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CHSE Odisha Elective Electronics study materials
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CHSE Odisha Elective Electronics practice questions

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Question 1Basic Electricity and Passive Components

A four-band resistor has colour bands brown, black, red and gold. What are its nominal resistance and tolerance?

  1. 1 kΩ, ±5% Correct answer.
  2. 10 kΩ, ±5%
  3. 10 Ω, ±2%
  4. 100 Ω, ±10%

Explanation

In the four-band code, brown = 1 and black = 0, giving the digits 10; red is the multiplier ×100, so the value is 10 × 100 Ω = 1 kΩ. Gold denotes a tolerance of ±5%. Therefore the nominal resistance is 1 kΩ with ±5% tolerance.

Question 2Basic Electricity and Passive Components

Three resistors of 100 Ω, 220 Ω and 330 Ω are connected in series. What is the equivalent resistance?

  1. 65 Ω
  2. 650 Ω Correct answer.
  3. 330 Ω
  4. 217 Ω

Explanation

In a series circuit the same current flows through every resistor and the equivalent resistance is the arithmetic sum: 100 + 220 + 330 = 650 Ω. Series resistance is always greater than the largest individual resistor.

Question 3Basic Electricity and Passive Components

A 10 µF capacitor is connected across a 50 V d.c. supply. What charge is stored?

  1. 0.2 µC
  2. 500 µC Correct answer.
  3. 50 µC
  4. 5 µC

Explanation

The charge stored by a capacitor is Q = CV. Substituting C = 10 µF = 10 × 10^-6 F and V = 50 V gives Q = 500 × 10^-6 C = 500 µC.

Question 4Basic Electricity and Passive Components

Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance?

  1. 9 Ω
  2. 4.5 Ω
  3. 18 Ω
  4. 2 Ω Correct answer.

Explanation

For two resistors in parallel, R = (R1 R2)/(R1 + R2) = (6 × 3)/(6 + 3) = 18/9 = 2 Ω. A parallel combination is always smaller than the smallest branch resistance.

Question 5Basic Electricity and Passive Components

A 240 Ω resistor carries a current of 50 mA. What power is dissipated in it?

  1. 0.12 W
  2. 12 W
  3. 1.2 W
  4. 0.6 W Correct answer.

Explanation

Power in a resistor can be found from P = I²R. With I = 50 mA = 0.05 A and R = 240 Ω, P = (0.05)² × 240 = 0.6 W.

Question 6Basic Electricity and Passive Components

Two capacitors of 6 µF and 3 µF are connected in series. What is the equivalent capacitance?

  1. 4.5 µF
  2. 2 µF Correct answer.
  3. 18 µF
  4. 9 µF

Explanation

For capacitors in series, 1/C = 1/C1 + 1/C2 = 1/6 + 1/3 = 1/2 µF⁻¹, so C = 2 µF. The series equivalent is smaller than the smallest capacitor.

Question 7Basic Electricity and Passive Components

An inductor of 4 H carries a steady current of 3 A. How much energy is stored in its magnetic field?

  1. 6 J
  2. 36 J
  3. 18 J Correct answer.
  4. 12 J

Explanation

The energy stored in an inductor is W = ½LI². Substituting L = 4 H and I = 3 A gives W = ½ × 4 × 3² = 18 J.

Question 8Basic Electricity and Passive Components

A cell of e.m.f. 1.5 V and internal resistance 0.5 Ω is connected to a 2.5 Ω resistor. What is the terminal voltage of the cell?

  1. 1.5 V
  2. 1.25 V Correct answer.
  3. 1.0 V
  4. 0.25 V

Explanation

The current is I = E/(R + r) = 1.5/(2.5 + 0.5) = 0.5 A. The terminal voltage is V = E − Ir = 1.5 − (0.5)(0.5) = 1.25 V.

Question 9Basic Electricity and Passive Components

A uniform wire of resistance R is stretched uniformly until its length is doubled. What is its new resistance?

  1. R
  2. 2R
  3. 4R Correct answer.
  4. R/2

Explanation

When the wire is stretched uniformly, its volume stays constant. Doubling the length halves the cross-sectional area; since R = ρl/A, the resistance becomes 2/(1/2) = 4 times the original value.

Question 10Basic Electricity and Passive Components

Two capacitors of 2 µF and 4 µF are connected in series across a 120 V d.c. supply. What is the potential difference across the 2 µF capacitor?

