Small Signal Amplifiers Questions and Answers

Q: Why does degeneration reduce distortion?

Series feedback makes the stage's gain depend on a resistor ratio rather than on the device's curved characteristic, so the non-linearity is divided down along with the gain. The amplitude at the base is actually reduced as a consequence, which is part of the same effect rather than a separate one. It does not change the current or filter anything. Feedback trades gain for every kind of accuracy at once, and linearity is one of the benefits.

Q: Which noise source dominates in a well-designed bipolar input stage?

At useful frequencies shot noise in the junction currents sets the floor, with thermal noise in the base spreading resistance contributing as well. Flicker noise matters only at low frequencies, divider thermal noise can be made small by design, and quantisation noise belongs to digital coding entirely. Identifying which mechanism dominates decides where design effort belongs, and in a bipolar front end it is the operating current. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Q: Why does the choice of operating current involve a noise trade-off?

Raising the current increases the shot noise but lowers the equivalent input noise resistance, so the optimum depends on the source impedance the stage works from. Neither extreme is universally better. Noise is certainly current-dependent. A parameter with an optimum rather than a direction is one where the source impedance has to be known before the design can be settled, which is why low-noise design begins with the source. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Q: What does an amplifier's equivalent input noise voltage let you do?

Referring the noise to the input removes the gain from the comparison, so two amplifiers of different gain can be judged on the same footing. Distortion, bias and bandwidth are all specified separately. Referring an imperfection to the input is a general technique, and it works because everything at the input is multiplied by the same gain as the signal.

Q: A stage has a gain of \( 100 \) and an output noise of \( 1 \ \mathrm{mV} \). Its equivalent input noise is

Referring the output figure back through the gain:\[ v_{n} = \frac{1 \ \mathrm{mV}}{100} = 10 \ \mathrm{\mu V} \]giving 10 microvolts. Multiplying rather than dividing gives 100 millivolts, which would be an absurdly noisy amplifier. Referring a quantity to the input always divides by the gain, and checking the answer for plausibility catches the direction error at once.

Q: A signal of \( 5 \ \mathrm{mV} \) is applied to a stage of gain \( -180 \). The output is

Multiplying amplitude by gain and noting the sign:\[ v_{o} = -180 \times 5 \ \mathrm{mV} = -0.9 \ \mathrm{V} \]so 0.9 volts, inverted. The magnitude alone is not the whole answer in a common-emitter stage, where the inversion matters as soon as the stage is placed in a feedback loop or cascaded. Carrying the sign through rather than taking magnitudes is a habit worth keeping.

Q: What happens to the analysis if the input is raised to fifty millivolts?

At twice the thermal voltage the exponential curvature is plainly significant and the output acquires substantial harmonic content, so the small-signal model no longer predicts the waveform. The gain does not rise usefully, the capacitors are unaffected by amplitude, and the input impedance change is a second-order effect. Knowing where a model stops being valid is more useful than knowing the model, because that boundary is where real circuits get designed.

71.

Why does degeneration reduce distortion?

A.It lowers the operating current and so the non-linearity
B.It removes the harmonics by filtering them at the emitter
C.It linearises the transfer characteristic through feedback
D.It reduces the signal amplitude reaching the device's base

Answer: C

Series feedback makes the stage's gain depend on a resistor ratio rather than on the device's curved characteristic, so the non-linearity is divided down along with the gain. The amplitude at the base is actually reduced as a consequence, which is part of the same effect rather than a separate one. It does not change the current or filter anything. Feedback trades gain for every kind of accuracy at once, and linearity is one of the benefits.

Discuss in Forum

72.

Which noise source dominates in a well-designed bipolar input stage?

A.Flicker noise in the collector load resistor
B.Thermal noise in the bias divider resistors
C.Quantisation noise arising in the coupling capacitors
D.Shot noise in the base and collector currents

Answer: D

At useful frequencies shot noise in the junction currents sets the floor, with thermal noise in the base spreading resistance contributing as well. Flicker noise matters only at low frequencies, divider thermal noise can be made small by design, and quantisation noise belongs to digital coding entirely. Identifying which mechanism dominates decides where design effort belongs, and in a bipolar front end it is the operating current. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Discuss in Forum

73.

Why does the choice of operating current involve a noise trade-off?

A.Shot noise and base resistance noise move oppositely with current
B.Higher current always produces proportionally more noise
C.Lower current always reduces the noise the stage generates
D.Noise is independent of current once the bias is stable

Answer: A

Raising the current increases the shot noise but lowers the equivalent input noise resistance, so the optimum depends on the source impedance the stage works from. Neither extreme is universally better. Noise is certainly current-dependent. A parameter with an optimum rather than a direction is one where the source impedance has to be known before the design can be settled, which is why low-noise design begins with the source. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Discuss in Forum

74.

What does an amplifier's equivalent input noise voltage let you do?

