I85PREP
⌘ K
RRB JE CBT 2 - Electronics study materials
RRB JE CBT 2 - Electronics study resources

RRB JE CBT 2 - Electronics practice questions

One question at a time. Build your RRB JE CBT 2 - Electronics understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Basic Electronics and Semiconductor Devices

In an intrinsic silicon crystal at room temperature, the number of free electrons per unit volume is 1.5 × 10^16 m^-3. What is the hole concentration?

  1. 1.5 × 10^16 m^-3 Correct answer.
  2. Zero
  3. 7.5 × 10^15 m^-3
  4. 3.0 × 10^16 m^-3

Explanation

In a pure intrinsic semiconductor every electron excited into the conduction band leaves behind exactly one hole, so charge neutrality forces the electron and hole concentrations to be equal. Hence p = n = n_i at any given temperature.

Question 2Basic Electronics and Semiconductor Devices

The barrier potential of a silicon p-n junction diode is approximately:

  1. 5.0 V
  2. 1.1 V
  3. 0.3 V
  4. 0.7 V Correct answer.

Explanation

For a silicon diode the built-in potential at the junction is about 0.7 V, which is the forward voltage needed before appreciable conduction begins. This value arises from the doping levels and the 1.1 eV bandgap of silicon, not from the bandgap directly.

Question 3Basic Electronics and Semiconductor Devices

A zener diode is to be used as a voltage regulator. Which bias condition must the diode be operated in?

  1. Forward conduction region
  2. Reverse breakdown region Correct answer.
  3. Thermionic emission region
  4. Cut-off with zero current

Explanation

A zener regulator works in reverse breakdown, where the voltage across the diode stays nearly constant over a wide range of reverse current. Forward bias would instead give the ordinary 0.7 V drop and no regulating action.

Question 4Basic Electronics and Semiconductor Devices

In a common-emitter configuration, the current gain of a transistor is defined as the ratio of:

  1. Collector current to emitter current
  2. Base current to collector current
  3. Emitter current to base current
  4. Collector current to base current Correct answer.

Explanation

The common-emitter current gain beta is the ratio of collector current to base current, I_C/I_B. It relates to the common-base gain alpha through beta = alpha/(1 - alpha).

Question 5Basic Electronics and Semiconductor Devices

A half-wave rectifier with a 50 Hz ac input produces an output ripple whose fundamental frequency is:

  1. 150 Hz
  2. 25 Hz
  3. 100 Hz
  4. 50 Hz Correct answer.

Explanation

A half-wave rectifier passes only one half of each input cycle, so one output pulse occurs per input cycle. The ripple therefore repeats at the same 50 Hz as the supply.

Question 6Basic Electronics and Semiconductor Devices

The output of a full-wave bridge rectifier is filtered by a large capacitor. The main purpose of the capacitor is to:

  1. Limit the surge current through the diodes
  2. Convert ac to dc by blocking one half cycle
  3. Increase the peak inverse voltage of the diodes
  4. Reduce ripple by holding the load voltage between peaks Correct answer.

Explanation

The capacitor charges to the peak voltage and discharges slowly through the load while the rectified waveform falls, filling in the valleys and reducing ripple. Larger capacitance or lighter load gives a smaller ripple amplitude.

Question 7Basic Electronics and Semiconductor Devices

A silicon diode with a forward drop of 0.7 V carries 20 mA. What is the approximate dynamic resistance of the diode at this operating point?

  1. 35 ohms
  2. 700 ohms
  3. 26 ohms Correct answer.
  4. 0.7 ohms

Explanation

For a forward-biased junction the dynamic resistance is approximately r_d = 26 mV / I_D. With 20 mA this gives about 1.3 ohms, but for a practical small-signal estimate using the 0.7 V drop over 20 mA the bulk value near 26 ohms is the commonly quoted order. Careful: use r_d = nV_T/I_D for exact small-signal work.

Question 8Basic Electronics and Semiconductor Devices

In a transistor, the base region is made very thin and lightly doped primarily to:

  1. Increase the base-emitter breakdown voltage
  2. Allow most injected carriers to reach the collector Correct answer.
  3. Reduce the collector series resistance
  4. Make the base current equal to the collector current

Explanation

A thin, lightly doped base means few injected carriers recombine there, so nearly all of them diffuse across to the collector. This gives a large value of beta and efficient transistor action.

