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Engineer Electronics practice questions

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Question 1Electronic Circuits and Devices

A silicon diode is forward-biased with a current of 10 mA. Using the Shockley equation with thermal voltage V_T = 26 mV and ideality factor n = 1, what is the dynamic (small-signal) resistance r_d of the diode?

  1. 0.26 Ω
  2. 2.6 Ω Correct answer.
  3. 26 Ω
  4. 260 Ω

Explanation

The small-signal resistance is r_d = n·V_T/I_D. With n = 1, V_T = 26 mV, and I_D = 10 mA, r_d = 0.026/0.010 = 2.6 Ω.

Question 2Electronic Circuits and Devices

In a common-emitter BJT amplifier biased in the forward-active region, which parameter most directly sets the small-signal transconductance g_m?

  1. The base spreading resistance r_bb'
  2. The collector quiescent current I_C Correct answer.
  3. The base-emitter junction capacitance
  4. The collector-emitter breakdown voltage

Explanation

Transconductance is g_m = I_C/V_T, so the DC collector bias current directly determines it. Capacitance and breakdown voltage do not set g_m, and r_bb' only affects high-frequency behavior.

Question 3Electronic Circuits and Devices

An ideal op-amp is configured as a non-inverting amplifier with R_f = 90 kΩ and R_in = 10 kΩ. What closed-loop voltage gain results?

  1. 11
  2. 0.1
  3. 10 Correct answer.
  4. 9

Explanation

For the non-inverting topology, A_v = 1 + R_f/R_in = 1 + 90/10 = 10. The feedback ratio sets the gain directly when loop gain is very high.

Question 4Electronic Circuits and Devices

A Zener diode with V_Z = 6.2 V is used as a shunt regulator feeding a load. For the regulator to hold the output near 6.2 V, what must be true of the operating conditions?

  1. The load current must exceed the Zener current at all times.
  2. The series resistor must be removed so the diode sees the full source voltage.
  3. The diode must be reverse-biased above its knee current with sufficient series resistance. Correct answer.
  4. The diode must be forward-biased with at least 0.7 V across it.

Explanation

Zener regulation requires reverse breakdown operation beyond the knee, with a series resistor to drop excess voltage and limit current. Without adequate reverse current the output sags; without the series resistor the diode can be destroyed.

Question 5Electronic Circuits and Devices

A MOSFET is specified with R_DS(on) = 50 mΩ. It carries a drain current of 4 A. What is the approximate conduction power dissipation?

  1. 0.8 W Correct answer.
  2. 2.0 W
  3. 0.2 W
  4. 8.0 W

Explanation

Conduction loss is P = I²·R_DS(on) = 16 × 0.05 = 0.8 W. This quadratic dependence on current is why conduction loss dominates at high load in switching converters.

Question 6Electronic Circuits and Devices

Which condition places an NPN bipolar transistor in saturation?

  1. The base-emitter junction is reverse-biased and the base-collector is forward-biased.
  2. The base-emitter junction is forward-biased and the collector current equals β·I_B.
  3. Both junctions are forward-biased and the collector current is limited by the external circuit. Correct answer.
  4. Both junctions are reverse-biased.

Explanation

Saturation occurs when both junctions are forward-biased and the external collector circuit, not β, limits collector current. The collector-emitter voltage then falls to a small value V_CE(sat).

Question 7Electronic Circuits and Devices

A capacitor of 10 µF is charged to 12 V. How much energy is stored?

  1. 120 µJ
  2. 72 µJ
  3. 1.44 mJ
  4. 720 µJ Correct answer.

Explanation

Stored energy is E = ½·C·V² = 0.5 × 10 µF × 144 = 720 µJ. The factor of one-half arises because voltage rises linearly as charge accumulates.

Question 8Electronic Circuits and Devices

In a series RLC circuit at resonance, which statement correctly describes the impedance and current?

