Semiconductors, p-n Junction Diode and Applications

Q: What is the difference between intrinsic and extrinsic semiconductors?

An intrinsic semiconductor is pure silicon or germanium, where every free electron leaves a hole, so ne = nh = ni. An extrinsic semiconductor is doped: pentavalent donors make n-type (electrons majority), trivalent acceptors make p-type (holes majority). Doping raises conductivity enormously while nenh = ni² still holds.

Q: What are the depletion region and the barrier potential of a p-n junction?

When the junction forms, electrons and holes diffuse across and recombine, leaving fixed negative ions on the p side and positive ions on the n side. This carrier-free layer, about 0.1 μm wide, is the depletion region. Its ions set up a potential difference, the barrier potential, with the n side higher, which stops further diffusion.

Q: Why does a p-n junction diode conduct in only one direction?

In forward bias the external voltage lowers the barrier to V₀ - V, so majority carriers cross and a current of milliamperes flows. In reverse bias the barrier rises to V₀ + V, so only a few minority carriers drift, giving microamperes. This huge difference in resistance lets the diode rectify alternating voltage.

Q: Why is the reverse current of a diode almost independent of voltage?

The reverse current is carried by minority carriers, whose number is fixed by temperature, not by the applied voltage. Even a small reverse voltage sweeps all of them across the junction, so raising the voltage further cannot increase the current, until breakdown. That is why it is called the reverse saturation current.

Q: How does a Zener diode regulate voltage?

A Zener diode is heavily doped and is used in reverse breakdown, where its voltage stays at VZ over a wide range of current. Placed in reverse bias across the load, with a series resistor, it absorbs changes: extra input voltage drops across the series resistor, while the load voltage stays at VZ.

Q: Why is a photodiode operated in reverse bias?

Light creates equal numbers of extra electrons and holes. In reverse bias the current is carried by minority carriers, which are few, so the fractional change in current caused by light is large and easy to measure. In forward bias the same extra carriers would be lost in the much larger majority carrier current.

Q: Which semiconductor questions are common in NEET?

NEET often asks which dopant makes n-type or p-type material, band gap order of carbon, silicon and germanium, forward and reverse bias of a diode from given potentials, output frequency of rectifiers, identifying Zener, LED, photodiode and solar cell characteristics, and simple numericals using the mass action law.

Q: How is the p-n junction diode tested in JEE Main?

JEE Main asks circuit problems where each diode must be judged on or off, Zener regulator currents and power, rectifier output frequency and average voltage, mass action law and conductivity numericals, LED or photodiode wavelength from the band gap, and reading of V-I characteristics, sometimes with dynamic resistance.

PhysicsElectronic DevicesFor NEET aspirants

Semiconductors are solids with a small band gap (less than about $3 eV$ ), so their conductivity can be controlled by temperature, light and doping. Doping gives n-type and p-type material, and joining the two makes a p-n junction diode, which conducts in forward bias and blocks in reverse bias. This page covers energy bands, intrinsic and extrinsic semiconductors, junction formation, the diode V-I characteristic, rectifiers, and the Zener diode, LED, photodiode and solar cell. Semiconductors and the p-n junction diode give sure questions in NEET and JEE Main every year.

On this page1Energy bands2Intrinsic3n-type and p-type4p-n junction5Bias and V-I6Rectifiers7Zener regulator8LED, photodiode, solar cell9Revision

Key Formulas - Quick Reference

★ Must learn Band gap: metal $E_{g} \approx 0$ (bands overlap); semiconductor $E_{g} < 3 eV$ (Si $1.1$ , Ge $0.7$ ); insulator $E_{g} > 3 eV$ (diamond $5.4$ )
★ Must learn Mass action law: $n_{e} n_{h} = n_{i}^{2}$ . Intrinsic: $n_{e} = n_{h} = n_{i}$ ; n-type: $n_{e} \approx N_{D}$ , $n_{h} = \frac{n _{i}^{2}}{N _{D}}$
Conductivity: $σ = e (n_{e} μ_{e} + n_{h} μ_{h}) = \frac{1}{ρ}$ ; current $I = I_{e} + I_{h}$
Intrinsic carriers: $n_{i} = n_{0} e^{- E_{g} /2 k_{B} T}$ , so $σ$ rises very fast with $T$ (resistance falls)
★ Must learn Barrier height: forward bias $V_{0} - V$ , reverse bias $V_{0} + V$ ; knee voltage Si $\approx 0.7 V$ , Ge $\approx 0.2$ to $0.3 V$
Diode equation: $I = I_{0} (e^{e V / η k_{B} T} - 1)$ ; dynamic resistance $r_{d} = \frac{Δ V}{Δ I}$
★ Must learn Rectifier output frequency: half-wave $f$ , full-wave (centre-tap or bridge) $2 f$ ; average output $\frac{V _{m}}{π}$ and $\frac{2 V _{m}}{π}$
★ Must learn Zener regulator: $I_{S} = \frac{V _{in} - V _{Z}}{R _{S}}$ , $I_{L} = \frac{V _{Z}}{R _{L}}$ , $I_{Z} = I_{S} - I_{L}$ ; $V_{o u t} = V_{Z}$
Photon and band gap: $λ (nm) = \frac{1240}{E _{g} ( eV )}$ (LED emits it; photodiode and solar cell need $λ \leq$ this)

1. Metals, Semiconductors and Insulators

Solids are sorted by resistivity $ρ$ (or conductivity $σ = \frac{1}{ρ}$ ), but the deeper test is the band gap and how resistance changes with temperature.

