Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits Application Questions | physics Class 12 CBSE

Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A student connects a p-n junction diode to a battery such that the p-side is at higher potential. She observes a current of several milliamperes. With reference to the forward biasing of a p-n junction, explain why current flows so readily in this configuration.

When p-side is at higher potential (forward bias), the applied voltage reduces the barrier potential from V0 to (V0 - V), narrowing the depletion region and lowering the resistance offered to majority carriers.
Majority carriers — electrons from n-side and holes from p-side — now have enough energy to cross the junction; minority carrier injection occurs, increasing minority carrier concentration near the boundary and driving diffusion currents.
The forward current is in the mA range because the combined hole diffusion current and electron diffusion current on either side of the junction are both significant; forward dynamic resistance is very low (few ohms).

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

In a practical circuit, an engineer replaces a silicon diode with a germanium diode in a small-signal detector. With reference to the V-I characteristics of both diodes, explain what difference in behaviour he should expect.

The threshold (cut-in) voltage is ~0.7 V for silicon and ~0.2 V for germanium; the germanium diode starts conducting at a much lower forward voltage, making it more sensitive for detecting small-amplitude signals.
For the same forward bias voltage (above 0.2 V but below 0.7 V), the Ge diode conducts while the Si diode does not, giving a significant current difference in this voltage range.
However, the reverse saturation current is slightly higher in Ge than in Si due to its smaller band gap (0.7 eV vs 1.1 eV), meaning Ge diodes have poorer reverse blocking behaviour and greater temperature sensitivity.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

A pure Si crystal has 5 × 10^28 atoms m^-3 and is doped with 1 ppm of pentavalent arsenic. Given ni = 1.5 × 10^16 m^-3, calculate the number of conduction electrons and holes after doping and comment on whether electrons or holes dominate.

Donor concentration ND = 1 ppm of 5×10^28 = 5×10^22 m^-3; since ND >> ni, thermally generated intrinsic electrons are negligible, so ne ≈ ND = 5×10^22 m^-3.
Using nenh = ni^2: nh = ni^2/ne = (1.5×10^16)^2/(5×10^22) = 2.25×10^32/5×10^22 ≈ 4.5×10^9 m^-3.
Electrons (5×10^22 m^-3) vastly outnumber holes (4.5×10^9 m^-3) by a factor of more than 10^13; this confirms an n-type semiconductor where electrons are overwhelmingly the majority carriers.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A technician measures the reverse current of a p-n junction diode at 25°C and finds it to be 1 μA. She then raises the temperature to 50°C and finds the reverse current has increased noticeably. With reference to the origin of reverse saturation current, explain this observation.

Reverse saturation current arises from minority carriers (electrons on p-side and holes on n-side) being swept across the junction by the electric field in the depletion region; its magnitude depends on minority carrier concentration.
As temperature increases, more electron-hole pairs are thermally generated (ni increases significantly with temperature), raising the minority carrier concentration on both sides.
With more minority carriers available to be swept across the junction, the reverse saturation current increases; this temperature sensitivity is a key difference between the ideal diode model and real diode behaviour.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A household adapter converts 230 V ac mains into 9 V dc for a mobile charger. Explain, with reference to the circuit components, how a full-wave rectifier with a capacitor filter achieves this dc output from ac input.

A step-down transformer first reduces 230 V ac to the required lower ac voltage; the centre-tap of the secondary provides two out-of-phase half-voltages for the two diodes.
The two diodes (D1 and D2) conduct alternately — D1 during positive half-cycles and D2 during negative half-cycles of the ac input — so current always flows through the load RL in the same direction, giving full-wave rectified pulsating dc at 100 Hz.
A large capacitor connected in parallel with RL charges to the peak voltage during each pulse and discharges slowly through RL between pulses; the time constant CRL is large enough to keep the output voltage nearly steady, producing smooth dc close to the peak rectified voltage.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

With reference to the energy band model, explain why adding phosphorus (a group V element) to silicon increases its conductivity, while pure silicon at room temperature is a poor conductor.

Pure silicon has a band gap of 1.1 eV; at room temperature only a tiny fraction of electrons (~10^16 m^-3 out of ~10^28 m^-3 atoms) are thermally excited to the conduction band, leaving most bonds intact and conductivity very low.
When phosphorus is added (pentavalent dopant), each P atom contributes one extra electron that sits in a donor energy level ED just 0.05 eV below the conduction band bottom EC.
This tiny energy gap means virtually all donor electrons are ionised and in the conduction band at room temperature, increasing electron concentration to ~10^22 m^-3 for 1 ppm doping — many orders of magnitude higher than the intrinsic value, dramatically increasing conductivity.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A student observes that connecting a diode in reverse bias to a battery of increasing voltage produces almost no current until suddenly a large current destroys the diode. With reference to breakdown voltage, explain what happened.