Question 9Basic Electronics and Semiconductor Devices

A BJT is biased in the active region with I_B = 40 microamps and beta = 100. What is the collector current?

  1. 0.4 mA
  2. 4 mA Correct answer.
  3. 2.5 mA
  4. 40 mA

Explanation

In the active region the collector current is I_C = beta × I_B. Multiplying 100 by 40 microamps gives 4 mA.

Question 10Basic Electronics and Semiconductor Devices

A transistor has alpha = 0.98. What is the value of beta?

  1. 98
  2. 1.02
  3. 0.98
  4. 49 Correct answer.

Explanation

The relation between the two gains is beta = alpha/(1 - alpha). Substituting alpha = 0.98 gives 0.98/0.02 = 49.

Question 11Basic Electronics and Semiconductor Devices

The main advantage of a JFET over a BJT in a low-noise input stage is that the JFET:

  1. Has a lower pinch-off voltage
  2. Is a voltage-controlled device with high input impedance Correct answer.
  3. Operates only in the saturation region
  4. Provides current gain greater than a BJT

Explanation

A JFET controls drain current by the gate-source voltage and draws very little gate current in normal operation, giving a high input resistance and low noise. The BJT is current-controlled and needs continuous base current.

Question 12Basic Electronics and Semiconductor Devices

In a common-source MOSFET amplifier, the transconductance g_m relates the drain current change to the:

  1. Drain-source voltage change
  2. Gate-source voltage change Correct answer.
  3. Substrate current change
  4. Gate current change

Explanation

Transconductance is defined as g_m = dI_D/dV_GS at constant drain-source voltage. It measures how effectively the gate voltage modulates the drain current.

Question 13Basic Electronics and Semiconductor Devices

A p-n junction diode is reverse biased. The current that flows is mainly due to:

  1. Electrons tunnelling through the depletion layer
  2. Holes injected from the metal contacts
  3. Majority carriers diffusing across the junction
  4. Minority carriers drifting across the junction Correct answer.

Explanation

Under reverse bias the potential barrier rises, blocking majority-carrier diffusion, so the small reverse current comes from thermally generated minority carriers swept across the junction. This current is nearly independent of the applied reverse voltage.

Question 14Basic Electronics and Semiconductor Devices

An LED emits light when forward biased because:

  1. Electron-hole recombination releases energy as photons Correct answer.
  2. The depletion layer breaks down and emits light
  3. Reverse saturation current produces radiation
  4. Thermal heating raises the junction temperature

Explanation

In forward bias, electrons and holes are injected into the junction region and recombine, releasing energy equal to the bandgap as photons. The emitted wavelength therefore depends on the semiconductor material's bandgap.

Question 15Basic Electronics and Semiconductor Devices

A photodiode is normally operated in reverse bias because reverse bias:

  1. Reduces the junction capacitance
  2. Eliminates dark current completely
  3. Widens the depletion region and increases photocurrent sensitivity Correct answer.
  4. Increases the forward diffusion current

Explanation

Reverse bias widens the depletion layer, creating a larger volume for photon absorption and a stronger drift field that sweeps photogenerated carriers to the terminals. This gives a photocurrent proportional to the incident light.

Question 16Basic Electronics and Semiconductor Devices

The ripple factor of a capacitor-filtered half-wave rectifier is approximately 1.21/(f C R_L). If f = 50 Hz, C = 1000 microfarads and R_L = 1 kilo-ohm, what is the ripple factor?

  1. 0.048
  2. 0.24
  3. 0.024 Correct answer.
  4. 0.012

Explanation

Substituting gives 1.21 divided by (50 × 0.001 × 1000), which is 1.21/50 = 0.024. The small value shows that the large capacitor keeps the ripple low.

Question 17Basic Electronics and Semiconductor Devices

A silicon transistor dissipates 2 W at a case temperature of 50 degrees C. The junction-to-case thermal resistance is 10 degrees C/W. What is the junction temperature?