  1. Impedance is purely resistive and current is maximum. Correct answer.
  2. Impedance is purely capacitive and current leads voltage by 90°.
  3. Impedance is purely inductive and current lags voltage by 90°.
  4. Impedance is maximum and current is minimum.

Explanation

At series resonance the inductive and capacitive reactances cancel, leaving only R. Impedance is therefore minimum and purely resistive, so current reaches its maximum and is in phase with the source.

Question 9Electronic Circuits and Devices

A common-source MOSFET amplifier with a resistive drain load is biased in saturation. What is the phase relationship between the gate input and the drain output at midband?

  1. Inverted (180°) Correct answer.
  2. Leading by 90°
  3. Lagging by 90°
  4. In phase (0°)

Explanation

An increase in gate voltage increases drain current, which increases the drop across the drain resistor and lowers the drain voltage. The output is therefore inverted relative to the input at midband.

Question 10Electronic Circuits and Devices

What is the primary purpose of a bypass capacitor connected across the emitter resistor in a common-emitter amplifier?

  1. To lower the input impedance at low frequency
  2. To protect the transistor from reverse breakdown
  3. To raise the AC voltage gain by reducing AC degeneration Correct answer.
  4. To increase the DC bias current

Explanation

The bypass capacitor provides a low-impedance AC path around R_E, removing negative feedback at signal frequencies. This raises the AC gain while leaving the DC bias point essentially unchanged.

Question 11Electronic Circuits and Devices

A full-wave bridge rectifier is fed from a 50 Hz AC source. What is the ripple frequency of the unfiltered output?

  1. 200 Hz
  2. 25 Hz
  3. 100 Hz Correct answer.
  4. 50 Hz

Explanation

A full-wave rectifier produces two output pulses per input cycle, so the ripple frequency is twice the line frequency. For a 50 Hz source that gives 100 Hz.

Question 12Electronic Circuits and Devices

A differential amplifier has a differential gain of 1000 and a common-mode gain of 0.5. What is the common-mode rejection ratio in dB?

  1. 66 dB Correct answer.
  2. 60 dB
  3. 46 dB
  4. 54 dB

Explanation

CMRR = A_d/A_cm = 1000/0.5 = 2000, and 20·log10(2000) ≈ 66 dB. Expressing it in decibels uses the 20·log form because it is a voltage ratio.

Question 13Electronic Circuits and Devices

Which biasing method for a BJT amplifier is most sensitive to variations in the transistor's current gain β?

  1. Fixed base bias with a single base resistor Correct answer.
  2. Collector-to-base feedback bias
  3. Voltage-divider bias with a stiff divider
  4. Emitter-stabilized bias with an emitter resistor

Explanation

Fixed base bias sets I_B from the supply and a resistor, so I_C = β·I_B varies directly with β. Emitter and voltage-divider schemes use feedback or stiff dividers to desensitize the operating point.

Question 14Electronic Circuits and Devices

An ideal inverting op-amp summer has input resistors R1 = 10 kΩ and R2 = 20 kΩ, with R_f = 40 kΩ. What output results for V1 = 1 V and V2 = 0.5 V?

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

Explanation

Each input contributes -R_f/R_i times its voltage: -(40/10)(1) - (40/20)(0.5) = -4 - 1 = -5 V. Wait-recomputing: -4 V plus -1 V gives -5 V, so the correct output is -5 V.

Question 15Electronic Circuits and Devices

What distinguishes a JFET from a depletion-mode MOSFET in terms of gate structure?

  1. A JFET gate is an insulated oxide layer, while a depletion MOSFET gate is a PN junction.
  2. Both devices require forward gate current to operate.
  3. A JFET gate is a reverse-biased PN junction, while a depletion MOSFET gate is an insulated oxide layer. Correct answer.
  4. Both devices use a Schottky gate contact.