Property	Metals	Semiconductors	Insulators
Resistivity $ρ$ in $Ω m$	$1 0^{- 2}$ to $1 0^{- 8}$	$1 0^{- 5}$ to $1 0^{6}$	$1 0^{11}$ to $1 0^{19}$
Conductivity $σ$ in $S m^{- 1}$	$1 0^{2}$ to $1 0^{8}$	$1 0^{5}$ to $1 0^{- 6}$	$1 0^{- 11}$ to $1 0^{- 19}$
Band gap $E_{g}$	$\approx 0$ (overlap)	$< 3 eV$	$> 3 eV$
Resistance as T rises	increases	decreases (fast)	stays very high
Examples	Cu, Al, Ag	Si, Ge, GaAs, CdS	diamond, glass, wood

Types of semiconductors. Elemental: Si, Ge. Compound inorganic: CdS, GaAs, CdSe, InP. Organic: anthracene, doped phthalocyanines. Organic polymers: polypyrrole, polyaniline, polythiophene. Almost all devices use Si, Ge or inorganic compounds.

1.1 Energy bands

In a crystal the outer electrons of neighbouring atoms interact, so each sharp atomic level spreads into a band of very closely spaced levels. The band holding the valence electrons is the valence band (top edge $E_{V}$ ); the next band is the conduction band (bottom edge $E_{C}$ ). The gap $E_{g} = E_{C} - E_{V}$ decides everything.

Figure 1: Band pictures on one energy scale (

18

units per eV). Metal: bands overlap,

E_{g} \approx 0

. Insulator (diamond):

E_{g} = 5.4 eV

, far too wide for thermal energy (

k T \approx 0.026 eV

300 K

). Semiconductor (Si):

E_{g} = 1.1 eV

, so a few electrons cross and leave equal holes behind.

Why Si and Ge have exactly a full band at 0 K. A crystal of $N$ atoms has $4 N$ outer electrons (two s and two p per atom) and $8 N$ outer states. At the actual atomic spacing these states split into two bands of $4 N$ states each. The lower band (valence band) takes all $4 N$ electrons and is completely full; the upper band (conduction band) is completely empty at $0 K$ .

Order of band gaps: C (5.4) > Si (1.1) > Ge (0.7 eV) > Sn (0). All four are group 14 with the same lattice; the gap falls down the group, so diamond is an insulator, Si and Ge are semiconductors and tin is a metal. Thermal energy at room temperature is only $k T \approx 0.026 eV$ , which is why a few electrons cross $1.1 eV$ but none cross $5.4 eV$ .

2. Intrinsic Semiconductors

A pure Si or Ge crystal has a diamond-like lattice: each atom shares one electron with each of its four neighbours, forming four covalent bonds. At $0 K$ every bond is intact, so a pure semiconductor is an insulator. Heat breaks a few bonds.

Figure 2: One broken bond gives one free electron and one hole, so

n_{e} = n_{h} = n_{i}

. The hole moves when a bound electron from site 2 jumps into it; the free electron is not involved. Total current

I = I_{e} + I_{h}

Hole: the vacancy left in a bond when an electron breaks free. It behaves as a particle of charge $+ e$ , although no positive particle moves.
Intrinsic semiconductor: pure; every free electron leaves one hole, so $n_{e} = n_{h} = n_{i}$ (intrinsic carrier concentration).
Current is carried by both: $I = I_{e} + I_{h}$ . Holes move towards the negative potential.
Equilibrium: generation of pairs (by heat) and recombination (electron meets hole) go on together; at equilibrium their rates are equal.

The number of pairs grows exponentially with temperature, $n_{i} = n_{0} e^{- E_{g} /2 k_{B} T}$ , so the conductivity $σ = e (n_{e} μ_{e} + n_{h} μ_{h})$ of a semiconductor rises sharply when it is heated. Room-temperature intrinsic conductivity is still far too low for devices, which is why we dope.

Key idea

Pure semiconductor:

n_{e} = n_{h} = n_{i}

; an insulator at

0 K

; resistance falls as temperature rises.

3. Extrinsic Semiconductors: n-type and p-type

Doping adds a tiny amount (a few parts per million) of a suitable impurity, the dopant, and raises the conductivity many times. The dopant atom must be about the size of Si or Ge so that it takes a lattice site without distorting the crystal. Group 15 atoms (5 valence electrons) and group 13 atoms (3 valence electrons) are used.

Figure 3: A donor adds a free electron and becomes a fixed

+

ion; an acceptor adds a hole and becomes a fixed

-

ion.

E_{D}

sits about

0.01 eV

(Ge) or

0.05 eV

(Si) below

E_{C}

E_{A}

just above

E_{V}

, so at room temperature almost every dopant is ionised. Each material stays electrically neutral.

	n-type	p-type
Dopant	pentavalent (donor): P, As, Sb	trivalent (acceptor): B, Al, In
What it adds	one nearly free electron	one hole
Majority / minority	electrons / holes: $n_{e} ≫ n_{h}$	holes / electrons: $n_{h} ≫ n_{e}$
Fixed ion left behind	donor ion, $+$	acceptor ion, $-$
Energy level	$E_{D}$ just below $E_{C}$	$E_{A}$ just above $E_{V}$
Net charge	neutral	neutral

Mass action law. In thermal equilibrium $n_{e} n_{h} = n_{i}^{2}$ for any doping. Adding donors raises $n_{e}$ , so more holes recombine and $n_{h}$ falls below $n_{i}$ : doping increases the majority carriers and reduces the minority carriers. The number of donor electrons is fixed by the doping level, not by temperature.

