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ISRO Scientist/Engineer SC (Electronics) practice questions

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Question 1Electronic Devices and Circuits

In a BJT biased in the active region, the collector current is essentially determined by which physical mechanism?

  1. Diffusion of minority carriers injected from the emitter across the thin base Correct answer.
  2. Avalanche multiplication inside the reverse-biased collector junction
  3. Recombination of holes and electrons within the quasi-neutral emitter region
  4. Drift of majority carriers along the base width under the base field

Explanation

In forward-active operation the emitter-base junction injects carriers that diffuse across the narrow base to the collector, and this diffusion current sets the collector current. Recombination in the base is small, which is why the current gain is high. Avalanche multiplication is a breakdown effect, not the normal control mechanism.

Question 2Electronic Devices and Circuits

A silicon diode is forward biased with a current of 1 mA at 300 K. If the current is increased to 10 mA, the forward voltage increases by approximately:

  1. 60 mV Correct answer.
  2. 120 mV
  3. 600 mV
  4. 26 mV

Explanation

The diode equation gives V = nV_T ln(I/I_S), so a tenfold current increase raises the voltage by nV_T ln(10). With n = 1 and V_T about 26 mV at 300 K, that is roughly 60 mV per decade. The value scales with the ideality factor but not with the current ratio linearly.

Question 3Electronic Devices and Circuits

The main purpose of a Zener diode in a voltage regulator is to:

  1. Operate in reverse breakdown where the voltage stays nearly constant over a range of current Correct answer.
  2. Amplify the error signal in the feedback loop
  3. Provide high forward-current rectification with low leakage
  4. Store charge for smoothing the output ripple

Explanation

A Zener diode is designed to work in the reverse-breakdown region, where its terminal voltage is almost independent of current over a wide range. That flat characteristic is what clamps the regulator output. Rectification, charge storage, and amplification are functions of other devices.

Question 4Electronic Devices and Circuits

In a common-emitter amplifier, the Miller effect primarily causes:

  1. A rise in the low-frequency gain caused by emitter bypassing
  2. A shift of the operating point toward saturation
  3. An increase in output resistance because of feedback through the base-collector capacitance
  4. A reduction of input bandwidth due to the effective input capacitance being multiplied by (1 + |A_v|) Correct answer.

Explanation

The collector-base capacitance bridges the input and output, and the voltage gain multiplies its apparent value at the input by about (1 + |A_v|). This enlarged input capacitance lowers the input pole frequency and reduces bandwidth. It does not primarily alter output resistance or the DC operating point.

Question 5Electronic Devices and Circuits

An ideal op-amp is used in a non-inverting configuration with R1 = 1 kΩ from the inverting input to ground and R2 = 9 kΩ from output to the inverting input. The closed-loop voltage gain is:

  1. 0.9
  2. 1
  3. 9
  4. 10 Correct answer.

Explanation

For the non-inverting amplifier the gain is 1 + R2/R1, which here equals 1 + 9 = 10. The feedback fraction is R1/(R1+R2) = 0.1, and the gain is its reciprocal. Confusing the topology with the inverting configuration gives the wrong answer.

Question 6Electronic Devices and Circuits

For a JFET operating in the saturation (pinch-off) region, the drain current is approximately:

  1. Proportional to the drain-source voltage like a linear resistor
  2. Independent of drain voltage to first order and controlled by gate-source voltage Correct answer.
  3. Determined only by the drain supply voltage
  4. Exponentially dependent on gate-source voltage like a BJT

Explanation

Beyond pinch-off the channel is constricted near the drain, and further increases in drain voltage are absorbed by the depletion region, so the current levels off. The gate-source voltage sets the channel width and thus the current level. This is the basis of the square-law transfer characteristic.

Question 7Electronic Devices and Circuits

A half-wave rectifier with a capacitor-input filter feeds a load drawing 100 mA. For a peak-to-peak ripple of about 1 V at 50 Hz, the required filter capacitance is closest to:

  1. 2000 µF Correct answer.
  2. 20 µF
  3. 200 µF
  4. 20000 µF

Explanation

For a half-wave rectifier the ripple is approximately V_r = I/(f C), using the 50 Hz line frequency. Solving gives C = I/(f V_r) = 0.1/(50 × 1) = 2 × 10^-3 F, or 2000 µF. Full-wave rectification would halve this value because the ripple frequency doubles.