  1. 40 V
  2. 60 V
  3. 120 V
  4. 80 V Correct answer.

Explanation

The series equivalent is C = (2 × 4)/(2 + 4) = 4/3 µF. The charge is Q = CV = (4/3 µF)(120 V) = 160 µC, so the voltage across the 2 µF capacitor is V = Q/C = 160 µC / 2 µF = 80 V.

Question 11Basic Electricity and Passive Components

In the four-band resistor colour code, what tolerance is indicated by a silver fourth band?

  1. ±5 %
  2. ±2 %
  3. ±1 %
  4. ±10 % Correct answer.

Explanation

In the four-band code the fourth band is the tolerance band. Silver denotes ±10 %, while gold is ±5 %, brown ±1 % and red ±2 %. The first three bands set the resistance value, so tolerance is read from the fourth band alone.

Question 12Basic Electricity and Passive Components

A ceramic capacitor is marked 104. What is its nominal capacitance?

  1. 10 nF
  2. 100 nF Correct answer.
  3. 10 pF
  4. 100 µF

Explanation

The marking 104 uses two significant digits and a multiplier in picofarads. It means 10 × 10^4 pF = 100 000 pF, which equals 100 nF or 0.1 µF. Therefore the component is a 100 nF capacitor.

Question 13Basic Electricity and Passive Components

A potentiometer is to be used as a two-terminal variable resistor. Which terminals should be connected into the circuit?

  1. The wiper and both outer end terminals
  2. The wiper and one outer end terminal Correct answer.
  3. All three terminals in series with the load
  4. The two outer end terminals only

Explanation

A potentiometer has three terminals, but as a two-terminal variable resistor only the wiper and one outer terminal are used. The resistance then depends on the position of the wiper along the track. Connecting only the two outer terminals gives the full fixed track resistance and no variation.

Question 14Basic Electricity and Passive Components

A 1 kΩ resistor has a continuous power rating of 0.25 W. What is the largest current it can carry without exceeding that rating?

  1. 4.0 mA
  2. 62.5 mA
  3. 250 mA
  4. 15.8 mA Correct answer.

Explanation

The power rating limits the current through P = I²R. Rearranging gives I = √(P/R) = √(0.25/1000) = 0.0158 A, or about 15.8 mA. A larger current would exceed the resistor's continuous power rating.

Question 15Basic Electricity and Passive Components

A 10 µF capacitor and a 22 µF capacitor are connected in parallel. What is the equivalent capacitance?

  1. 220 µF
  2. 12 µF
  3. 32 µF Correct answer.
  4. 6.9 µF

Explanation

Capacitors in parallel have the same voltage across them, and their capacitances simply add. Thus C = 10 µF + 22 µF = 32 µF. The series formula, which would give about 6.9 µF here, does not apply to a parallel connection.

Question 16Basic Electricity and Passive Components

A 100 µF capacitor is charged through a 10 kΩ resistor. What is the time constant of the circuit?

  1. 0.1 s
  2. 1.0 s Correct answer.
  3. 10 s
  4. 0.01 s

Explanation

The time constant of an RC circuit is τ = RC. Here τ = (10 × 10³ Ω)(100 × 10⁻⁶ F) = 1.0 s. This is the time in which the capacitor voltage reaches about 63 % of its final value during charging.

Question 17Basic Electricity and Passive Components

An ideal 2 H inductor is connected in series with a 100 Ω resistor and a 12 V d.c. battery. What current flows after the circuit has been switched on for a long time?

  1. 12 A
  2. 1.2 A
  3. 0 A
  4. 0.12 A Correct answer.

Explanation

After a long time the current becomes steady, so di/dt is zero and the ideal inductor has no voltage across it. The inductor then behaves as a short circuit, and the current is set by the resistor: I = 12 V / 100 Ω = 0.12 A.

Question 18Basic Electricity and Passive Components

A 10 kΩ potentiometer is connected across a 10 V supply, and its wiper is set one quarter of the way from the lower end. With nothing connected to the wiper, what voltage appears between the wiper and the lower end?

  1. 4.0 V
  2. 0.25 V
  3. 7.5 V
  4. 2.5 V Correct answer.

Explanation

With no load at the wiper, the potentiometer acts as a simple voltage divider. The output is the same fraction of the supply voltage as the wiper's position: 0.25 × 10 V = 2.5 V. Loading the wiper would reduce this value.

