of the following cases damage the transistor?
a) when VCE is increased too far
b) when VCE is decreased too far
c) when VBE is increased too far
d) when VBE is decreased too far

Answer: a [Reason:] When VCE is increased too far, collector base junction completely breaks down and due to this avalanche breakdown, collector current increases rapidly. This is not shown in the characteristic. In this case, the transistor is damaged.

6. When the collector junction is reverse biased and emitter junction is forward biased, the operating region of the transistor is called_________
a) inverted region
b) active region
c) cut off region
d) cut in region

Answer: b [Reason:] In the active region, for small values of base current, the effect of collector voltage over collector current is small while for large base currents this effect increases. The shape of characteristic here is same as that of CB transistors.

7. The small amount of current which flows even when base current IB=0 is called_________
a) IBEO
b) ICBO
c) ICEO
d) IC

Answer: c [Reason:] In the cut off region, a small amount of collector current flows even when base current IB is zero. This is called ICEO. Since the main current is also zero, the transistor is said to be cut off.

8. A change in 700mV in base emitter voltage causes a change of 200µA in the base current. Determine the dynamic input resistance.
a) 2kΩ
b) 10kΩ
c) 3kΩ
d) 3.5kΩ

Answer: c [Reason:] ro=∆VBE/∆IB =700m/200µ=3.5kΩ.

9. The change in collector emitter voltage from 6V to 9V causes increase in collector current from 6mA to 6.3mA. Determine the dynamic output resistance.
a) 20kΩ
b) 10kΩ
c) 50kΩ
d) 60kΩ

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

10. Which of the following points locates the quiescent point?
a) (IC, VCB)
b) (IE, VCE)
c) (IE, VCB)
d) (IC, VCE)

Answer: a [Reason:] The quiescent point is best located between the cut off and saturation point. IE= VEE/RE, VCB=VCC-ICRL. It is denoted by ‘Q’.

## Set 4

1. The collector current (IC) that is obtained in a collector to base biased transistor is_________
a) (VCC-VBE)/RB
b) (VCC+VBE)/RB
c) (VCE-VBE)/RB
d) (VCE+VBE)/RB

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 – RC(IC-IB)
b) VCC – RC(IC+IB)
c) VCC + RC(IC+IB)
d) VCC + RC(IC-IB)

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. 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.

4. The thermal runway is avoided in a collector to base bias because_________
a) of its independence of β
b) of the positive feedback produced by the base resistor
c) of the negative feedback produced by the base 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 base resistor in a collector to base bias. The IC which is responsible for the damage is reduced by decreased output signal.

5. 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 base current increases with decreasing collector current and the thermal runway too.

6. The demerit of a collector to base bias is_________
a) its need of high resistance values
b) its dependence on β
c) its independence on β
d) the positive feedback produced by the base resistor

Answer: a [Reason:] When the stability factor S=1, the collector resistor value should be very large when compared to the base resistor. So, when RC is large we need to provide large power supply which increases the cost. At the same time, as the base resistor is small we need to provide small power supply.

7. The negative feedback does good for DC signal by_________
a) decreasing the gain
b) increasing the gain
c) stabilising the operating point
d) increasing the stability factor

Answer: c [Reason:] The resistor RB can provide negative feedback for both AC and DC signals. The negative feedback for DC signal is done good as it can provide stable operating point. On the other side, the negative feedback is badly done for AC signal by decreasing the voltage gain.

8. In the circuit, transistor has β =60, VBE=0.7V. Find the collector to emitter voltage drop VCE. a) 5V
b) 3V
c) 8V
d) 6V

Answer: d [Reason:] We know, IC=(VCC-VBE)/RB By putting the values, we have IC=5.9mA. IE=IC/α. So, IE=5.99mA. VCE= VCC-RC(IC+IB). We have VCE=6V.

9. In the circuit shown below, β =100 and VBE=0.7V. The Zener diode has a breakdown voltage of 6V. Find the operating point. a) (6.7V, 5.3mA)
b) (5.7V, 5.3mA)
c) (6.7V, 5mA)
d) (6V, 5mA)

Answer; a [Reason:] We know, by KVL -12+(IC+IB)1K+6+VBE=0 We have IE=5.3. IC= αIE=5.24mA. From another loop, -12+IEIK+VBE=0 We have, VCE=12-5.3m*1000=6.7V. Hence the Q point is (6.7V, 5.3mA).

10. When the β value is large for a given transistor, the IC and VCE values are given by_________
a) (VCC-VBE)/RB, VCC-RCIC
b) (VCC+VBE)/RB, VCC-RC(IC+IB)
c) (VCC+VBE)/RB, VCC+RC(IC+IB)
d) (VCC+VBE)/RB, VCC+RC(IC-IB)

Answer: a [Reason:] The base current IB is zero when β value is large. So, the VCE changes to VCC-RCIC. The collector current IC is changed to (VCC-VBE)/RB from β(VCC-VBE)/(1+ β)RE+ RB.

## Set 5

1. The AC current gain in a common base configuration is_________
a) -∆IC/∆IE
b) ∆IC/∆IE
c) ∆IE/∆IC
d) -∆IE/∆IC

Answer: a [Reason:] The AC current gain is denoted by αac. The ratio of change in collector current to the change in emitter current at constant collector base voltage is defined as current amplification factor.

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

Answer: d [Reason:] For all practical purposes, αac=αdc=α and practical values in commercial transistors range from 0.9-0.99. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.

3. A transistor has an IC of 100mA and IB of 0.5mA. What is the value of αdc?
a) 0.787
b) 0.995
c) 0.543
d) 0.659

Answer: b [Reason:] Emitter current IE=IC+IB =100+0.5=100.5mA αdc=IC/IE=100/100.5=0.995.

4. 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.

5. The emitter current IE in a transistor is 3mA. If the leakage current ICBO is 5µA and α=0.98, calculate the collector and base current.
a) 3.64mA and 35µA
b) 2.945mA and 55µA
c) 3.64mA and 33µA
d) 5.89mA and 65µA

Answer: b [Reason:] IC=αIE + ICBO =0.98*3+0.005=2.945mA. IE=IC+IB . So, IB=3-2.495=0.055mA=55µA.

6. Determine the value of emitter current and collector current of a transistor having α=0.98 and collector to base leakage current ICBO=4µA. The base current is 50µA.
a) 1.5mA
b) 3.7mA
c) 2.7mA
d) 4.5mA

Answer: c [Reason:] Given, IB=50µA=0.05mA ICBO=4µA=0.004Ma IC=α/(1- α)IB+1/(1- α)ICBO=2.45+0.2=2.65Ma IE=IC+IB=2.65+0.05=2.7mA.

7. The negative sign in the formula of amplification factor indicates_________
a) that IE flows into transistor while IC flows out it
b) that IC flows into transistor while IE flows out it
c) that IB flows into transistor while IC flows out it
d) that IC flows into transistor while IB flows out it

Answer: a [Reason:] When no signal is applied, the ratio of collector current to emitter current is called dc alpha, αdc of a transistor. αdc=-IC/IE. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.

8. 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.

9. A transistor has an IE of 0.9mA and amplification factor of 0.98. What will be the IC?
a) 0.745mA
b) 0.564mA
c) 0.236mA
d) 0.882mA