Questions: Semiconductor Electronics

Q1. In an n-type silicon, which statement is true?

Answer (English): The correct option says holes are the minority carriers and the dopants are pentavalent. The donor's extra electron is only weakly bound, so electrons dominate and holes are scarce. Trivalent atoms do the opposite job: they accept an electron and give p-type material.

Hinglish: Sahi statement yeh hai ki holes minority carriers hain aur pentavalent atoms dopants hain. Pentavalent donor ka extra electron kam energy par free ho jaata hai, isliye electrons majority ban jaate hain aur holes minority. Trivalent atoms acceptors hote hain jo p-type banate hain, n-type nahi.

Q2. Which statement from the n-type question is true for a p-type semiconductor?

Answer (English): For p-type silicon the correct option names holes as majority carriers and trivalent atoms as dopants. Every trivalent acceptor leaves one bond unsatisfied, and that gap acts as a hole, so holes outnumber electrons.

Hinglish: p-type silicon ke liye sahi statement yeh hai ki holes majority carriers hain aur trivalent atoms dopants hain. Har trivalent acceptor ek bond adhura chhodta hai jo hole ban jaata hai, isliye holes electrons se zyada ho jaate hain.

Q3. Carbon, silicon and germanium all have four valence electrons. Which relation between their band gaps is true?

Answer (English): The gaps are about 5.4 eV for carbon, 1.1 eV for silicon and 0.72 eV for germanium, so the order is carbon greatest, then silicon, then germanium. The relation is E g of carbon greater than E g of silicon greater than E g of germanium, which is why diamond behaves as an insulator while silicon and germanium are semiconductors.

Hinglish: Gaps lagbhag carbon 5.4 eV, silicon 1.1 eV aur germanium 0.72 eV hain, isliye order carbon > silicon > germanium hai. Isi wajah se diamond insulator jaisa behave karta hai jabki silicon aur germanium semiconductors hain.

Q4. In an unbiased p-n junction, why do holes diffuse from the p-region to the n-region?

Answer (English): Holes diffuse from p to n because the concentration of holes is much higher in the p-region than in the n-region, so there is a concentration gradient across the junction. The diffusion is driven by that gradient and not by any applied voltage or by attraction from free electrons.

Hinglish: Holes p se n ki taraf isliye diffuse karte hain kyunki p-region mein holes ki concentration n-region se bahut zyada hoti hai, isliye junction par concentration gradient hota hai. Yeh diffusion usi gradient ki wajah se hota hai, kisi applied voltage ya free electrons ke attraction se nahi.

Q5. If a p-n junction is forward biased, what happens to its potential barrier?

Answer (English): The potential barrier is lowered. The applied voltage opposes the built-in barrier field, so the effective barrier becomes V0 - V. The depletion region also becomes narrower, which lets more carriers cross and the current rises.

Hinglish: Potential barrier kam ho jaata hai. Battery ka voltage built-in field ke khilaaf kaam karta hai, isliye barrier ghatkar V0 - V ho jaata hai. Depletion region bhi patli hoti hai, jisse zyada carriers cross karte hain aur current badh jaata hai.

Q6. An ac input of 50 Hz is fed to a half-wave rectifier. What is the output frequency, and what would it be for a full-wave rectifier?

Answer (English): With half-wave rectification one pulse appears for each input cycle, so the output frequency is still 50 Hz. Full-wave rectification produces a pulse in every half cycle, which doubles the output frequency to 100 Hz.

Hinglish: Half-wave rectifier har input cycle mein ek pulse deta hai, isliye output frequency 50 Hz hi rehti hai. Full-wave rectifier har half cycle mein pulse deta hai, isliye output frequency double, yaani 100 Hz hoti hai.

Q7. A pure silicon crystal has 5 × 10 to the 28 atoms per cubic metre and is doped with 1 ppm of arsenic. Taking n i = 1.5 × 10 to the 16 per cubic metre, find the electron and hole concentrations.

Answer (English): One ppm of 5 × 10 to the 28 is 5 × 10 to the 22 donor atoms per cubic metre, and each donor gives one electron, so n e is about 5 × 10 to the 22 per cubic metre. Using n e n h = n i squared, n h = (2.25 × 10 to the 32)/(5 × 10 to the 22), which is about 4.5 × 10 to the 9 per cubic metre. The holes are many orders of magnitude fewer than the electrons, as expected in n-type material.

