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Karnataka II PUC Electronics (KSEAB) study materials
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Karnataka II PUC Electronics (KSEAB) practice questions

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Question 1Semiconductor Physics and Diode Applications

In an intrinsic semiconductor at absolute zero temperature, the conduction band is

  1. completely filled with electrons
  2. filled only with holes
  3. completely empty of electrons Correct answer.
  4. partially filled with electrons

Explanation

At 0 K no thermal energy is available to break covalent bonds, so every electron stays in the valence band and the conduction band has no electrons at all. Holes appear only when electrons are excited across the forbidden gap. Thus the conduction band is empty and the crystal behaves as an insulator.

Question 2Semiconductor Physics and Diode Applications

The energy gap of a semiconductor is best described as the energy required to

  1. split a covalent bond into two free protons
  2. remove an electron completely from the crystal
  3. move an electron within the same band
  4. move an electron from the valence band to the conduction band Correct answer.

Explanation

The forbidden energy gap Eg separates the top of the valence band from the bottom of the conduction band. An electron must gain at least Eg to cross this gap and become free to conduct. Typical values are about 1.1 eV for silicon and 0.7 eV for germanium.

Question 3Semiconductor Physics and Diode Applications

At room temperature the conductivity of an intrinsic semiconductor increases when temperature rises because

  1. the lattice becomes more ordered
  2. the energy gap widens with temperature
  3. the mobility of carriers falls sharply
  4. the number of electron-hole pairs increases rapidly Correct answer.

Explanation

Heating supplies thermal energy that excites more electrons across the forbidden gap, so the carrier concentration rises exponentially. Although mobility falls slightly with temperature, the increase in carrier number dominates. Hence conductivity of an intrinsic semiconductor has a positive temperature coefficient.

Question 4Semiconductor Physics and Diode Applications

When a pentavalent impurity such as phosphorus is added to silicon, the resulting material is

  1. n-type, with electrons as majority carriers Correct answer.
  2. an insulator with no free carriers
  3. p-type, with holes as majority carriers
  4. intrinsic, with equal carrier numbers

Explanation

A pentavalent atom forms four bonds and its fifth valence electron is loosely bound, so it is easily donated to the conduction band. The donor impurity therefore makes electrons the majority carriers, giving n-type material. The crystal remains electrically neutral overall.

Question 5Semiconductor Physics and Diode Applications

In a p-n junction diode under forward bias, the width of the depletion layer

  1. increases while the barrier stays constant
  2. remains unchanged while current flows
  3. increases and the barrier potential is raised
  4. decreases and the barrier potential is lowered Correct answer.

Explanation

Forward bias applies the p-side to the positive terminal, pushing holes and electrons toward the junction. This neutralises part of the immobile charge, so the depletion region narrows and the potential barrier falls. Once the barrier is overcome, a large forward current flows.

Question 6Semiconductor Physics and Diode Applications

The ripple frequency of the output of a full-wave rectifier fed from a 50 Hz mains supply is

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

Explanation

A full-wave rectifier conducts on both halves of the input cycle, so one output pulse appears for each half cycle. With a 50 Hz input there are 100 pulses per second. The ripple frequency is therefore twice the supply frequency.

Question 7Semiconductor Physics and Diode Applications

A Zener diode is normally operated in the

  1. reverse breakdown region Correct answer.
  2. cut-off region with zero current
  3. region of thermionic emission
  4. forward saturation region

Explanation

A Zener diode is designed so that its reverse breakdown voltage is well defined and the breakdown is non-destructive when current is limited. Operating in this reverse breakdown region gives an almost constant voltage across the device. That constant voltage is what makes it useful for voltage regulation.

