Hello dear learner. Welcome to HSLC GURU. In this lesson we present the complete English-medium question-and-answer set, key concepts, derivations, formula sheet and additional important questions of the fourteenth chapter — Semiconductor Electronics: Materials, Devices and Simple Circuits — of the ASSEB (Assam State School Education Board) Class 12 Physics syllabus. The microchip in your smartphone, the LED on your study lamp, the solar panel on your rooftop, the diode that converts AC to DC inside every adapter — every one of these devices is built on the physics of semiconductors. Mastering this chapter is therefore essential not only for your board examination but also for any future engineering or electronics-oriented career.
Summary
A semiconductor is a crystalline solid whose electrical conductivity lies between that of a metal and an insulator. According to the band theory of solids, the close packing of atoms in a crystal splits the discrete atomic energy levels into closely-spaced energy bands. The highest band that is normally completely filled at $T=0$ K is called the valence band (VB), the next higher band is the conduction band (CB), and the gap separating them is the energy gap $E_g$. In a conductor the two bands either overlap or the conduction band is partly filled, so $E_g \approx 0$. In an insulator $E_g > 3$ eV, so almost no electron can cross at room temperature. In a semiconductor $E_g \sim 1$ eV (e.g., $E_g(\text{Ge})\approx 0.72$ eV, $E_g(\text{Si})\approx 1.1$ eV); thermal energy can excite a small fraction of valence electrons across the gap, leaving behind holes in the valence band.
A pure tetravalent crystal of Si or Ge is called an intrinsic semiconductor, in which the number of free electrons equals the number of holes: $n_e = n_h = n_i$. By doping the crystal with a small fraction of impurity atoms one obtains an extrinsic semiconductor. A pentavalent dopant (P, As, Sb) donates an extra electron to the lattice and produces an n-type semiconductor (majority carriers — electrons; minority carriers — holes). A trivalent dopant (B, Al, In) creates a hole and produces a p-type semiconductor (majority carriers — holes; minority carriers — electrons). At thermal equilibrium the product of the two carrier concentrations is fixed by the law of mass action, $n_e n_h = n_i^2$, regardless of the doping level.
When a p-type and an n-type region are formed in the same single crystal we get a p-n junction. Diffusion of majority carriers across the junction leaves behind a thin region of immobile ionised impurity atoms — the depletion region — across which a built-in barrier potential $V_b$ (about $0.3$ V for Ge and $0.7$ V for Si) opposes further diffusion. Under forward bias the external EMF lowers $V_b$, the depletion layer narrows and a large current flows; under reverse bias the depletion layer widens and only a tiny reverse-saturation (leakage) current due to minority carriers flows until breakdown occurs at a critical reverse voltage. This unidirectional behaviour is summarised by the diode I-V characteristic.
The unidirectional property is exploited in the half-wave rectifier (a single diode passes only one half of every input cycle) and in the full-wave rectifier (centre-tap or bridge configuration — both half-cycles appear with the same polarity at the load). A Zener diode, heavily doped to produce a sharp reverse breakdown, is used as a voltage regulator. Special-purpose diodes include the photodiode (light-controlled current — used in detectors), the light-emitting diode (LED) (electroluminescent recombination — used in indicators and displays) and the solar cell (photovoltaic action — converts sunlight directly into electrical energy).
Key Formulas and Quantities
| Quantity | Formula / Value | Remark |
|---|---|---|
| Energy gap (insulator) | $E_g > 3$ eV | Diamond $\sim 6$ eV |
| Energy gap (semiconductor) | $E_g \sim 1$ eV | Si: $1.1$ eV · Ge: $0.72$ eV |
| Energy gap (conductor) | $E_g \approx 0$ (overlap) | VB and CB overlap |
| Intrinsic carrier relation | $n_e = n_h = n_i$ | Pure crystal |
| Mass-action law | $n_e n_h = n_i^2$ | Holds in extrinsic case |
| Conductivity | $\sigma = e(n_e\mu_e + n_h\mu_h)$ | $\mu$ = mobility |
| Resistivity | $\rho = 1/\sigma$ | SI: $\Omega\cdot\text{m}$ |
| Drift current | $I = neA v_d$ | $v_d$ = drift speed |
| Diode equation | $I = I_0\!\left(e^{eV/kT}-1\right)$ | Shockley relation |
| Knee/cut-in voltage | Si $\approx 0.7$ V; Ge $\approx 0.3$ V | Forward turn-on |
| Output frequency (HW rectifier) | $f_{\text{out}} = f_{\text{in}}$ | $50$ Hz $\to 50$ Hz |
| Output frequency (FW rectifier) | $f_{\text{out}} = 2f_{\text{in}}$ | $50$ Hz $\to 100$ Hz |
| Rectifier efficiency (HW) | $\eta_{\max} \approx 40.6\%$ | For ideal diode |
| Rectifier efficiency (FW) | $\eta_{\max} \approx 81.2\%$ | For ideal diode |
| Photon energy condition (LED, photodiode) | $h\nu \ge E_g$ | $\lambda \le hc/E_g$ |
| Solar-cell band gap (optimum) | $1.0$ eV $\le E_g \le 1.8$ eV | Match solar spectrum |
Key Concepts at a Glance
1. Energy bands in solids
Isolated atoms have sharp discrete energy levels. When $N$ atoms come together to form a crystal, each level splits into $N$ closely spaced sub-levels — together they look like a continuous band. The two outermost (highest-energy) bands of interest are the valence band and the conduction band, separated by a forbidden gap of width $E_g$.