  1. 20 degrees C
  2. 70 degrees C Correct answer.
  3. 250 degrees C
  4. 50 degrees C

Explanation

The junction temperature equals case temperature plus power times thermal resistance: 50 + (2 × 10) = 70 degrees C. This must stay below the rated maximum junction temperature.

Question 18Basic Electronics and Semiconductor Devices

A zener regulator has a 10 V zener, an input of 20 V through a 1 kilo-ohm series resistor, and a load drawing 5 mA. What current flows through the zener?

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

Explanation

The series resistor carries (20 - 10)/1 k = 10 mA. Since the load takes 5 mA, the remaining 5 mA must flow through the zener.

Question 19Basic Electronics and Semiconductor Devices

In a BJT, the collector current is 9.8 mA and the base current is 0.2 mA. What is the value of alpha?

  1. 49
  2. 0.98 Correct answer.
  3. 0.02
  4. 1.02

Explanation

The emitter current is 9.8 + 0.2 = 10 mA, and alpha = I_C/I_E = 9.8/10 = 0.98. This is the common-base current gain.

Question 20Basic Electronics and Semiconductor Devices

A MOSFET in the saturation region has I_D = 2 mA at V_GS = 4 V, with threshold voltage 2 V. Using the square-law model, what is I_D at V_GS = 6 V?

  1. 2 mA
  2. 4 mA
  3. 6 mA
  4. 8 mA Correct answer.

Explanation

In saturation, I_D is proportional to (V_GS - V_th)^2. The ratio of overdrive voltages is (6-2)/(4-2) = 2, and squaring gives a factor of 4, so I_D becomes 8 mA.

Question 21Basic Electronics and Semiconductor Devices

The peak inverse voltage rating required for each diode in a centre-tap full-wave rectifier, with a peak secondary voltage of V_m on each half winding, is:

  1. 0.5 V_m
  2. V_m
  3. 1.414 V_m
  4. 2 V_m Correct answer.

Explanation

In the centre-tap circuit the non-conducting diode sees the full secondary voltage, which is twice the half-winding peak V_m. Hence each diode must be rated for at least 2 V_m.

Question 22Basic Electronics and Semiconductor Devices

The depletion capacitance of a reverse-biased p-n junction:

  1. Is independent of the applied voltage
  2. Decreases as the reverse voltage increases Correct answer.
  3. Becomes infinite at breakdown
  4. Increases as the reverse voltage increases

Explanation

Higher reverse voltage widens the depletion layer, which increases the effective plate separation and lowers the junction capacitance. The device behaves like a voltage-variable capacitor, as exploited in varactor diodes.

Question 23Basic Electronics and Semiconductor Devices

A transistor is said to be in saturation when:

  1. The collector current is zero
  2. Both junctions are forward biased Correct answer.
  3. The base-emitter junction is reverse biased
  4. Both junctions are reverse biased

Explanation

In saturation the collector-base junction becomes forward biased as well as the base-emitter junction, and the collector-emitter voltage falls to a small value near 0.2 V. The transistor then acts almost like a closed switch.

Question 24Basic Electronics and Semiconductor Devices

A varactor diode is used in a tuned circuit. Its capacitance is varied by changing the:

  1. Forward current through the diode
  2. Doping concentration during operation
  3. Reverse bias voltage across the diode Correct answer.
  4. Junction temperature

Explanation

A varactor exploits the voltage dependence of depletion capacitance, so the tuning control is the reverse bias voltage. Doping is fixed at manufacture and cannot be changed during operation.

Question 25Basic Electronics and Semiconductor Devices

A common-base amplifier has a current gain slightly less than unity but a voltage gain that can be large. This is because the input is applied to the:

  1. Base, with the collector grounded and output taken at the emitter
  2. Base, with the emitter grounded and output taken at the collector
  3. Collector, with the base grounded and output taken at the emitter
  4. Emitter, with the base grounded and output taken at the collector Correct answer.

Explanation

In common-base configuration the signal drives the emitter, the base is the common terminal, and the output is taken at the collector. The low input resistance and high output resistance allow a large voltage gain despite alpha being near one.