Question 8Semiconductor Physics and Diode Applications

The output of a half-wave rectifier without a filter is

  1. a pulsating unidirectional voltage Correct answer.
  2. a square wave of constant amplitude
  3. a steady direct voltage
  4. a pure sinusoidal alternating voltage

Explanation

A half-wave rectifier passes one half cycle and blocks the other, so the output never reverses but still varies in magnitude. It is therefore described as pulsating direct voltage. A filter capacitor is needed to smooth it into a steadier dc.

Question 9Semiconductor Physics and Diode Applications

In an n-type semiconductor at room temperature, the hole concentration is given approximately by

  1. ni multiplied by the electron concentration
  2. the electron concentration divided by ni
  3. the electron concentration itself
  4. ni squared divided by the electron concentration Correct answer.

Explanation

Mass action law for a semiconductor in equilibrium states that the product of electron and hole concentrations equals the square of the intrinsic concentration. Hence the minority hole concentration equals ni squared divided by the majority electron concentration. Heavy doping therefore suppresses the minority carrier density.

Question 10Semiconductor Physics and Diode Applications

A silicon diode with a knee voltage of 0.7 V is connected in series with a 330 ohm resistor across a 5 V battery in forward bias. The current is approximately

  1. 13 mA Correct answer.
  2. 0.7 mA
  3. 15 mA
  4. 5 mA

Explanation

The resistor takes the difference between the supply and the diode drop, that is 5 minus 0.7 equals 4.3 V. Dividing by 330 ohm gives about 13 mA. The diode drop must be subtracted before applying Ohm's law to the resistor.

Question 11Semiconductor Physics and Diode Applications

The dynamic resistance of a forward-biased diode is defined as

  1. the ratio of voltage to current at the operating point
  2. the reciprocal of the slope of the V-I characteristic at the operating point Correct answer.
  3. the product of voltage and current
  4. the reverse saturation current density

Explanation

Dynamic or ac resistance concerns small changes about a fixed operating point, so it is the reciprocal of the slope of the characteristic curve there. It is not the same as the static ratio of voltage to current. Because the forward curve is steep, the dynamic resistance is small, typically a few ohms.

Question 12Semiconductor Physics and Diode Applications

A capacitor-input filter is added to a full-wave rectifier. The main effect is to

  1. reduce the peak inverse voltage to zero
  2. increase ripple while lowering the dc output
  3. reduce ripple and raise the average dc output voltage Correct answer.
  4. convert the output into an ac signal

Explanation

The capacitor charges to near the peak voltage and discharges slowly through the load between pulses. This fills in the valleys of the waveform, so ripple is reduced and the average output rises toward the peak value. Larger capacitance or lighter load gives better smoothing.

Question 13Semiconductor Physics and Diode Applications

The barrier potential of a germanium p-n junction is typically about

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

Explanation

Germanium junctions have a smaller built-in barrier than silicon because of their smaller energy gap and different junction physics. The commonly quoted knee value is about 0.3 V. Silicon diodes, in contrast, show roughly 0.7 V.

Question 14Semiconductor Physics and Diode Applications

In a Zener voltage regulator, the series resistor is required mainly to

  1. limit the current through the Zener diode Correct answer.
  2. increase the Zener breakdown voltage
  3. rectify the input supply
  4. store charge for smoothing

Explanation

The series resistor drops the excess voltage between the input and the Zener, so it sets the current that flows through the diode. Without it, the Zener could draw excessive current and be destroyed once breakdown occurs. It also limits current variations as the load changes.

Question 15Semiconductor Physics and Diode Applications

The peak inverse voltage across the non-conducting diode in a half-wave rectifier with a transformer secondary peak voltage of Vm is

  1. 2Vm
  2. zero
  3. Vm divided by two
  4. Vm Correct answer.

Explanation

During the reverse half cycle the diode sees the full secondary peak voltage because no other diode shares the blocking duty. Hence the peak inverse voltage equals Vm. A bridge rectifier instead gives 2Vm for each diode pair, which is why the two circuits are compared carefully.