- Conductor: VB and CB overlap, or CB is partially filled — large free-electron density, $E_g\approx 0$.
- Insulator: VB completely filled, CB empty, $E_g > 3$ eV — no carriers at room temperature.
- Semiconductor: $E_g \sim 1$ eV — at $T>0$ K thermal excitation produces a few electron-hole pairs.
2. Intrinsic and extrinsic semiconductors
An intrinsic semiconductor is the pure tetravalent element Si (or Ge). Each atom forms four covalent bonds with its neighbours. Thermal energy breaks a few bonds, releasing an electron into the CB and leaving a hole in the VB; both contribute to conduction. As the temperature rises $n_i$ increases very rapidly, so the resistivity of a pure semiconductor decreases with rising temperature — the opposite of a metal.
Adding a controlled, very small fraction (one part in $\sim 10^6$ to $10^8$) of an impurity is called doping. A pentavalent dopant (P, As, Sb) has one electron more than is needed for the four bonds; this loosely-bound electron sits on a donor level just below the CB and is easily excited into the CB at room temperature, producing an n-type semiconductor. A trivalent dopant (B, Al, In, Ga) has one electron short; the unfilled bond constitutes a hole bound to an acceptor level just above the VB, producing a p-type semiconductor.
| Property | n-type | p-type |
|---|---|---|
| Dopant valency | Pentavalent (P, As, Sb) | Trivalent (B, Al, In, Ga) |
| Impurity name | Donor | Acceptor |
| Majority carriers | Electrons | Holes |
| Minority carriers | Holes | Electrons |
| Impurity level | Just below CB | Just above VB |
| Net charge | Neutral | Neutral |
3. p-n junction and depletion region
When p- and n-regions are produced in the same crystal, electrons diffuse from the n-side to the p-side and holes diffuse the other way. They leave behind immobile, ionised donor and acceptor atoms — positively charged on the n-side and negatively charged on the p-side. This thin charged layer is the depletion region. The associated electric field gives rise to the barrier potential $V_b$ which finally stops further diffusion at equilibrium.
- Depletion-region width $\approx 10^{-6}$ m for Si.
- $V_b \approx 0.7$ V for Si, $\approx 0.3$ V for Ge.
- The depletion region behaves like a charged capacitor with a built-in field directed from n to p.
4. Forward and reverse bias
If an external battery is connected so that the p-side becomes positive with respect to the n-side, the diode is forward biased. The external field opposes the built-in field, the depletion region narrows, the barrier potential reduces and a large current of majority carriers flows. Above the knee voltage the current rises almost exponentially with $V$, and the diode behaves as a low-resistance closed switch.
If the polarity is reversed (p-side negative, n-side positive) the diode is reverse biased. The external field adds to the built-in field, the depletion region widens, the barrier height increases and only a small reverse-saturation current (of minority carriers) flows. When the reverse voltage exceeds a critical value $V_z$ — the breakdown voltage — the current rises very sharply (avalanche or Zener breakdown).
Visual Aids
Figure 1 — Energy band diagrams (conductor / insulator / semiconductor)
Figure 2 — p-n junction and depletion region
Figure 3 — Forward and reverse I-V characteristic
Figure 4 — Half-wave rectifier circuit and output
Figure 5 — Full-wave (centre-tap) rectifier
Figure 6 — Zener diode as voltage regulator
Textbook (NCERT-pattern) Question and Answer
Q1. In an n-type silicon, which of the following statement is true?
(a) Electrons are majority carriers and trivalent atoms are dopants.
(b) Electrons are minority carriers and pentavalent atoms are dopants.
(c) Holes are minority carriers and pentavalent atoms are dopants.
(d) Holes are majority carriers and trivalent atoms are dopants.
Answer: Option (c) is correct. In an n-type silicon, the dopant is pentavalent (e.g., P, As, Sb), so electrons are the majority carriers and holes are the minority carriers.
Q2. Which of the statements given below is true for p-type semiconductors?
(a) Holes are majority carriers and trivalent atoms are dopants.
(b) Holes are minority carriers and pentavalent atoms are dopants.
(c) Electrons are majority carriers and trivalent atoms are dopants.
(d) Electrons are minority carriers and pentavalent atoms are dopants.
