I85PREP
⌘ K
Tamil Nadu HSE Second Year Basic Electronics Engineering study materials
Tamil Nadu HSE Second Year Basic Electronics Engineering study resources

Tamil Nadu HSE Second Year Basic Electronics Engineering practice questions

One question at a time. Build your Tamil Nadu HSE Second Year Basic Electronics Engineering 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 1Semiconductor Devices and PN Junction

In an intrinsic semiconductor such as pure silicon at room temperature, what is the relationship between the number of free electrons and the number of holes?

  1. There are far more holes than free electrons
  2. There are far more free electrons than holes
  3. Both are zero because no charge carriers exist
  4. The number of free electrons equals the number of holes Correct answer.

Explanation

In an intrinsic crystal, every electron that breaks free from a covalent bond leaves behind exactly one vacancy, called a hole. Therefore the two carrier concentrations are always equal, and their common value is the intrinsic carrier concentration ni.

Question 2Semiconductor Devices and PN Junction

A silicon crystal is doped with phosphorus atoms, each of which has five valence electrons. What type of material results and which carrier is the majority carrier?

  1. p-type material with holes as majority carriers
  2. n-type material with holes as majority carriers
  3. n-type material with electrons as majority carriers Correct answer.
  4. intrinsic material with equal carriers

Explanation

A pentavalent impurity such as phosphorus contributes an extra electron that is only loosely bound and becomes free easily. Such impurities are donors, the material becomes n-type, and electrons are the majority carriers.

Question 3Semiconductor Devices and PN Junction

Which statement correctly describes the depletion region of an unbiased p-n junction?

  1. It is depleted of mobile carriers and contains fixed immobile ions Correct answer.
  2. It is a region rich in mobile majority carriers
  3. It contains only minority carriers in large numbers
  4. It is a thin layer of insulating oxide between the two materials

Explanation

Diffusion of carriers across the junction leaves behind acceptor and donor ions that are fixed in the crystal lattice. This region is therefore free of mobile carriers and holds an internal electric field that opposes further diffusion.

Question 4Semiconductor Devices and PN Junction

In a forward-biased silicon diode, the applied voltage is gradually increased from zero. At approximately what voltage does the current begin to rise sharply?

  1. 0.7 V Correct answer.
  2. 0.1 V
  3. 5.0 V
  4. 12 V

Explanation

For a silicon diode the barrier potential is about 0.7 V, so the knee of the forward characteristic occurs near this value. Beyond it the depletion region narrows and conduction increases rapidly.

Question 5Semiconductor Devices and PN Junction

What is the effect of increasing the reverse bias voltage across a p-n junction, up to the breakdown limit, on the width of the depletion layer?

  1. The depletion layer width remains unchanged
  2. The depletion layer width increases Correct answer.
  3. The depletion layer disappears completely
  4. The depletion layer width decreases

Explanation

Reverse bias adds to the internal barrier potential, pulling majority carriers away from the junction. This exposes more fixed ions on both sides, so the depletion layer widens with increasing reverse voltage.

Question 6Semiconductor Devices and PN Junction

Which property makes a zener diode suitable for voltage regulation in a reverse-biased circuit?

  1. Its forward current is exactly zero at all voltages
  2. Its reverse voltage stays nearly constant over a range of reverse currents in breakdown Correct answer.
  3. Its capacitance changes with temperature
  4. Its forward resistance is extremely high

Explanation

In the zener breakdown region, a large change in reverse current produces only a small change in terminal voltage. This near-constant breakdown voltage is what allows the device to hold a reference level across a load.

Question 7Semiconductor Devices and PN Junction

Which semiconductor material is most commonly used for light-emitting diodes because of its direct band gap?

  1. Selenium
  2. Intrinsic silicon
  3. Germanium
  4. Gallium arsenide Correct answer.

Explanation

Gallium arsenide has a direct band gap, so electron-hole recombination releases energy mainly as photons. Silicon and germanium have indirect band gaps and release energy mostly as heat, making them poor light emitters.

Question 8Semiconductor Devices and PN Junction

A p-n junction diode is connected in series with a 200 ohm resistor across a 5 V supply, forward biased, with a barrier potential of 0.7 V. What is the approximate current in the circuit?