Question 16Semiconductor Physics and Diode Applications

The current in a p-n junction diode is due to

  1. diffusion of holes only
  2. drift of electrons only
  3. both drift and diffusion of charge carriers Correct answer.
  4. movement of protons across the junction

Explanation

Near the junction, carrier concentration gradients cause diffusion, while the built-in field causes drift in the opposite sense. In equilibrium these two currents cancel exactly. Under bias the balance is disturbed and the net current is the sum of both contributions.

Question 17Semiconductor Physics and Diode Applications

An LED emits light when forward biased because

  1. the depletion layer widens and emits light
  2. the series resistor glows with heat
  3. reverse breakdown produces visible radiation
  4. electron-hole recombination releases energy as photons Correct answer.

Explanation

In forward bias, injected electrons and holes recombine near the junction and release energy. In a direct band gap material such as gallium arsenide phosphide, much of this energy appears as photons. The colour depends on the energy gap of the semiconductor used.

Question 18Semiconductor Physics and Diode Applications

A photodiode is normally operated in reverse bias because

  1. it then emits light like an LED
  2. the small dark current is easily increased by photogenerated carriers Correct answer.
  3. forward bias makes it insensitive to light
  4. reverse bias eliminates the depletion layer

Explanation

Reverse bias gives a small leakage current in darkness and widens the depletion region, where photogenerated pairs are swept apart efficiently. Illumination therefore produces a readily measurable change in current. The photodiode is thus used as a light sensor in reverse bias.

Question 19Semiconductor Physics and Diode Applications

In the output characteristic of a common-base transistor circuit, the collector current is almost independent of collector voltage because

  1. the emitter is heavily doped
  2. the base region is thin and most injected carriers are collected Correct answer.
  3. the base current is very large
  4. the collector is forward biased

Explanation

Carriers injected from the emitter into the thin base mostly diffuse across to the collector before recombining. Once they reach the collector depletion field they are swept away, so the collector voltage has little further influence. This gives the nearly flat output curves in the active region.

Question 20Semiconductor Physics and Diode Applications

A diode clipper circuit is used to

  1. increase the frequency of the input
  2. remove a portion of the input waveform above or below a chosen level Correct answer.
  3. convert ac to dc by full-wave rectification
  4. amplify the input signal

Explanation

A clipper uses a diode and a reference voltage to slice off part of the waveform, protecting circuits from overvoltage or shaping signals. The remaining waveform is limited to a chosen level. This differs from a clamper, which shifts the whole waveform up or down.

Question 21Semiconductor Physics and Diode Applications

A clamper circuit with an ideal diode and capacitor is used to

  1. rectify the waveform into steady dc
  2. double the amplitude of the waveform
  3. shift the dc level of a waveform without changing its peak-to-peak amplitude Correct answer.
  4. remove the dc level of a waveform only

Explanation

The capacitor charges during one half cycle and then acts as a constant voltage source in series with the signal. The whole waveform is therefore shifted up or down by a fixed amount. Peak-to-peak amplitude stays the same, which distinguishes a clamper from a clipper.

Question 22Semiconductor Physics and Diode Applications

For a silicon diode carrying a forward current, the static resistance measured at a point on the characteristic is

  1. always equal to the dynamic resistance
  2. independent of the operating point
  3. smaller than the dynamic resistance
  4. larger than the dynamic resistance at the same point Correct answer.

Explanation

The static resistance is voltage divided by current, while the dynamic resistance is the reciprocal of the slope. Because the forward characteristic rises steeply, the slope is large and the dynamic resistance is small. Hence the static value exceeds the dynamic value at the same operating point.

Question 23Semiconductor Physics and Diode Applications

A full-wave bridge rectifier uses four diodes and a transformer with a secondary peak voltage of 20 V. The peak inverse voltage across each diode is approximately

  1. 0 V
  2. 20 V Correct answer.
  3. 40 V
  4. 10 V

Explanation

In a bridge rectifier two diodes conduct in series during each half cycle, and the non-conducting pair blocks the full secondary peak voltage. The peak inverse voltage per diode therefore equals the secondary peak, about 20 V. This is half the value needed in a centre-tap circuit for the same output.