Answer: Option (a) is correct. In p-type silicon, trivalent dopants (B, Al, In, Ga) provide acceptor levels — holes are the majority carriers and electrons are the minority carriers.
Q3. Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to $(E_g)_C$, $(E_g)_{Si}$ and $(E_g)_{Ge}$. Which of the following statement is true?
(a) $(E_g)_{Si} < (E_g)_{Ge} < (E_g)_C$
(b) $(E_g)_C < (E_g)_{Ge} > (E_g)_{Si}$
(c) $(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$
(d) $(E_g)_C = (E_g)_{Si} = (E_g)_{Ge}$
Answer: Option (c) is correct. The energy gaps are approximately $E_g(C) \approx 5.4$ eV (insulator), $E_g(Si) \approx 1.1$ eV (semiconductor) and $E_g(Ge) \approx 0.72$ eV (semiconductor). Therefore $(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$.
Q4. In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) all of the above.
Answer: Option (c) is correct. Holes diffuse because of the concentration gradient: their concentration is much larger in the p-region than in the n-region, and diffusion always proceeds from higher to lower concentration.
Q5. When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) none of the above.
Answer: Option (c) is correct. Forward biasing reduces the height of the potential barrier and narrows the depletion region, allowing a large majority-carrier current to flow.
Q6. In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency?
Answer: A half-wave rectifier passes only one half (positive) of each input cycle, so the output has the same frequency as the input: $f_{\text{HW}} = 50$ Hz. A full-wave rectifier inverts the negative half-cycles into positive ones, doubling the number of pulses per second: $f_{\text{FW}} = 2 \times 50 = 100$ Hz.
Q7. The number of silicon atoms per $\text{m}^3$ is $5 \times 10^{28}$. This is doped simultaneously with $5 \times 10^{22}$ atoms per $\text{m}^3$ of arsenic and $5 \times 10^{20}$ per $\text{m}^3$ atoms of indium. Calculate the number of electrons and holes. Given that $n_i = 1.5 \times 10^{16}\,\text{m}^{-3}$. Is the material n-type or p-type?
Answer: Arsenic (pentavalent) donates electrons; indium (trivalent) creates holes. The net donor density is
$$N_D – N_A = 5\times10^{22} – 5\times10^{20} \approx 4.95\times10^{22}\,\text{m}^{-3}.$$
Since $N_D – N_A \gg n_i$, the electron concentration is
$$n_e \approx N_D – N_A = 4.95\times 10^{22}\,\text{m}^{-3}.$$
By the law of mass action $n_e n_h = n_i^2$,
$$n_h = \frac{n_i^2}{n_e} = \frac{(1.5\times10^{16})^2}{4.95\times10^{22}} \approx 4.55\times10^{9}\,\text{m}^{-3}.$$
Because $n_e \gg n_h$, the resulting material is n-type.
Q8. In an intrinsic semiconductor the energy gap $E_g$ is 1.2 eV. Its hole mobility is much smaller than electron mobility and is independent of temperature. What is the ratio between conductivity at 600 K and at 300 K? Assume that the temperature dependence of intrinsic carrier concentration is given by $n_i = n_0\exp\!\left(-\dfrac{E_g}{2k_BT}\right)$ where $n_0$ is a constant.
Answer: Conductivity is proportional to $n_i$ (since mobility is independent of $T$). Therefore
$$\frac{\sigma_{600}}{\sigma_{300}} = \frac{n_i(600)}{n_i(300)} = \exp\!\left[\frac{E_g}{2k_B}\!\left(\frac{1}{300}-\frac{1}{600}\right)\right].$$
With $E_g = 1.2$ eV and $k_B = 8.617\times10^{-5}$ eV/K,
$$\frac{E_g}{2k_B}\!\left(\frac{1}{300}-\frac{1}{600}\right) = \frac{1.2}{2\times8.617\times10^{-5}}\times\frac{1}{600} \approx 11.6.$$
Hence $\sigma_{600}/\sigma_{300} = e^{11.6} \approx 1.09\times 10^{5}$. The conductivity at 600 K is about $10^5$ times that at 300 K.
Q9. In a p-n junction diode, the current $I$ can be expressed as $I = I_0\!\left[\exp\!\left(\dfrac{eV}{2k_BT}\right) – 1\right]$ where $I_0$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, $I$ is the current through the diode, $k_B$ is the Boltzmann constant ($8.6\times10^{-5}$ eV/K) and $T$ is the absolute temperature. If for a given diode $I_0 = 5\times 10^{-12}$ A and $T = 300$ K, then (a) what will be the forward current at a forward voltage of 0.6 V? (b) What will be the increase in current if the voltage across the diode is increased to 0.7 V?
Answer: At $T = 300$ K, $2k_B T = 2 \times 8.6\times 10^{-5}\times 300 = 0.0516$ eV.