Both dopants present. Donors and acceptors cancel: $n_{e} \approx N_{D} - N_{A}$ if $N_{D} > N_{A}$ (n-type), and $n_{h} \approx N_{A} - N_{D}$ if $N_{A} > N_{D}$ (p-type).

Quick Recall: tap to check

Is n-type silicon negatively charged?

No. Every extra electron is balanced by a fixed

+

donor ion; the material is neutral.

Which dopants make p-type silicon?

Trivalent (group 13) atoms: boron, aluminium, indium.

n-type sample:

n_{e} = 1 0^{22} m^{- 3}

n_{i} = 1 0^{16} m^{- 3}

. Find

n_{h}

n_{h}

= \frac{n _{i}^{2}}{n _{e}}

= \frac{1 0 ^{32}}{1 0 ^{22}}

= 1 0^{10} m^{- 3}

4. The p-n Junction

A p-n junction is made in one crystal: part of a p-type wafer is converted to n-type by adding donors. (Two separate slabs pressed together cannot work: no surface is smooth on the atomic scale of $2$ to $3 \overset{˚}{A}$ , so the contact is a break for charge carriers.) Two processes then act across the junction.

Figure 4: Diffusion leaves fixed

-

ions on the p side and

+

ions on the n side. Their field (from n to p, largest at the junction) drives a drift current opposite to diffusion; at equilibrium the two are equal. The n side is at a higher potential by the barrier

V_{0}

; layer width about

0.1 μ m

(drawn for equal doping).

Diffusion current

Majority carriers move from high to low concentration: holes p to n, electrons n to p. Large at first, it builds up the fixed ions.

Drift current

Minority carriers are pushed by the junction field (n to p): electrons p to n, holes n to p. It grows as the layer builds up.

Depletion layer: the region around the junction emptied of free carriers, holding only fixed ions ( $-$ on the p side, $+$ on the n side). Its width is about $0.1 μ m$ .
Equilibrium: diffusion current $=$ drift current, so no net current flows in an unbiased junction.
Barrier potential $V_{0}$ : the n side is at a higher potential than the p side. It opposes further diffusion of majority carriers.

JEE Advanced

Field in the depletion layer. The average field is $\frac{V _{0}}{W}$ ; for $V_{0} = 0.5 V$ and $W = 0.1 μ m$ it is $5 \times 1 0^{6} V m^{- 1}$ . For a sharp junction the field is triangular (Figure 4), so its peak at the junction is $E_{ma x} = \frac{2 V _{0}}{W}$ .

Crossing the barrier. An electron going from n to p (or a hole from p to n) loses energy $e V_{0}$ : it crosses only if its kinetic energy exceeds $e V_{0}$ , and leaves with $K - e V_{0}$ . A minority electron going from p to n gains $e V_{0}$ . Heavier doping gives a thinner layer and a stronger field; under bias the width varies as the square root of the barrier height, $V_{0} - V$ (forward) or $V_{0} + V$ (reverse).

Key idea

Diffusion builds fixed ions; their field drives drift the other way. At balance: no net current, a depletion layer about

0.1 μ m

wide and a barrier

V_{0}

(n side higher).

5. Junction Diode: Biasing and V-I Characteristic

A semiconductor diode is a p-n junction with metal contacts at both ends. In its symbol the arrow points from p to n, the direction of conventional current in forward bias. The external voltage drops almost entirely across the depletion layer, because the layer has no free carriers and so the highest resistance.

Figure 5: Forward bias lowers the barrier to

V_{0} - V

and narrows the depletion layer; reverse bias raises it to

V_{0} + V

and widens it. Drawn for

V_{0} = 0.7 V

with

0.35 V

forward and

0.7 V

reverse; the widths follow

W \propto barrier

(dashed: no bias).

	Forward bias	Reverse bias
Connection	p to $+$ , n to $-$	p to $-$ , n to $+$
Barrier height	$V_{0} - V$ (lowered)	$V_{0} + V$ (raised)
Depletion layer	narrower	wider
Current carriers	majority carriers cross (minority carrier injection)	only minority carriers drift
Size of current	mA, rises exponentially after the knee	a few $μ A$ , nearly independent of $V$
Resistance	low (about $10 Ω$ )	very high (about $1 0^{7} Ω$ )

To trace the characteristic, the diode is fed from a battery through a potentiometer; a milliammeter reads the forward current and a microammeter the reverse current. The graph (Figure 6) has three regions:

Forward, below the knee: almost no current until the threshold (cut-in) voltage, about $0.7 V$ for Si and $0.2$ to $0.3 V$ for Ge.
Forward, above the knee: current rises exponentially; a small rise in $V$ gives a large rise in $I$ .
Reverse: a tiny reverse saturation current, limited by the number of minority carriers (so it depends on temperature, not on voltage), until the breakdown voltage $V_{b r}$ , where it shoots up. An ordinary diode is never used beyond breakdown; too much current in either direction overheats and destroys it.