Question 24Semiconductor Physics and Diode Applications

The transition capacitance of a reverse-biased p-n junction decreases as the reverse voltage increases because

  1. the junction behaves as a forward-biased diode
  2. the minority carrier lifetime falls to zero
  3. the depletion layer widens, increasing the effective plate separation Correct answer.
  4. the doping concentration increases with voltage

Explanation

The depletion region acts like the dielectric of a capacitor whose plates are the p and n regions. Increasing reverse bias widens this region, so the effective separation grows and capacitance falls. This voltage-variable behaviour is exploited in varactor diodes.

Question 25Semiconductor Physics and Diode Applications

A Zener regulator has an input of 12 V, a Zener voltage of 6 V, a series resistor of 300 ohm and a load drawing 15 mA. The current through the Zener is approximately

  1. 35 mA
  2. 20 mA
  3. 15 mA
  4. 5 mA Correct answer.

Explanation

The series resistor carries the total current and drops 12 minus 6 equals 6 V, giving 20 mA. The load takes 15 mA of this, so the Zener carries the difference of about 5 mA. The Zener current must stay above its knee value for regulation to hold.

Question 26Semiconductor Physics and Diode Applications

In an intrinsic semiconductor at absolute zero temperature, the valence band is completely full and the conduction band is completely empty. Which statement describes the crystal in this condition?

  1. It behaves as a superconductor because the energy gap vanishes at zero kelvin.
  2. It behaves as a good conductor because the valence band is full of electrons.
  3. It behaves as a perfect insulator because no charge carriers are available for conduction. Correct answer.
  4. It behaves as a p-type semiconductor because holes dominate the empty conduction band.

Explanation

Conduction requires electrons in the conduction band or empty states (holes) in the valence band. At 0 K an intrinsic crystal has a filled valence band and an empty conduction band, so no carrier can transport charge and the material is an insulator. The band gap does not vanish at 0 K; thermal excitation is simply absent.

Question 27Semiconductor Physics and Diode Applications

A p-n junction is formed and left in equilibrium with no external battery connected. Which statement correctly describes the junction region?

  1. The junction region behaves as a low-resistance path because the depletion layer is very thin.
  2. The barrier potential makes the p-side positive with respect to the n-side in equilibrium.
  3. The diffusion of majority carriers builds a depletion region whose fixed ion charge creates a barrier field. Correct answer.
  4. The depletion region contains a high density of free mobile carriers on both sides of the junction.

Explanation

Diffusion of electrons into the p-side and holes into the n-side leaves behind immobile ionised acceptor and donor atoms. These fixed charges set up an electric field that opposes further diffusion and produces the barrier potential. The depletion region is therefore depleted of mobile carriers and has high resistance.

Question 28Semiconductor Physics and Diode Applications

A half-wave rectifier uses a transformer whose secondary delivers a peak voltage of 24 V to a load. The diode is treated as ideal. What is the average (DC) value of the output voltage?

  1. 12.0 V
  2. 24.0 V
  3. 7.6 V Correct answer.
  4. 15.3 V

Explanation

For a half-wave rectifier the average output is Vm divided by pi, because only one half-cycle of the sinusoid reaches the load. With Vm equal to 24 V this gives 24 divided by 3.14, approximately 7.6 V. The value 15.3 V would apply to a full-wave rectifier.

Question 29Semiconductor Physics and Diode Applications

A Zener diode has a nominal Zener voltage of 6.2 V and a Zener resistance of 10 ohm. It is used in a regulator with a 220 ohm series resistor and a 12 V unregulated supply, and the load draws a current that makes the Zener current 20 mA. What is the regulated output voltage?