(a) For $V = 0.6$ V:
$$\frac{eV}{2k_BT} = \frac{0.6}{0.0516} \approx 11.63 \;\Rightarrow\; e^{11.63} \approx 1.12\times 10^{5}.$$
$$I_1 = I_0(e^{11.63}-1) \approx 5\times 10^{-12}\times 1.12\times 10^{5} \approx 5.6\times 10^{-7}\,\text{A}.$$
(b) For $V = 0.7$ V:
$$\frac{eV}{2k_BT} = \frac{0.7}{0.0516} \approx 13.57 \;\Rightarrow\; e^{13.57} \approx 7.83\times 10^{5}.$$
$$I_2 \approx 5\times 10^{-12}\times 7.83\times 10^{5} \approx 3.9\times 10^{-6}\,\text{A}.$$
Increase in current: $I_2 – I_1 \approx 3.9\times 10^{-6} – 5.6\times 10^{-7} \approx 3.34\times 10^{-6}$ A. The forward current increases roughly seven-fold for an additional 0.1 V.
Q10. The number of densities in a pure semiconductor is $n_i$. Suppose the number of electrons is $n_e$ and the number of holes is $n_h$. Show that the conductivity of the semiconductor is given by $\sigma = e(n_e\mu_e + n_h\mu_h)$.
Answer: When an electric field $E$ is applied, electrons drift with velocity $v_e = \mu_e E$ (opposite to $E$) and holes with $v_h = \mu_h E$ (along $E$). The total current density is
$$J = J_e + J_h = e n_e v_e + e n_h v_h = e(n_e\mu_e + n_h\mu_h)E.$$
By Ohm’s law $J = \sigma E$, so
$$\sigma = e\,(n_e\mu_e + n_h\mu_h).$$
Additional Important Questions and Answers
Very Short Answer (1 mark)
Q. Define an intrinsic semiconductor.
Answer: A pure semiconductor (free of any impurity) in which the number of conduction electrons equals the number of holes is called an intrinsic semiconductor. Examples: pure Si and pure Ge with $n_e = n_h = n_i$.
Q. Why is silicon preferred over germanium for fabricating semiconductor devices?
Answer: Silicon has a larger band gap ($1.1$ eV) than germanium ($0.72$ eV). At normal operating temperatures the leakage (reverse) current in silicon is much smaller than in germanium, and the silicon-oxide layer formed naturally on its surface allows easy fabrication of integrated circuits.
Q. What is the law of mass action for a semiconductor?
Answer: At thermal equilibrium the product of the concentrations of free electrons and holes in any semiconductor is equal to the square of the intrinsic carrier concentration: $n_e n_h = n_i^2$.
Q. What is the order of magnitude of the depletion-region width and barrier potential in a p-n junction?
Answer: Width $\approx 10^{-6}$ m; barrier potential $\approx 0.7$ V for Si, $\approx 0.3$ V for Ge.
Q. What is reverse saturation current?
Answer: The small, almost constant current that flows through a reverse-biased p-n junction due to the drift of thermally-generated minority carriers across the junction is called the reverse-saturation current. It increases slightly with temperature.
Q. State one use of a solar cell.
Answer: A solar cell directly converts solar energy into electrical energy. It is used in calculators, satellites, water pumps, street lights and rooftop solar power plants — especially in remote areas where conventional electricity supply is unavailable.
Q. Write full forms of LCD, LED and CRT.
Answer: LCD — Liquid Crystal Display; LED — Light Emitting Diode; CRT — Cathode Ray Tube.
Q. What is the optimum band-gap range for solar-cell material?
Answer: Approximately $1.0$ eV to $1.8$ eV — chosen so that a large fraction of the solar spectrum is absorbed and converted efficiently. Silicon ($1.1$ eV) lies in this range.
Q. Resistivity of a semiconductor lies in the range $10^{-5}$ to $10^{6}\,\Omega\cdot\text{m}$. What is the corresponding range of conductivity?
Answer: Conductivity $\sigma = 1/\rho$, so the range is $10^{-6}$ to $10^{5}$ S/m.
Q. Name two categories of integrated circuits (ICs).
Answer: (i) Linear ICs (e.g., operational amplifiers) and (ii) Digital ICs (e.g., logic gates, memories, microprocessors).
Short Answer (2-3 marks)
Q. How will you dope a pure silicon crystal to obtain (i) an n-type and (ii) a p-type semiconductor?
Answer: (i) For an n-type semiconductor, dope pure silicon with a small fraction of pentavalent atoms such as phosphorus (P), arsenic (As) or antimony (Sb). Four of the dopant’s five valence electrons form covalent bonds with neighbouring Si atoms; the fifth electron is loosely bound and easily released into the conduction band, so electrons become the majority carriers. (ii) For a p-type semiconductor, dope pure silicon with trivalent atoms such as boron (B), aluminium (Al), gallium (Ga) or indium (In). The dopant has only three valence electrons; one bond remains incomplete (a hole), and these holes act as the majority carriers.
