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CBSE Class 12 Physics · 11 questions · 26 marks
This chapter explains how the electrical behaviour of silicon and germanium can be engineered by adding tiny amounts of impurities, turning them into the diodes and transistors that form the backbone of modern electronic circuits. It builds from the energy-band picture of solids up to how a p-n junction rectifies alternating current and how a transistor amplifies signals.
In a pure (intrinsic) semiconductor at room temperature, current is carried by:
Answer
Equal numbers of electrons and holes — in an intrinsic semiconductor, every electron that is thermally excited into the conduction band leaves behind a hole in the valence band, so electrons and holes are always generated in equal numbers and both contribute to conduction.
Adding a small amount of pentavalent impurity (like phosphorus) to pure silicon produces:
Answer
n-type semiconductor with electrons as majority carriers — a pentavalent atom has five valence electrons, but silicon needs only four for its covalent bonds, so the fifth electron is loosely bound and easily freed, becoming a majority charge carrier and giving the material n-type (negative carrier) character.
When a p-n junction diode is connected in reverse bias, the width of the depletion region:
Answer
Increases — reverse bias pulls majority carriers further away from the junction on both sides (electrons in the n-region toward the positive terminal, holes in the p-region toward the negative terminal), widening the region depleted of mobile carriers and increasing the barrier to current flow.
In a common-emitter transistor amplifier, if the base current changes by 20 μA and the collector current changes by 3 mA, the current amplification factor β is:
Answer
150 — β = ΔI_C/ΔI_B = (3×10⁻³ A)/(20×10⁻⁶ A) = 3000/20 = 150, meaning a small change in base current produces a collector current change 150 times as large, which is the basis of transistor current amplification.
Assertion (A): A Zener diode is always operated in reverse bias when used as a voltage regulator. Reason (R): In reverse breakdown, the voltage across a Zener diode stays nearly constant over a wide range of currents through it.
Answer
Both A and R are true and R is the correct explanation of A — the whole purpose of a voltage regulator is to hold output voltage steady; the Zener diode's reverse breakdown region is specially engineered (through heavy doping) to keep voltage nearly fixed even as current varies considerably, which is exactly the property exploited in reverse-bias regulator circuits.
Distinguish between an intrinsic and an extrinsic semiconductor, giving one example of how each is obtained.
Answer
An intrinsic semiconductor is a pure semiconductor material (like pure silicon or germanium) with no significant impurities, where electron and hole concentrations are equal and depend only on thermal excitation across the energy gap. An extrinsic semiconductor is obtained by deliberately adding a small, controlled amount of impurity (doping) to an intrinsic semiconductor — for example, doping pure silicon with a pentavalent element like phosphorus produces an n-type extrinsic semiconductor with a much higher electron concentration than a comparable intrinsic sample.
With the help of a labelled circuit description, explain how a p-n junction diode works as a half-wave rectifier.
Answer
The AC input voltage is connected across the diode in series with a load resistor. During the half-cycle when the p-side of the junction is at a higher potential than the n-side, the diode is forward biased, its resistance is low, and current flows through the load, reproducing that half of the input waveform across the load. During the other half-cycle, the diode becomes reverse biased, its resistance becomes very high, and essentially no current flows through the load. The result is that only alternate half-cycles of the input AC appear across the load, giving a pulsating output that is unidirectional (always the same polarity) though not smooth, hence the name half-wave rectifier — only half the input wave is 'used' in each full cycle.
Why is the base region of a transistor made thin and only lightly doped compared to the emitter and collector regions?
Answer
The base is made thin and lightly doped so that only a small fraction of the electrons (or holes) injected from the heavily doped emitter recombine with majority carriers in the base; most of them are able to diffuse across the thin base and reach the collector. This design ensures the collector current is nearly as large as the emitter current, with only a small base current needed to sustain the process, which is what makes a large current amplification factor (β) possible.
Explain, with the help of energy band diagrams described in words, how conductors, insulators, and semiconductors differ in terms of the valence band, conduction band, and forbidden energy gap. Explain why the conductivity of a pure semiconductor increases with increasing temperature, unlike that of a metal.
