Bipolar Junction Transistors Questions and Answers

Q: In the following circuit, if RL = RC = 10 kΩ, then the value of Vo will be :

The output is 2·5 V.With \( V_{in} = 0 \) there is no forward bias on the base-emitter junction, so the transistor is cut off and draws no collector current. \( R_{C} \) and \( R_{L} \) then form a simple voltage divider across the 5 V supply:\[ \begin{aligned} V_{o} &= 5 \times \frac{R_{L}}{R_{C} + R_{L}} \\ &= 5 \times \frac{10}{20} = 2.5\ \text{V} \end{aligned} \]Zero would be the output of a saturated transistor, which needs base drive. The full 5 V would appear only with no load connected, and 4·55 V or 1·0 V do not follow from equal 10 kΩ resistors.

Q: Ebers-Moll model of BJT has two coupled back to back diodes in parallel with controlled current sources.Which of the following is/are correct for switching applications ?(1) If two diodes are both in forward bias, then the BJT is in a low impedance state.(2) If two diodes are both in reverse bias, then BJT is in a high impedance state.(3) If two diodes are both in forward bias, then the BJT is in high impedance state.(4) If two diodes are both in reverse bias, then BJT is in low impedance state.

The correct statements are 1 and 2.In the Ebers-Moll model the BJT is two back-to-back diodes, base-emitter and base-collector, with controlled current sources. When both junctions are forward biased the transistor is in saturation: it conducts heavily with only a small collector-emitter voltage, a low impedance state, like a closed switch (1). When both are reverse biased it is in cut-off: almost no current flows, a high impedance state, like an open switch (2).Statements 3 and 4 swap these states, so every option containing either of them is wrong.

Q: Determine the values of current amplification factor (\(\beta\)) of two different transistors having \(\alpha_{1} = 0.99\) and \(\alpha_{2} = 0.98\) :

The current amplification factors are 99 and 49 respectively.\[ \beta = \frac{\alpha}{1 - \alpha} \]\[ \begin{aligned} \beta_{1} &= \frac{0.99}{0.01} = 99 \\ \beta_{2} &= \frac{0.98}{0.02} = 49 \end{aligned} \]A small fall in alpha from 0.99 to 0.98 halves beta, because \(1 - \alpha\) doubles. '99 and 98' wrongly scales beta in proportion to alpha, and the options with 99 second reverse the two transistors.

Q: A Si n-p-n bipolar transistor has the following parameters collector current Ic = 6 mA, common emitter current gain factor hFE = 120, operational temperature T = 300°K. Determine the mutual conductance of small signal transistor.

The mutual conductance is about 0.23 mho.For a bipolar transistor the small-signal transconductance depends only on the collector current and the thermal voltage:\[ \begin{aligned} V_{T} &= \frac{kT}{q} \approx 25.9\ \text{mV at } 300\ \text{K} \\ g_{m} &= \frac{I_{C}}{V_{T}} = \frac{6 \times 10^{-3}}{25.9 \times 10^{-3}} \\ &\approx 0.23\ \text{S (mho)} \end{aligned} \]The current gain hFE = 120 is not needed for gm; it would be used to find rπ = β/gm, about 520 Ω. 0.02 mho is ten times too small, and 2.3 or 4.33 mho would need currents of tens of milliamperes.

Q: For effective working as an amplifier Base-emitter and Base-collector junctions of a BJT should be biased as :

For amplification the junctions are biased forward and reverse respectively.In the active region the base-emitter junction is forward biased, so the emitter injects carriers into the thin base, and the base-collector junction is reverse biased, so its field sweeps those carriers into the collector. The collector current is then controlled by the base current and nearly independent of the collector voltage, which is what makes the transistor an amplifier.Both junctions forward biased is saturation, both reverse biased is cut-off (the two states used in switching), and reverse-forward is the inverse active mode, which gives very poor gain.

Q: What does the hFE value represent in the common-emitter configuration?