Q. Distinguish between intrinsic and extrinsic semiconductors.
Answer:
| Intrinsic semiconductor | Extrinsic semiconductor |
|---|---|
| Pure form of Si or Ge. | Doped with a controlled impurity. |
| $n_e = n_h = n_i$. | $n_e \neq n_h$ in general. |
| Conductivity is small and depends only on temperature. | Conductivity is much larger and is controlled by doping. |
| No donor / acceptor levels. | Has donor (n-type) or acceptor (p-type) impurity levels. |
| Used mainly for fundamental study. | Used to fabricate diodes, transistors, ICs, solar cells. |
Q. Explain how a depletion region is formed at a p-n junction.
Answer: When p- and n-type regions are produced inside the same single crystal, electrons from the n-region (where their concentration is high) diffuse to the p-region, and holes diffuse the other way because of their concentration gradients. Each electron that crosses leaves behind an immobile, positively charged donor ion on the n-side; each hole that crosses leaves behind an immobile, negatively charged acceptor ion on the p-side. A thin layer near the junction is therefore depleted of mobile charges and contains only fixed ionised impurity atoms — this is the depletion region. The space-charge produces a built-in electric field directed from n to p that opposes further diffusion; at equilibrium the diffusion current is exactly cancelled by this drift current and no net current flows.
Q. Differentiate forward bias and reverse bias of a p-n junction diode.
Answer:
| Property | Forward bias | Reverse bias |
|---|---|---|
| Polarity | p $\to$ +ve, n $\to$ −ve | p $\to$ −ve, n $\to$ +ve |
| External field | Opposes built-in field | Aids built-in field |
| Depletion region | Narrows | Widens |
| Barrier potential | Decreases | Increases |
| Resistance | Low ($\sim 10^2\,\Omega$) | Very high ($\sim 10^6\,\Omega$) |
| Current | Large (mA, majority carriers) | Very small ($\mu$A, minority carriers) |
Q. Draw and explain the I-V characteristic of a p-n junction diode.
Answer: See Figure 3 above. In the forward region the current is negligible until the forward voltage reaches the cut-in (knee) voltage ($\approx 0.7$ V for Si and $\approx 0.3$ V for Ge); beyond this the current rises almost exponentially as predicted by $I = I_0(e^{eV/k_BT}-1)$. In the reverse region only a small, nearly constant reverse-saturation current flows due to minority carriers, until the reverse voltage reaches the breakdown value $V_z$, beyond which the current rises very sharply. A diode therefore acts as a one-way switch — closed in forward bias and open in reverse bias up to $V_z$.
Q. Explain the working of a Zener diode as a voltage regulator.
Answer: A Zener diode is a heavily-doped p-n junction designed to operate in the reverse-breakdown region without damage. The unregulated DC input is applied through a series resistor $R_s$ to the Zener diode connected in reverse bias (see Figure 6). The load $R_L$ is connected in parallel with the Zener. As long as the input voltage is greater than the Zener voltage $V_z$, breakdown occurs and the voltage across the diode is clamped at $V_z$, regardless of small variations in input voltage or load current. The series resistor $R_s$ absorbs the difference $(V_{\text{in}}-V_z)$. Thus the load receives a steady output $V_{\text{out}} = V_z$.
Q. What is a photodiode? How does it differ from an ordinary diode?
Answer: A photodiode is a special-purpose p-n junction diode designed to detect light. It is operated in reverse bias and a transparent window is provided so that light can fall on the depletion region. Photons of energy $h\nu \ge E_g$ generate extra electron-hole pairs which contribute to the reverse current; thus the reverse current increases with light intensity. Unlike an ordinary diode (which is opaque-encapsulated and used as a rectifier or switch), a photodiode acts as a light-controlled resistor and is used in optical detectors, light meters, fibre-optic receivers and burglar alarms.
Q. How does a light-emitting diode (LED) work? List two advantages over an incandescent lamp.
Answer: An LED is a heavily-doped p-n junction made from a direct-band-gap material (e.g., GaAs, GaP, GaN, AlGaAs). When forward biased, electrons injected into the p-region recombine with holes; the energy released ($h\nu \approx E_g$) is emitted as a photon (electroluminescence). The colour of light depends on the band gap of the semiconductor used. Advantages over incandescent lamps: (i) very high luminous efficiency — most of the input energy becomes light, very little heat; (ii) very long life ($\sim 10^5$ hours), low operating voltage, fast switching, mechanical robustness and small size.
Q. Describe the principle of a solar cell.
Answer: A solar cell is essentially a large-area p-n junction in which the top layer is made very thin so that incident sunlight reaches the depletion region. Photons of energy $h\nu \ge E_g$ generate electron-hole pairs in or near the depletion region. The built-in field then sweeps the electrons to the n-side and holes to the p-side, producing a photo-EMF (photovoltaic effect). When the cell is connected to an external load, a current flows. Suitable materials have $E_g$ between about $1.0$ eV and $1.8$ eV — Si and GaAs are typical.
