Bipolar Junction Transistors Questions and Answers

Q: Why is a transistor's base made very thin?

The base must be thin compared with the diffusion length, so carriers injected from the emitter cross it and reach the collector before recombining - which is what makes the collector current nearly equal to the emitter current and the base current small. A thin base actually raises the base resistance, and capacitance and breakdown voltage are affected in ways that constrain rather than motivate the design. Transistor action is entirely a matter of carriers surviving the base, and every geometric choice serves that.

Q: Why is a transistor's base lightly doped compared with its emitter?

Light base doping means few majority carriers there to recombine with the injected ones, so the emitter injection efficiency is high and almost all the emitter current becomes collector current. The raised bulk resistance is an unwanted consequence rather than the aim, and the two remaining effects are undesirable. Doping asymmetry is what makes the emitter inject far more than the base does back into it, and that asymmetry is why the device is not symmetrical.

Q: Which junction biases distinguish the active region?

The active region needs the emitter junction forward biased to inject carriers and the collector junction reverse biased to collect them. Both forward is saturation, both reverse is cut-off, and the reversed arrangement is the inverse active region, which works but with a much poorer gain because the doping is wrong for it. Naming the four regions by their two junction biases is the compact way to hold them, and it makes the bias conditions in a circuit immediately readable.

Q: What bias condition defines saturation?

In saturation both junctions are forward biased, the collector can no longer take all the carriers offered, and the collector-emitter voltage falls to a few tenths of a volt. The collector current then depends on the external circuit rather than on the base current, which is precisely what a switching application wants. Recognising saturation from the junction biases rather than from a current ratio is more reliable, because the ratio is what breaks down there.

Q: With no base drive at all, only leakage flows. Which region is the device in?

Cut-off has both junctions reverse biased, so only leakage flows and the device is effectively open. Both forward is saturation and the mixed case is the active region. Equal base and collector currents would mean a beta of one, which describes no useful region at all. Since a switch spends its life in cut-off or saturation, these two regions matter far more in digital work than the active region between them.

Q: What does the current gain alpha describe?

Alpha is the ratio of collector to emitter current, so it is a little less than unity - typically 0.98 to 0.998. Collector over base is beta, and emitter over base is gamma, one more than beta. Because alpha is so close to one, small changes in it produce large changes in beta, which is why beta varies so widely between nominally identical devices and why no good design depends on its exact value.

Q: A transistor has \( \alpha = 0.98 \). Its beta is

Converting between the two gains:\[ \beta = \frac{\alpha}{1-\alpha} = \frac{0.98}{0.02} = 49 \]so 49. Ninety-eight mistakes alpha's decimals for the answer, and fifty would follow from an alpha of 0.9804. The steepness of this relation is the important part: raising alpha from 0.98 to 0.99 doubles beta to 99, which is why manufacturing spreads in alpha appear as enormous spreads in beta.

Q: A transistor has \( \beta = 99 \). Its alpha is

Inverting the relation:\[ \alpha = \frac{\beta}{1+\beta} = \frac{99}{100} = 0.99 \]giving 0.99. The value 1.01 is impossible, since alpha must be less than one for a device that cannot deliver more collector current than emitter current. Checking that a computed alpha lies below unity is a quick sanity test, and it catches the commonest algebraic slip in this pair of formulae.

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Assertion-Reason type questions : Q. 41 to 50 :

The following items consist of two statements, one labelled the ‘Assertion (A)’ and the other labelled the ‘Reason (R)’. You are to examine these two statements and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly.

Codes :
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true and (R) is false.
(D) (A) is false and (R) is true.

Assertion (A) : A BJT is said to be operating in a saturation region if both the junctions are forward biased.

Reason (R) : In saturation region collector doping concentration is reduced.

A.Both (A) and (R) are true and (R) is the correct explanation of (A).
B.Both (A) and (R) are true, but (R) is not the correct explanation of (A).
C.(A) is true and (R) is false.
D.(A) is false and (R) is true.

Exam Year(s)

UGC NET - 2012,

Answer: C

The assertion is true and the reason is false, so option (C) is right. A bipolar junction transistor works in saturation when both the emitter-base junction and the collector-base junction are forward biased, and the collector current is then limited by the circuit and not by the base current, which makes the assertion correct. The reason, that the collector doping concentration is reduced in saturation, is wrong, because the doping of a transistor is fixed at manufacture and does not change with the bias. What does change in saturation is the amount of charge stored in the base and collector. Options (A) and (B) accept the reason, and option (D) rejects the assertion.

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Why is a transistor's base made very thin?

A.So most injected carriers reach the collector
B.So its resistance is as low as the design allows
C.So its capacitance to the emitter stays small
D.So the breakdown voltage of the device is raised

Answer: A

The base must be thin compared with the diffusion length, so carriers injected from the emitter cross it and reach the collector before recombining - which is what makes the collector current nearly equal to the emitter current and the base current small. A thin base actually raises the base resistance, and capacitance and breakdown voltage are affected in ways that constrain rather than motivate the design. Transistor action is entirely a matter of carriers surviving the base, and every geometric choice serves that.