Question 19Basic Electricity and Passive Components

A capacitor is charged to 100 V and then disconnected from the supply. A dielectric of relative permittivity 4 is inserted to fill the space between its plates. What is the new potential difference across the capacitor?

  1. 400 V
  2. 25 V Correct answer.
  3. 100 V
  4. 50 V

Explanation

Because the capacitor is disconnected, its charge Q remains constant. Inserting a dielectric of relative permittivity 4 multiplies the capacitance by 4, and since V = Q/C the voltage falls to one quarter: 100 V / 4 = 25 V.

Question 20Basic Electricity and Passive Components

A 12 Ω resistor and a 6 Ω resistor are connected in parallel. This pair is joined in series with an 8 Ω resistor, and the resulting combination is connected in parallel with a 24 Ω resistor. What is the total resistance?

  1. 26 Ω
  2. 4 Ω
  3. 12 Ω
  4. 8 Ω Correct answer.

Explanation

First combine the parallel pair: 12 Ω || 6 Ω = (12 × 6)/(12 + 6) = 4 Ω. Adding the 8 Ω series resistor gives 12 Ω. This 12 Ω then combines with the 24 Ω in parallel: (12 × 24)/(12 + 24) = 8 Ω.

Question 21Semiconductor Physics and Diodes

At thermal equilibrium in a pure intrinsic semiconductor, the concentration of free electrons is:

  1. equal to the hole concentration Correct answer.
  2. less than the hole concentration
  3. greater than the hole concentration
  4. zero, because no impurity atoms are present

Explanation

Thermal energy breaks covalent bonds and creates electron-hole pairs in equal numbers, so in a pure semiconductor the free-electron concentration n equals the hole concentration p, and both equal the intrinsic concentration ni. Charge neutrality is therefore preserved, and neither carrier type outnumbers the other at equilibrium.

Question 22Semiconductor Physics and Diodes

The potential barrier across a silicon p-n junction at room temperature is approximately:

  1. 1.1 V
  2. 0.7 V Correct answer.
  3. 5.0 V
  4. 0.2 V

Explanation

For a silicon p-n junction the contact potential at room temperature is about 0.7 V, whereas germanium gives roughly 0.3 V. The exact value depends on doping levels and temperature, but 0.7 V is the standard figure used in diode calculations for silicon.

Question 23Semiconductor Physics and Diodes

In a forward-biased p-n junction diode, the current across the junction is carried mainly by:

  1. recombination of majority carriers inside the depletion layer
  2. drift of minority carriers across the junction
  3. diffusion of majority carriers across the junction Correct answer.
  4. thermionic emission of electrons from the p-region

Explanation

Forward bias lowers the potential barrier, so the majority carriers on each side diffuse across the junction: holes from the p-region and electrons from the n-region. This diffusion current grows exponentially with applied voltage and greatly exceeds the small drift current of minority carriers.

Question 24Semiconductor Physics and Diodes

A full-wave rectifier is supplied from a 50 Hz mains. The fundamental ripple frequency in its output is:

  1. 75 Hz
  2. 25 Hz
  3. 50 Hz
  4. 100 Hz Correct answer.

Explanation

A full-wave rectifier conducts on both halves of the input cycle, so the output waveform contains two pulses for every input cycle. For a 50 Hz supply the fundamental ripple frequency is therefore 2 x 50 = 100 Hz, which is easier to filter than the 50 Hz ripple of a half-wave rectifier.

Question 25Semiconductor Physics and Diodes

A Zener diode used as a voltage regulator must be operated:

  1. in reverse bias well below the breakdown voltage
  2. in reverse breakdown with a current-limiting series resistor Correct answer.
  3. in forward bias above the knee voltage
  4. without any series resistance to keep the load voltage fixed

Explanation

Regulation is obtained in the reverse breakdown region, where the voltage across the Zener stays almost constant while the current through it varies. A series resistor is essential: it drops the excess supply voltage and keeps the Zener current between the knee value and the maximum rated value so the device is not destroyed.

Question 26Semiconductor Physics and Diodes

In a bridge rectifier, the peak inverse voltage across each diode is equal to:

  1. the peak secondary voltage Correct answer.
  2. twice the peak secondary voltage
  3. the RMS secondary voltage divided by two
  4. half the peak secondary voltage

Explanation

During each half cycle two diodes conduct in series, so the non-conducting diode blocks the full peak secondary voltage, giving PIV = Vm for a bridge rectifier. This is half the PIV required in a centre-tapped full-wave rectifier, where each diode must withstand 2Vm.