Long Answer (5 marks)
Q. With circuit diagram and waveforms, explain the working of a half-wave rectifier. Why is its efficiency low?
Answer: See Figure 4. A step-down transformer feeds the AC mains to a single diode connected in series with the load $R_L$. During the positive half-cycle of the secondary voltage, the diode is forward-biased; it conducts and a current flows through $R_L$ in one direction. During the negative half-cycle the diode is reverse-biased and behaves as an open circuit; no current flows through $R_L$. Thus, only the positive halves of the input appear across the load. The output is a unidirectional but pulsating voltage, with the same frequency as the input. Because half of every cycle is wasted, the maximum theoretical rectification efficiency is only $\eta_{\max} \approx 40.6\%$, the ripple factor is high ($\approx 1.21$), and the transformer utilisation factor is poor — therefore half-wave rectifiers are used only in low-power applications (e.g., charging small batteries).
Q. With diagram, explain the principle and working of a full-wave rectifier (centre-tap type).
Answer: See Figure 5. The secondary of a centre-tapped transformer feeds two diodes $D_1$ and $D_2$ whose cathodes are joined and connected through the load $R_L$ back to the centre tap. When the upper end of the secondary is positive (positive half-cycle), $D_1$ is forward-biased and conducts; the current flows through $R_L$ from top to bottom. During the negative half-cycle the lower end becomes positive, $D_2$ conducts, and the current again flows through $R_L$ in the same direction. Therefore both halves of every input cycle contribute to the output. The output frequency is twice the input frequency, the maximum efficiency is $\approx 81.2\%$, the ripple factor reduces to $\approx 0.482$, and the smoothing capacitor required to obtain steady DC is much smaller than for a half-wave rectifier. (A bridge rectifier with four diodes does the same job without needing a centre tap and is widely used in modern adapters.)
Q. On the basis of band theory of solids, distinguish between conductors, insulators and semiconductors. Draw labelled energy band diagrams.
Answer: See Figure 1.
- Conductor: The conduction band is partially filled, or it overlaps the valence band; the energy gap $E_g\approx 0$. A vast number of free electrons are available and the conductivity is very high ($\sigma \sim 10^7$ S/m).
- Insulator: The valence band is full, the conduction band is empty, and the forbidden gap $E_g > 3$ eV (e.g., diamond $\sim 6$ eV). At room temperature thermal energy ($\sim 0.026$ eV) is far too small to push an electron across the gap, so almost no carriers are present and the conductivity is extremely small ($\sigma \sim 10^{-12}$ S/m).
- Semiconductor: The gap is small ($E_g \sim 1$ eV — Si: $1.1$, Ge: $0.72$). Thermal excitation produces a measurable concentration $n_i$ of electron-hole pairs, and conductivity lies between that of metals and insulators ($\sigma \sim 10^{-6}$ to $10^4$ S/m). Conductivity rises sharply with temperature, opposite to a metal.
Q. Compare half-wave and full-wave rectifiers in tabular form.
| Property | Half-wave | Full-wave |
|---|---|---|
| Number of diodes | 1 | 2 (centre-tap) / 4 (bridge) |
| Use of input cycle | One half only | Both halves |
| Output frequency | $f_{\text{in}}$ | $2 f_{\text{in}}$ |
| Maximum efficiency | $\approx 40.6\%$ | $\approx 81.2\%$ |
| Ripple factor | $\approx 1.21$ | $\approx 0.48$ |
| Filter requirement | Large | Small |
| Application | Low-power chargers | Most DC power supplies |
Q. State and prove the law of mass action $n_e n_h = n_i^2$.
Answer: In any non-degenerate semiconductor at thermal equilibrium, the rate at which electron-hole pairs are generated thermally equals the rate at which they recombine. The recombination rate is proportional to the product $n_e n_h$, while the generation rate depends only on temperature. Therefore the product is fixed at a given temperature: $n_e n_h = $ constant. For an intrinsic semiconductor $n_e = n_h = n_i$, so the constant equals $n_i^2$. Hence in any (intrinsic or extrinsic) semiconductor at the same temperature, $n_e n_h = n_i^2$. This is the law of mass action. Its practical consequence is that doping increases one carrier population while reducing the other so that the product remains $n_i^2$.
Multiple Choice / Objective Questions
- The forbidden energy gap of an insulator is —
(a) $\sim 1$ eV (b) $\sim 0$ eV (c) $> 3$ eV (d) $\sim 0.5$ eV.
Answer: (c). - The dopant used to obtain a p-type semiconductor is —
(a) Phosphorus (b) Arsenic (c) Antimony (d) Boron.
Answer: (d). - In an n-type semiconductor the Fermi level lies —
(a) Inside the valence band (b) Just below the conduction band (c) At the centre of the gap (d) Inside the conduction band.