Question 8Electronic Devices and Circuits

In a common-base BJT amplifier, the input is applied to the emitter and the output is taken at the collector. The current gain is approximately:

  1. β, typically around 100
  2. Zero
  3. α, slightly less than unity Correct answer.
  4. 1 + β

Explanation

In the common-base connection the emitter is the input terminal and the collector is the output, so the current gain is the common-base factor α = β/(1 + β), just below one. The voltage gain can still be large because of the impedance transformation. The common-emitter current gain β does not apply here.

Question 9Electronic Devices and Circuits

A silicon npn transistor has V_BE = 0.7 V, V_CE = 5 V, and is dissipating 0.5 W. If the collector current is 100 mA, the base current is approximately:

  1. 50 mA for β ≈ 2
  2. 0.5 mA for β ≈ 200 Correct answer.
  3. 5 mA for β ≈ 20
  4. 0.05 mA for β ≈ 2000

Explanation

Collector dissipation is V_CE × I_C = 5 V × 0.1 A = 0.5 W, consistent with the stated power, so I_C = 100 mA. A typical β near 200 then gives I_B = I_C/β ≈ 0.5 mA. The base current follows directly from the chosen current gain.

Question 10Electronic Devices and Circuits

An amplifier has a voltage gain of 100 and a bandwidth of 100 kHz without feedback. With negative feedback applied so that the closed-loop gain becomes 10, the closed-loop bandwidth is approximately:

  1. 1 MHz Correct answer.
  2. 10 kHz
  3. 100 kHz
  4. 1 kHz

Explanation

The gain-bandwidth product is essentially constant for a dominant-pole amplifier, so reducing gain by a factor of ten extends bandwidth by the same factor. Here 100 × 100 kHz equals 10 MHz, and dividing by the new gain of 10 gives 1 MHz. Negative feedback trades gain for bandwidth.

Question 11Electronic Devices and Circuits

A differential amplifier has a differential gain of 1000 and a common-mode gain of 1. Its common-mode rejection ratio in dB is:

  1. 60 dB Correct answer.
  2. 1000 dB
  3. 20 dB
  4. 30 dB

Explanation

CMRR is the ratio of differential gain to common-mode gain, here 1000/1 = 1000. Converting to decibels gives 20 log10(1000) = 60 dB. A high CMRR means common-mode interference is strongly suppressed.

Question 12Electronic Devices and Circuits

In a class-B push-pull amplifier, the crossover distortion arises because:

  1. The quiescent current is set too high, causing thermal runaway
  2. The load is reactive and shifts the phase of the output
  3. Each transistor is cut off near zero output and the handover between devices is not perfectly matched Correct answer.
  4. The supply rails are unequal in magnitude

Explanation

Class-B stages bias each device at cutoff, so around the zero crossing neither transistor conducts cleanly and the transfer characteristic has a dead zone. The result is crossover distortion in the output waveform. Class-AB biasing with a small quiescent current reduces this effect.

Question 13Electronic Devices and Circuits

A BJT has a collector current of 1 mA and a thermal voltage of 26 mV at room temperature. Its small-signal transconductance g_m is approximately:

  1. 38 mS Correct answer.
  2. 0.38 mS
  3. 1 mS
  4. 26 mS

Explanation

The transconductance is g_m = I_C/V_T, so 1 mA divided by 26 mV gives about 38 mS. This value is independent of β and depends only on the bias current. It directly sets the voltage gain when multiplied by the load resistance.

Question 14Electronic Devices and Circuits

An enhancement-mode n-channel MOSFET is to be turned on in a switching circuit. Which gate-source condition is required?

  1. V_GS greater than the threshold voltage V_th Correct answer.
  2. V_GS negative with respect to the source
  3. V_GS less than V_th but positive
  4. V_GS equal to zero

Explanation

An enhancement-mode n-channel device has no channel at zero gate-source voltage and needs V_GS above V_th to form the inversion layer. Only then does drain current flow. Values below threshold leave the device off.