  1. 21.5 mA Correct answer.
  2. 25.0 mA
  3. 28.5 mA
  4. 3.5 mA

Explanation

The resistor drops the supply voltage minus the diode barrier potential, that is 5 V minus 0.7 V equals 4.3 V. Dividing 4.3 V by 200 ohm gives 0.0215 A, or about 21.5 mA.

Question 9Semiconductor Devices and PN Junction

In a p-n junction, what is the direction of the built-in electric field at equilibrium?

  1. From the p-side toward the n-side
  2. It is zero everywhere at equilibrium
  3. Parallel to the junction plane
  4. From the n-side toward the p-side Correct answer.

Explanation

Positive donor ions remain on the n-side and negative acceptor ions on the p-side of the depletion layer. The resulting field therefore points from the positive n-side to the negative p-side.

Question 10Semiconductor Devices and PN Junction

A germanium diode has a barrier potential of about 0.3 V. If it is forward biased in series with a 100 ohm resistor from a 3 V supply, what current flows?

  1. 30 mA
  2. 3 mA
  3. 27 mA Correct answer.
  4. 33 mA

Explanation

The voltage across the resistor is 3 V minus 0.3 V, which is 2.7 V. Using Ohm's law, 2.7 V divided by 100 ohm gives 0.027 A, or 27 mA.

Question 11Semiconductor Devices and PN Junction

Why does the reverse saturation current of a p-n junction diode increase rapidly with temperature?

  1. The barrier potential increases sharply with temperature
  2. The depletion layer width becomes zero
  3. The doping concentration rises with temperature
  4. The number of thermally generated minority carriers increases Correct answer.

Explanation

Reverse current is carried by minority carriers, which are created thermally. As temperature rises, more covalent bonds break, so the minority carrier concentration and hence the reverse saturation current grow.

Question 12Semiconductor Devices and PN Junction

A half-wave rectifier uses a single diode and a load. What is the main function of the diode in this circuit?

  1. To amplify the input voltage
  2. To store charge for the next cycle
  3. To allow current in one direction only Correct answer.
  4. To change the frequency of the supply

Explanation

A diode conducts in forward bias and blocks in reverse bias, so it passes only one half of the alternating input. The load therefore receives a pulsating unidirectional voltage.

Question 13Semiconductor Devices and PN Junction

What happens to the barrier potential of a p-n junction when the doping concentration on both sides is increased?

  1. The barrier potential increases slightly Correct answer.
  2. The barrier potential becomes negative
  3. The barrier potential falls to zero
  4. The barrier potential is unaffected by doping

Explanation

The barrier potential depends on the ratio of acceptor to donor concentrations through the term involving the logarithm of their product. Heavier doping raises this value, so the built-in potential increases modestly.

Question 14Semiconductor Devices and PN Junction

A zener diode with a breakdown voltage of 6 V is used with a series resistor of 300 ohm from a 12 V supply to regulate a load. If the load draws negligible current, what current flows through the zener?

  1. 60 mA
  2. 6 mA
  3. 20 mA Correct answer.
  4. 40 mA

Explanation

The series resistor drops the difference between the supply and the zener voltage, that is 12 V minus 6 V equals 6 V. Dividing 6 V by 300 ohm gives 0.02 A, or 20 mA.

Question 15Semiconductor Devices and PN Junction

Which of the following best explains why a photodiode is operated in reverse bias?

  1. Reverse bias increases the depletion layer width and the light-generated current Correct answer.
  2. Reverse bias reduces the depletion layer capacitance to zero
  3. Reverse bias makes the diode emit light
  4. Reverse bias eliminates all leakage current

Explanation

In reverse bias the depletion region widens, giving a larger volume in which photons can create electron-hole pairs. The resulting photocurrent is proportional to illumination and is easily measured.

Question 16Semiconductor Devices and PN Junction

In a full-wave bridge rectifier, how many diodes conduct during each half-cycle of the input?

  1. One diode
  2. Four diodes
  3. Two diodes Correct answer.
  4. Three diodes

Explanation

In a bridge rectifier, the input half-cycle forward biases one diagonal pair of diodes while the other pair is reverse biased. Thus exactly two diodes conduct at any instant.

Question 17Semiconductor Devices and PN Junction

What is the main reason a varactor diode is used in tuned circuits?

  1. Its forward resistance varies with current
  2. It stores charge indefinitely
  3. It emits light when forward biased
  4. Its junction capacitance varies with reverse bias voltage Correct answer.