Answer: (b). - The barrier potential of a Si diode at room temperature is approximately —
(a) $0.1$ V (b) $0.3$ V (c) $0.7$ V (d) $1.1$ V.
Answer: (c). - For a full-wave rectifier whose input frequency is $50$ Hz, the output frequency is —
(a) $25$ Hz (b) $50$ Hz (c) $100$ Hz (d) $200$ Hz.
Answer: (c). - A Zener diode is operated in —
(a) Forward bias (b) Reverse breakdown (c) Cut-off (d) Saturation.
Answer: (b). - Solar cells convert —
(a) Electrical to light energy (b) Light to electrical energy (c) Heat to electrical energy (d) Sound to electrical energy.
Answer: (b). - The photon energy required to operate a photodiode satisfies —
(a) $h\nu < E_g$ (b) $h\nu = 0$ (c) $h\nu \ge E_g$ (d) $h\nu = 2E_g$.
Answer: (c). - The depletion region of an unbiased p-n junction contains —
(a) Only mobile electrons (b) Only mobile holes (c) Only immobile ionised impurities (d) Both mobile electrons and holes.
Answer: (c). - The mass-action law for a semiconductor is —
(a) $n_e + n_h = n_i$ (b) $n_e – n_h = n_i$ (c) $n_e n_h = n_i^2$ (d) $n_e/n_h = n_i$.
Answer: (c).
Fill in the Blanks
- In an n-type semiconductor, the majority carriers are __________. (electrons)
- The energy gap of pure germanium is approximately __________ eV. (0.72)
- A p-type semiconductor is obtained by adding a __________ valent impurity. (tri)
- The output frequency of a half-wave rectifier with $50$ Hz input is __________ Hz. (50)
- A Zener diode is operated in the __________ region. (reverse-breakdown)
- An LED emits light when it is __________ biased. (forward)
- A photodiode is operated in __________ bias for detection. (reverse)
- The full form of LCD is __________. (Liquid Crystal Display)
- The barrier potential of a Si diode is approximately __________ V. (0.7)
- The expression for the conductivity of a semiconductor is $\sigma = $ __________. ($e(n_e\mu_e + n_h\mu_h)$)
True / False
- A semiconductor is electrically neutral whether it is intrinsic or extrinsic. True.
- Adding a pentavalent impurity to silicon produces a p-type material. False — it produces an n-type material.
- In reverse bias the depletion region of a p-n junction widens. True.
- A half-wave rectifier produces an output frequency twice that of the input. False — only equal to the input.
- The barrier potential of a Si diode is larger than that of a Ge diode. True.
- A Zener diode is mainly used as a voltage amplifier. False — it is used as a voltage regulator.
- The conductivity of an intrinsic semiconductor decreases with rising temperature. False — it increases.
- An LED converts electrical energy into light energy by electroluminescence. True.
- A solar cell requires an external bias to deliver an output current. False — it generates its own EMF (photovoltaic effect).
- In an extrinsic semiconductor the law $n_e n_h = n_i^2$ no longer holds. False — it always holds at thermal equilibrium.
Numerical Practice
N1. Pure germanium at $300$ K has $n_i = 2.4\times 10^{19}\,\text{m}^{-3}$. If it is doped with $4\times 10^{23}$ donor atoms/m$^3$, find $n_e$ and $n_h$.
Solution: Since $N_D \gg n_i$, $n_e \approx N_D = 4\times 10^{23}\,\text{m}^{-3}$. From the law of mass action,
$$n_h = \frac{n_i^2}{n_e} = \frac{(2.4\times 10^{19})^2}{4\times 10^{23}} = 1.44\times 10^{15}\,\text{m}^{-3}.$$
N2. A silicon diode has a reverse-saturation current $I_0 = 10^{-9}$ A at $T = 300$ K. Find the forward current at $V = 0.5$ V (use $kT/e = 0.026$ V and the ideal diode equation).
$$I = I_0(e^{V/(kT/e)} – 1) = 10^{-9}\!\left(e^{0.5/0.026} – 1\right) = 10^{-9}\times e^{19.23} \approx 2.25\times 10^{-1}\,\text{A}.$$
(About $0.225$ A — well within the operating range of a normal Si diode.)
N3. The energy gap of a semiconductor is $1.5$ eV. Find the maximum wavelength of light that can produce electron-hole pairs in it.
$$\lambda_{\max} = \frac{hc}{E_g} = \frac{1240\,\text{eV}\cdot\text{nm}}{1.5\,\text{eV}} \approx 826.7\,\text{nm}.$$
(Light of wavelength shorter than $\sim 827$ nm — i.e., in the near-infrared and visible range — can excite electrons across the gap.)