Figure 6: Note the two scales: mA and

0

1 V

forward,

μ A

and

0

100 V

reverse. Forward curve drawn from

I \propto e^{e V / η k T}

(

η = 1.5

300 K

): it gives

7.6 mA

0.70 V

and

100 mA

0.80 V

. Reverse current stays near

1 μ A

until breakdown at

V_{b r}

Dynamic resistance

r_{d} = \frac{Δ V}{Δ I}

, the inverse slope of the V-I curve at the working point. It is small in forward bias and huge in reverse bias, which is exactly why a diode rectifies.

Exam Trick

Forward biased means p higher than n. Compare the potentials at the two ends: if the anode (p, the triangle's base) is more positive than the cathode (n, the bar) by at least the knee voltage, the diode is ON. It does not matter whether the numbers are positive or negative: p at $- 2 V$ and n at $- 5 V$ is forward biased. In circuits, replace an ON diode by a $0.7 V$ drop (or a plain wire if "ideal") and an OFF diode by a break.

Quick Recall: tap to check

Barrier height with a forward bias

V

? With a reverse bias

V

V_{0} - V

and

V_{0} + V

Why does the reverse current hardly change with reverse voltage?

It is set by the few minority carriers available; even a small voltage sweeps all of them across.

Which meter measures the reverse current, and why?

A microammeter: the reverse current is only a few

μ A

6. Diode as a Rectifier

A rectifier turns alternating voltage into one-way (pulsating) voltage, using the diode's one-way conduction. A transformer first sets the ac voltage; the diode's breakdown voltage must be well above the peak reverse voltage it will face.

6.1 Half-wave rectifier

Figure 7: Half-wave rectifier. D conducts only when A is positive, so the load gets one half-sine pulse per cycle: output frequency

=

input frequency (

50 Hz

in,

50 Hz

out). The diode must withstand a reverse peak of

V_{m}

6.2 Full-wave rectifier (centre-tap)

Two diodes feed a common load from the two ends of a centre-tap secondary. The ends A and B are $18 0^{\circ}$ out of phase, so the diodes take turns, and the load current flows the same way in both halves.

Figure 8: Full-wave (centre-tap) rectifier. A and B are

18 0^{\circ}

out of phase, so

D_{1}

and

D_{2}

conduct in alternate halves and the load current never reverses. Output frequency

= 2 \times

input (

50 Hz

in,

100 Hz

out); each diode uses only half the secondary.

6.3 Bridge rectifier

Four diodes give full-wave output without a centre tap; in each half cycle two opposite diodes conduct in series with the load.

Figure 9: Bridge rectifier (orange = conducting). Positive half:

D_{1}

D_{3}

; negative half:

D_{2}

D_{4}

. Two diodes are always in series with

R_{L}

, current through

R_{L}

is always downward, and the output frequency is

2 f

	Half-wave	Full-wave (centre-tap)	Bridge
Diodes	1	2	4
Output frequency	$f$ (50 Hz for 50 Hz)	$2 f$ (100 Hz)	$2 f$ (100 Hz)
Average (dc) output	$\frac{V _{m}}{π} = 0.318 V_{m}$	$\frac{2 V _{m}}{π} = 0.637 V_{m}$	$\frac{2 V _{m}}{π}$
rms output	$\frac{V _{m}}{2}$	$\frac{V _{m}}{2}$	$\frac{V _{m}}{2}$
Maximum efficiency	$40.6%$	$81.2%$	$81.2%$
Ripple factor	1.21	0.48	0.48
Peak inverse voltage per diode	$V_{m}$	$2 V_{m}$	$V_{m}$

Here $V_{m}$ is the peak voltage reaching the load (for the centre-tap circuit, the peak of each half of the secondary). With real Si diodes subtract $0.7 V$ per conducting diode from the peak ( $1.4 V$ in a bridge).

6.4 Filter: from pulses to steady dc

The rectified output is one-way but pulsating. A large capacitor across the load charges to the peak on each pulse and discharges slowly through $R_{L}$ between pulses, so the output stays close to $V_{m}$ with a small ripple. The larger the time constant $R_{L} C$ , the smaller the ripple. (An inductor in series with $R_{L}$ also filters.) The capacitor-input filter is the most common in power supplies.

Figure 10: Capacitor filter, computed for

R_{L} C = 4 T

with ideal diodes. The output stays near

V_{m}

and dips by only

Δ V = 0.10 V_{m}

each half cycle; a larger

R_{L} C

gives a smaller ripple (estimate

Δ V \approx \frac{V _{m}}{2 f R _{L} C} = 0.125 V_{m}

here).

Exam Trick

Count the pulses. Half-wave: one pulse per cycle, output (and ripple) frequency $f$ . Full-wave or bridge: two pulses per cycle, $2 f$ . So a $50 Hz$ supply gives $50 Hz$ and $100 Hz$ outputs. For the peak inverse voltage, remember the odd one out: the centre-tap diode faces $2 V_{m}$ .

Key idea

Half-wave:

f

\frac{V _{m}}{π}

. Full-wave and bridge:

2 f

\frac{2 V _{m}}{π}

. A capacitor across the load smooths the output; larger

R_{L} C

, smaller ripple.

7. Zener Diode and Voltage Regulator

A Zener diode (named after C. Zener) is designed to work in reverse breakdown. Both sides are heavily doped, so the depletion layer is very thin (below $1 0^{- 6} m$ ) and even about $5 V$ of reverse bias creates a field of about $5 \times 1 0^{6} V m^{- 1}$ . At $V_{Z}$ this field pulls valence electrons out of their bonds (field ionisation or internal field emission, needing about $1 0^{6} V m^{- 1}$ ), and the reverse current rises sharply.