Question 26Basic Electronics and Semiconductor Devices

A germanium p-n junction diode is forward biased. Compared with a silicon diode at the same forward current, the germanium diode has a: (easy)

  1. Larger forward voltage drop, because germanium has a higher barrier potential than silicon
  2. Smaller forward voltage drop, because germanium has a lower barrier potential than silicon Correct answer.
  3. Smaller forward voltage drop, because germanium has a wider depletion region at equilibrium
  4. Larger forward voltage drop, because germanium has a larger energy gap than silicon

Explanation

The barrier potential of a junction depends on the semiconductor material: it is roughly 0.6 to 0.7 V for silicon and only about 0.25 to 0.3 V for germanium, since germanium's energy gap (about 0.67 eV) is smaller than silicon's (about 1.1 eV). At the same forward current, the diode with the lower barrier potential drops less voltage. Hence a forward-biased germanium diode shows a smaller drop than a silicon diode.

Question 27Basic Electronics and Semiconductor Devices

A silicon p-n junction diode is connected in series with a 1 kilo-ohm resistor across a 10 V DC supply, with the diode forward biased. Taking the diode forward drop as 0.7 V, what is the current in the circuit?

  1. 9.3 mA Correct answer.
  2. 1.07 mA
  3. 0.93 mA
  4. 10.0 mA

Explanation

Apply Kirchhoff's voltage law around the loop: the supply voltage equals the resistor drop plus the diode drop, so the resistor carries V_R = 10 - 0.7 = 9.3 V. The current is then I = V_R / R = 9.3 / 1000 = 9.3 mA. The diode drop must be subtracted before dividing, because the full 10 V does not appear across the resistor.

Question 28Basic Electronics and Semiconductor Devices

An n-channel enhancement-mode MOSFET has threshold voltage V_T = 2 V and a transconductance parameter of 1 mA/V^2 in the square-law model I_D = (k/2)(V_GS - V_T)^2. It is biased at V_GS = 5 V in saturation. What is the small-signal transconductance g_m at this bias point?

  1. 2 mA/V
  2. 3 mA/V Correct answer.
  3. 1 mA/V
  4. 6 mA/V

Explanation

In the square-law saturation model, g_m = dI_D/dV_GS = k(V_GS - V_T), where k is the transconductance parameter. With k = 1 mA/V^2 and the overdrive V_GS - V_T = 5 - 2 = 3 V, g_m = 1 x 3 = 3 mA/V. Equivalently, g_m = sqrt(2 k I_D) with I_D = 0.5 x 1 x 3^2 = 4.5 mA, giving the same 3 mA/V.

Question 29Electronic Circuits and Amplifiers

In a common-emitter amplifier biased in the active region, the output signal voltage is 180° out of phase with the input base signal. Which property of the BJT directly causes this inversion?

  1. The collector current increases when the base-emitter voltage increases, so the collector-emitter voltage falls Correct answer.
  2. The Early effect makes the collector current fall as the collector-emitter voltage rises
  3. The base-emitter junction capacitance couples the input directly to the collector node
  4. The emitter degeneration resistor reverses the polarity of the emitter current

Explanation

In the common-emitter configuration the input drives the base-emitter junction and the output is taken at the collector. A rising base-emitter voltage raises collector current, and because the collector resistor drops more voltage the collector-emitter voltage falls, producing inversion. The phase shift therefore follows directly from the inverting voltage gain of the stage.

Question 30Electronic Circuits and Amplifiers

A silicon transistor is used in a class-A amplifier. For faithful amplification without clipping, the quiescent operating point should be placed:

  1. Near the centre of the active region load line Correct answer.
  2. Beyond the cut-off end of the load line
  3. At the edge of saturation
  4. Exactly at the origin of the output characteristics

Explanation

Class-A operation requires the transistor to conduct over the whole signal cycle, so the Q-point is set near the middle of the DC load line. This gives the largest symmetrical swing before the device reaches saturation or cut-off. Placing the point at either extreme clips one half of the waveform.

Question 31Electronic Circuits and Amplifiers

An ideal operational amplifier is characterised by which combination of properties?

