BJT Biasing and Stability Questions and Answers

Q: Why must a transistor amplifier be biased?

Bias establishes a quiescent point so that the whole signal excursion stays inside the active region, with neither peak driven into cut-off or saturation. Limiting current, matching impedance and protecting the junction are all real concerns addressed by other parts of the design. Since a transistor conducts in one direction only, an alternating signal needs a standing current to ride on, and providing that is the whole purpose of a bias network.

Q: Where on the load line does the quiescent point usually sit for a linear amplifier?

The midpoint gives the largest symmetrical swing before either extreme is reached, which is what a linear stage wants. A point near saturation clips the negative peaks first, and one near cut-off clips the positive peaks; both are used deliberately in class B and class C stages but not in a class A amplifier. Choosing an operating point is always a matter of which excursion the design has to accommodate. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Q: What does the direct-current load line represent?

The load line is the locus of collector current and collector-emitter voltage that the supply and the resistances allow, drawn on the output characteristics. Where it crosses the curve for the actual base current is the operating point. Frequency response, input range and the input characteristic are all plotted elsewhere. Superimposing what the circuit permits on what the device can do is the graphical method's whole idea, and the intersection is the answer.

Q: What are the two intercepts of the direct-current load line?

Setting the current to zero gives the supply voltage on the voltage axis, and setting the voltage to zero gives the supply divided by the total series resistance on the current axis. Those two points fix the line. The quiescent point lies on it rather than defining it. Finding a straight line from its two intercepts is quicker than plotting it point by point, and both intercepts here follow from setting one variable to zero. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Q: With a base current giving \( I_{C} = 2 \ \mathrm{mA} \), that stage's collector-emitter voltage is

Subtracting both resistive drops from the supply:\[ \begin{aligned} V_{CE} &= 15 - 2\mathrm{m}(3000+1000) \\ &= 15 - 8 = 7 \ \mathrm{V} \end{aligned} \]so 7 volts, close enough to mid-supply for a good swing. Nine volts counts only the collector resistor. Because the same current flows in both resistors, they can be added before multiplying, which halves the arithmetic and the chances of an error.

Q: What is a stability factor?

The stability factor measures how much the collector current moves for a given change in leakage current, so a small value means a stable bias. It can be defined with respect to beta or to the base-emitter voltage as well, each isolating one drifting parameter. The current ratio is beta, and the other options describe unrelated quantities. A figure defined as a sensitivity is the natural way to compare bias arrangements, since it is the sensitivity rather than the nominal point that distinguishes them.

Q: What value of stability factor is ideal?

Unity is the best achievable: the collector current then changes only by the leakage change itself, with no multiplication. Zero is impossible, since the leakage current is part of the collector current by definition. A factor equal to one plus beta is the worst case and is exactly what fixed bias delivers. Knowing both the ideal and the worst case gives a scale against which any arrangement can be placed. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Direction for [ Question No: 1 To 1 ] :

Directions : In the following questions, match List - I and List - II and select the correct answer using the codes given below the lists :

List - I	List - II
a. Stability factor	i. Large bandwidth
b. Loop gain \(\ge\) 1	ii. Voltage divider bias
c. Lock range	iii. Oscillators
d. Negative feedback	iv. PLL

Codes : a b c d

A.ii iv iii i
B.ii iii iv i
C.iii iv ii i
D.i ii iii iv

Exam Year(s)

UGC NET - 2012,

Answer: B

The matching is: a stability factor with voltage divider bias (ii), b loop gain of at least 1 with oscillators (iii), c lock range with PLL (iv), d negative feedback with large bandwidth (i), the code ii iii iv i (option B).

A voltage divider bias gives a low stability factor, a loop gain of unity or more is the Barkhausen condition for oscillation, the lock range is a property of a phase-locked loop, and negative feedback widens the bandwidth. Options A, C and D mix these pairs.

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Why must a transistor amplifier be biased?

A.To keep the device in the active region throughout the signal
B.To limit the current the device draws from the supply
C.To match the input impedance to the driving source
D.To prevent the base-emitter junction from breaking down

Answer: A

Bias establishes a quiescent point so that the whole signal excursion stays inside the active region, with neither peak driven into cut-off or saturation. Limiting current, matching impedance and protecting the junction are all real concerns addressed by other parts of the design. Since a transistor conducts in one direction only, an alternating signal needs a standing current to ride on, and providing that is the whole purpose of a bias network.

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Where on the load line does the quiescent point usually sit for a linear amplifier?