Figure 11: Zener diode (drawn for

V_{Z} = 6 V

). Beyond breakdown,

I_{Z}

can change over a wide range while the voltage stays at

V_{Z}

. In the regulator, extra input voltage drops across

R_{S}

I_{S} = \frac{V _{in} - V _{Z}}{R _{S}}

I_{L} = \frac{V _{Z}}{R _{L}}

I_{Z} = I_{S} - I_{L}

★ Must learn Zener property: in breakdown the voltage across the diode stays at $V_{Z}$ while the current through it changes over a wide range. Connected in reverse bias across the load, it holds the output at

V_{Z}

How regulation works. The unregulated dc (filtered rectifier output) feeds the Zener through a series resistor $R_{S}$ . If $V_{in}$ rises, $I_{S}$ and $I_{Z}$ rise and the extra voltage drops across $R_{S}$ ; if $V_{in}$ falls, $I_{S}$ and $I_{Z}$ fall. Either way $V_{o u t}$ stays $V_{Z}$ . Choose $R_{S}$ so that $I_{Z}$ is well above $I_{L}$ (a common design choice is $I_{Z} \approx 5 I_{L}$ ) and the Zener power $V_{Z} I_{Z}$ stays within its rating. Before using $V_{o u t} = V_{Z}$ , check that the Zener is actually in breakdown: without it, the divider voltage $V_{in} \frac{R _{L}}{R _{S} + R _{L}}$ must be at least $V_{Z}$ .

Zener breakdown

Heavily doped, thin depletion layer; low $V_{Z}$ (below about $6 V$ ). Caused by the strong field tearing electrons out of bonds.

Avalanche breakdown

Lightly doped, wide layer; higher breakdown voltage. Minority carriers gain speed and knock out more electrons by collision, in a chain.

8. Optoelectronic Devices: Photodiode, LED and Solar Cell

In these junction devices, photons create or are created by electron-hole pairs. The key number is $λ (nm) = \frac{1240}{E _{g} ( eV )}$ .

Figure 12: (a) Photodiode: reverse current

\propto

light intensity, almost independent of the reverse voltage. (b) Solar cell (drawn for

I_{sc} = 30 mA

V_{oc} = 0.60 V

): it lies in the 4th quadrant because it delivers current. Maximum power

14.9 mW

0.52 V

(shaded rectangle).

	Photodiode	LED	Solar cell
Bias	reverse	forward	none
Process	light ( $h ν > E_{g}$ ) makes e-h pairs near the junction; the field separates them	injected carriers recombine at the junction and emit photons	generation, separation (junction field), collection at the contacts
Output	reverse current proportional to light intensity	light with $h ν \approx E_{g}$ (nearly monochromatic)	emf; current to a load
I-V graph	3rd quadrant	like a diode, higher knee	4th quadrant
Use	detecting optical signals	displays, remote controls, lighting	power for satellites, calculators, grids

8.1 Photodiode

A photodiode has a transparent window over the junction and is reverse biased. Why not forward bias? In reverse bias the current is carried by minority carriers, and light adds the same number of carriers $Δ n = Δ p$ to both types. Since $n ≫ p$ in (say) n-type material, the fractional change $\frac{Δ p}{p}$ is far larger than $\frac{Δ n}{n}$ , so the change in current is easy to measure.

8.2 Light emitting diode (LED)

A heavily doped junction, forward biased, in a transparent cover. Excess minority carriers recombine near the junction and release photons of energy about $E_{g}$ (radiative recombination, strong in GaAs and GaAs-GaP).
Visible light needs $E_{g} \geq 1.8 eV$ (visible spans about $1.8$ to $3 eV$ ). $GaA s_{0.6} P_{0.4}$ ( $E_{g} \approx 1.9 eV$ ) gives red; GaAs ( $1.4 eV$ ) gives infrared (remote controls).
Knee voltages are higher than a Si diode and differ with colour; reverse breakdown is low, about $5 V$ . Light output rises with current up to a maximum, then falls.
Advantages over filament lamps: low voltage and power, no warm-up, fast on-off switching, long life and ruggedness, nearly monochromatic light, with a bandwidth of only $100$ to $500 \overset{˚}{A}$ . White LEDs (a blue InGaN chip with a yellow phosphor) now replace filament lamps in lighting.

8.3 Solar cell

A solar cell is a large-area p-n junction with no external bias. A p-Si wafer (about $300 μ m$ ) carries a thin n-Si layer (about $0.3 μ m$ ) on top; a metal finger grid (under $15%$ of the area) is the front contact and a metal coating the back contact. Light makes pairs near the junction, the field sends electrons to n and holes to p, so p becomes positive: a photovoltage. Its graph lies in the 4th quadrant because the cell supplies current rather than drawing it; it is marked by the open-circuit voltage $V_{oc}$ and short-circuit current $I_{sc}$ .

Best band gap about $1.5 eV$ (range $1.0$ to $1.8 eV$ ), near the peak of the solar spectrum. Si ( $1.1$ ), GaAs ( $1.43$ ), CdTe ( $1.45$ ), $CuInS e_{2}$ ( $1.04 eV$ ) are used; GaAs beats Si despite its larger gap because it absorbs light more strongly.
Too large a gap (CdS, $2.4 eV$ ) wastes most of sunlight; too small (PbS, $0.4 eV$ ) absorbs light in the top layer, far from the junction, so the pairs are not separated.
Other criteria: high optical absorption (about $1 0^{4} cm^{- 1}$ ), good conductivity, available raw material, low cost. Any light with $h ν > E_{g}$ works, not only sunlight.