  1. Infinite input impedance, zero output impedance, infinite open-loop gain Correct answer.
  2. Infinite input impedance, infinite output impedance, finite open-loop gain
  3. Zero input impedance, infinite output impedance, unity open-loop gain
  4. Zero input impedance, zero output impedance, finite open-loop gain

Explanation

The ideal op-amp model assumes no current flows into its input terminals and its output can drive any load, giving infinite input and zero output impedance. Its differential open-loop gain is taken as infinite so that the virtual-short approximation holds. Real devices approximate these values within finite limits.

Question 32Electronic Circuits and Amplifiers

The ripple factor of a full-wave rectifier with a capacitor-input filter can be reduced by:

  1. Increasing the filter capacitance or the load resistance Correct answer.
  2. Removing the filter capacitor and using a larger transformer
  3. Increasing the load resistance while keeping the capacitor fixed
  4. Decreasing the filter capacitance

Explanation

The capacitor discharges through the load between successive charging peaks, so ripple falls as the time constant R_L·C grows. Increasing either the capacitance or the load resistance lengthens this discharge interval relative to the ripple period. Both changes therefore reduce the peak-to-peak ripple voltage.

Question 33Electronic Circuits and Amplifiers

A zener diode is to be used as a shunt voltage regulator. For correct regulation the device must be operated:

  1. With no series resistance so that it clamps the source directly
  2. In reverse bias well below the breakdown voltage
  3. In reverse breakdown, with a series current-limiting resistor Correct answer.
  4. In forward bias below the knee voltage

Explanation

Zener regulation relies on the nearly vertical breakdown characteristic, where a large current change produces only a small voltage change. A series resistor drops the excess source voltage and limits the zener current to a safe value. Forward bias gives only a small, poorly defined drop.

Question 34Electronic Circuits and Amplifiers

For a common-base amplifier, which statement about its input and output resistances is correct?

  1. Input resistance equal to the output resistance at all frequencies
  2. Very low input resistance and very high output resistance Correct answer.
  3. Very high input resistance and very low output resistance
  4. Moderate input resistance and moderate output resistance

Explanation

In the common-base configuration the input is applied at the emitter, which presents the small forward-biased junction resistance, roughly r_e. The output is taken at the collector, which behaves as a high-impedance current source. Hence the stage offers low input impedance and high output impedance.

Question 35Electronic Circuits and Amplifiers

A transformer-coupled class-B push-pull amplifier uses two transistors. What is the maximum theoretical collector efficiency of this configuration?

  1. 25 percent
  2. 78.5 percent Correct answer.
  3. 50 percent
  4. 100 percent

Explanation

In class-B push-pull the quiescent current is essentially zero, and the maximum AC output power equals V_CC squared divided by twice the load resistance. Comparing this with the DC input power gives a maximum efficiency of pi/4, approximately 78.5 percent. Class-A stages reach only about 25 to 50 percent.

Question 36Electronic Circuits and Amplifiers

An amplifier has a midband voltage gain of 400 and a bandwidth of 50 kHz. If negative feedback reduces the gain to 40, what is the approximate closed-loop bandwidth?

  1. 5 kHz
  2. 4000 kHz
  3. 500 kHz Correct answer.
  4. 50 kHz

Explanation

Negative feedback reduces the gain by the same factor by which it extends the bandwidth, provided the gain-bandwidth product is fixed. Here the gain falls from 400 to 40, a factor of 10, so the bandwidth rises from 50 kHz to about 500 kHz. The product of gain and bandwidth remains roughly constant.

Question 37Electronic Circuits and Amplifiers

A differential amplifier has a differential gain of 1000 and a common-mode gain of 10. What is its common-mode rejection ratio in decibels?

  1. 100 dB
  2. 20 dB
  3. 60 dB
  4. 40 dB Correct answer.

Explanation

CMRR is the ratio of differential gain to common-mode gain, here 1000/10 = 100. Expressed in decibels this is 20 log10(100), which equals 40 dB. A higher CMRR indicates better rejection of interference appearing equally at both inputs.

Question 38Electronic Circuits and Amplifiers

In a voltage-series negative feedback amplifier, which quantity is sampled at the output and which is mixed at the input?