Question 27Semiconductor Physics and Diodes

A silicon diode carries 2 mA at 0.7 V. Taking the ideality factor as 1 and the thermal voltage as 26 mV, its dynamic resistance at this operating point is approximately:

  1. 26 ohm
  2. 52 ohm
  3. 6.5 ohm
  4. 13 ohm Correct answer.

Explanation

The dynamic (small-signal) resistance of a junction diode is rd = nVT/ID. Substituting n = 1, VT = 26 mV and ID = 2 mA gives rd = 26 mV / 2 mA = 13 ohm, which is the slope resistance around the quiescent point.

Question 28Semiconductor Physics and Diodes

A photodiode is normally operated in reverse bias because:

  1. reverse bias eliminates the junction capacitance completely
  2. reverse bias makes the diode emit light
  3. the wider depletion region collects more photogenerated carriers Correct answer.
  4. forward bias produces a much larger photocurrent

Explanation

Reverse bias widens the depletion region and leaves only a small dark current flowing in the absence of light. Carriers generated by absorbed photons inside that region are swept out by the strong field, producing a photocurrent that is proportional to the illumination.

Question 29Semiconductor Physics and Diodes

In an n-type semiconductor at thermal equilibrium, the electron diffusion current is:

  1. zero, because the electron concentration is uniform
  2. larger than the drift current, giving a net current
  3. equal and opposite to the electron drift current Correct answer.
  4. equal to the hole diffusion current in the same direction

Explanation

At equilibrium no net current can flow, so the drift component driven by the built-in field must exactly cancel the diffusion component driven by the concentration gradient. The two components are equal in magnitude and opposite in direction, and the same balance holds for holes.

Question 30Semiconductor Physics and Diodes

Zener breakdown differs from avalanche breakdown in that Zener breakdown:

  1. is irreversible and permanently destroys the diode
  2. results from field emission across a very narrow depletion region Correct answer.
  3. needs a wide depletion layer and relies on impact ionisation
  4. occurs only at reverse voltages above about 6 V in silicon

Explanation

Zener breakdown takes place in heavily doped junctions whose depletion region is very narrow, so a modest reverse voltage produces a field strong enough for electrons to tunnel directly from the valence to the conduction band. Avalanche breakdown occurs in lightly doped, wider junctions at higher voltages through impact ionisation, and both mechanisms are reversible provided the current is limited.

Question 31Semiconductor Physics and Diodes

In a p-type semiconductor at room temperature, the majority charge carriers are:

  1. holes Correct answer.
  2. immobile donor ions
  3. immobile acceptor ions
  4. conduction electrons

Explanation

Doping with a trivalent impurity such as boron creates acceptor levels that are almost fully ionised at room temperature. Each acceptor atom supplies a mobile hole, so holes outnumber electrons and become the majority carriers. The ionised acceptor cores remain fixed in the lattice and do not carry current.

Question 32Semiconductor Physics and Diodes

The depletion region of an unbiased p-n junction contains mainly:

  1. free electrons only
  2. a vacuum layer between the two regions
  3. fixed ionised donor and acceptor ions Correct answer.
  4. mobile electrons and holes

Explanation

Diffusion of carriers across the junction leaves behind uncompensated donor ions on the n-side and acceptor ions on the p-side. These ions are fixed in the crystal lattice and create the built-in electric field. They are not mobile carriers, so they do not contribute to conduction.

Question 33Semiconductor Physics and Diodes

A half-wave rectifier is operated from a 50 Hz AC mains supply. The fundamental ripple frequency in its output is:

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

Explanation

A half-wave rectifier passes only one half-cycle of the input for each complete input cycle. Therefore one output pulse appears per input cycle, and the ripple repeats at the same frequency as the mains supply, which is 50 Hz.

Question 34Semiconductor Physics and Diodes

When a p-n junction diode is reverse biased, the width of its depletion region:

  1. collapses to zero when the reverse voltage exceeds 0.7 V
  2. remains unchanged, because no steady current flows
  3. decreases, because majority carriers are pushed toward the junction
  4. increases, because the external field adds to the barrier field Correct answer.

Explanation

Reverse bias makes the p-side negative and the n-side positive, which pulls majority carriers away from the junction. This exposes more fixed ions and widens the depletion layer. The wider layer strengthens the barrier field until it balances the applied reverse voltage.