Discuss in Forum

Why is a transistor's base lightly doped compared with its emitter?

A.To increase the base-emitter junction's capacitance
B.To keep recombination in the base low
C.To raise the base's own bulk resistance
D.To lower the collector-base breakdown voltage

Answer: B

Light base doping means few majority carriers there to recombine with the injected ones, so the emitter injection efficiency is high and almost all the emitter current becomes collector current. The raised bulk resistance is an unwanted consequence rather than the aim, and the two remaining effects are undesirable. Doping asymmetry is what makes the emitter inject far more than the base does back into it, and that asymmetry is why the device is not symmetrical.

Discuss in Forum

Which junction biases distinguish the active region?

A.Both junctions reverse biased
B.Emitter reverse, collector forward
C.Emitter forward, collector reverse
D.Both junctions forward biased

Answer: C

The active region needs the emitter junction forward biased to inject carriers and the collector junction reverse biased to collect them. Both forward is saturation, both reverse is cut-off, and the reversed arrangement is the inverse active region, which works but with a much poorer gain because the doping is wrong for it. Naming the four regions by their two junction biases is the compact way to hold them, and it makes the bias conditions in a circuit immediately readable.

Discuss in Forum

What bias condition defines saturation?

A.Emitter forward and collector reverse biased
B.Both junctions reverse biased
C.Emitter reverse and collector forward biased
D.Both junctions forward biased

Answer: D

In saturation both junctions are forward biased, the collector can no longer take all the carriers offered, and the collector-emitter voltage falls to a few tenths of a volt. The collector current then depends on the external circuit rather than on the base current, which is precisely what a switching application wants. Recognising saturation from the junction biases rather than from a current ratio is more reliable, because the ratio is what breaks down there.

Discuss in Forum

With no base drive at all, only leakage flows. Which region is the device in?

A.Both junctions reverse biased
B.Both junctions forward biased
C.Emitter forward and collector reverse biased
D.The base current exactly equal to the collector current

Answer: A

Cut-off has both junctions reverse biased, so only leakage flows and the device is effectively open. Both forward is saturation and the mixed case is the active region. Equal base and collector currents would mean a beta of one, which describes no useful region at all. Since a switch spends its life in cut-off or saturation, these two regions matter far more in digital work than the active region between them.

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What does the current gain alpha describe?

A.Base current divided by collector current
B.Collector current divided by emitter current
C.Collector current divided by base current
D.Emitter current divided by base current

Answer: B

Alpha is the ratio of collector to emitter current, so it is a little less than unity - typically 0.98 to 0.998. Collector over base is beta, and emitter over base is gamma, one more than beta. Because alpha is so close to one, small changes in it produce large changes in beta, which is why beta varies so widely between nominally identical devices and why no good design depends on its exact value.

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A transistor has \( \alpha = 0.98 \). Its beta is

A.0.98
B.50
C.49
D.98

Answer: C

Converting between the two gains:

\[ \beta = \frac{\alpha}{1-\alpha} = \frac{0.98}{0.02} = 49 \]

so 49. Ninety-eight mistakes alpha's decimals for the answer, and fifty would follow from an alpha of 0.9804. The steepness of this relation is the important part: raising alpha from 0.98 to 0.99 doubles beta to 99, which is why manufacturing spreads in alpha appear as enormous spreads in beta.

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A transistor has \( \beta = 99 \). Its alpha is

A.0.98
B.0.90
C.1.01
D.0.99

Answer: D

Inverting the relation:

\[ \alpha = \frac{\beta}{1+\beta} = \frac{99}{100} = 0.99 \]

giving 0.99. The value 1.01 is impossible, since alpha must be less than one for a device that cannot deliver more collector current than emitter current. Checking that a computed alpha lies below unity is a quick sanity test, and it catches the commonest algebraic slip in this pair of formulae.

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10.

A transistor carries \( I_{B} = 20 \ \mathrm{\mu A} \) with \( \beta = 100 \). Its collector and emitter currents are

A.2 mA and 2.02 mA
B.2 mA and 1.98 mA
C.20 mA and 20.2 mA
D.0.2 mA and 0.22 mA

Answer: A

Applying beta and then Kirchhoff's current law at the device:

\[ \begin{aligned} I_{C} &= 100 \times 20\mathrm{\mu} = 2 \ \mathrm{mA} \\ I_{E} &= 2 + 0.02 = 2.02 \ \mathrm{mA} \end{aligned} \]

so 2 mA and 2.02 mA. The emitter current is the largest of the three, since it supplies both the others - a check worth making, because an answer with the emitter current smallest has a sign error somewhere.

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Bipolar Junction Transistors Questions and Answers

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