Answer
In a conductor, the valence band and conduction band overlap, or the conduction band is already partially filled with electrons even at absolute zero, so electrons can move freely under an applied field with no energy gap to cross; this gives conductors their very high conductivity. In an insulator, the valence band is completely full and the conduction band is completely empty, separated by a large forbidden energy gap (Eg > 3 eV); at ordinary temperatures, essentially no electrons have enough thermal energy to jump this gap, so almost no conduction occurs. In a semiconductor, the arrangement is similar to an insulator but the forbidden gap is much smaller (Eg ≈ 1 eV); at room temperature, a small but significant number of electrons acquire enough thermal energy to jump from the valence band to the conduction band, leaving holes behind, so limited conduction is possible. As temperature increases further, more electrons gain enough thermal energy to cross the gap, increasing the number of free electrons and holes, so the conductivity of a semiconductor increases with temperature. In a metal, conductivity instead decreases with temperature, because the number of free electrons is already essentially fixed (no gap to cross), and rising temperature only increases the vibration of the lattice ions, causing more frequent collisions that impede electron flow (increased resistance).
(a) Draw and explain the input and output characteristics of an n-p-n transistor in common-emitter configuration (describe in words). (b) A common-emitter amplifier has a base current of 25 μA and β = 100. Calculate the collector current and, if the collector supply voltage is 12 V with a 2 kΩ collector resistor, find the collector-emitter voltage assuming negligible saturation voltage.
Answer
(a) The input characteristic is a plot of base current I_B against base-emitter voltage V_BE at constant collector-emitter voltage V_CE; it resembles the forward characteristic of a p-n junction diode, since the base-emitter junction is forward biased, showing very small I_B until V_BE crosses a threshold (~0.6-0.7 V for silicon), after which I_B rises steeply. The output characteristic is a plot of collector current I_C against V_CE for different fixed values of I_B; for small V_CE, I_C rises rapidly with V_CE (active-to-saturation transition), but beyond a certain V_CE, I_C becomes nearly constant (flat/active region) and depends mainly on I_B rather than V_CE, which is the region used for linear amplification. (b) I_C = β × I_B = 100 × 25×10⁻⁶ = 2.5×10⁻³ A = 2.5 mA. Voltage drop across collector resistor = I_C × R_C = 2.5×10⁻³ × 2000 = 5 V. V_CE = V_CC - I_C R_C = 12 - 5 = 7 V (assuming negligible saturation/base-emitter drop is absorbed elsewhere in the circuit as stated).
Read the passage and answer the questions that follow: A student builds a full-wave bridge rectifier circuit using four identical silicon diodes to convert household AC mains into DC to charge a small battery. The AC input has a peak voltage of 12 V, and each diode has a forward voltage drop of about 0.7 V. In a bridge rectifier, current always flows through two diodes in series with the load during each half-cycle. (a) Explain briefly why two diodes conduct in series during each half-cycle in a bridge rectifier, unlike a simple half-wave rectifier. (b) Calculate the approximate peak output voltage across the load, accounting for the diode drops. (c) State one advantage of a full-wave bridge rectifier over a half-wave rectifier. (d) Explain why silicon diodes are generally preferred over germanium diodes in such practical circuits.
Answer
(a) In a bridge rectifier, four diodes are arranged so that for each half-cycle of the AC input, current is directed through the load in the same direction via a different pair of diodes conducting in series (one pair for the positive half-cycle, the other pair for the negative half-cycle), which is what allows both halves of the AC cycle to be used, unlike the single diode of a half-wave rectifier. (b) Two diodes conduct in series, so the total voltage drop is 2 × 0.7 = 1.4 V. Peak output voltage ≈ 12 - 1.4 = 10.6 V across the load. (c) A full-wave bridge rectifier utilises both halves of the AC input cycle (unlike a half-wave rectifier, which wastes one half), giving a higher average output voltage, less pulsation (ripple), and more efficient use of the transformer and input power. (d) Silicon diodes are preferred because silicon has a larger forbidden energy gap (1.1 eV vs 0.7 eV for germanium), giving silicon diodes much lower reverse saturation current, greater stability with temperature changes, and the ability to withstand higher temperatures before being damaged, making them more reliable in everyday practical circuits.
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