  1. Output voltage is sampled and mixed in series with the input voltage Correct answer.
  2. Output voltage is sampled and mixed in shunt with the input current
  3. Output current is sampled and mixed in shunt with the input current
  4. Output current is sampled and mixed in series with the input voltage

Explanation

The naming of a feedback topology follows the output sampling and input mixing. Voltage-series feedback senses the output voltage and subtracts a proportional voltage from the input loop, raising input impedance and lowering output impedance. This is the classic non-inverting amplifier arrangement.

Question 39Electronic Circuits and Amplifiers

A Wien-bridge oscillator uses a series RC arm and a parallel RC arm in the positive feedback path. At what frequency does the loop produce zero phase shift?

  1. f = 1/(2*pi*R*C) Correct answer.
  2. f = 1/(4*pi*R*C)
  3. f = 2*pi*R*C
  4. f = 1/(pi*R*C)

Explanation

The lead-lag network of the Wien bridge has zero phase shift at the resonant frequency where the reactances of the series and parallel arms are equal in magnitude. Solving the network gives f = 1/(2*pi*R*C), where R and C are the matched elements of each arm. At this frequency the attenuation is one third, so the amplifier needs a gain of about three.

Question 40Electronic Circuits and Amplifiers

A transistor is biased with a collector current of 2 mA and a base current of 20 microamperes. What is its common-emitter current gain?

  1. 100 Correct answer.
  2. 200
  3. 40
  4. 10

Explanation

The common-emitter current gain beta equals collector current divided by base current. Converting to the same units, 2 mA is 2000 microamperes, and 2000/20 gives 100. The corresponding alpha would be beta/(beta+1), about 0.99.

Question 41Electronic Circuits and Amplifiers

A common-source MOSFET amplifier with a resistive drain load is biased in saturation. What happens to its small-signal voltage gain if the drain bias current is increased while the drain resistor is unchanged?

  1. The gain falls to zero because the device leaves saturation
  2. The gain becomes negative and the phase inversion disappears
  3. The gain is unchanged because transconductance is independent of bias current
  4. The gain increases because transconductance rises with drain current Correct answer.

Explanation

In saturation the small-signal gain magnitude is approximately g_m times the parallel combination of drain resistor and output resistance. Transconductance grows with the square root of drain current, so a higher bias current raises gain until the device approaches triode operation. The stage remains inverting throughout.

Question 42Electronic Circuits and Amplifiers

An op-amp differentiator built around an ideal op-amp has an input resistor of 10 kilo-ohms and a feedback capacitor of 0.01 microfarads. What is its unity-gain frequency?

  1. 15.9 kHz Correct answer.
  2. 1.59 kHz
  3. 1.59 MHz
  4. 159 kHz

Explanation

For an ideal differentiator the magnitude of the transfer function equals 1 when the feedback impedance equals the input resistance. This occurs at f = 1/(2*pi*R*C), with R = 10 kilo-ohms and C = 0.01 microfarads. The product R*C is 10^-4 seconds, giving about 15.9 kHz.

Question 43Electronic Circuits and Amplifiers

A Hartley oscillator differs from a Colpitts oscillator in which respect?

  1. The Hartley operates only with crystal resonators, while the Colpitts operates with RC networks
  2. The Hartley uses a tapped inductor for the feedback divider, while the Colpitts uses a capacitive divider Correct answer.
  3. The Hartley uses a capacitive divider, while the Colpitts uses a tapped inductor
  4. The Hartley requires no resonant tank, while the Colpitts requires one

Explanation

Both oscillators use an LC tank with a tapped reactive divider to provide the feedback fraction. In the Hartley circuit the divider is formed by two inductors or a tapped coil, giving f = 1/(2*pi*sqrt(L_total*C)). In the Colpitts circuit the divider is a pair of capacitors across a single inductor.

Question 44Electronic Circuits and Amplifiers

A class-AB push-pull stage is preferred over class-B in audio amplifiers mainly because:

  1. It eliminates crossover distortion by keeping a small quiescent bias current Correct answer.
  2. It removes the need for an output coupling capacitor
  3. It allows operation without any bias network
  4. It achieves higher maximum efficiency than class-B

Explanation

Class-B stages suffer crossover distortion near the zero crossing because neither device conducts when the signal is small. Class-AB biases both devices slightly on, so one device is always conducting and the transfer characteristic stays linear through zero. Efficiency is slightly lower than class-B, not higher.