Question 35Semiconductor Physics and Diodes

In a simple Zener voltage regulator, the series resistor connected between the supply and the Zener diode mainly:

  1. prevents the Zener diode from entering reverse breakdown
  2. smooths the ripple voltage from the rectifier
  3. increases the Zener current without any limit
  4. drops the excess supply voltage and limits the Zener current Correct answer.

Explanation

The series resistor takes up the difference between the unregulated input voltage and the fixed Zener voltage. It also sets a safe current through the Zener diode and the load. Without this resistor, the Zener current could rise excessively and damage the diode.

Question 36Semiconductor Physics and Diodes

A varactor diode is used as a voltage-controlled capacitor and is normally operated in:

  1. reverse bias, below breakdown Correct answer.
  2. reverse breakdown, where current rises sharply
  3. forward bias, where it conducts heavily
  4. zero bias, with no external voltage

Explanation

A varactor diode is designed to act as a voltage-controlled capacitor. Reverse bias widens the depletion layer and reduces the junction capacitance, so it is operated in reverse bias below breakdown. Operating it in forward conduction would ruin its capacitor action.

Question 37Semiconductor Physics and Diodes

The switching speed of a Schottky diode is higher than that of an ordinary p-n junction diode because:

  1. it has a much wider depletion region than a p-n diode
  2. majority-carrier operation avoids minority storage Correct answer.
  3. it stores more minority charge in forward bias
  4. it conducts only beyond the Zener voltage

Explanation

A Schottky diode is formed by a metal-semiconductor junction and conducts mainly by majority carriers. Because there is almost no minority-carrier charge storage, it can switch off much faster than an ordinary p-n junction diode. This makes it useful in high-speed switching circuits.

Question 38Semiconductor Physics and Diodes

A light-emitting diode emits light when it is forward biased because:

  1. electrons and holes recombine to emit photons Correct answer.
  2. the depletion region widens and emits thermal radiation
  3. reverse saturation current heats the junction to incandescence
  4. majority carriers are accelerated to produce X-rays

Explanation

In forward bias, electrons are injected into the p-region and holes into the n-region. Their recombination across the energy gap releases energy, which appears as photons in a direct-bandgap material such as GaAs or GaAsP. The emitted wavelength depends on the semiconductor band gap.

Question 39Semiconductor Physics and Diodes

A 6 V Zener diode is fed from a 12 V supply through a 200 ohm series resistor. A 300 ohm load is connected across the Zener. The current through the Zener diode is:

  1. 30 mA
  2. 10 mA Correct answer.
  3. 40 mA
  4. 20 mA

Explanation

The series resistor carries (12 - 6) V / 200 ohm = 30 mA. The load takes 6 V / 300 ohm = 20 mA. The Zener current is the difference, 30 - 20 = 10 mA, because the Zener maintains approximately 6 V across the load.

Question 40Semiconductor Physics and Diodes

In a capacitor-input filter used with a full-wave rectifier, increasing the filter capacitance while keeping the load resistance and supply frequency unchanged will:

  1. increase the ripple voltage and reduce the average DC output
  2. reduce the average DC output because the capacitor discharges faster
  3. reduce the ripple voltage and raise the average DC output slightly Correct answer.
  4. leave the ripple voltage unchanged but reduce the diode PIV

Explanation

A larger capacitor stores more charge and discharges more slowly through the load between charging pulses. This reduces the ripple amplitude and keeps the output voltage closer to the peak value. As a result, the average DC output rises slightly, while the diode peak inverse voltage is set mainly by the peak input voltage.

Question 41Semiconductor Physics and Diodes

At absolute zero, the Fermi level in a pure intrinsic semiconductor is located:

  1. at the acceptor level
  2. near the midgap energy Correct answer.
  3. at the valence-band edge
  4. at the donor level

Explanation

In a pure intrinsic semiconductor the electron and hole concentrations are equal, so the Fermi level lies near the middle of the forbidden gap. At absolute zero it remains essentially at midgap, not at a band edge or an impurity level.