Explanation

The JFET controls channel width with a reverse-biased gate-channel junction, so gate current is tiny. A depletion-mode MOSFET uses a metal-oxide-semiconductor gate that is insulated from the channel.

Question 16Electronic Circuits and Devices

A transistor operates with a junction temperature of 125 °C and dissipates 2 W. The junction-to-case thermal resistance is 5 °C/W and the case-to-ambient is 10 °C/W. What ambient temperature is implied?

  1. 85 °C
  2. 95 °C
  3. 105 °C
  4. 65 °C Correct answer.

Explanation

Total thermal resistance is 5 + 10 = 15 °C/W, so the rise is 2 W × 15 °C/W = 30 °C. Subtracting from 125 °C gives an ambient of 95 °C.

Question 17Electronic Circuits and Devices

In a CMOS inverter, both transistors are momentarily on during the input transition. What is the main consequence?

  1. Increased dynamic power loss from crowbar current Correct answer.
  2. Increased noise margin
  3. Lower output capacitance
  4. Reduced propagation delay

Explanation

During a slow input transition both the PMOS and NMOS conduct, creating a short-circuit path from supply to ground. This crowbar current adds to dynamic power, which is why fast input edges are preferred.

Question 18Electronic Circuits and Devices

An ideal op-amp integrator has R = 10 kΩ and C = 0.1 µF. What is the magnitude of its gain at 1 kHz?

  1. 0.159
  2. 159
  3. 1.59
  4. 15.9 Correct answer.

Explanation

Integrator gain magnitude is 1/(2πfRC). With RC = 1 ms and f = 1 kHz, the denominator is 2π × 1 ≈ 6.28, giving a gain of about 0.159. Rechecking: 1/(2π×1000×0.001) = 1/6.28 ≈ 0.159.

Question 19Electronic Circuits and Devices

What is the primary function of a varactor diode in a tuned circuit?

  1. Providing voltage-controlled capacitance Correct answer.
  2. Rectifying high-frequency signals
  3. Storing charge as an electrolytic element
  4. Emitting light when forward-biased

Explanation

A varactor is operated reverse-biased so its depletion capacitance varies with applied voltage. This voltage-controlled capacitance tunes oscillators and filters electronically.

Question 20Electronic Circuits and Devices

A common-collector (emitter-follower) amplifier is used as a buffer. Which set of properties best describes it?

  1. Voltage gain near unity and low output impedance Correct answer.
  2. Voltage gain near unity and high output impedance
  3. High voltage gain and high output impedance
  4. High voltage gain and low input impedance

Explanation

The emitter follower has a voltage gain slightly below one, a high input impedance, and a low output impedance. That combination makes it useful for driving low-impedance loads without loading the source.

Question 21Electronic Circuits and Devices

A half-wave rectifier with a smoothing capacitor feeds a load drawing 50 mA. If the capacitor is 1000 µF and the ripple is approximated by I·t/C with t = 10 ms, what ripple voltage results?

  1. 50 mV
  2. 0.5 V Correct answer.
  3. 0.05 V
  4. 5 V

Explanation

Ripple is approximately ΔV = I·t/C = 0.05 A × 0.01 s / 0.001 F = 0.5 V. Larger capacitance or smaller load current reduces this ripple proportionally.

Question 22Electronic Circuits and Devices

Which mechanism primarily limits the high-frequency gain of a bipolar transistor amplifier?

  1. The thermal resistance of the package
  2. The Miller-multiplied base-collector capacitance Correct answer.
  3. The DC current gain β
  4. The collector-emitter saturation voltage

Explanation

The base-collector capacitance appears multiplied by the voltage gain at the input (Miller effect), forming a dominant pole. This capacitance charging limits the amplifier bandwidth at high frequency.

Question 23Electronic Circuits and Devices

A diode has a forward voltage of 0.7 V at 5 mA. What is the approximate power dissipated in the diode?

  1. 0.35 mW
  2. 35 mW
  3. 350 mW
  4. 3.5 mW Correct answer.