Exam Trick

Bias memory line: "LED lights Forward, Photo Reverses, Sun needs Nothing". Then use $1240$ : an LED with $E_{g} = 2.5 eV$ emits about $\frac{1240}{2.5} = 496 nm$ (blue-green); a photodiode with $E_{g} = 2.0 eV$ responds only to $λ < 620 nm$ .

Quick Recall: tap to check

Which of the photodiode, LED and solar cell is forward biased?

Only the LED. The photodiode is reverse biased and the solar cell has no bias.

Least band gap for a visible LED?

About

1.8 eV

(red end, about

700 nm

Why is the solar cell graph in the 4th quadrant?

It delivers current to the load: voltage positive, current negative in the diode convention.

Key idea

Photodiode: reverse bias, current

\propto

light. LED: forward bias,

λ \approx \frac{1240}{E _{g}}

nm. Solar cell: no bias, 4th quadrant, best

E_{g} \approx 1.5 eV

9. Numbers to Remember and Revision

Quantity	Value
Band gap: C, Si, Ge, Sn	$5.4$ , $1.1$ , $0.7$ , $0 eV$
Lattice spacing of C, Si, Ge	$3.56$ , $5.43$ , $5.66 \overset{˚}{A}$
Donor ionisation energy	Ge $\approx 0.01 eV$ , Si $\approx 0.05 eV$
$n_{i}$ of Si at 300 K	$1.5 \times 1 0^{16} m^{- 3}$ (Si has $5 \times 1 0^{28}$ atoms per $m^{3}$ )
Depletion layer width	about $0.1 μ m$ ( $1 0^{- 7} m$ )
Knee voltage	Si $0.7 V$ , Ge $0.2$ to $0.3 V$
Forward / reverse current	mA / $μ A$
Zener layer and field	below $1 0^{- 6} m$ ; about $5 \times 1 0^{6} V m^{- 1}$ at $5 V$
LED materials	$GaA s_{0.6} P_{0.4}$ red ( $1.9 eV$ ), GaAs infrared ( $1.4 eV$ )
History	vacuum tubes: diode 2, triode 3, tetrode 4, pentode 5 electrodes; transistor invented 1947 (announced 1948); galena (PbS) point-contact radio detector

Use the flowchart for circuit problems, then the mind map for a last revision.

Figure 13: Problem-solving flowchart. Decide ON or OFF for each diode first, replace it by its model (

0.7 V

source or open switch;

V_{Z}

source for a Zener in breakdown), solve, then check that the assumed current direction is right.

Figure 14: Mind map for revision: every exam fact of this concept on one screen.

10. Solved Examples

Solved Example 1

A pure Si crystal has

5 \times 1 0^{28}

atoms

m^{- 3}

. It is doped with 1 ppm of arsenic. Find the electron and hole concentrations. (

n_{i} = 1.5 \times 1 0^{16} m^{- 3}

)

Solution:

Donors: $N_{D}$ $= 5 \times 1 0^{28} \times 1 0^{- 6}$ $= 5 \times 1 0^{22} m^{- 3}$ . Thermal electrons ( $\sim 1 0^{16}$ ) are negligible, so $n_{e} \approx N_{D}$ .

$n_{h}$ $= \frac{n _{i}^{2}}{n _{e}}$ $= \frac{2.25 \times 1 0 ^{32}}{5 \times 1 0 ^{22}}$ .

Answer: $n_{e} = 5 \times 1 0^{22} m^{- 3}$ , $n_{h} = 4.5 \times 1 0^{9} m^{- 3}$ .

Solved Example 2

Si (

5 \times 1 0^{28}

atoms

m^{- 3}

) is doped at the same time with

5 \times 1 0^{22} m^{- 3}

arsenic and

5 \times 1 0^{20} m^{- 3}

indium. Find

n_{e}

and

n_{h}

. Is it n-type or p-type? (

n_{i} = 1.5 \times 1 0^{16} m^{- 3}

)

Solution:

Acceptors cancel an equal number of donors: $n_{e} \approx N_{D} - N_{A}$ $= 5 \times 1 0^{22} - 5 \times 1 0^{20}$ $= 4.95 \times 1 0^{22} m^{- 3}$ .

$n_{h}$ $= \frac{n _{i}^{2}}{n _{e}}$ $= \frac{2.25 \times 1 0 ^{32}}{4.95 \times 1 0 ^{22}}$ $= 4.5 \times 1 0^{9} m^{- 3}$ .

Answer: $n_{e} \approx 4.95 \times 1 0^{22} m^{- 3}$ , $n_{h} \approx 4.5 \times 1 0^{9} m^{- 3}$ ; n-type.

Solved Example 3

An intrinsic semiconductor has

E_{g} = 1.2 eV

. Its mobilities do not depend on temperature. Find

\frac{σ _{600 K}}{σ _{300 K}}

, given

n_{i} = n_{0} e^{- E_{g} /2 k_{B} T}

(

k_{B} = 8.62 \times 1 0^{- 5} eV K^{- 1}

Solution:

With constant mobilities, $σ \propto n_{i}$ , so $\frac{σ _{2}}{σ _{1}}$ $= exp [\frac{E _{g}}{2 k _{B}} (\frac{1}{T _{1}} - \frac{1}{T _{2}})]$ .