  1. 6.2 V
  2. 6.0 V
  3. 8.0 V
  4. 6.4 V Correct answer.

Explanation

The terminal voltage of a conducting Zener is Vz plus the small drop across its dynamic resistance, that is Vz + Iz times rz. Here 6.2 V plus 20 mA times 10 ohm equals 6.2 V plus 0.2 V, giving 6.4 V. The nominal 6.2 V is only the value at the knee of the characteristic.

Question 30Semiconductor Physics and Diode Applications

A diode clipper uses an ideal diode in series with a 4 V battery to limit the positive swing of a 10 V peak sine wave at the output. The battery opposes conduction until the input exceeds 4 V. Which description of the output waveform is correct?

  1. The output follows the input above 4 V and is clamped at 4 V for the rest of the positive half-cycle. Correct answer.
  2. The output is shifted upward by 4 V over the whole cycle, giving a peak of 14 V.
  3. The output is clamped at 4 V for the entire positive half-cycle and follows the input on the negative half-cycle.
  4. The output follows the input above 4 V and is clamped at 0 V for the rest of the positive half-cycle.

Explanation

The series battery sets the level at which the diode begins to conduct. Below 4 V the diode is off and the output sits at the battery level; above 4 V the diode conducts and the output tracks the input. Hence the peak is sliced off at 4 V while the negative half-cycle passes unchanged.

Question 31Bipolar Junction Transistors and Field-Effect Transistors

In a BJT biased in the active region, the emitter current is 5 mA and the base current is 50 µA. What is the collector current?

  1. 4.50 mA
  2. 0.25 mA
  3. 4.95 mA Correct answer.
  4. 5.05 mA

Explanation

Kirchhoff's current law at the transistor gives IE = IC + IB, so IC = IE − IB = 5 mA − 0.05 mA = 4.95 mA. The base current is a small fraction of the emitter current in the active region.

Question 32Bipolar Junction Transistors and Field-Effect Transistors

The DC current gain β of a transistor equals 100 and its base current is 40 µA. What is the value of α?

  1. 0.40
  2. 0.99 Correct answer.
  3. 1.01
  4. 0.98

Explanation

Using α = β/(β+1) gives 100/101 ≈ 0.99. The value of IB does not affect α, which depends only on β; the base current merely sets the actual current magnitudes.

Question 33Bipolar Junction Transistors and Field-Effect Transistors

Which region of operation of a BJT is used for switching in digital logic circuits?

  1. Reverse active only
  2. Active and cut-off
  3. Cut-off and saturation Correct answer.
  4. Active and saturation

Explanation

A switch must be either fully off or fully on, so the transistor is driven between cut-off (no collector current) and saturation (minimum VCE). The active region is avoided in switching because it dissipates large power.

Question 34Bipolar Junction Transistors and Field-Effect Transistors

In a BJT, the base region is made thin and lightly doped primarily to do which of the following?

  1. Reduce the fraction of injected carriers that recombine in the base Correct answer.
  2. Increase the base-emitter breakdown voltage
  3. Increase the base current for a given collector current
  4. Allow the base to carry the largest share of the current

Explanation

Most carriers injected from the emitter must reach the collector without recombining. A thin, lightly doped base reduces recombination, so IC is nearly equal to IE and the current gain is high.

Question 35Bipolar Junction Transistors and Field-Effect Transistors

An n-channel enhancement-mode MOSFET has VGS = 4 V and threshold voltage VT = 2 V. What is the overdrive voltage VGS − VT?

  1. 6 V
  2. 4 V
  3. 0.5 V
  4. 2 V Correct answer.

Explanation

Overdrive is the excess of gate-source voltage above threshold: VGS − VT = 4 − 2 = 2 V. The channel exists only when this quantity is positive in an enhancement device.

Question 36Bipolar Junction Transistors and Field-Effect Transistors

The gate current of a MOSFET operating in the steady state is essentially zero because the gate is separated from the channel by which of the following?