Explanation

Diode power is P = V_F × I_F = 0.7 V × 5 mA = 3.5 mW. This dissipation must be checked against the device's thermal rating.

Question 24Electronic Circuits and Devices

In a class-B push-pull output stage, what is the purpose of the small forward bias applied to the bases of the two transistors?

  1. To lower the supply voltage requirement
  2. To reduce crossover distortion by keeping both devices slightly conducting Correct answer.
  3. To increase the maximum output swing
  4. To increase the output impedance

Explanation

Class-B stages suffer crossover distortion near zero crossing because neither device conducts there. A small forward bias moves the stage toward class AB, keeping both devices slightly on and smoothing the transition.

Question 25Electronic Circuits and Devices

A photodiode is operated in photoconductive mode. Which bias condition and resulting characteristic are correct?

  1. Forward-biased, producing a large forward current
  2. Reverse-biased, producing a photocurrent proportional to illumination Correct answer.
  3. Unbiased, producing zero current regardless of light
  4. Reverse-biased, producing a voltage independent of light

Explanation

In photoconductive mode the photodiode is reverse-biased, widening the depletion region and reducing junction capacitance. Incident light generates carriers, and the resulting photocurrent is proportional to illumination.

Question 26Electronic Circuits and Devices

A silicon pn-junction diode is reverse biased by 20 V at room temperature. Which statement describes the dominant current mechanism?

  1. Zero current, because the depletion region blocks all carrier motion
  2. A diffusion current of majority carriers that doubles for every 20 mV of reverse bias
  3. A small leakage current of thermally generated minority carriers that is nearly independent of reverse voltage Correct answer.
  4. A large drift current of majority carriers proportional to the applied reverse voltage

Explanation

Under reverse bias the depletion region widens and majority-carrier diffusion is suppressed, leaving a small reverse saturation current carried by thermally generated minority carriers drifting across the junction. This current is set mainly by temperature and material, not by the magnitude of the reverse voltage, until breakdown is approached.

Question 27Electronic Circuits and Devices

A zener diode with a nominal 6.2 V rating is to be used as a simple shunt regulator. Which operating condition makes the zener regulate effectively?

  1. Reverse biased in breakdown with a current between I_z(min) and I_z(max) Correct answer.
  2. Reverse biased with the load drawing more current than the zener can carry
  3. Reverse biased below its breakdown voltage
  4. Forward biased with at least 0.7 V across it

Explanation

A zener regulates by operating in controlled reverse breakdown, where a small change in current produces only a small change in voltage. The series resistor must be sized so the zener current stays above the knee current and below the rated maximum over the full load range.

Question 28Electronic Circuits and Devices

An npn transistor is biased with V_B = 2.0 V, V_E = 1.3 V and V_C = 5.0 V. Which statement about its operating region is correct?

  1. It is in cutoff because the collector voltage is below the supply rail
  2. It is in forward active mode because the base-emitter junction is forward biased and the collector-base junction is reverse biased Correct answer.
  3. It is in saturation because V_CE is greater than 0.2 V
  4. It is in reverse active mode because the collector junction is forward biased

Explanation

Here V_BE is about 0.7 V, so the emitter junction is forward biased, while V_C exceeds V_B by 3 V, so the collector junction is reverse biased. That combination defines forward active operation, which is the normal region for linear amplification.

Question 29Electronic Circuits and Devices

A MOSFET is used as a switch driving a relay coil. Why is a freewheeling diode placed across the coil?

  1. To clamp the inductive kick when the MOSFET turns off and protect it from overvoltage Correct answer.
  2. To prevent the coil from conducting when the MOSFET is on
  3. To increase the coil current during the on-time
  4. To speed up the turn-on of the MOSFET by charging its gate

Explanation

An inductive load stores energy in its magnetic field, and interrupting the coil current forces a large voltage spike that can exceed the MOSFET's drain-source rating. The freewheeling diode gives that current a path to circulate and decay, clamping the drain voltage to roughly the supply plus a diode drop.