A.Determine the highest frequency the stage will pass
B.Compare stages independently of their gain
C.Predict the distortion at a given output amplitude
D.Compute the bias current needed for a stated gain

Answer: B

Referring the noise to the input removes the gain from the comparison, so two amplifiers of different gain can be judged on the same footing. Distortion, bias and bandwidth are all specified separately. Referring an imperfection to the input is a general technique, and it works because everything at the input is multiplied by the same gain as the signal.

Discuss in Forum

75.

Why is a stage's noise figure worse when driven from a very low impedance?

A.The shot noise in the base current rises with source conductance
B.The coupling capacitor's reactance becomes comparable to the source
C.The amplifier's own noise voltage dominates the source's
D.The stage's input impedance loads the source too heavily

Answer: C

A low source resistance generates little thermal noise of its own, so the amplifier's fixed input noise voltage becomes a large fraction of the total and the ratio degrades. Loading and capacitor reactance affect the signal rather than the noise ratio. Noise figure is always a comparison against what the source itself contributes, which is why the same amplifier can look excellent on one source and poor on another. Framing the question as which quantity the change acts upon makes the answer follow without calculation. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Discuss in Forum

76.

A stage has a gain of \( 100 \) and an output noise of \( 1 \ \mathrm{mV} \). Its equivalent input noise is

A.1 millivolt
B.100 millivolts
C.10 millivolts
D.10 microvolts

Answer: D

Referring the output figure back through the gain:

\[ v_{n} = \frac{1 \ \mathrm{mV}}{100} = 10 \ \mathrm{\mu V} \]

giving 10 microvolts. Multiplying rather than dividing gives 100 millivolts, which would be an absurdly noisy amplifier. Referring a quantity to the input always divides by the gain, and checking the answer for plausibility catches the direction error at once.

Discuss in Forum

77.

A signal of \( 5 \ \mathrm{mV} \) is applied to a stage of gain \( -180 \). The output is

A.0.9 V in antiphase
B.0.9 V in phase
C.36 V
D.27.8 mV

Answer: A

Multiplying amplitude by gain and noting the sign:

\[ v_{o} = -180 \times 5 \ \mathrm{mV} = -0.9 \ \mathrm{V} \]

so 0.9 volts, inverted. The magnitude alone is not the whole answer in a common-emitter stage, where the inversion matters as soon as the stage is placed in a feedback loop or cascaded. Carrying the sign through rather than taking magnitudes is a habit worth keeping.

Discuss in Forum

78.

Would a five millivolt input still count as a small signal at the base?

A.Only if the emitter resistor is left unbypassed
B.Yes, since it is well below the thermal voltage
C.No, because it exceeds one millivolt at the input
D.No, because the resulting output approaches a volt

Answer: B

Five millivolts is a fifth of the twenty-six millivolt thermal voltage, so the exponential is well approximated by its tangent and the small-signal analysis holds. The output amplitude is irrelevant to the question, which concerns the excursion at the device. A one-millivolt threshold is arbitrary. Judging smallness against the thermal voltage rather than against a fixed figure is what makes the criterion meaningful. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Discuss in Forum

79.

What happens to the analysis if the input is raised to fifty millivolts?

A.The coupling capacitors begin to limit the response
B.The input impedance falls by roughly the same factor
C.Harmonic distortion becomes significant
D.The gain rises because the device is driven harder

Answer: C

At twice the thermal voltage the exponential curvature is plainly significant and the output acquires substantial harmonic content, so the small-signal model no longer predicts the waveform. The gain does not rise usefully, the capacitors are unaffected by amplitude, and the input impedance change is a second-order effect. Knowing where a model stops being valid is more useful than knowing the model, because that boundary is where real circuits get designed.

Discuss in Forum

80.

Which specification would you improve by raising the operating current?

A.The stage's tolerance of a low supply voltage
B.The size of the emitter bypass capacitor required
C.The input impedance the stage presents to its source
D.The gain available from a given collector resistor

Answer: D

Higher current means a smaller intrinsic emitter resistance and so a larger gain from the same collector resistor. It uses more of the supply in the collector resistor's drop, demands a larger bypass capacitor because the emitter resistance is lower, and reduces the input impedance. One change improving one specification while worsening three is the normal case, and the design question is only which of the four matters most. Framing the question as which quantity the change acts upon makes the answer follow without calculation. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Small Signal Amplifiers Questions and Answers || Small Signal Amplifiers MCQ Questions and Answers || Small Signal Amplifiers GK Questions and Answers || Small Signal Amplifiers GK MCQ Questions || Small Signal Amplifiers Multiple Choice Questions and Answers || Small Signal Amplifiers GK || GK on Small Signal Amplifiers || Analog Electronics Questions and Answers || Analog Electronics MCQ Questions and Answers || Analog Electronics GK Questions and Answers || GK on Analog Electronics

Small Signal Amplifiers Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications