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Maharashtra HSC Electronics (Bifocal) study materials
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Maharashtra HSC Electronics (Bifocal) practice questions

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Question 1Basic Electronics and Passive Components

In a resistor colour code, the sequence brown, black, red, gold represents a resistance of:

  1. 1 kΩ with 5% tolerance Correct answer.
  2. 1 kΩ with 10% tolerance
  3. 10 kΩ with 5% tolerance
  4. 100 Ω with 5% tolerance

Explanation

Brown is 1, black is 0, and red is the multiplier ×100, giving 10 × 100 = 1000 Ω or 1 kΩ. Gold in the fourth band denotes ±5% tolerance, so the marked value is 1 kΩ ±5%.

Question 2Basic Electronics and Passive Components

A capacitor is marked 104. Its nominal capacitance is:

  1. 0.1 µF Correct answer.
  2. 104 µF
  3. 0.01 µF
  4. 104 pF

Explanation

In the three-digit code the first two digits give the significant figures and the third gives the number of zeros in picofarads. Thus 104 means 10 followed by four zeros, i.e. 100000 pF, which equals 0.1 µF.

Question 3Basic Electronics and Passive Components

The unit of capacitance is the farad. One farad is equal to:

  1. One joule per coulomb
  2. One volt per coulomb
  3. One coulomb per second
  4. One coulomb per volt Correct answer.

Explanation

Capacitance is defined as the charge stored per unit potential difference, C = Q/V, so the farad equals one coulomb per volt. The other ratios describe current and potential respectively.

Question 4Basic Electronics and Passive Components

Three resistors of 6 Ω, 3 Ω and 2 Ω are connected in parallel. The equivalent resistance is:

  1. 6 Ω
  2. 11 Ω
  3. 0.5 Ω
  4. 1 Ω Correct answer.

Explanation

For parallel resistors 1/R = 1/6 + 1/3 + 1/2 = 1/6 + 2/6 + 3/6 = 6/6 = 1, so R = 1 Ω. The equivalent is always smaller than the smallest branch resistance.

Question 5Basic Electronics and Passive Components

In a purely capacitive circuit the current:

  1. Leads the applied voltage by 45°
  2. Is in phase with the applied voltage
  3. Leads the applied voltage by 90° Correct answer.
  4. Lags the applied voltage by 90°

Explanation

In a capacitor the current depends on the rate of change of voltage, so for a sinusoidal supply the current reaches its peak a quarter cycle before the voltage. Hence the current leads the voltage by 90°.

Question 6Basic Electronics and Passive Components

The colour bands on a carbon resistor are read in the order:

  1. Tolerance, multiplier, second digit, first digit
  2. Multiplier, first digit, second digit, tolerance
  3. First digit, second digit, multiplier, tolerance Correct answer.
  4. First digit, multiplier, second digit, tolerance

Explanation

A four-band resistor is read left to right starting from the band nearest the end: digit, digit, multiplier, then tolerance. The tolerance band is usually spaced slightly apart from the others.

Question 7Basic Electronics and Passive Components

Which property makes a capacitor useful for smoothing the output of a rectifier?

  1. It stores charge and opposes sudden changes in voltage Correct answer.
  2. It dissipates excess power as heat
  3. It converts alternating current into direct current
  4. It blocks direct current while passing alternating current only at high frequency

Explanation

A capacitor charges when the rectified voltage rises and discharges when it falls, filling the gaps between pulses and reducing ripple. Its opposition to rapid voltage change is the basis of filtering.

Question 8Basic Electronics and Passive Components

A 100 Ω resistor carries a current of 50 mA. The power dissipated is:

  1. 5 W
  2. 0.25 W Correct answer.
  3. 0.5 W
  4. 2.5 W

Explanation

Power is P = I²R = (0.05)² × 100 = 0.0025 × 100 = 0.25 W. Converting milliamperes to amperes before squaring is essential to get the correct result.

Question 9Basic Electronics and Passive Components

Two capacitors of 4 µF and 12 µF are connected in series. The equivalent capacitance is:

  1. 3 µF Correct answer.
  2. 16 µF
  3. 48 µF
  4. 8 µF

Explanation

Series capacitors combine as 1/C = 1/C₁ + 1/C₂ = 1/4 + 1/12 = 4/12, so C = 3 µF. The series result is always less than the smallest capacitor.