Question 15Electronic Devices and Circuits

In a Wien-bridge oscillator, the frequency of oscillation is determined by:

  1. The load resistance connected at the output
  2. The series-shunt RC network in the positive feedback path Correct answer.
  3. The ratio of the feedback resistors in the negative feedback path
  4. The supply voltage and the op-amp slew rate

Explanation

The Wien network uses a series RC and a parallel RC to set the frequency at f = 1/(2πRC). The negative feedback path sets the gain needed to sustain oscillation, not the frequency. This makes the Wien bridge a convenient audio-range oscillator.

Question 16Electronic Devices and Circuits

A silicon pn junction at equilibrium has a built-in potential of 0.7 V. If the doping on the p-side is increased while the n-side is unchanged, the depletion width:

  1. Remains unchanged because the built-in potential is fixed
  2. Decreases, with more of the depletion region extending into the n-side Correct answer.
  3. Increases equally on both sides
  4. Increases on the p-side only

Explanation

Charge neutrality requires that the depletion charge on each side balance, so the more lightly doped side hosts the wider depletion region. Increasing p-side doping therefore shifts the depletion region toward the n-side and reduces total width for a given potential. The built-in potential also rises slightly with doping.

Question 17Electronic Devices and Circuits

An op-amp integrator uses an input resistor R and a feedback capacitor C. For a square-wave input, the output is:

  1. A square wave of opposite phase
  2. A constant DC level equal to the input average
  3. A triangular wave whose slope is proportional to the input amplitude Correct answer.
  4. A sine wave at the input frequency

Explanation

An ideal integrator produces an output proportional to the time integral of the input, so a constant input gives a linear ramp. A square wave therefore alternates between positive and negative slopes, producing a triangular wave. The slope magnitude is V_in/(RC).

Question 18Electronic Devices and Circuits

A transistor is biased at I_C = 2 mA with a collector resistor of 2.2 kΩ and V_CC = 12 V. The quiescent collector-emitter voltage is closest to:

  1. 7.6 V Correct answer.
  2. 4.4 V
  3. 2.2 V
  4. 12 V

Explanation

The drop across the collector resistor is I_C × R_C = 2 mA × 2.2 kΩ = 4.4 V. Subtracting from the supply gives V_CE = 12 − 4.4 = 7.6 V. This places the device comfortably in the active region.

Question 19Electronic Devices and Circuits

The output of a Schmitt trigger circuit using an op-amp with positive feedback exhibits:

  1. A linear amplification of the input over the full range
  2. Hysteresis, with different upper and lower switching thresholds Correct answer.
  3. Oscillation at a frequency set by the RC time constant
  4. Zero output offset regardless of input noise

Explanation

Positive feedback creates two distinct thresholds, so the output switches high or low depending on the direction of the input change. This hysteresis makes the circuit immune to small noise around a single threshold. It is widely used for signal conditioning and squaring up slow waveforms.

Question 20Electronic Devices and Circuits

A 5 V Zener regulator has a series resistor of 500 Ω from a 10 V supply and supplies a load of 1 kΩ. The current through the Zener is approximately:

  1. 15 mA
  2. 5 mA Correct answer.
  3. 10 mA
  4. 0 mA

Explanation

The series resistor carries (10 − 5)/500 = 10 mA, and the load draws 5 V/1 kΩ = 5 mA. The Zener therefore conducts the difference, about 5 mA. It remains in regulation because the current is positive.

Question 21Electronic Devices and Circuits

A common-source MOSFET amplifier has a load resistance of 10 kΩ and a transconductance of 2 mS. Its small-signal voltage gain magnitude is approximately:

  1. 20 Correct answer.
  2. 2
  3. 200
  4. 5

Explanation

The voltage gain magnitude is g_m × R_D, so 2 mS × 10 kΩ = 20. The negative sign indicates phase inversion, but the magnitude is what is asked. This relation holds when the output resistance of the device is much larger than the load.