$\frac{E _{g}}{2 k _{B}}$ $= \frac{1.2}{2 \times 8.62 \times 1 0 ^{- 5}}$ $= 6961 K$ ; $\frac{1}{300} - \frac{1}{600}$ $= \frac{1}{600} K^{- 1}$ ; exponent $= 11.6$ .

Answer: $\frac{σ _{600}}{σ _{300}}$ $= e^{11.6} \approx 1.1 \times 1 0^{5}$ . Doubling the temperature raises the conductivity about a hundred thousand times.

Solved Example 4

A Si diode passes

I = 10 mA

0.7 V

I = 20 mA

0.8 V

, and

I = - 1 μ A

- 10 V

. Find the forward resistance near

15 mA

and the reverse resistance at

- 10 V

Solution:

(a) Treat the curve between $10$ and $20 mA$ as straight: $r_{f b}$ $= \frac{Δ V}{Δ I}$ $= \frac{0.1 V}{10 mA}$ $= 10 Ω$ .

(b) $r_{r b}$ $= \frac{V}{I}$ $= \frac{10 V}{1 μ A}$ $= 1.0 \times 1 0^{7} Ω$ .

Answer: about $10 Ω$ forward and $1 0^{7} Ω$ reverse. The million-fold difference is what makes the diode a one-way valve.

Solved Example 5

A diode obeys

I = I_{0} (e^{e V /2 k_{B} T} - 1)

with

I_{0} = 5 \times 1 0^{- 12} A

300 K

(

k_{B} = 8.6 \times 1 0^{- 5} eV K^{- 1}

). Find (a)

I

0.6 V

, (b) the rise in current from

0.6

0.7 V

, (c) the dynamic resistance over this step, (d) the current as the reverse bias changes from

1

2 V

Solution:

$k_{B} T = 8.6 \times 1 0^{- 5} \times 300 = 0.0258 eV$ , so $2 k_{B} T = 0.0516 eV$ .

(a) $\frac{0.6}{0.0516} = 11.63$ : $I$ $= 5 \times 1 0^{- 12} \times e^{11.63}$ $= 5.6 \times 1 0^{- 7} A$ .

(b) $\frac{0.7}{0.0516} = 13.57$ : $I = 3.9 \times 1 0^{- 6} A$ , a rise of $3.3 \times 1 0^{- 6} A$ (about 7 times).

(d) For $V = - 1$ or $- 2 V$ the exponential is negligible, so $I \approx - I_{0} = - 5 \times 1 0^{- 12} A$ in both cases.

Answer: (a) $0.56 μ A$ ; (b) $3.3 μ A$ ; (c) $30 k Ω$ ; (d) $- 5 \times 1 0^{- 12} A$ , unchanged. With $e^{e V / k_{B} T}$ (no factor 2) the same steps give $0.063 A$ and $3.0 A$ and $r_{d} \approx 0.034 Ω$ ; read the formula given in the question.

Solved Example 6

A Zener-regulated supply uses

V_{Z} = 6.0 V

. The load current is

4.0 mA

and the unregulated input is

10.0 V

. Choose the series resistor

R_{S}

Solution:

For good regulation keep $I_{Z}$ well above $I_{L}$ ; take $I_{Z} = 5 I_{L} = 20 mA$ . Then $I_{S} = 20 + 4 = 24 mA$ .

Drop across $R_{S}$ : $10.0 - 6.0 = 4.0 V$ , so $R_{S} = \frac{4.0}{24 \times 1 0 ^{- 3}} = 167 Ω$ .

Answer: $R_{S} \approx 167 Ω$ . A standard $150 Ω$ resistor also works: it gives $I_{S} = 26.7 mA$ and $I_{Z} = 22.7 mA$ ; the exact value matters less than keeping $I_{Z} ≫ I_{L}$ .

Solved Example 7

Which diode is forward biased?
(A) p at

- 10 V

, n at

0 V

(B) p at

5 V

, n at

10 V

- 2 V

, n at

- 5 V

(D) p at

0 V

, n at

2 V

Solution:

Answer: (C). Forward bias needs the p side at a higher potential than the n side. Only in (C) is $V_{p} - V_{n} = - 2 - (- 5) = + 3 V$ positive; the other three have $V_{p} < V_{n}$ and are reverse biased.

Solved Example 8

A Si diode (

0.7 V

drop) is connected in series with a

1 k Ω

resistor across a

5 V

battery, p side towards

+

. Find the current. What is it if the diode is reversed, and if the diode is ideal?

Solution:

Forward biased: $I = \frac{5 - 0.7}{1000} = 4.3 \times 1 0^{- 3} A$ .

Reversed: the diode is OFF, only the reverse saturation current (a few $μ A$ or less) flows, so $I \approx 0$ .

Ideal diode (no drop): $I = \frac{5}{1000} = 5 mA$ .

Answer: $4.3 mA$ ; about zero when reversed; $5 mA$ if ideal.

Solved Example 9

10 V

Zener is fed from

15 V

through

R_{S} = 250 Ω

and supplies a

1 k Ω

load. Find

I_{S}

I_{L}

I_{Z}

and the power in the Zener. What happens if the input rises to

20 V

Solution:

Check breakdown: without the Zener, the load would get $15 \times \frac{1000}{1250}$ $= 12 V > 10 V$ , so the Zener conducts and $V_{o u t} = 10 V$ .