Question 30Electronic Circuits and Devices

A common-emitter amplifier has a voltage gain magnitude of 50 and a bandwidth of 100 kHz. If negative feedback reduces the gain to 10, what is the approximate closed-loop bandwidth, assuming a single dominant pole?

  1. 500 kHz Correct answer.
  2. 100 kHz
  3. 5 MHz
  4. 20 kHz

Explanation

For a single-pole amplifier the gain-bandwidth product is constant, so reducing gain by a factor of five extends bandwidth by the same factor. The product here is 50 × 100 kHz = 5 MHz, and dividing by the new gain of 10 gives about 500 kHz.

Question 31Electronic Circuits and Devices

An op-amp is configured as a non-inverting amplifier with R_f = 90 kΩ and R_1 = 10 kΩ. A 0.5 V input is applied. What is the output voltage, assuming ideal behaviour?

  1. 5.0 V Correct answer.
  2. 4.5 V
  3. 10 V
  4. 0.5 V

Explanation

The non-inverting gain is 1 + R_f/R_1, which here equals 1 + 9 = 10. Multiplying the 0.5 V input by that gain gives 5.0 V at the output.

Question 32Electronic Circuits and Devices

A comparator built from an op-amp has no hysteresis and its input sits close to the reference with superimposed noise. What is the likely observed problem?

  1. The input offset voltage is automatically cancelled by the noise
  2. The output oscillates or chatters between its high and low states as the noise crosses the threshold Correct answer.
  3. The slew rate increases in proportion to the noise amplitude
  4. The output settles permanently at the midpoint of the supply rails

Explanation

Without hysteresis, a single threshold exists, so noise superimposed on a slowly moving input repeatedly crosses that threshold and toggles the output. Adding positive feedback creates two thresholds and a dead band that rejects small noise excursions.

Question 33Electronic Circuits and Devices

A crystal oscillator is preferred over an RC phase-shift oscillator in a digital clock circuit. Which property of the crystal explains this choice?

  1. It provides gain without any external active device
  2. It generates a square wave directly at the output
  3. It has a very high Q, giving a precise and stable series-resonant frequency Correct answer.
  4. It eliminates the need for any supply decoupling

Explanation

A quartz crystal behaves as a high-Q resonator whose resonant frequency is set by its physical dimensions and is only weakly temperature dependent. That high Q gives the low phase noise and tight frequency tolerance needed for clock generation, far better than an RC network.

Question 34Electronic Circuits and Devices

A bridge rectifier with a large smoothing capacitor feeds a light load. Which statement about the diode current is correct?

  1. The diode current is a smooth DC equal to the load current
  2. The diodes conduct continuously over the whole AC cycle
  3. The diodes conduct in short, high-amplitude pulses near the peak of each half-cycle Correct answer.
  4. The diodes conduct only during the negative half-cycle

Explanation

The capacitor holds the output near the peak and only recharges when the rectified input exceeds the capacitor voltage. Conduction is therefore confined to brief intervals around each peak, so the diodes carry narrow pulses whose average equals the load current.

Question 35Electronic Circuits and Devices

A bipolar transistor dissipates 0.8 W while its case is held at 60 °C. The junction-to-case thermal resistance is 50 °C/W and the case-to-ambient path is negligible. What is the approximate junction temperature?

  1. 100 °C Correct answer.
  2. 60 °C
  3. 160 °C
  4. 40 °C

Explanation

The junction temperature equals the case temperature plus the temperature rise across the junction-to-case path, which is power times thermal resistance. That rise is 0.8 W × 50 °C/W = 40 °C, so the junction sits at about 100 °C.

Question 36Electronic Circuits and Devices

A two-stage amplifier has a first stage with gain 20 and a second stage with gain 40, both referred to the same source and load conditions. What is the overall gain in decibels?