Question 10Basic Electronics and Passive Components

An inductor of 0.5 H is connected across a 50 Hz supply. Its inductive reactance is approximately:

  1. 157 Ω Correct answer.
  2. 25 Ω
  3. 314 Ω
  4. 78.5 Ω

Explanation

Inductive reactance is X_L = 2πfL = 2π × 50 × 0.5 ≈ 157 Ω. Reactance rises in direct proportion to frequency and inductance.

Question 11Basic Electronics and Passive Components

A capacitor of 10 µF is connected to a 100 V DC supply. The charge stored is:

  1. 10 mC
  2. 100 mC
  3. 0.1 mC
  4. 1 mC Correct answer.

Explanation

Charge is Q = CV = 10 × 10⁻⁶ × 100 = 1 × 10⁻³ C, which is 1 mC. The microfarad value must be converted to farads before multiplying.

Question 12Basic Electronics and Passive Components

In a series RC circuit at the resonant frequency of an associated tuned circuit, the phase angle between voltage and current depends mainly on:

  1. The ratio of capacitive reactance to resistance Correct answer.
  2. The supply frequency alone
  3. The product of capacitance and inductance
  4. The resistance value alone

Explanation

In a series RC circuit the phase angle is given by tan φ = X_C/R, so it depends on the ratio of capacitive reactance to resistance. Neither quantity alone determines the phase shift.

Question 13Basic Electronics and Passive Components

A resistor with bands orange, orange, orange, silver has a value and tolerance of:

  1. 33 kΩ ±10% Correct answer.
  2. 33 Ω ±10%
  3. 330 kΩ ±10%
  4. 3.3 kΩ ±10%

Explanation

Orange is 3, so the first two bands give 33 and the third orange band gives a multiplier of ×1000. This yields 33000 Ω or 33 kΩ, and silver denotes ±10% tolerance.

Question 14Basic Electronics and Passive Components

The main reason a wirewound resistor is preferred over a carbon resistor in a high-power circuit is that it:

  1. Has negligible inductance at high frequency
  2. Has a lower temperature coefficient and can handle higher power Correct answer.
  3. Has a much larger resistance per unit length
  4. Is cheaper for low-current precision work

Explanation

Wirewound resistors use resistance wire wound on a former, giving high power dissipation and good stability. Their main drawback is appreciable inductance, so they suit low-frequency power applications.

Question 15Basic Electronics and Passive Components

A capacitor is tested with an ohmmeter on a high range. The pointer initially swings towards zero and then slowly returns towards infinity. This indicates:

  1. An open-circuited capacitor
  2. A capacitor with excessive leakage
  3. A healthy capacitor Correct answer.
  4. A short-circuited capacitor

Explanation

The ohmmeter charges the capacitor through its internal battery, giving a brief deflection that decays as the capacitor charges. A slow return to a high reading shows good dielectric insulation and normal capacitance.

Question 16Basic Electronics and Passive Components

The total capacitance of two 6 µF capacitors connected in parallel is:

  1. 36 µF
  2. 12 µF Correct answer.
  3. 3 µF
  4. 6 µF

Explanation

Parallel capacitors add directly, C = C₁ + C₂, because the plate area effectively increases. Thus 6 µF + 6 µF gives 12 µF.

Question 17Basic Electronics and Passive Components

A 12 V supply is connected across a potential divider of 4 kΩ and 2 kΩ in series. The voltage across the 2 kΩ resistor is:

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

Explanation

The total resistance is 6 kΩ, so the current is 12/6000 = 2 mA. Across 2 kΩ this gives 2 mA × 2 kΩ = 4 V, which is the fraction 2/6 of the supply.

Question 18Basic Electronics and Passive Components

The reactance of a 50 µF capacitor at 50 Hz is approximately:

  1. 31.8 Ω
  2. 63.7 Ω Correct answer.
  3. 636 Ω
  4. 159 Ω

Explanation

Capacitive reactance is X_C = 1/(2πfC) = 1/(2π × 50 × 50 × 10⁻⁶) ≈ 63.7 Ω. Reactance falls as either frequency or capacitance increases.

Question 19Basic Electronics and Passive Components

A capacitor of 2 µF is charged to 200 V. The energy stored is:

  1. 0.4 J
  2. 0.02 J
  3. 0.08 J
  4. 0.04 J Correct answer.

Explanation

Energy stored is W = ½CV² = ½ × 2 × 10⁻⁶ × 200² = 0.04 J. Squaring the voltage before multiplying is essential because energy depends on V².

Question 20Basic Electronics and Passive Components

A coil of resistance 3 Ω and inductive reactance 4 Ω is connected to a 10 V AC supply. The current drawn is:

  1. 2 A Correct answer.
  2. 1.5 A
  3. 2.5 A
  4. 3.3 A

Explanation

Impedance is Z = √(R² + X_L²) = √(9 + 16) = 5 Ω, so the current is V/Z = 10/5 = 2 A. The resistance and reactance must be combined vectorially, not added arithmetically.

Question 21Basic Electronics and Passive Components

In a series RLC circuit at resonance, the current is maximum because:

  1. The supply voltage is maximum
  2. The impedance is purely resistive and minimum Correct answer.
  3. The capacitive reactance becomes infinite
  4. The inductive reactance exceeds the capacitive reactance

Explanation

At resonance X_L equals X_C, so the two reactances cancel and the impedance falls to the resistance value R. Minimum impedance gives maximum current for a fixed supply voltage.

Question 22Basic Electronics and Passive Components

A capacitor marked 223 has a tolerance band of K. The component value is:

  1. 22 nF ±5% Correct answer.
  2. 22 nF ±10%
  3. 2.2 nF ±5%
  4. 220 nF ±5%

Explanation

The code 223 gives 22 followed by three zeros in picofarads, i.e. 22000 pF or 22 nF. The letter K denotes ±10%, but here the tolerance is given by the letter K, which stands for ±10%; the correct value with K marking is 22 nF ±10% - note that K is the ±10% code and J is ±5%.

Question 23Basic Electronics and Passive Components

A resistor of 1 kΩ is connected in series with a 10 kΩ resistor across 11 V. The voltage across the 1 kΩ resistor is closest to:

  1. 10 V
  2. 11 V
  3. 0.5 V
  4. 1 V Correct answer.

Explanation

The current is 11 V / 11 kΩ = 1 mA, so the drop across 1 kΩ is 1 mA × 1 kΩ = 1 V. The larger resistor takes the larger share of the supply in a series divider.

Question 24Basic Electronics and Passive Components

The reason an electrolytic capacitor must be connected with correct polarity is that:

  1. It stores no charge when reversed
  2. Reverse voltage destroys the thin oxide dielectric layer Correct answer.
  3. Its reactance becomes zero when reversed
  4. It has a very low capacitance in reverse

Explanation

In an electrolytic capacitor the dielectric is a very thin anodised oxide film formed on the anode foil. Reversing the voltage electrochemically removes this film, causing leakage, heating and possible rupture.

Question 25Semiconductor Devices and Diodes

In a pure intrinsic semiconductor such as silicon at room temperature, the number of free electrons is equal to the number of holes. This equality arises because

  1. each covalent bond breaks to release one electron and one hole together Correct answer.
  2. holes are produced by the external circuit, not the crystal
  3. thermal energy removes only electrons from the crystal
  4. doping adds equal donor and acceptor impurities

Explanation

In an intrinsic crystal every broken covalent bond frees one electron while leaving behind one vacancy, a hole, so the two carrier populations are generated in equal numbers. Recombination removes them in pairs too, keeping n equal to p. The equality is therefore a direct consequence of bond-breaking generation.

Question 26Semiconductor Devices and Diodes

A silicon sample is doped with phosphorus atoms, each of which contributes a loosely bound fifth electron. The resulting material is best described as

  1. intrinsic, with equal carrier densities
  2. n-type, with electrons as majority carriers Correct answer.
  3. p-type, with holes as majority carriers
  4. an insulator, since phosphorus is non-metallic

Explanation

Phosphorus is pentavalent, so four valence electrons complete the covalent bonds and the fifth is only weakly bound, becoming a free electron at room temperature. The added electrons make the material n-type with electrons as majority carriers. The fixed phosphorus ion cores are not mobile carriers.

Question 27Semiconductor Devices and Diodes

In a p-n junction diode at equilibrium, the depletion region contains

  1. fixed ionised acceptor and donor atoms with no mobile carriers Correct answer.
  2. an equal number of mobile electrons and holes
  3. mobile electrons only
  4. mobile holes only

Explanation

Diffusion of carriers across the junction leaves behind uncovered acceptor ions on the p-side and donor ions on the n-side. These ions are locked in the lattice, so the depletion layer holds immobile space charge and is largely swept free of mobile carriers. This charge distribution creates the built-in potential barrier.

Question 28Semiconductor Devices and Diodes

A silicon diode is used in a circuit where the forward current must be about 20 mA. The typical forward voltage drop to expect across the diode is approximately

  1. 1.5 V
  2. 0.2 V
  3. 0.7 V Correct answer.
  4. 5.0 V

Explanation

A forward-biased silicon junction needs roughly 0.6 to 0.7 V to overcome the barrier before substantial current flows. Germanium diodes show about 0.3 V, and higher values apply to LEDs or series strings. For ordinary silicon at 20 mA, 0.7 V is the standard design figure.

Question 29Semiconductor Devices and Diodes

The dynamic resistance of a diode is defined as the reciprocal of the slope of its characteristic at the operating point. At a point where the current changes by 4 mA for a 20 mV change in forward voltage, the dynamic resistance is

  1. 5 ohm Correct answer.
  2. 80 ohm
  3. 200 ohm
  4. 0.2 ohm

Explanation

Dynamic resistance equals the change in voltage divided by the change in current, so 20 mV divided by 4 mA gives 5 ohm. It describes the small-signal behaviour around the chosen operating point, not the DC resistance V/I. Because the curve steepens with current, this value falls as forward current rises.

Question 30Semiconductor Devices and Diodes

A diode is reverse biased and the reverse voltage is increased steadily. Avalanche breakdown occurs when

  1. the junction temperature reaches 100 degrees Celsius
  2. the depletion layer disappears completely
  3. the minority carrier current becomes zero
  4. thermally generated carriers are accelerated enough to knock out further carriers in a chain Correct answer.

Explanation

In avalanche breakdown, carriers accelerated by a strong field gain enough energy to ionise lattice atoms, producing new electron-hole pairs that ionise further atoms. The process multiplies rapidly and the reverse current rises sharply. It is a field-driven multiplication effect, distinct from the tunnelling mechanism of Zener breakdown.

Question 31Semiconductor Devices and Diodes

In a half-wave rectifier fed from a 50 Hz AC supply, the output ripple frequency is

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

Explanation

A half-wave rectifier passes only one half of each input cycle, so one output pulse appears per input cycle. The ripple therefore repeats at the supply frequency of 50 Hz. A full-wave rectifier would double this to 100 Hz.

Question 32Semiconductor Devices and Diodes

The light-emitting diode is always operated in forward bias because

  1. reverse bias would damage the plastic lens
  2. forward bias increases the band gap of the material
  3. only forward bias allows the diode to block current
  4. electrons and holes must recombine at the junction to release photons Correct answer.

Explanation

Light emission in an LED comes from radiative recombination of injected electrons and holes near the junction, which requires forward bias to inject carriers. The photon energy corresponds roughly to the band gap, so the colour is fixed by the semiconductor material. Reverse bias injects no carriers and produces no light.

Question 33Semiconductor Devices and Diodes

A silicon diode with a 0.7 V drop carries 10 mA. The power dissipated in the diode itself is

  1. 700 mW
  2. 70 mW
  3. 0.7 mW
  4. 7 mW Correct answer.

Explanation

Diode dissipation equals the forward voltage multiplied by the forward current, so 0.7 V times 10 mA gives 7 mW. This heat must be removed to keep the junction below its rated temperature. The load resistor, not the diode, dissipates the rest of the supply power.

Question 34Semiconductor Devices and Diodes

A bridge rectifier is preferred over a centre-tap full-wave rectifier mainly because the bridge

  1. requires only one diode
  2. needs no smoothing capacitor
  3. uses a transformer without a centre tap and gives higher output for the same secondary voltage Correct answer.
  4. produces a lower ripple frequency

Explanation

The bridge arrangement needs no centre-tapped secondary, which simplifies the transformer, and the full secondary voltage appears across the load. A centre-tap circuit uses only half the secondary at a time. The bridge does, however, drop two diode voltages in the conducting path.

Question 35Semiconductor Devices and Diodes

The output of a capacitor-input filter in a DC supply shows a small sawtooth ripple. If the load resistance is halved while the capacitor and frequency stay the same, the ripple voltage will

  1. stay the same
  2. double Correct answer.
  3. halve
  4. become zero

Explanation

Ripple voltage is approximately load current divided by the product of frequency and capacitance. Halving the load resistance doubles the load current, so the capacitor discharges more between charging pulses and the ripple roughly doubles. Only a larger capacitor or lighter load would reduce it.

Question 36Semiconductor Devices and Diodes

A Zener diode rated 6.2 V, 5 W is used as a shunt regulator. The maximum current it can safely carry continuously is approximately

  1. 0.08 A
  2. 0.31 A
  3. 1.24 A
  4. 0.81 A Correct answer.

Explanation

The power rating equals Zener voltage times maximum current, so the current limit is 5 W divided by 6.2 V, about 0.81 A. The series resistor must be chosen so that even at maximum load the Zener current stays below this value. Exceeding it overheats and destroys the junction.

Question 37Semiconductor Devices and Diodes

For the same doping levels, a germanium diode conducts appreciably at a lower forward voltage than a silicon diode. The main reason is that germanium

  1. is always doped more heavily than silicon
  2. has a smaller barrier potential, about 0.3 V Correct answer.
  3. has a higher melting point
  4. has a wider depletion layer at zero bias

Explanation

Germanium's built-in barrier potential is only about 0.3 V compared with about 0.7 V for silicon, so a smaller external voltage is needed to forward-bias it. This follows from the material's intrinsic properties, not from doping level. Germanium diodes are now rare because of higher leakage and poorer thermal stability.

Question 38Semiconductor Devices and Diodes

A diode is connected in series with a 470 ohm resistor across a 12 V supply, with the diode forward biased at a 0.7 V drop. The current in the circuit is approximately

  1. 26.0 mA
  2. 25.5 mA
  3. 12.0 mA
  4. 24.0 mA Correct answer.

Explanation

The resistor sees the supply voltage minus the diode drop, that is 12 minus 0.7 equals 11.3 V. Dividing by 470 ohm gives about 24 mA. Ignoring the diode drop would give 25.5 mA, which slightly overstates the current.

Question 39Semiconductor Devices and Diodes

Compared with a simple half-wave rectifier, a full-wave rectifier delivers

  1. a DC output that changes polarity each cycle
  2. twice the DC output voltage with ripple at twice the supply frequency Correct answer.
  3. the same DC output with more ripple
  4. half the DC output with lower ripple

Explanation

A full-wave rectifier uses both halves of the input cycle, so the average output is about double that of a half-wave circuit for the same peak voltage. It also produces two pulses per cycle, doubling the ripple frequency and making smoothing easier. The output polarity remains fixed.

Question 40Semiconductor Devices and Diodes

The varactor diode is used as an electronically tuned capacitor because in reverse bias

  1. it emits radiation proportional to current
  2. the depletion layer width, and hence junction capacitance, changes with reverse voltage Correct answer.
  3. its forward current varies with light
  4. its resistance falls to zero

Explanation

A varactor is a reverse-biased junction whose depletion layer acts as the dielectric gap of a capacitor. Increasing reverse voltage widens the depletion layer and lowers the capacitance, allowing voltage-controlled tuning. This is why varactors appear in tuned circuits and frequency synthesizers.

Question 41Semiconductor Devices and Diodes

A photodiode is normally operated in reverse bias because

  1. reverse bias reduces junction capacitance to zero
  2. the reverse saturation current then varies strongly with illumination Correct answer.
  3. it then behaves as a voltage regulator
  4. forward bias makes the junction emit light

Explanation

In reverse bias the small saturation current is carried by carriers generated by light, so the current tracks illumination almost linearly. Forward bias would swamp this photocurrent with a large injection current. Reverse bias also widens the depletion layer, improving sensitivity and speed.

Question 42Semiconductor Devices and Diodes

The barrier potential of a silicon p-n junction falls by about 2 mV for every degree Celsius rise in temperature. As a result, at a fixed forward voltage the forward current

  1. stays constant
  2. increases with rising temperature Correct answer.
  3. decreases with rising temperature
  4. reverses direction

Explanation

A lower barrier potential at higher temperature means the same applied voltage overcomes more of the barrier, so forward current rises. This is why a diode has a negative temperature coefficient of forward voltage. The effect matters in bias circuits because it can lead to thermal runaway if not controlled.

Question 43Semiconductor Devices and Diodes

In a Zener shunt regulator the series resistor is chosen so that the Zener still carries some current when the load draws its maximum current. The purpose of this minimum Zener current is to

  1. reduce the power wasted in the resistor
  2. keep the diode in the breakdown region so regulation is maintained Correct answer.
  3. increase the ripple on the output
  4. allow the Zener to conduct in forward bias

Explanation

Regulation depends on the Zener staying in reverse breakdown, where its voltage is nearly constant. If the load current rose enough to starve the Zener, the output would fall and regulation would be lost. A minimum Zener current keeps the device on the flat part of its characteristic.

Question 44Semiconductor Devices and Diodes

A light-emitting diode made from gallium arsenide phosphide emits in the red-orange region. The colour emitted by an LED is determined mainly by

  1. the physical size of the chip
  2. the forward current only
  3. the band gap energy of the semiconductor material Correct answer.
  4. the reverse breakdown voltage

Explanation

Each recombining electron-hole pair releases a photon whose energy is close to the band gap, so the wavelength and hence colour are set by the material. Gallium arsenide phosphide has a band gap giving red-orange light, while gallium nitride gives blue. Current mainly controls brightness, not colour.

Question 45Semiconductor Devices and Diodes

A diode is reverse biased with a voltage well below its breakdown rating. The small current that still flows is mainly

  1. thermally generated minority carriers drifting across the junction Correct answer.
  2. zero at all temperatures
  3. the same as the forward current
  4. majority carriers diffusing from p to n

Explanation

Below breakdown, the reverse current is the saturation current carried by minority carriers thermally generated near the junction and swept across by the field. It is nearly independent of reverse voltage but doubles roughly every 10 degrees Celsius. It is far smaller than the forward current.

Question 46Semiconductor Devices and Diodes

Two silicon diodes are connected in series, both forward biased, and carry 15 mA. The total voltage drop across the pair is approximately

  1. 2.1 V
  2. 0.7 V
  3. 1.4 V Correct answer.
  4. 3.5 V

Explanation

Series diodes each drop about 0.7 V at normal forward current, so the combined drop is roughly 1.4 V. The same current flows through both because they are in series. This principle is used in diode-based voltage references and in level shifting.

Question 47Semiconductor Devices and Diodes

In a bridge rectifier feeding a capacitor-input filter, the peak inverse voltage across each diode is approximately equal to

  1. half the peak secondary voltage
  2. twice the peak secondary voltage
  3. zero
  4. the peak secondary voltage Correct answer.

Explanation

In the bridge circuit each non-conducting diode sees the full peak secondary voltage across it during the opposite half-cycle. This is half the value required by a centre-tap full-wave rectifier, where each diode must withstand twice the peak of one half winding. The lower stress is a further advantage of the bridge.

Question 48Semiconductor Devices and Diodes

A Zener diode and a forward-biased silicon diode are connected in series in the same branch, with the Zener reverse biased at its rated 5.1 V. The total branch drop is approximately

  1. 5.1 V
  2. 10.2 V
  3. 5.8 V Correct answer.
  4. 4.4 V

Explanation

The drops add because the devices are in series: 5.1 V across the reverse-biased Zener plus about 0.7 V across the forward-biased diode gives roughly 5.8 V. Such combinations are used to trim reference voltages. Current must be limited to keep the Zener within its rating.

Question 49Semiconductor Devices and Diodes

A diode whose reverse saturation current is 2 microampere at 25 degrees Celsius is heated to 55 degrees Celsius. The reverse current becomes approximately

  1. 8 microampere
  2. 2 microampere
  3. 16 microampere Correct answer.
  4. 4 microampere

Explanation

Reverse saturation current roughly doubles for every 10 degrees Celsius rise, and a 30 degree rise contains three such intervals. Doubling three times multiplies the current by eight, giving about 16 microampere. This strong temperature dependence is why germanium diodes leak badly when warm.

Question 50Semiconductor Devices and Diodes

A p-n junction diode is made from silicon and is connected in a circuit. Which property of the depletion region is responsible for the diode conducting current easily in forward bias but hardly at all in reverse bias?

  1. The depletion region narrows in forward bias, so the barrier is lowered, and widens in reverse bias, so the barrier is raised. Correct answer.
  2. The depletion region widens in forward bias and narrows in reverse bias.
  3. The depletion region has the same width in both biases, but its resistance changes with current direction.
  4. The depletion region disappears completely in reverse bias, blocking the flow of majority carriers.

Explanation

Forward bias applies a field that opposes the built-in barrier, so the depletion layer becomes thin and majority carriers cross easily. Reverse bias adds to the barrier, widening the depletion layer and leaving only a tiny leakage current of minority carriers. This asymmetry in barrier width and height is what gives the diode its one-way conduction property.