N4. A full-wave bridge rectifier supplies a load of $R_L = 1\,\text{k}\Omega$ from a sinusoidal source whose peak voltage is $V_m = 12$ V. Assuming ideal diodes, find (i) the peak load current and (ii) the average (DC) load current.
$$I_m = \frac{V_m}{R_L} = \frac{12}{1000} = 12\,\text{mA}; \qquad I_{\text{dc}} = \frac{2I_m}{\pi} \approx \frac{2\times 12}{3.1416} \approx 7.64\,\text{mA}.$$
N5. A Zener diode regulator has $V_z = 6$ V, $V_{\text{in}} = 10$ V, $R_s = 100\,\Omega$ and $R_L = 500\,\Omega$. Find the load current and the current through the Zener.
$$I_L = \frac{V_z}{R_L} = \frac{6}{500} = 12\,\text{mA}; \qquad I_s = \frac{V_{\text{in}}-V_z}{R_s} = \frac{4}{100} = 40\,\text{mA}; \qquad I_z = I_s – I_L = 28\,\text{mA}.$$
Glossary / Key Terms
| Term | Meaning |
|---|---|
| Valence band (VB) | Highest energy band normally filled with valence electrons at $T=0$ K. |
| Conduction band (CB) | Next higher band; electrons in it are free to conduct. |
| Energy gap $E_g$ | Forbidden energy region between the VB and CB. |
| Hole | Vacancy of an electron in the valence band; behaves like a positive charge $+e$. |
| Intrinsic semiconductor | Pure semiconductor with $n_e = n_h = n_i$. |
| Doping | Deliberate addition of a small amount of impurity to alter conductivity. |
| Donor | Pentavalent dopant — provides extra electron in n-type material. |
| Acceptor | Trivalent dopant — provides a hole in p-type material. |
| Majority carrier | Carrier with the larger concentration in an extrinsic semiconductor. |
| Minority carrier | Carrier with the smaller concentration in an extrinsic semiconductor. |
| Mobility $\mu$ | Drift velocity per unit electric field, $v_d = \mu E$. |
| p-n junction | Single crystal containing adjacent p- and n-regions. |
| Depletion region | Thin layer near the junction containing only ionised, immobile impurity atoms. |
| Barrier potential $V_b$ | Built-in potential difference across the depletion region. |
| Forward bias | p-side made positive — depletion region narrows, diode conducts. |
| Reverse bias | p-side made negative — depletion region widens, only leakage flows. |
| Reverse-saturation current | Small minority-carrier current under reverse bias. |
| Knee voltage | Forward voltage at which the diode begins to conduct strongly. |
| Breakdown voltage | Critical reverse voltage at which diode current rises abruptly. |
| Rectifier | Circuit that converts AC to pulsating DC using diodes. |
| Half-wave rectifier | Single-diode rectifier — passes only one half of every cycle. |
| Full-wave rectifier | Two-diode (centre-tap) or four-diode (bridge) rectifier — passes both halves. |
| Ripple factor | Ratio of AC component to DC component in the rectified output. |
| Zener diode | Heavily-doped diode operated in reverse breakdown — used as a voltage regulator. |
| Photodiode | Reverse-biased diode whose current depends on incident light intensity. |
| LED | Forward-biased diode that emits light by electroluminescence. |
| Solar cell | Large-area p-n junction that produces an EMF when illuminated by sunlight. |
| IC | Integrated Circuit — many devices fabricated on a single semiconductor chip. |
Quick Recap
- Solids are classified by band theory: conductors ($E_g\approx 0$), insulators ($E_g>3$ eV), semiconductors ($E_g\sim 1$ eV).
- Intrinsic Si and Ge: $n_e = n_h = n_i$. Mass-action law: $n_e n_h = n_i^2$.
- Doping: pentavalent $\to$ n-type (electrons majority); trivalent $\to$ p-type (holes majority).
- p-n junction: diffusion forms a depletion region with built-in field and barrier potential ($\approx 0.7$ V Si, $0.3$ V Ge).
- Forward bias: barrier reduces, current flows. Reverse bias: barrier increases, only leakage flows.
- Half-wave rectifier (1 diode): $f_{\text{out}}=f_{\text{in}}$, $\eta_{\max}\approx 40.6\%$.
- Full-wave rectifier (2 or 4 diodes): $f_{\text{out}}=2f_{\text{in}}$, $\eta_{\max}\approx 81.2\%$.
- Zener diode — reverse-breakdown — voltage regulator.
- Photodiode — reverse-biased optical detector; LED — forward-biased emitter; Solar cell — photovoltaic cell.
- Always remember: $h\nu \ge E_g$ for any optoelectronic action.
That completes the full English-medium question-answer set, conceptual notes, illustrative figures and additional practice for ASSEB Class 12 Physics Chapter 14 — Semiconductor Electronics: Materials, Devices and Simple Circuits. Revise the energy-band picture and the law of mass action thoroughly — almost every conceptual question can be tackled by combining these two ideas with the diode I-V characteristic. Wishing you the very best for your board examination.