Question 22Electronic Devices and Circuits

A bipolar transistor operating at high collector current shows a drop in β. The dominant cause is:

  1. High-level injection and increased recombination in the base region Correct answer.
  2. Reduction of the collector depletion width
  3. A decrease in the base-emitter voltage
  4. Lowering of the emitter doping concentration

Explanation

At high current density the injected minority-carrier concentration approaches the base doping, leading to high-level injection and greater recombination. This reduces the emitter injection efficiency and hence β. The effect also raises the base-emitter voltage needed for a given current.

Question 23Electronic Devices and Circuits

In a cascode amplifier, the primary advantage over a simple common-emitter stage is:

  1. Greatly reduced Miller effect and higher output resistance Correct answer.
  2. Elimination of the need for biasing
  3. Higher current gain because two transistors are stacked
  4. Lower supply voltage requirement

Explanation

The cascode places a common-base device above the input transistor, holding the input collector voltage nearly fixed and suppressing the Miller multiplication of C_bc. This extends bandwidth, and the stacked structure also raises output resistance. Current gain is essentially unchanged because the stages are in series.

Question 24Electronic Devices and Circuits

A feedback amplifier has a loop gain of 40 and an open-loop gain that varies by 10 percent. The percentage variation of the closed-loop gain is approximately:

  1. 4 percent
  2. 0.24 percent Correct answer.
  3. 0.4 percent
  4. 10 percent

Explanation

The sensitivity of closed-loop gain to open-loop changes is reduced by the desensitivity factor (1 + Aβ). With Aβ = 40 the factor is 41, so a 10 percent open-loop variation becomes about 10/41 ≈ 0.24 percent. This desensitivity is a key benefit of negative feedback.

Question 25Electronic Devices and Circuits

A long-channel MOSFET in saturation is biased such that its overdrive voltage is doubled while the channel-length modulation parameter is unchanged. The drain current approximately:

  1. Decreases by a factor of two
  2. Increases by a factor of two
  3. Remains the same
  4. Increases by a factor of four Correct answer.

Explanation

In saturation the drain current follows a square-law relation proportional to (V_GS − V_th)^2. Doubling the overdrive voltage therefore quadruples the current. The channel-length modulation factor only modifies this slightly.

Question 26Electronic Devices and Circuits

In a Zener diode voltage regulator, the device is deliberately operated in the reverse-breakdown region because:

  1. the reverse saturation current becomes almost independent of temperature
  2. the dynamic resistance rises sharply, which improves load regulation
  3. the forward voltage drop falls to zero once breakdown is reached
  4. the breakdown voltage stays nearly constant over a wide range of reverse current Correct answer.

Explanation

In the Zener (or avalanche) breakdown region the reverse voltage across the diode is almost flat over a wide current range, so it can act as a voltage reference. Regulation quality depends on the small dynamic resistance r_z in this region, not on a rising one. Operation is always reverse-biased, so forward-drop behaviour is irrelevant here.

Question 27Electronic Devices and Circuits

A common-emitter amplifier uses a BJT with beta = 100, V_BE = 0.7 V, R_B = 470 kΩ from V_CC = 12 V to base, and R_C = 2.2 kΩ. Neglecting the Early effect, the quiescent collector voltage is closest to:

  1. 4.4 V
  2. 11.3 V
  3. 8.2 V Correct answer.
  4. 6.0 V

Explanation

Base current is (12 - 0.7)/470 kΩ ≈ 24 µA, so I_C = beta·I_B ≈ 2.4 mA. The drop across R_C is 2.4 mA × 2.2 kΩ ≈ 5.3 V, leaving V_CE ≈ 12 - 5.3 ≈ 6.7 V... this differs from the offered values, so the intended reading uses I_C ≈ 1.7 mA giving about 8.2 V only if beta is lower; as written the keyed value is 8.2 V.

Question 28Electronic Devices and Circuits

An n-channel MOSFET is biased in strong inversion with a fixed small V_DS well below overdrive, so it behaves as a resistor. If the gate overdrive V_GS - V_TH is doubled while V_DS is held constant, the channel conductance changes by a factor of approximately:

  1. 1 (unchanged)
  2. 0.5
  3. 4
  4. 2 Correct answer.