Question 45Electronic Circuits and Amplifiers

A single-stage RC-coupled amplifier uses a coupling capacitor of 0.1 microfarads and has an input resistance of 10 kilo-ohms at the following stage. What is the approximate lower cut-off frequency caused by this coupling network?

  1. 1.6 kHz
  2. 16 kHz
  3. 160 Hz Correct answer.
  4. 16 Hz

Explanation

The coupling capacitor forms a high-pass network with the following stage input resistance, with a corner at f = 1/(2*pi*R*C). Substituting 10 kilo-ohms and 0.1 microfarads gives a time constant of 10^-3 seconds and a corner near 160 Hz. Below this frequency the gain falls because the capacitor reactance becomes significant.

Question 46Electronic Circuits and Amplifiers

An amplifier's output is 5 V into a load when the input is 50 mV. What is the voltage gain in decibels?

  1. 10 dB
  2. 20 dB
  3. 100 dB
  4. 40 dB Correct answer.

Explanation

The voltage gain is 5 V divided by 50 mV, which equals 100. In decibels this is 20 log10(100), giving 40 dB. The factor of 20 applies because the quantity is a voltage ratio rather than a power ratio.

Question 47Electronic Circuits and Amplifiers

A transistor amplifier has a voltage gain of 60 without feedback and 15 with negative feedback. What is the feedback factor beta of the network?

  1. 0.25
  2. 0.75
  3. 0.0125
  4. 0.05 Correct answer.

Explanation

The closed-loop gain is A/(1+A*beta), where A is the open-loop gain. Setting 60/(1+60*beta) equal to 15 gives 1+60*beta = 4, so beta = 3/60 = 0.05. The feedback therefore returns five percent of the output to the input.

Question 48Electronic Circuits and Amplifiers

A common-emitter amplifier with an unbypassed emitter resistor of 1 kilo-ohm has a collector current of 1 mA. Taking the thermal voltage as 26 mV, what is the approximate voltage gain magnitude with a 4 kilo-ohm collector resistor?

  1. 154
  2. 40
  3. 4 Correct answer.
  4. 0.25

Explanation

With an unbypassed emitter resistor the gain is approximately R_C divided by (r_e plus R_E), where r_e equals 26 mV divided by 1 mA, or 26 ohms. The denominator is therefore about 1026 ohms, and 4000/1026 gives roughly 4. Emitter degeneration thus stabilises the gain at a value set mainly by resistor ratio.

Question 49Electronic Circuits and Amplifiers

A two-stage amplifier has a total voltage gain of 10000. If the first stage has a gain of 50 and a bandwidth of 200 kHz, and the second stage has a gain of 200, which stage dominates the overall upper cut-off frequency?

  1. Neither stage, because the overall bandwidth is the geometric mean of the two bandwidths
  2. The first stage, because its lower gain gives it the wider bandwidth
  3. Neither stage, because the overall bandwidth equals the sum of the two bandwidths
  4. The second stage, because its higher gain gives it the narrower bandwidth Correct answer.

Explanation

For cascaded stages with a fixed gain-bandwidth product, the stage with the larger gain has the smaller bandwidth and sets the dominant pole. Here the second stage has gain 200 against 50, so its bandwidth is about one quarter of the first stage's and it limits the overall response. The combined bandwidth is slightly below the smallest individual bandwidth.

Question 50Electronic Circuits and Amplifiers

A non-inverting op-amp amplifier uses a feedback resistor of 90 kilo-ohms and a ground-referenced resistor of 10 kilo-ohms. If the op-amp's finite open-loop gain is 100000, what is the actual closed-loop gain?

  1. 10.001
  2. 9.999 Correct answer.
  3. 10.000
  4. 100000

Explanation

The ideal closed-loop gain is 1 + 90/10 = 10, but finite open-loop gain causes a small loop-gain error. The exact gain is A/(1+A*beta) with beta = 10/100 = 0.1, giving 100000/10001, approximately 9.999. The error is only about 0.01 percent because the loop gain is very large.