Bipolar Junction Transistors Questions and Answers

Q: What information does the measurement of dc beta provide about a bipolar junction transistor ?

The dc beta tells about the current gain characteristics of the transistor.The dc beta, \(\beta_{dc} = h_{FE} = I_{C}/I_{B}\), is the ratio of dc collector current to dc base current in the common-emitter configuration. It shows how much the transistor amplifies current and how that gain varies with collector current and temperature, which is needed in biasing and switching design.The voltage drop across the collector resistor depends on the circuit. The gain-bandwidth product (fT) is a high-frequency property found from ac measurements. The base-emitter threshold voltage, about 0.6-0.7 V for silicon, is a junction property, not a current gain.

Q: The phase difference between the input and output voltages in a common base arrangement is .........

The phase difference in a common base amplifier is 0°.In the common base configuration the input is applied to the emitter and the output taken from the collector, with the base common. When the emitter voltage rises (for an npn transistor), the emitter-base forward bias falls, collector current falls, the drop across the collector resistor falls, and the collector voltage rises. Input and output therefore move together, in phase.A phase shift of 180° belongs to the common emitter amplifier, where a rise in base voltage raises collector current and lowers the collector voltage. 90° and 270° are not the phase relations of any of the three basic configurations at mid-band frequencies.

121.

What information does the measurement of dc beta provide about a bipolar junction transistor ?

A.Voltage drop across the collector resistor
B.Gain bandwidth product
C.Current gain characteristics
D.Base-emitter voltage threshold

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The dc beta tells about the current gain characteristics of the transistor.

The dc beta, \(\beta_{dc} = h_{FE} = I_{C}/I_{B}\), is the ratio of dc collector current to dc base current in the common-emitter configuration. It shows how much the transistor amplifies current and how that gain varies with collector current and temperature, which is needed in biasing and switching design.

The voltage drop across the collector resistor depends on the circuit. The gain-bandwidth product (f_T) is a high-frequency property found from ac measurements. The base-emitter threshold voltage, about 0.6-0.7 V for silicon, is a junction property, not a current gain.

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122.

The phase difference between the input and output voltages in a common base arrangement is .........

A.180°
B.90°
C.270°
D.0°

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The phase difference in a common base amplifier is 0°.

In the common base configuration the input is applied to the emitter and the output taken from the collector, with the base common. When the emitter voltage rises (for an npn transistor), the emitter-base forward bias falls, collector current falls, the drop across the collector resistor falls, and the collector voltage rises. Input and output therefore move together, in phase.

A phase shift of 180° belongs to the common emitter amplifier, where a rise in base voltage raises collector current and lowers the collector voltage. 90° and 270° are not the phase relations of any of the three basic configurations at mid-band frequencies.

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123.

Arrange the following configurations of BJT in the increasing order of their input impedance.

I. CE II. CB
III. CC IV. Darlington pair

A.III, II, IV, I
B.I, II, III, IV
C.IV, III, II, I
D.II, I, III, IV

Exam Year(s)

UGC NET - 2012,

Answer: D

The input impedance of the common base configuration is the lowest, the common emitter is higher, the common collector is higher still, and the Darlington pair, which is two transistors in cascade, gives the highest input impedance of all, so the increasing order is CB, CE, CC and Darlington, which is II, I, III, IV. In figures, the common base stage has an input impedance of a few tens of ohms, the common emitter about a kilo-ohm or so, and the common collector and the Darlington pair hundreds of kilo-ohms or more. The other sequences place the common collector or the Darlington pair before the common base. The answer is II, I, III, IV.

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124.

In an npn transistor, the expression for avalanche multiplication factor is given by

Where, \(V_{CB}\) is collector to base voltage and \(BV_{CBO}\) is maximum reverse biasing voltage which may be applied before breakdown between the collector and base terminals.

A.\(M = \dfrac{1}{(1 + V_{CB}/BV_{CBO})^{n}}\)
B.\(M = (1 + V_{CB}/BV_{CBO})^{n}\)
C.\(M = (1 - V_{CB}/BV_{CBO})^{n}\)
D.\(M = \dfrac{1}{[1 - (V_{CB}/BV_{CBO})^{n}]}\)

Exam Year(s)

UGC NET - 2013,

Answer: D

The avalanche multiplication factor of an npn transistor is given by Miller's empirical formula \(M = \dfrac{1}{1 - (V_{CB}/BV_{CBO})^{n}}\), the fourth option. As \(V_{CB}\) approaches the breakdown voltage \(BV_{CBO}\), the denominator tends to zero and M grows without limit, which is the onset of avalanche breakdown.

The first option has a plus sign in the denominator, so M would fall below 1 and never rise. The second and third options give M as a power of a bracket and do not become infinite at \(V_{CB} = BV_{CBO}\), so they cannot represent breakdown.

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125.

Consider the following four common type of transistors :

1. Point Contact Transistor
2. Bipolar Junction Transistor
3. MOS Field Effect Transistor
4. Junction Field Effect Transistor Correct arrangement of these transistors in the increasing order of input impedance is

A.1, 2, 4, 3
B.1, 2, 3, 4
C.2, 1, 3, 4
D.2, 1, 4, 3

Exam Year(s)

UGC NET - 2013,

Answer: D

The key gives the order 2, 1, 4, 3, the fourth option. The BJT has the lowest input impedance, since its base-emitter junction is forward biased, then the point contact transistor, then the junction field effect transistor with a reverse biased gate junction, and the MOSFET has the highest because its gate is insulated by an oxide layer.

The other options place the MOSFET or the JFET lower in the sequence, which does not fit the key.

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