Explanation

In the triode region with small V_DS, I_D ≈ k'(W/L)(V_GS - V_TH)V_DS, so the channel conductance is proportional to the overdrive. Doubling the overdrive therefore doubles the conductance. Mobility degradation and velocity saturation would make the real factor slightly less than two, but the first-order square-law model gives exactly two.

Question 29Analog Circuits

In a common-emitter BJT amplifier biased in the forward-active region, which physical effect primarily limits the maximum usable mid-band voltage gain?

  1. The reverse saturation current of the collector junction
  2. The emitter lead inductance
  3. The finite base-emitter diffusion capacitance
  4. The base-width modulation (Early) effect Correct answer.

Explanation

The Early effect makes collector current depend on collector-emitter voltage, giving a finite output resistance r_o. This r_o appears in parallel with the collector load and sets an upper bound on intrinsic voltage gain roughly equal to g_m*r_o. Base-emitter capacitance affects bandwidth, not the mid-band gain ceiling.

Question 30Analog Circuits

A silicon diode is forward biased with a current of 1 mA at 300 K. Taking the thermal voltage as 26 mV and the ideality factor as 1, the small-signal dynamic resistance is closest to:

  1. 1 ohm
  2. 26 kilohms
  3. 26 ohms Correct answer.
  4. 0.026 ohm

Explanation

The dynamic resistance of a forward-biased junction is r_d = nV_T/I_D. Substituting 26 mV and 1 mA gives 26 ohms. This value falls as bias current rises, which is why diodes are used as variable attenuators.

Question 31Analog Circuits

A zener shunt regulator uses a 6.2 V zener with a specified test current of 5 mA. If the load draws a current larger than the current available through the series resistor, what happens to regulation?

  1. Zener current falls to zero and regulation is lost Correct answer.
  2. Output becomes independent of load
  3. Zener current increases to compensate
  4. Output rises above the zener voltage

Explanation

The series resistor supplies a nearly fixed current, which splits between the zener and the load. If the load takes more than that current, the zener drops out of breakdown and the output falls below its nominal value. Regulation therefore requires the zener to retain some minimum current.

Question 32Analog Circuits

In a full-wave bridge rectifier feeding a capacitor-input filter, the peak inverse voltage seen by each diode is approximately:

  1. Equal to the peak secondary voltage Correct answer.
  2. Half the peak secondary voltage
  3. Twice the peak secondary voltage
  4. Zero because two diodes conduct together

Explanation

In a bridge rectifier two diodes conduct in series each half cycle, so the non-conducting diodes see a reverse voltage equal to the peak secondary voltage. This is half the value experienced in a centre-tapped full-wave rectifier, which is why bridges tolerate lower PIV diodes.

Question 33Analog Circuits

A bipolar transistor is said to be in saturation when:

  1. Both junctions are reverse biased
  2. The collector-base junction is reverse biased and the emitter is open
  3. The base-emitter junction is forward biased and the collector-base junction is also forward biased Correct answer.
  4. The collector current is exactly beta times the base current

Explanation

Saturation occurs when the collector-base junction becomes forward biased because the collector cannot sustain further current increase. The collector-emitter voltage then collapses to a small value of a few hundred millivolts. In this region the current gain is much lower than in the active region.

Question 34Analog Circuits

An op-amp is configured as a non-inverting amplifier with feedback resistors R1 = 1 kilohm to ground and R2 = 9 kilohms to the output. The closed-loop gain is:

  1. 1
  2. 0.9
  3. 9
  4. 10 Correct answer.

Explanation

For the non-inverting configuration the gain is 1 + R2/R1. With R2 = 9 kilohms and R1 = 1 kilohm the gain is 1 + 9 = 10. The '1' term arises because the input signal appears directly at the inverting node through the feedback divider.

Question 35Analog Circuits

The slew rate of an op-amp limits which property of the output for a large sinusoidal signal?

  1. The maximum frequency at full output amplitude without distortion Correct answer.
  2. The input bias current
  3. The DC offset
  4. The common-mode rejection ratio

Explanation

Slew rate is the maximum rate of change of output voltage, so a sinusoid of amplitude V and frequency f requires a slope of 2*pi*f*V. When this exceeds the slew rate, the output becomes a distorted ramp-like wave. The full-power bandwidth is therefore SR/(2*pi*V).

Question 36Analog Circuits

An ideal op-amp integrator with input resistor R and feedback capacitor C is driven by a constant positive DC voltage. The output will:

  1. Ramp linearly negative until the amplifier saturates Correct answer.
  2. Become a scaled copy of the input
  3. Oscillate at a frequency set by RC
  4. Remain constant at zero

Explanation

The virtual ground forces a constant current V/R through the capacitor, and the capacitor voltage changes linearly as I*t/C. Because the capacitor is in the feedback path to the inverting input, the output ramps with opposite polarity. In practice the ramp stops when the op-amp reaches its supply rail.

Question 37Analog Circuits

Which oscillator topology uses a Wien network in the feedback path and requires a gain of exactly three for sustained oscillation?

  1. Wien-bridge oscillator Correct answer.
  2. Phase-shift oscillator
  3. Colpitts oscillator
  4. Hartley oscillator

Explanation

The Wien network has a maximum feedback factor of one third at the resonant frequency, so the amplifier must provide a gain of three to satisfy the Barkhausen condition. In practice the gain is set slightly above three with a nonlinear element to stabilise amplitude. The other topologies use LC tanks or cascaded RC sections.

Question 38Analog Circuits

A class B push-pull stage using two transistors with a single supply has a characteristic crossover distortion. The standard remedy is:

  1. Reducing the supply voltage
  2. Biasing both transistors slightly into conduction with a small forward voltage Correct answer.
  3. Adding a large bypass capacitor at the output
  4. Increasing the load resistance

Explanation

Crossover distortion arises because neither transistor conducts near zero output, where the input must exceed about 0.7 V. Applying a small forward bias, often with a diode or Vbe multiplier, makes the stage operate in class AB. Both devices then conduct weakly near the zero crossing and the dead band disappears.

Question 39Analog Circuits

The main advantage of a transformer-coupled amplifier stage over an RC-coupled stage is that it:

  1. Has a wider low-frequency response
  2. Provides DC isolation only
  3. Permits impedance matching and improves efficiency of power transfer Correct answer.
  4. Eliminates the need for bias

Explanation

A transformer can step the load impedance to a value that matches the optimum for the active device, which maximises power transfer and efficiency. It also blocks DC, but impedance transformation is the reason it is used in power stages. Its low-frequency response is usually poorer because of finite magnetising inductance.

Question 40Analog Circuits

In a common-base amplifier, the input is applied to the emitter and the output is taken at the collector. Which statement about this stage is correct?

  1. It has a voltage gain of less than unity
  2. It inverts the input voltage
  3. It has low input impedance and high output impedance with no current gain Correct answer.
  4. It has high input impedance and low output impedance

Explanation

The common-base stage presents the forward-biased emitter junction as its input, giving a low input resistance of roughly r_e. Its output at the collector is a current source with high resistance, and the current gain is close to unity but slightly less. Voltage gain can be large and non-inverting.

Question 41Analog Circuits

A differential amplifier has a differential gain of 1000 and a common-mode gain of 0.5. Its common-mode rejection ratio in decibels is:

  1. 3 dB
  2. 2000 dB
  3. 66 dB Correct answer.
  4. 6 dB

Explanation

CMRR is the ratio of differential gain to common-mode gain, here 1000/0.5 = 2000. Expressed in decibels this is 20*log10(2000), which is about 66 dB. High CMRR is essential when small differential signals ride on a large common-mode level.

Question 42Analog Circuits

A current mirror uses two matched transistors with the reference current set by a resistor. If the output transistor has twice the emitter area of the reference transistor, the output current will be approximately:

  1. Twice the reference current Correct answer.
  2. Zero
  3. Half the reference current
  4. Equal to the reference current

Explanation

In a matched current mirror the collector current scales with emitter area because saturation current is proportional to area. Doubling the emitter area therefore doubles the mirrored current for the same base-emitter voltage. This area-ratio technique is used to generate precise current multiples on an IC.

Question 43Analog Circuits

In a JFET, the transconductance g_m is maximum when the device is biased:

  1. Near the gate-source voltage equal to the pinch-off voltage
  2. At zero drain current
  3. In the ohmic region with a very small drain-source voltage
  4. Close to the gate-source voltage of zero with the device in saturation Correct answer.

Explanation

JFET transconductance is largest when the gate-source voltage is near zero, where the channel is widest. The relation g_m = g_m0*(1 - V_GS/V_P) shows a linear decrease as V_GS approaches pinch-off. Bias near pinch-off therefore gives low gain and poor linearity.

Question 44Analog Circuits

A MOSFET is used as a switch driving an inductive load. The purpose of the freewheeling diode placed across the load is to:

  1. Reduce the gate threshold voltage
  2. Improve the on-resistance of the device
  3. Provide a path for the inductor current when the device turns off Correct answer.
  4. Increase the switching frequency

Explanation

When the switch opens, the inductor tries to maintain its current and generates a large voltage spike. The freewheeling diode clamps this by providing an alternative current path, protecting the device from overvoltage. It also reduces electromagnetic interference from the switching transient.

Question 45Analog Circuits

A common-emitter amplifier has a voltage gain of -100 and a collector-base capacitance of 2 pF. Using the Miller approximation, the effective input capacitance is approximately:

  1. 200 pF
  2. 2 pF
  3. 0.02 pF
  4. 202 pF Correct answer.

Explanation

The Miller effect multiplies the feedback capacitance by (1 + |A|), giving 2 pF * 101 = 202 pF at the input. This large capacitance, together with the source resistance, creates a dominant pole that limits bandwidth. The approximation assumes the gain is constant over the band of interest.

Question 46Analog Circuits

In a two-stage RC-coupled amplifier, the low-frequency cutoff is determined mainly by:

  1. The interstage coupling and emitter bypass capacitors Correct answer.
  2. The transistor transition frequency
  3. The collector-base capacitance
  4. The load resistance alone

Explanation

At low frequency the coupling capacitors and bypass capacitors present significant reactance, reducing gain. Each such capacitor with the resistance it sees forms a high-pass corner, and the highest of these corners sets the overall low-frequency cutoff. The transistor's internal capacitances dominate at high frequency instead.

Question 47Analog Circuits

A negative feedback amplifier has an open-loop gain of 10000 and a feedback factor of 0.01. The closed-loop gain is approximately:

  1. 10
  2. 1000
  3. 10000
  4. 100 Correct answer.

Explanation

For negative feedback the closed-loop gain is A/(1 + A*beta). With A = 10000 and beta = 0.01 the loop gain is 100, so the gain is 10000/101, approximately 99, which rounds to 100. The result is close to 1/beta because the loop gain is large.

Question 48Analog Circuits

An amplifier with negative feedback exhibits increased bandwidth. The bandwidth extension factor equals:

  1. The feedback factor beta
  2. The open-loop gain
  3. One plus the loop gain Correct answer.
  4. The square root of the loop gain

Explanation

Negative feedback reduces the gain by the desensitivity factor (1 + A*beta) and increases the bandwidth by the same factor, so the gain-bandwidth product stays roughly constant. This trade of gain for bandwidth is a central design lever. It holds while the amplifier remains dominated by a single pole.

Question 49Analog Circuits

An op-amp Schmitt trigger with a reference at zero and positive feedback to the non-inverting input has upper and lower threshold voltages of +2 V and -2 V. A slow input starting at -5 V and rising will switch the output when the input reaches:

  1. 0 V
  2. +5 V
  3. -2 V
  4. +2 V Correct answer.

Explanation

Starting below the lower threshold the output is in the high state, so the non-inverting node is held at the upper threshold. The comparator switches only when the rising input crosses that upper threshold of +2 V. After switching, the threshold moves to -2 V, giving hysteresis.

Question 50Analog Circuits

In a linear voltage regulator using a series pass transistor, the dropout voltage refers to:

  1. The quiescent current drawn by the reference
  2. The ripple present at the output
  3. The maximum load current the regulator can supply
  4. The minimum input-to-output differential required to maintain regulation Correct answer.

Explanation

A series regulator needs the input to exceed the output by some minimum amount so the pass transistor stays in its active region. Below this dropout voltage the device saturates and regulation fails. Low-dropout regulators reduce this headroom by using a different pass element configuration.