A.At either end, whichever suits the load
B.Near the middle of the line
C.Close to the saturation end
D.Close to the cut-off end

Answer: B

The midpoint gives the largest symmetrical swing before either extreme is reached, which is what a linear stage wants. A point near saturation clips the negative peaks first, and one near cut-off clips the positive peaks; both are used deliberately in class B and class C stages but not in a class A amplifier. Choosing an operating point is always a matter of which excursion the design has to accommodate. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

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What does the direct-current load line represent?

A.The signal excursion the input can accommodate
B.The relation between base current and base voltage
C.The collector current and voltage the bias circuit permits
D.The gain of the stage plotted against frequency

Answer: C

The load line is the locus of collector current and collector-emitter voltage that the supply and the resistances allow, drawn on the output characteristics. Where it crosses the curve for the actual base current is the operating point. Frequency response, input range and the input characteristic are all plotted elsewhere. Superimposing what the circuit permits on what the device can do is the graphical method's whole idea, and the intersection is the answer.

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What are the two intercepts of the direct-current load line?

A.The base current and the collector current at saturation
B.The saturation voltage and the maximum rated current
C.The quiescent voltage and the quiescent current
D.The supply voltage and the supply divided by the total resistance

Answer: D

Setting the current to zero gives the supply voltage on the voltage axis, and setting the voltage to zero gives the supply divided by the total series resistance on the current axis. Those two points fix the line. The quiescent point lies on it rather than defining it. Finding a straight line from its two intercepts is quicker than plotting it point by point, and both intercepts here follow from setting one variable to zero. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

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How does the alternating load line differ from the direct one?

A.It reflects the load seen only by the signal
B.It has the same slope but a different intercept
C.It is always steeper because the supply is higher
D.It passes through the origin rather than the supply

Answer: A

The signal sees any coupled load in parallel with the collector resistor and any bypassed emitter resistor as a short, so its load line is steeper and pivots about the quiescent point. Same slope with a shifted intercept describes a change of supply. It does not pass through the origin. Two load lines through one operating point, one for direct conditions and one for the signal, is the standard picture, and the swing available is read from the steeper one.

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A stage has \( V_{CC} = 15 \ \mathrm{V} \), \( R_{C} = 3 \ \mathrm{k\Omega} \) and \( R_{E} = 1 \ \mathrm{k\Omega} \). The saturation current is

A.1.25 mA
B.3.75 mA
C.5 mA
D.15 mA

Answer: B

At saturation the whole supply appears across the two resistors:

\[ I_{C(sat)} = \frac{15}{3000+1000} = 3.75 \ \mathrm{mA} \]

giving 3.75 mA. Five milliamps uses the collector resistor alone and forgets the emitter, which is the commonest slip. Including every resistance in the collector-emitter path is what makes the intercept come out, and it is worth listing them explicitly before dividing.

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With a base current giving \( I_{C} = 2 \ \mathrm{mA} \), that stage's collector-emitter voltage is

A.15 V
B.6 V
C.7 V
D.9 V

Answer: C

Subtracting both resistive drops from the supply:

\[ \begin{aligned} V_{CE} &= 15 - 2\mathrm{m}(3000+1000) \\ &= 15 - 8 = 7 \ \mathrm{V} \end{aligned} \]

so 7 volts, close enough to mid-supply for a good swing. Nine volts counts only the collector resistor. Because the same current flows in both resistors, they can be added before multiplying, which halves the arithmetic and the chances of an error.

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What is a stability factor?

A.The ratio of collector current to base current at bias
B.The change in gain per degree of temperature rise
C.The fraction of the supply appearing across the device
D.The change in collector current per unit change in leakage

Answer: D

The stability factor measures how much the collector current moves for a given change in leakage current, so a small value means a stable bias. It can be defined with respect to beta or to the base-emitter voltage as well, each isolating one drifting parameter. The current ratio is beta, and the other options describe unrelated quantities. A figure defined as a sensitivity is the natural way to compare bias arrangements, since it is the sensitivity rather than the nominal point that distinguishes them.

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10.

What value of stability factor is ideal?

A.One
B.Zero
C.Equal to beta
D.Equal to one plus beta

Answer: A

Unity is the best achievable: the collector current then changes only by the leakage change itself, with no multiplication. Zero is impossible, since the leakage current is part of the collector current by definition. A factor equal to one plus beta is the worst case and is exactly what fixed bias delivers. Knowing both the ideal and the worst case gives a scale against which any arrangement can be placed. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Discuss in Forum

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BJT Biasing and Stability Questions and Answers

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