\(h_{FE}\) is the forward current transfer ratio.In h-parameter notation the first subscript names the parameter (i input impedance, r reverse voltage ratio, f forward current ratio, o output admittance) and the second the configuration (e common emitter). Capital subscripts mean the DC value, so\[ h_{FE} = \frac{I_{C}}{I_{B}} = \beta_{dc} \]the DC current gain of the transistor in common emitter.The reverse transfer parameter is \(h_{re}\), a voltage ratio rather than a current ratio. Output behaviour is described by \(h_{oe}\), an admittance. Voltage gain is a circuit quantity that depends on the load, not an h-parameter of the transistor.

Q: In any circuit connections, what is the biasing direction for the emitter and collector of a transistor ?

For a transistor working as an amplifier, the emitter is forward biased and the collector is reverse biased.This is the active region. Forward bias on the emitter-base junction injects carriers from the emitter into the thin base; reverse bias on the collector-base junction sets up a field that sweeps almost all of them into the collector, giving \(I_{C} \approx \beta I_{B}\). This holds whatever the configuration, CE, CB or CC.Both junctions forward biased is saturation, where the transistor acts as a closed switch. Both reverse biased is cut-off, an open switch. Emitter reverse and collector forward is the inverse-active mode, which works but with very low gain and is not used for normal amplification.

Q: Which of the following transistor amplifier configuration is often used in impedance-matching application ?

The configuration is the emitter follower.The emitter follower (common-collector amplifier) has a high input impedance and a low output impedance, with voltage gain close to 1. It can therefore connect a high-impedance source to a low-impedance load without loading the source, which is the definition of an impedance-matching buffer.The common-base stage has a low input impedance; the common-emitter stage is used for voltage gain, with moderate impedances; the common-source FET amplifier is the analogue of common-emitter, not of the follower.

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.

11.

In the following circuit, if R_L = R_C = 10 kΩ, then the value of V_o will be :

Transistor Q1 with its base fed from Vin = 0 through RB, emitter grounded, collector joined to VCC = +5 V through RC, and the output Vo taken at the collector, which is loaded by RL to ground

A.4·55 V
B.2·5 V
C.1·0 V
D.Zero

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The output is 2·5 V.

With \( V_{in} = 0 \) there is no forward bias on the base-emitter junction, so the transistor is cut off and draws no collector current. \( R_{C} \) and \( R_{L} \) then form a simple voltage divider across the 5 V supply:

\[ \begin{aligned} V_{o} &= 5 \times \frac{R_{L}}{R_{C} + R_{L}} \\ &= 5 \times \frac{10}{20} = 2.5\ \text{V} \end{aligned} \]

Zero would be the output of a saturated transistor, which needs base drive. The full 5 V would appear only with no load connected, and 4·55 V or 1·0 V do not follow from equal 10 kΩ resistors.

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12.

Ebers-Moll model of BJT has two coupled back to back diodes in parallel with controlled current sources.

Which of the following is/are correct for switching applications ?

(1) If two diodes are both in forward bias, then the BJT is in a low impedance state.
(2) If two diodes are both in reverse bias, then BJT is in a high impedance state.
(3) If two diodes are both in forward bias, then the BJT is in high impedance state.
(4) If two diodes are both in reverse bias, then BJT is in low impedance state.

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

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The correct statements are 1 and 2.

In the Ebers-Moll model the BJT is two back-to-back diodes, base-emitter and base-collector, with controlled current sources. When both junctions are forward biased the transistor is in saturation: it conducts heavily with only a small collector-emitter voltage, a low impedance state, like a closed switch (1). When both are reverse biased it is in cut-off: almost no current flows, a high impedance state, like an open switch (2).

Statements 3 and 4 swap these states, so every option containing either of them is wrong.

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13.

Determine the values of current amplification factor (\(\beta\)) of two different transistors having \(\alpha_{1} = 0.99\) and \(\alpha_{2} = 0.98\) :

A.99 and 98 respectively
B.49 and 99 respectively
C.99 and 49 respectively
D.98 and 99 respectively

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The current amplification factors are 99 and 49 respectively.

\[ \beta = \frac{\alpha}{1 - \alpha} \]

\[ \begin{aligned} \beta_{1} &= \frac{0.99}{0.01} = 99 \\ \beta_{2} &= \frac{0.98}{0.02} = 49 \end{aligned} \]

A small fall in alpha from 0.99 to 0.98 halves beta, because \(1 - \alpha\) doubles. '99 and 98' wrongly scales beta in proportion to alpha, and the options with 99 second reverse the two transistors.