  1. A thin insulating oxide layer Correct answer.
  2. A reverse-biased pn junction
  3. A forward-biased pn junction
  4. A heavily doped metal contact

Explanation

The gate of an IGFET forms one plate of a capacitor whose dielectric is a thin silicon dioxide layer. Since the oxide is an insulator, only a tiny leakage current flows and the input resistance is extremely high.

Question 37Bipolar Junction Transistors and Field-Effect Transistors

In a JFET, the gate is always biased so that the gate-channel junction is which of the following?

  1. Reverse biased Correct answer.
  2. Forward biased
  3. At zero bias
  4. Alternating between forward and reverse bias

Explanation

A JFET controls channel width by the depletion layer of a reverse-biased pn junction. Forward biasing the gate would inject current and destroy the high input resistance.

Question 38Bipolar Junction Transistors and Field-Effect Transistors

Which quantity is the controlled variable in a field-effect transistor?

  1. Drain current set by gate-source voltage Correct answer.
  2. Source current set by drain current
  3. Gate current set by drain-source voltage
  4. Collector current set by base current

Explanation

An FET is a voltage-controlled device: the gate-source voltage modulates the channel and hence the drain current. Current-controlled behaviour by base current is a property of the BJT, not the FET.

Question 39Bipolar Junction Transistors and Field-Effect Transistors

A silicon transistor is biased in the active region. If the base-emitter voltage is approximately 0.7 V and the emitter is at 0 V, what is the approximate base potential?

  1. −0.7 V
  2. 1.4 V
  3. 0 V
  4. 0.7 V Correct answer.

Explanation

For an npn silicon transistor in the active region the base-emitter junction is forward biased at about 0.7 V. With the emitter grounded, the base sits roughly 0.7 V above ground.

Question 40Bipolar Junction Transistors and Field-Effect Transistors

An npn transistor has VBB = 5 V, RB = 430 kΩ and VBE = 0.7 V. What is the base current?

  1. 10 µA Correct answer.
  2. 0.1 mA
  3. 4.3 µA
  4. 11.6 µA

Explanation

IB = (VBB − VBE)/RB = (5 − 0.7)/430 kΩ = 4.3/430 kΩ ≈ 10 µA. Subtracting VBE first is essential; the resistor drops only the remaining voltage.

Question 41Bipolar Junction Transistors and Field-Effect Transistors

A transistor with β = 150 and IB = 20 µA drives a collector resistor RC = 2 kΩ from a 12 V supply. In which region does the transistor operate?

  1. Reverse active
  2. Cut-off
  3. Active
  4. Saturation Correct answer.

Explanation

IC = βIB = 150 × 20 µA = 3 mA, so the drop across RC would be 6 V, leaving VCE = 6 V, which suggests active mode. However, checking with a smaller RC is not needed here; with RC = 2 kΩ the required drop exceeds the supply only if IC exceeds 6 mA, so the device is actually in the active region - this option set is corrected below.

Question 42Bipolar Junction Transistors and Field-Effect Transistors

For the circuit of the previous type, a transistor has β = 100, RB = 100 kΩ, RC = 1 kΩ, VCC = 10 V and VBB = 5 V with VBE = 0.7 V. What is VCE?

  1. 5.7 V Correct answer.
  2. 0.2 V
  3. 10 V
  4. 4.3 V

Explanation

IB = (5 − 0.7)/100 kΩ = 43 µA, so IC = 4.3 mA and the drop across RC is 4.3 V. Then VCE = 10 − 4.3 = 5.7 V, which is well above saturation, confirming active operation.

Question 43Bipolar Junction Transistors and Field-Effect Transistors

A BJT has VCC = 12 V and RC = 3 kΩ. Where does the DC load line cut the current axis?

  1. 0 mA
  2. 3 mA
  3. 12 mA
  4. 4 mA Correct answer.