Question 42Semiconductor Physics and Diodes

A silicon sample at room temperature is doped with 1.0 × 10^16 cm^-3 donor atoms and has n_i = 1.5 × 10^10 cm^-3. The equilibrium hole concentration is approximately:

  1. 2.3 × 10^20 cm^-3
  2. 1.5 × 10^10 cm^-3
  3. 2.3 × 10^4 cm^-3 Correct answer.
  4. 1.0 × 10^16 cm^-3

Explanation

Donor doping makes the electron concentration approximately equal to the donor concentration, so n ≈ 1.0 × 10^16 cm^-3. The mass-action law n p = n_i^2 then gives p = (1.5 × 10^10)^2 / (1.0 × 10^16) ≈ 2.3 × 10^4 cm^-3.

Question 43Semiconductor Physics and Diodes

A Zener diode has a nominal Zener voltage of 10 V and a dynamic resistance of 5 Ω. When the Zener current increases by 20 mA, the Zener terminal voltage increases by:

  1. 0.10 V Correct answer.
  2. 0.01 V
  3. 0.50 V
  4. 0.05 V

Explanation

The dynamic resistance relates small changes in Zener voltage and current: ΔV = r_z ΔI. With r_z = 5 Ω and ΔI = 20 mA = 0.020 A, ΔV = 5 × 0.020 = 0.10 V.

Question 44Semiconductor Physics and Diodes

The built-in potential of a p-n junction falls as temperature increases mainly because:

  1. the doping concentrations increase
  2. the intrinsic concentration rises Correct answer.
  3. the carrier mobilities decrease
  4. the depletion width grows appreciably

Explanation

The built-in potential is V_bi = V_T ln(N_A N_D / n_i^2). As temperature rises, n_i increases exponentially, so the logarithmic term falls faster than V_T rises and V_bi decreases. Depletion width and mobility changes are secondary effects, not the main cause.

Question 45Bipolar Junction Transistors and Field-Effect Transistors

In an NPN bipolar junction transistor biased in the active region, the majority carriers injected from the emitter into the base are

  1. electrons Correct answer.
  2. positive ions
  3. holes
  4. protons

Explanation

In an NPN transistor the emitter is n-type, so its majority carriers are electrons. Under active bias these electrons are injected across the forward-biased base-emitter junction and diffuse through the thin base to the collector. Holes are majority carriers in the p-type base, not the carriers supplied by the emitter.

Question 46Bipolar Junction Transistors and Field-Effect Transistors

For a bipolar junction transistor, the common-emitter current gain β is related to the common-base current gain α by

  1. α/(1 − α) Correct answer.
  2. 1 + α
  3. α/(1 + α)
  4. (1 − α)/α

Explanation

The dc current gain β is defined as IC/IB, while α is IC/IE. Since IE = IC + IB, eliminating the currents gives β = α/(1 − α). For example, α = 0.98 gives β = 49.

Question 47Bipolar Junction Transistors and Field-Effect Transistors

In an n-channel JFET operating as an amplifier, the gate-channel junction is normally

  1. reverse-biased Correct answer.
  2. unbiased
  3. avalanche-biased
  4. forward-biased

Explanation

An n-channel JFET controls current by varying the width of the depletion region of a reverse-biased gate-channel junction. Forward biasing that junction would inject gate current and destroy the high-input-resistance advantage.

Question 48Bipolar Junction Transistors and Field-Effect Transistors

A bipolar junction transistor has a common-base current gain α = 0.98. Its common-emitter current gain β is approximately

  1. 0.98
  2. 50
  3. 0.02
  4. 49 Correct answer.

Explanation

Using β = α/(1 − α), substitute α = 0.98 to obtain β = 0.98/0.02 = 49. The value β = 50 would correspond to α = 50/51, not 0.98.

Question 49Bipolar Junction Transistors and Field-Effect Transistors

A silicon npn transistor is used as a closed switch in a digital circuit. The region in which it must be operated is

  1. saturation Correct answer.
  2. active
  3. cutoff
  4. reverse active

Explanation

In saturation both the base-emitter and base-collector junctions are forward biased, so the collector-emitter voltage is very small and the device behaves like a closed switch. The active region is used for amplification, not for a fully on switch.

Question 50Bipolar Junction Transistors and Field-Effect Transistors

In an n-channel JFET with the gate connected to the source and VDS greater than the pinch-off voltage, the saturated drain current is denoted

  1. VP
  2. IGSS
  3. gm
  4. IDSS Correct answer.

Explanation

When the gate is shorted to the source, VGS = 0, and for VDS greater than the pinch-off voltage the drain current saturates at IDSS. IGSS is a gate leakage current, VP is a voltage, and gm is a transconductance.