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14.

A Si n-p-n bipolar transistor has the following parameters collector current I_c = 6 mA, common emitter current gain factor h_FE = 120, operational temperature T = 300°K. Determine the mutual conductance of small signal transistor.

A.0.23 mho
B.0.02 mho
C.2.3 mho
D.4.33 mho

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The mutual conductance is about 0.23 mho.

For a bipolar transistor the small-signal transconductance depends only on the collector current and the thermal voltage:

\[ \begin{aligned} V_{T} &= \frac{kT}{q} \approx 25.9\ \text{mV at } 300\ \text{K} \\ g_{m} &= \frac{I_{C}}{V_{T}} = \frac{6 \times 10^{-3}}{25.9 \times 10^{-3}} \\ &\approx 0.23\ \text{S (mho)} \end{aligned} \]

The current gain h_FE = 120 is not needed for g_m; it would be used to find r_π = β/g_m, about 520 Ω. 0.02 mho is ten times too small, and 2.3 or 4.33 mho would need currents of tens of milliamperes.

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15.

For effective working as an amplifier Base-emitter and Base-collector junctions of a BJT should be biased as :

A.Forward and forward respectively
B.Reverse and forward respectively
C.Reverse and reverse respectively
D.Forward and reverse respectively

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

For amplification the junctions are biased forward and reverse respectively.

In the active region the base-emitter junction is forward biased, so the emitter injects carriers into the thin base, and the base-collector junction is reverse biased, so its field sweeps those carriers into the collector. The collector current is then controlled by the base current and nearly independent of the collector voltage, which is what makes the transistor an amplifier.

Both junctions forward biased is saturation, both reverse biased is cut-off (the two states used in switching), and reverse-forward is the inverse active mode, which gives very poor gain.

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16.

What does the h_FE value represent in the common-emitter configuration?

A.Reverse current transfer ratio
B.Output impedance
C.Voltage gain
D.Forward current transfer ratio

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

\(h_{FE}\) is the forward current transfer ratio.

In h-parameter notation the first subscript names the parameter (i input impedance, r reverse voltage ratio, f forward current ratio, o output admittance) and the second the configuration (e common emitter). Capital subscripts mean the DC value, so

\[ h_{FE} = \frac{I_{C}}{I_{B}} = \beta_{dc} \]

the DC current gain of the transistor in common emitter.

The reverse transfer parameter is \(h_{re}\), a voltage ratio rather than a current ratio. Output behaviour is described by \(h_{oe}\), an admittance. Voltage gain is a circuit quantity that depends on the load, not an h-parameter of the transistor.

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17.

In any circuit connections, what is the biasing direction for the emitter and collector of a transistor ?

A.Both are forward biased
B.Both are reverse biased
C.Emitter is forward, and collector is reverse biased
D.Emitter is reverse, and collector is forward biased

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

For a transistor working as an amplifier, the emitter is forward biased and the collector is reverse biased.

This is the active region. Forward bias on the emitter-base junction injects carriers from the emitter into the thin base; reverse bias on the collector-base junction sets up a field that sweeps almost all of them into the collector, giving \(I_{C} \approx \beta I_{B}\). This holds whatever the configuration, CE, CB or CC.

Both junctions forward biased is saturation, where the transistor acts as a closed switch. Both reverse biased is cut-off, an open switch. Emitter reverse and collector forward is the inverse-active mode, which works but with very low gain and is not used for normal amplification.

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18.

Which of the following transistor amplifier configuration is often used in impedance-matching application ?

A.The common-base amplifier
B.The emitter follower
C.The common-emitter amplifier
D.The common-source amplifier

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The configuration is the emitter follower.

The emitter follower (common-collector amplifier) has a high input impedance and a low output impedance, with voltage gain close to 1. It can therefore connect a high-impedance source to a low-impedance load without loading the source, which is the definition of an impedance-matching buffer.

The common-base stage has a low input impedance; the common-emitter stage is used for voltage gain, with moderate impedances; the common-source FET amplifier is the analogue of common-emitter, not of the follower.

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19.

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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20.

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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