As reverse bias increases, the diode carries only the small reverse saturation current (~μA), which remains nearly constant because it is limited by minority carrier concentration, not by voltage.
At the breakdown voltage Vbr, the reverse current suddenly increases sharply; even a slight further increase in voltage causes a very large change in current, as the junction loses its rectifying property.
If the student had not limited the current externally using a resistor below the diode's rated value, the large current caused overheating and permanently destroyed the p-n junction; a protective resistor in series with the diode would have prevented this.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

In a radio receiver circuit, a galena crystal (PbS) with a metal point contact was historically used to detect radio waves. Relate this to the semiconductor properties discussed in the chapter.

Galena (lead sulphide, PbS) is a naturally occurring semiconductor; the metal-semiconductor contact forms a junction analogous to a p-n junction diode with rectifying properties.
This junction allows current predominantly in one direction (like a forward-biased diode), so the high-frequency alternating radio signal is rectified, extracting the audio information carried on the radio wave.
This is one of the earliest practical applications of semiconductor properties — well before the formal understanding of semiconductors — illustrating that the same principles of unidirectional conduction and junction rectification apply to naturally occurring semiconductor materials.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

An engineer must choose between a half-wave and a full-wave rectifier for a regulated dc power supply that must deliver steady output with minimum ripple. Which circuit should she choose and why, with reference to output frequency and filter capacitor requirements?

She should choose the full-wave rectifier, which produces output at 100 Hz for a 50 Hz input (double that of half-wave), meaning the gaps between output pulses are shorter and less voltage droop occurs.
The higher output frequency means the capacitor filter needs to maintain the voltage for only half the time between pulses compared to half-wave, requiring a smaller capacitance to achieve the same ripple level.
Full-wave rectification also utilises both half-cycles of the input, giving a higher average output voltage and more efficient use of the transformer, making it the standard choice for power supply design.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

With reference to the p-n junction, a student claims that two p-n junctions can be made by pressing three semiconductor slabs (p-n-p) together. Critically evaluate this claim.

The claim is incorrect; as stated in the chapter, any slab surface — no matter how polished — has roughness much larger than the inter-atomic crystal spacing (~2 to 3 Å).
Pressing slabs together cannot achieve continuous atomic-level contact across the entire interface, so the pressed interface would behave as a physical discontinuity rather than a true semiconductor junction.
True junctions require in-situ formation within a single crystal wafer, such as by diffusion or ion implantation of dopants, which ensures continuous atomic lattice and proper formation of the depletion region.

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Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits | Application Questions

A student connects a p-n junction diode to a battery such that the p-side is at higher potential. She observes a current of several milliamperes. With reference to the forward biasing of a p-n junction, explain why current flows so readily in this configuration.

In a practical circuit, an engineer replaces a silicon diode with a germanium diode in a small-signal detector. With reference to the V-I characteristics of both diodes, explain what difference in behaviour he should expect.

A pure Si crystal has 5 × 10^28 atoms m^-3 and is doped with 1 ppm of pentavalent arsenic. Given ni = 1.5 × 10^16 m^-3, calculate the number of conduction electrons and holes after doping and comment on whether electrons or holes dominate.

A technician measures the reverse current of a p-n junction diode at 25°C and finds it to be 1 μA. She then raises the temperature to 50°C and finds the reverse current has increased noticeably. With reference to the origin of reverse saturation current, explain this observation.

A household adapter converts 230 V ac mains into 9 V dc for a mobile charger. Explain, with reference to the circuit components, how a full-wave rectifier with a capacitor filter achieves this dc output from ac input.

With reference to the energy band model, explain why adding phosphorus (a group V element) to silicon increases its conductivity, while pure silicon at room temperature is a poor conductor.

A student observes that connecting a diode in reverse bias to a battery of increasing voltage produces almost no current until suddenly a large current destroys the diode. With reference to breakdown voltage, explain what happened.

In a radio receiver circuit, a galena crystal (PbS) with a metal point contact was historically used to detect radio waves. Relate this to the semiconductor properties discussed in the chapter.

An engineer must choose between a half-wave and a full-wave rectifier for a regulated dc power supply that must deliver steady output with minimum ripple. Which circuit should she choose and why, with reference to output frequency and filter capacitor requirements?

With reference to the p-n junction, a student claims that two p-n junctions can be made by pressing three semiconductor slabs (p-n-p) together. Critically evaluate this claim.

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