Hinglish: 5 × 10²⁸ ka 1 ppm yaani 5 × 10²² donor atoms per m³, aur har donor ek electron deta hai, isliye ne ≈ 5 × 10²² m⁻³. ne nh = ni² se nh = (2.25 × 10³²)/(5 × 10²²) ≈ 4.5 × 10⁹ m⁻³. n-type material mein holes electrons se kai orders kam hote hain, wahi yahan dikh raha hai.

Q8. How does the conductivity of an intrinsic semiconductor change with temperature? Explain and show the variation in a plot.

Answer (English): The conductivity increases with temperature. Heating creates more electron-hole pairs because more electrons can cross the energy gap, so the carrier concentration rises sharply. The plot has temperature on the x-axis and conductivity on the y-axis, and rises with increasing slope. This is opposite to a metal, where resistance grows because the lattice scatters electrons more at higher temperature.

Hinglish: Temperature badhne par conductivity badhti hai. Garam karne par zyada electrons energy gap cross karte hain, isliye zyada electron-hole pairs bante hain aur carrier concentration tezi se badhti hai. Plot mein temperature x-axis par aur conductivity y-axis par hoti hai, aur curve badhte dhalaan ke saath upar jaata hai. Metal mein iska ulta hota hai, kyunki wahan lattice scattering badhne se resistance badhta hai.

Q9. Explain the roles of diffusion current and drift current in the formation of the depletion layer of a p-n junction.

Answer (English): Diffusion carries electrons from the n-side to the p-side and holes from the p-side to the n-side, because of the concentration differences. This leaves fixed positive donor ions on the n-side and fixed negative acceptor ions on the p-side, building the space-charge layers that make the depletion region. That space charge sets up an electric field pointing from n to p, which drives electrons back to the n-side and holes back to the p-side as the drift current. Diffusion starts large and drift small; as the space charge grows, drift catches up, and the junction reaches equilibrium when the two currents are equal.

Hinglish: Diffusion concentration ke farq se electrons ko n se p aur holes ko p se n le jaata hai. Isse n-side par fixed positive donor ions aur p-side par fixed negative acceptor ions reh jaate hain, jo space-charge layers bana kar depletion region banate hain. Yeh space charge n se p ki taraf electric field banata hai, jo electrons ko wapas n-side aur holes ko wapas p-side bhejta hai, wahi drift current hai. Shuru mein diffusion bada aur drift chhota hota hai; space charge badhne par drift bhi badhta hai, aur jab dono barabar ho jaate hain to junction equilibrium mein aa jaata hai.

Q10. Explain the property of a p-n junction that makes it suitable for rectifying alternating voltages, and differentiate between a half-wave and a full-wave rectifier.

Answer (English): A p-n junction conducts only when it is forward biased, because then the barrier is lowered and the resistance is small. In reverse bias the barrier is raised, only a tiny saturation current flows, and the resistance is very large. Applied to an alternating voltage this means current passes in one half cycle and not in the other, which is rectification. A half-wave rectifier uses one diode and gives one pulse per input cycle, so the output frequency equals the input frequency. A full-wave rectifier uses two diodes with a centre-tapped transformer and gives a pulse in every half cycle, so its output frequency is twice the input frequency and its output is smoother.

Hinglish: p-n junction sirf forward bias mein conduct karta hai, kyunki tab barrier kam aur resistance chhoti hoti hai; reverse bias mein barrier badh jaata hai aur sirf chhota saturation current bahta hai. Alternating voltage par iska matlab hai ki current ek half cycle mein guzarta hai aur doosre mein nahi, yahi rectification hai. Half-wave rectifier ek diode se har cycle mein ek pulse deta hai, isliye output frequency input ke barabar hoti hai. Full-wave rectifier centre-tapped transformer aur do diodes se har half cycle mein pulse deta hai, isliye output frequency double aur output zyada smooth hota hai.

To check how much of this stuck, take the five-question quiz below. Each answer explains the reasoning, so read the explanation even when you get it right.

Quiz: the five question check for this chapter is set from the same topics. Attempt it after the questions above, then revisit any topic page where you lost a mark.