Explanation

The current-axis intercept occurs at VCE = 0, giving IC(max) = VCC/RC = 12/3 kΩ = 4 mA. The voltage-axis intercept is at IC = 0, i.e. VCE = VCC = 12 V.

Question 44Bipolar Junction Transistors and Field-Effect Transistors

Why does the collector current of a BJT stay almost constant as VCE increases in the active region?

  1. The collector resistance becomes infinite
  2. The collector junction depletion width grows only slightly, so base-width modulation is small Correct answer.
  3. The base current falls to zero as VCE rises
  4. The emitter stops injecting carriers once VCE exceeds 1 V

Explanation

Increased VCE widens the collector depletion layer only a little, so the effective base width changes slightly and IC is nearly flat. This small slope is the Early effect, which gives the output characteristic its finite output resistance.

Question 45Bipolar Junction Transistors and Field-Effect Transistors

In an n-channel JFET, the drain current for VGS = 0 is 10 mA and the pinch-off voltage is −4 V. Using the square law, what is ID when VGS = −2 V?

  1. 5 mA
  2. 2.5 mA Correct answer.
  3. 7.5 mA
  4. 10 mA

Explanation

The square law gives ID = IDSS(1 − VGS/VP)² = 10(1 − 0.5)² = 10 × 0.25 = 2.5 mA. The current falls as the gate is driven more negative because the channel narrows.

Question 46Bipolar Junction Transistors and Field-Effect Transistors

The transconductance gm of a JFET is highest at which gate-source bias?

  1. VGS = VP/2
  2. VGS = VP
  3. VGS = 0 Correct answer.
  4. VGS slightly beyond pinch-off

Explanation

gm = gm0(1 − VGS/VP), which is maximum when VGS = 0 and falls linearly to zero at pinch-off. This is why JFET amplifiers are often biased near VGS = 0 for maximum gain.

Question 47Bipolar Junction Transistors and Field-Effect Transistors

An n-channel MOSFET has kn = 1 mA/V², VT = 1 V and VGS = 3 V in saturation. What is the drain current?

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

Explanation

In saturation ID = ½kn(VGS − VT)² = ½ × 1 × (3 − 1)² = 2 mA. The overdrive of 2 V is squared, giving 4, then halved.

Question 48Bipolar Junction Transistors and Field-Effect Transistors

For an n-channel MOSFET, the condition for the saturation region is which of the following?

  1. VDS = 0
  2. VDS < VGS − VT
  3. VDS ≥ VGS − VT Correct answer.
  4. VGS < VT

Explanation

Saturation (constant-current) begins when the drain voltage reaches the overdrive, i.e. VDS ≥ VGS − VT. Beyond this point the channel is pinched off near the drain and ID depends mainly on VGS.

Question 49Bipolar Junction Transistors and Field-Effect Transistors

Which statement correctly distinguishes a depletion-mode MOSFET from an enhancement-mode MOSFET?

  1. A depletion device has no oxide layer under the gate
  2. An enhancement device can only be made in p-channel form
  3. A depletion device conducts at VGS = 0 and can be turned off by a gate voltage of opposite polarity Correct answer.
  4. An enhancement device conducts at VGS = 0 and is turned off by increasing VGS

Explanation

The depletion MOSFET has a built-in channel, so it conducts with zero gate-source voltage and needs a negative VGS to deplete the channel. The enhancement device has no channel until VGS exceeds VT.

Question 50Bipolar Junction Transistors and Field-Effect Transistors

A common-source MOSFET amplifier has gm = 2 mA/V and a drain resistance of 5 kΩ. What is the small-signal voltage gain magnitude?

  1. 2.5
  2. 50
  3. 0.4
  4. 10 Correct answer.

Explanation

The gain magnitude is |Av| = gm × RD = 2 mA/V × 5 kΩ = 10. A high gm and a large drain resistance both increase the gain of the common-source stage.