  1. 26 dB
  2. 600 dB
  3. 58 dB Correct answer.
  4. 800 dB

Explanation

Cascaded voltage gains multiply, so the overall gain is 20 × 40 = 800. Converting to decibels gives 20 log10(800), which is approximately 58 dB, matching the sum of 26 dB and 32 dB for the individual stages.

Question 37Analog Electronics

In a common-emitter BJT amplifier biased in the forward-active region, which statement correctly describes the phase relationship between the small-signal base-emitter voltage and the collector-emitter voltage?

  1. They are inverted, differing by 180 degrees. Correct answer.
  2. They are in phase by 0 degrees.
  3. They differ by 45 degrees depending on beta.
  4. They differ by 90 degrees with collector leading.

Explanation

In the forward-active region an increase in base-emitter voltage raises collector current, which increases the drop across the collector resistor and therefore lowers the collector-emitter voltage. This produces a 180-degree phase inversion between input and output. The effect follows directly from the inverting voltage gain of the common-emitter stage.

Question 38Analog Electronics

A silicon diode with a forward voltage of 0.7 V is in series with a 1 kΩ resistor across a 5 V supply. What is the approximate current through the resistor?

  1. 7.0 mA
  2. 0.7 mA
  3. 4.3 mA Correct answer.
  4. 5.0 mA

Explanation

The resistor sees the supply voltage minus the diode drop, so 5 V minus 0.7 V equals 4.3 V across 1 kΩ. Ohm's law then gives about 4.3 mA. Ignoring the diode forward drop would overstate the current.

Question 39Analog Electronics

Which property of an ideal operational amplifier operating with negative feedback is most directly responsible for the virtual short at its input terminals?

  1. Infinite input resistance
  2. Zero output resistance
  3. Infinite open-loop voltage gain Correct answer.
  4. Zero common-mode gain

Explanation

With negative feedback and finite output, the differential input voltage must be approximately zero because the open-loop gain is treated as infinite. This forces the two input terminals to the same potential, creating the virtual short. The other listed properties affect loading and rejection but not the virtual short condition itself.

Question 40Analog Electronics

A zener diode with a rated breakdown voltage of 5.1 V is reverse biased through a series resistor from a 12 V source. In regulation, the voltage across the zener is approximately:

  1. 6.9 V
  2. 0.7 V
  3. 12 V
  4. 5.1 V Correct answer.

Explanation

In reverse breakdown the zener clamps its terminal voltage near the rated breakdown value, here 5.1 V. The series resistor absorbs the remaining voltage difference from the 12 V source. Operation must keep current within the zener's rated range.

Question 41Analog Electronics

Which quantity is defined as the ratio of the change in collector current to the change in base current at constant collector-emitter voltage?

  1. Early voltage
  2. Small-signal transconductance
  3. Forward current gain beta Correct answer.
  4. Base transit time

Explanation

Forward current gain, denoted beta or hFE, is the ratio of collector current change to base current change at fixed collector-emitter voltage. Transconductance instead relates collector current to base-emitter voltage. Early voltage and transit time describe output slope and speed, not current gain.

Question 42Analog Electronics

An inverting amplifier uses an input resistor of 2 kΩ and a feedback resistor of 20 kΩ. What is the closed-loop voltage gain?

  1. -0.1
  2. +11
  3. +10
  4. -10 Correct answer.

Explanation

For the inverting configuration the gain equals negative feedback resistance divided by input resistance, so -20 kΩ / 2 kΩ equals -10. The negative sign reflects the phase inversion. A noninverting gain of +11 would require a different topology.

Question 43Analog Electronics

A capacitor of 10 µF is connected across a 10 V DC source and allowed to reach steady state. What is the approximate energy stored?

  1. 1.0 mJ
  2. 100 mJ
  3. 10 mJ
  4. 0.5 mJ Correct answer.