Explanation

A varactor is operated in reverse bias so that the depletion layer acts as the dielectric of a small capacitor. Changing the reverse voltage changes the layer width and hence the capacitance, allowing electronic tuning.

Question 18Semiconductor Devices and PN Junction

A silicon diode is reverse biased with 20 V. Which statement about the small current that flows is correct?

  1. It is carried mainly by majority carriers and grows with reverse voltage
  2. It equals the forward current at the same voltage
  3. It flows only after breakdown occurs
  4. It is carried by minority carriers and is nearly independent of reverse voltage Correct answer.

Explanation

The small reverse current is the reverse saturation current, produced by thermally generated minority carriers. It is almost constant over a wide range of reverse voltage until breakdown is reached.

Question 19Semiconductor Devices and PN Junction

A diode is tested and found to have a barrier potential of about 0.3 V. Which material is most likely used?

  1. Gallium arsenide
  2. Silicon
  3. Silicon carbide
  4. Germanium Correct answer.

Explanation

Germanium diodes have a barrier potential of roughly 0.3 V, compared with about 0.7 V for silicon. Gallium arsenide and silicon carbide have even higher values.

Question 20Semiconductor Devices and PN Junction

A silicon diode is used in a half-wave rectifier with a 10 V peak input. What is the approximate peak output voltage across the load, taking the barrier potential as 0.7 V?

  1. 10.7 V
  2. 5.0 V
  3. 9.3 V Correct answer.
  4. 10.0 V

Explanation

During conduction the diode drops about 0.7 V, so the peak load voltage is the input peak minus this drop. That gives 10 V minus 0.7 V, or about 9.3 V.

Question 21Semiconductor Devices and PN Junction

A p-n junction is formed and left at thermal equilibrium with no applied voltage. Which condition holds across the junction?

  1. Both diffusion and drift currents are zero
  2. The drift current exceeds the diffusion current
  3. The diffusion current exceeds the drift current
  4. The diffusion current and drift current are equal and opposite Correct answer.

Explanation

At equilibrium the net current across the junction must be zero. The diffusion of majority carriers is exactly balanced by the drift of minority carriers in the built-in field, so the two currents are equal in magnitude and opposite in direction.

Question 22Semiconductor Devices and PN Junction

The barrier potential of a silicon p-n junction is approximately 0.7 V. What does this value physically represent?

  1. The energy needed to break a covalent bond in silicon
  2. The voltage at which the diode begins to emit light
  3. The potential difference across the depletion layer that opposes further diffusion Correct answer.
  4. The reverse breakdown voltage of the diode

Explanation

The barrier potential is the built-in potential difference created by the fixed ions in the depletion layer. It acts as a potential hill that majority carriers must overcome to diffuse across the junction.

Question 23Semiconductor Devices and PN Junction

In a zener regulator, the input voltage rises while the load resistance is held constant. Assuming the zener remains in breakdown, what happens to the zener current and the voltage across the load?

  1. Zener current stays fixed and load voltage increases sharply
  2. Zener current falls to zero and load voltage falls to zero
  3. Zener current decreases and load voltage increases sharply
  4. Zener current increases and load voltage stays nearly constant Correct answer.

Explanation

The extra voltage appears across the series resistor, so the total current through it rises. Because the load resistance is fixed and the load voltage is clamped near the zener voltage, the additional current flows through the zener.

Question 24Semiconductor Devices and PN Junction

Two p-n junction diodes are connected in parallel, both forward biased directly across a supply, and one has a slightly lower barrier potential. Why is this connection generally avoided?

  1. The total current becomes exactly zero
  2. Both diodes immediately go into reverse breakdown
  3. The diode with the lower barrier potential hogs most of the current and may overheat Correct answer.
  4. The barrier potentials add together and block all current

Explanation

Diodes have a steep exponential forward characteristic, so a small difference in barrier potential causes a large difference in current. The lower-threshold device conducts most of the current and can exceed its rating, which is why series resistors or matched devices are used.

Question 25Bipolar Junction Transistors and Biasing

In an npn transistor operating in the active region, which carrier type forms the collector current?

  1. Holes drifting from collector to base
  2. Electrons injected from emitter to collector Correct answer.
  3. Holes injected from base to emitter
  4. Electrons recombining only in the base

Explanation

In an npn device the emitter is heavily doped n-type, so electrons are injected across the thin base into the collector. The base current is only the small fraction that recombines in the base. Hence the collector current is carried mainly by electrons.

Question 26Bipolar Junction Transistors and Biasing

For a silicon transistor biased in the active region, the base-emitter voltage is approximately

  1. 1.4 V
  2. 5.0 V
  3. 0.2 V
  4. 0.7 V Correct answer.

Explanation

The forward-biased base-emitter junction behaves like a silicon diode, so the drop is about 0.7 V. Germanium devices would show roughly 0.3 V instead. This value is used in bias calculations for silicon transistors.

Question 27Bipolar Junction Transistors and Biasing

The relation between the transistor currents in any region of operation is

  1. IE = IC − IB
  2. IE = IB + IC Correct answer.
  3. IB = IE + IC
  4. IC = IE + IB

Explanation

The emitter current is the sum of the collector and base currents because the emitter supplies both. This holds regardless of whether the device is active, saturated or cut off. It follows directly from Kirchhoff's current law at the transistor terminals.

Question 28Bipolar Junction Transistors and Biasing

A transistor has β = 100 and base current IB = 40 µA. Its collector current is

  1. 40 mA
  2. 4.0 mA Correct answer.
  3. 0.4 mA
  4. 1.0 mA

Explanation

In the active region IC = β × IB, so 100 × 40 µA = 4000 µA = 4 mA. The emitter current would then be about 4.04 mA. This direct proportionality is the basis of transistor current amplification.

Question 29Bipolar Junction Transistors and Biasing

The output characteristics of a BJT in common-emitter configuration are plotted as

  1. IC against IB for constant VCE
  2. IB against VBE for constant VCE
  3. VCE against VBE for constant IC
  4. IC against VCE for constant IB Correct answer.

Explanation

The common-emitter output characteristics show collector current on the vertical axis against collector-emitter voltage on the horizontal axis, drawn for several fixed base currents. The input characteristics instead plot IB against VBE. Keeping IB constant shows how the output behaves for a given drive.

Question 30Bipolar Junction Transistors and Biasing

The DC load line of a common-emitter amplifier is drawn between which two points on the output characteristics?

  1. IB = 0 and IB = IB(max)
  2. VBE = 0 and VBE = 0.7 V
  3. VCE = VCC and IC = VCC/RC Correct answer.
  4. VCE = 0 and IC = 0

Explanation

The load line comes from VCE = VCC − IC·RC. Setting IC to zero gives the intercept VCE = VCC, and setting VCE to zero gives IC = VCC/RC. The quiescent point must lie somewhere on this straight line.

Question 31Bipolar Junction Transistors and Biasing

A transistor with β = 150 has collector current 6 mA in the active region. Its base current is

  1. 900 µA
  2. 400 µA
  3. 40 µA Correct answer.
  4. 90 µA

Explanation

Rearranging IC = β·IB gives IB = IC/β = 6 mA / 150 = 0.04 mA, which is 40 µA. The emitter current is then about 6.04 mA. This small base current is what makes the device a current-controlled amplifier.

Question 32Bipolar Junction Transistors and Biasing

In a common-emitter output characteristic, the nearly horizontal portion of the curves at high VCE indicates that the collector current

  1. Rises steeply with VCE
  2. Is almost independent of VCE Correct answer.
  3. Falls to zero
  4. Equals the base current

Explanation

Beyond the knee, further increases in VCE barely change IC because the base region has already swept most injected carriers into the collector. This flat region is the active region used for amplification. The slight upward slope reflects the Early effect.

Question 33Bipolar Junction Transistors and Biasing

Why is a transistor biased at the centre of the DC load line in a class-A amplifier?

  1. To obtain maximum symmetrical swing without clipping Correct answer.
  2. To minimise the base-emitter voltage
  3. To make the collector current zero
  4. To keep the transistor always in cut-off

Explanation

Centring the quiescent point allows the collector voltage and current to swing equally in both directions before hitting saturation or cut-off. This maximises undistorted output amplitude. Off-centre biasing clips the waveform on one side earlier.

Question 34Bipolar Junction Transistors and Biasing

In fixed-bias (base-resistor) biasing, the base current is set by

  1. β·VCC/RB
  2. (VCC − VBE)/RB Correct answer.
  3. VBE/RC
  4. VCC/(RB + RC)

Explanation

Applying Kirchhoff's voltage law to the base loop gives VCC = IB·RB + VBE, so IB = (VCC − VBE)/RB. The collector resistor does not appear in the base loop. Because IB is fixed, the operating point shifts strongly if β changes.

Question 35Bipolar Junction Transistors and Biasing

The main drawback of fixed-bias with a single base resistor is that

  1. It requires two separate supply voltages
  2. The operating point drifts badly when β changes or temperature rises Correct answer.
  3. It cannot be used with npn transistors
  4. The base resistor dissipates excessive power

Explanation

Since IB is fixed by RB, any change in β directly changes IC and shifts the quiescent point, possibly into saturation. Temperature rise also increases leakage and β, worsening the drift. This poor stability is why voltage-divider bias is preferred.

Question 36Bipolar Junction Transistors and Biasing

In collector-to-base bias, the base resistor is connected between

  1. Collector and emitter
  2. Base and emitter
  3. Base and collector Correct answer.
  4. Base and ground

Explanation

Feedback bias returns the base resistor from the collector to the base, so the base current depends on VCE. If IC rises, VCE falls and IB is reduced, giving partial stabilisation. This negative feedback improves stability over fixed bias.

Question 37Bipolar Junction Transistors and Biasing

In voltage-divider bias, the base voltage is approximately set by the divider because

  1. The transistor is always in saturation
  2. The emitter resistor is very large
  3. The base current is small compared with the divider current Correct answer.
  4. The collector resistor carries no current

Explanation

When the current through the divider resistors greatly exceeds IB, the base voltage is fixed almost entirely by the resistor ratio. The emitter voltage then follows VBE below it, making IC nearly independent of β. This is the most widely used stable bias arrangement.

Question 38Bipolar Junction Transistors and Biasing

A silicon transistor uses voltage-divider bias with VB = 2.7 V and RE = 1 kΩ. The emitter current is approximately

  1. 0.7 mA
  2. 2.0 mA Correct answer.
  3. 3.4 mA
  4. 2.7 mA

Explanation

The emitter voltage is VE = VB − VBE = 2.7 − 0.7 = 2.0 V. Then IE = VE/RE = 2.0 V / 1 kΩ = 2 mA. Since IC ≈ IE, the collector current is also about 2 mA.

Question 39Bipolar Junction Transistors and Biasing

Emitter-resistor bias (with a resistor in the emitter lead) improves stability because

  1. Negative feedback reduces the effect of β and temperature changes Correct answer.
  2. It increases the base-emitter voltage directly
  3. It forces the transistor into saturation
  4. It removes the need for any supply voltage

Explanation

If IC tries to rise, the drop across RE increases, reducing VBE and therefore pulling IC back down. This self-correcting action is negative feedback. The result is an operating point much less sensitive to β spread and thermal drift.

Question 40Bipolar Junction Transistors and Biasing

The stability factor S of a bias circuit is defined as

  1. dIC/dβ at constant ICO
  2. dVB/dIB at constant VCE
  3. dIC/dICO at constant β and VBE Correct answer.
  4. dIB/dVBE at constant IC

Explanation

The stability factor expresses how strongly collector current changes when the reverse saturation current ICO changes, with β and VBE held constant. A smaller value of S means better thermal stability. Ideal bias circuits aim for the lowest practical S.

Question 41Bipolar Junction Transistors and Biasing

In a common-emitter amplifier, the phase relationship between input base signal and output collector signal is

  1. 90° out of phase
  2. 180° out of phase Correct answer.
  3. 45° out of phase
  4. In phase

Explanation

A rising base input increases collector current, which increases the drop across RC and therefore lowers VCE. The output voltage thus moves opposite to the input, giving 180° inversion. Common-base and common-collector stages do not invert.

Question 42Bipolar Junction Transistors and Biasing

A transistor is found to have VCE ≈ 0.2 V, IC limited only by RC, and both junctions forward biased. The device is operating in

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

Explanation

In saturation the collector-emitter voltage falls to a small value (about 0.2 V for silicon) and the collector current is set by the external resistor rather than by β. Both junctions are forward biased. Further increases in base current cannot raise IC appreciably.

Question 43Bipolar Junction Transistors and Biasing

A common-emitter amplifier has VCC = 12 V, RC = 2 kΩ and quiescent IC = 2 mA. The quiescent collector-emitter voltage is

  1. 12 V
  2. 10 V
  3. 4 V
  4. 8 V Correct answer.

Explanation

The drop across RC is IC·RC = 2 mA × 2 kΩ = 4 V. Subtracting from the supply gives VCEQ = 12 − 4 = 8 V. This point should lie near the middle of the load line for maximum swing.

Question 44Bipolar Junction Transistors and Biasing

Thermal runaway in a transistor is best prevented by

  1. Increasing the collector supply voltage
  2. Operating the device in class-A at maximum current
  3. Removing the emitter resistor
  4. Using a bias circuit with a low stability factor and adequate heat sinking Correct answer.

Explanation

Thermal runaway occurs when rising temperature increases IC, which raises dissipation and temperature further. A low stability factor plus heat sinking breaks this loop by limiting the current rise. Emitter degeneration also helps by reducing VBE as current grows.

Question 45Bipolar Junction Transistors and Biasing

A silicon npn transistor is biased with VBB = 3 V, RB = 200 kΩ and RE = 1 kΩ, with the base resistor feeding the base and RE in the emitter lead. Taking VBE = 0.7 V, the emitter current is approximately

  1. 1.15 mA Correct answer.
  2. 2.3 mA
  3. 11.5 mA
  4. 0.7 mA

Explanation

The base loop gives IB ≈ (VBB − VBE)/(RB + (1+β)RE), but with a large β the emitter current is approximately (VBB − VBE)/RE when βRE dominates. Here (3 − 0.7)/1 kΩ ≈ 2.3 mA is the upper bound; including the 200 kΩ base resistance with β = 100 gives about 1.15 mA. The exact value depends on β, so the calculation must state it.

Question 46Bipolar Junction Transistors and Biasing

The DC load line of a circuit with VCC = 15 V and RC = 3 kΩ intersects the current axis at

  1. 45 mA
  2. 3 mA
  3. 15 mA
  4. 5 mA Correct answer.

Explanation

The current-axis intercept occurs when VCE = 0, so IC = VCC/RC = 15 V / 3 kΩ = 5 mA. The voltage-axis intercept is VCC = 15 V. The quiescent point must lie between these two intercepts.

Question 47Bipolar Junction Transistors and Biasing

Compared with fixed bias, voltage-divider bias gives an operating point that is far less sensitive to β because

  1. The collector resistor is removed from the circuit
  2. The transistor is deliberately driven into saturation
  3. Base current is made larger than the divider current
  4. The base is held near a fixed potential and emitter feedback sets IC ≈ (VB − VBE)/RE Correct answer.

Explanation

The stiff divider clamps VB, and the emitter resistor converts that voltage into a current almost independent of β. Only a small residual dependence remains through IB. This is why the design rule keeps divider current much larger than base current.

Question 48Bipolar Junction Transistors and Biasing

For a BJT, the relation between α and β is correctly given by

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

Explanation

Since α = β/(1+β), rearranging yields β = α/(1 − α). For α = 0.98 this gives β = 49, consistent with typical values. The two gain parameters describe the same device in common-base and common-emitter configurations.

Question 49Field Effect Transistors and Amplifiers

In an n-channel JFET, the channel is made of which material, and what are the majority carriers that flow through it?

  1. p-type semiconductor; holes
  2. n-type semiconductor; electrons Correct answer.
  3. metal oxide layer; free electrons only
  4. intrinsic semiconductor; both electrons and holes equally

Explanation

An n-channel JFET uses an n-type semiconductor bar as the channel, and the source injects electrons into it. Because the channel is n-type, electrons are the majority carriers and current flows from drain to source.

Question 50Field Effect Transistors and Amplifiers

For a JFET, the parameter denoted g_m (transconductance) is defined as the ratio of which two quantities?

  1. Change in gate current to change in gate-source voltage
  2. Change in drain-source voltage to change in drain current at constant gate voltage
  3. Drain current to gate-source voltage at zero drain current
  4. Change in drain current to change in gate-source voltage at constant drain-source voltage Correct answer.

Explanation

Transconductance g_m = ΔI_D / ΔV_GS with V_DS held constant, and it is expressed in siemens or milliamperes per volt. It measures how effectively the gate voltage controls drain current. The reciprocal quantity ΔV_DS/ΔI_D is the drain resistance r_d, not g_m.