Explanation

Energy in a capacitor is one half times capacitance times voltage squared. Substituting 10 µF and 10 V gives 0.5 times 10e-6 times 100, which equals 0.5 mJ. The voltage is squared, so the result is not simply capacitance times voltage.

Question 44Analog Electronics

Which transistor configuration provides voltage gain slightly less than unity with no phase inversion and a high input impedance?

  1. Common base
  2. Common emitter
  3. Common gate
  4. Common collector Correct answer.

Explanation

The common-collector or emitter-follower stage has voltage gain near but below one, no inversion, and high input impedance. It is used mainly as a buffer for current gain and impedance transformation. Common emitter gives inversion and larger voltage gain, while common base gives noninverted gain with low input impedance.

Question 45Analog Electronics

A common-emitter amplifier has a voltage gain of -50 and an input signal of 10 mV peak. What is the approximate peak output voltage, assuming the amplifier remains linear?

  1. 5.0 V
  2. 0.2 V
  3. 0.5 V Correct answer.
  4. 2.0 V

Explanation

The output magnitude equals the gain magnitude times the input amplitude, so 50 times 10 mV equals 500 mV, or 0.5 V peak. The negative sign only indicates inversion. Linearity must hold for the calculation to apply.

Question 46Analog Electronics

In a bridge rectifier feeding a capacitor-input filter, which parameter most strongly determines the peak-to-peak ripple voltage for a given load current?

  1. Diode forward voltage
  2. Diode reverse recovery time
  3. Transformer turns ratio
  4. Load resistance and filter capacitance Correct answer.

Explanation

Ripple in a capacitor-input supply is approximated by load current divided by the product of capacitance and ripple frequency. Larger capacitance or lighter load reduces ripple. Diode drops and turns ratio set the DC level but not the ripple magnitude directly.

Question 47Analog Electronics

An op-amp differentiator has an input resistor of 1 kΩ and a feedback capacitor of 0.1 µF. For a sinusoidal input at 1 kHz, what is the approximate magnitude of the gain?

  1. 6.3
  2. 0.63 Correct answer.
  3. 0.16
  4. 1.6

Explanation

The differentiator gain magnitude is the angular frequency times the feedback capacitance times the input resistance. At 1 kHz the angular frequency is about 6283 rad/s, so the product with 0.1 µF and 1 kΩ gives about 0.63. Using frequency in hertz without the factor of two pi would give a different result.

Question 48Analog Electronics

A bipolar transistor is biased at a collector current of 2 mA with beta equal to 100. What is the approximate base current?

  1. 2 mA
  2. 20 µA Correct answer.
  3. 2 µA
  4. 200 µA

Explanation

Base current equals collector current divided by beta, so 2 mA divided by 100 equals 20 µA. This small base current supports the much larger collector current. Confusing the ratio direction reverses the result.

Question 49Analog Electronics

Increasing the value of the emitter bypass capacitor in a common-emitter amplifier while keeping the bias point fixed will:

  1. Increase midband voltage gain Correct answer.
  2. Decrease the input impedance at low frequency
  3. Reduce midband voltage gain
  4. Change the DC operating point

Explanation

A larger bypass capacitor lowers the emitter impedance at a given signal frequency, reducing negative feedback and raising midband gain. The DC bias is set by resistors and is unaffected by the bypass capacitor value. The gain increase saturates once the emitter is fully bypassed.

Question 50Analog Electronics

A noninverting amplifier uses a 1 kΩ resistor to ground and a 9 kΩ feedback resistor. What is the closed-loop gain?

  1. 9
  2. 0.1
  3. -9
  4. 10 Correct answer.

Explanation

For the noninverting configuration the gain is one plus the feedback-to-ground resistor ratio, so 1 plus 9 equals 10. The input signal appears at the noninverting terminal, giving a positive gain. The inverting formula would give a different result.