$I_{S} = \frac{15 - 10}{250} = 20 mA$ ; $I_{L} = \frac{10}{1000} = 10 mA$ ; $I_{Z} = 20 - 10 = 10 mA$ ; $P_{Z} = 10 \times 0.010 = 0.1 W$ .

At $20 V$ : $I_{S} = \frac{10}{250} = 40 mA$ , $I_{L}$ still $10 mA$ , so $I_{Z} = 30 mA$ ( $0.3 W$ ).

Answer: $20$ , $10$ , $10 mA$ and $0.1 W$ ; at $20 V$ the Zener takes the extra $20 mA$ and the load voltage stays $10 V$ .

Solved Example 10

A red LED is made of

GaA s_{0.6} P_{0.4}

with

E_{g} = 1.9 eV

. The wavelength it emits is nearly
(A)

450 nm

(B)

550 nm

(C)

650 nm

(D)

850 nm

Solution:

Answer: (C). An LED emits photons of energy about $E_{g}$ : $λ = \frac{1240}{1.9} = 653 nm$ , which is red. (A) and (B) need $2.8$ and $2.3 eV$ ; (D) is infrared, like a GaAs LED ( $1.4 eV$ , about $890 nm$ ).

Solved Example 11

An alternating voltage of peak

10 V

and frequency

50 Hz

is applied to (a) a half-wave and (b) a bridge rectifier with ideal diodes. Find the output frequency and the average output voltage in each case.

Solution:

(a) One pulse per cycle: $f_{o u t} = 50 Hz$ ; $V_{d c}$ $= \frac{V _{m}}{π}$ $= \frac{10}{π}$ $= 3.18 V$ .

(b) Two pulses per cycle: $f_{o u t} = 100 Hz$ ; $V_{d c} = \frac{2 V _{m}}{π} = 6.37 V$ .

Answer: (a) $50 Hz$ , $3.18 V$ ; (b) $100 Hz$ , $6.37 V$ . With Si diodes the bridge peak falls to $10 - 1.4 = 8.6 V$ and $V_{d c}$ to $5.47 V$ .

Solved Example 12

Pure Si has

n_{i} = 1.5 \times 1 0^{16} m^{- 3}

μ_{e} = 0.135

and

μ_{h} = 0.048 m^{2} V^{- 1} s^{- 1}

. Find its conductivity. By what factor does it rise after doping with

5 \times 1 0^{22}

donors

m^{- 3}

Solution:

Pure: $σ$ $= e n_{i} (μ_{e} + μ_{h})$ $= 1.6 \times 1 0^{- 19} \times 1.5 \times 1 0^{16} \times 0.183$ $= 4.4 \times 1 0^{- 4} S m^{- 1}$ .

Doped: holes are negligible ( $4.5 \times 1 0^{9} m^{- 3}$ ), so $σ \approx e N_{D} μ_{e}$ $= 1.6 \times 1 0^{- 19} \times 5 \times 1 0^{22} \times 0.135$ $= 1.08 \times 1 0^{3} S m^{- 1}$ .

Answer: $4.4 \times 1 0^{- 4} S m^{- 1}$ ; it rises about $2.5 \times 1 0^{6}$ times for only 1 ppm of dopant.

Practice Questions

In n-type silicon: (a) electrons are majority carriers and trivalent atoms are the dopants (b) electrons are minority carriers and pentavalent atoms are the dopants (c) holes are minority carriers and pentavalent atoms are the dopants (d) holes are majority carriers and trivalent atoms are the dopants.Answer: (c)
Which option of the question above is true for p-type silicon?Answer: (d)
C, Si and Ge have band gaps $(E_{g})_{C}$ , $(E_{g})_{S i}$ , $(E_{g})_{G e}$ . Which is true? (a) $(E_{g})_{S i} < (E_{g})_{G e} < (E_{g})_{C}$ (b) $(E_{g})_{C} < (E_{g})_{G e} > (E_{g})_{S i}$ (c) $(E_{g})_{C} > (E_{g})_{S i} > (E_{g})_{G e}$ (d) all equal.Answer: (c): $5.4 > 1.1 > 0.7 eV$
In an unbiased p-n junction holes diffuse from p to n because (a) free electrons in n attract them (b) the potential difference drives them (c) hole concentration is higher in p (d) all of these.Answer: (c): diffusion follows the concentration gradient
Forward bias applied to a p-n junction (a) raises the barrier (b) makes the majority carrier current zero (c) lowers the barrier (d) none of these.Answer: (c)
Input frequency $50 Hz$ . Find the output frequency of a half-wave and of a full-wave rectifier.Answer: $50 Hz$ and $100 Hz$
Can a photodiode made from a semiconductor with $E_{g} = 2.8 eV$ detect light of wavelength $6000 nm$ ?Answer: No: the photon energy is $\frac{1240}{6000} = 0.21 eV < 2.8 eV$ ; it detects only $λ < 443 nm$

Common Mistakes to Avoid

Watch out

Calling n-type material negatively charged. Every free electron is balanced by a fixed $+$ donor ion; n-type and p-type are both neutral.
Thinking a hole is a positive particle that travels. A hole moves because bound electrons jump into it; the free electron is not involved.
Using $n_{e} = N_{D}$ when acceptors are also present. Use the net doping $N_{D} - N_{A}$ , then $n_{h} = \frac{n _{i}^{2}}{n _{e}}$ .
Mixing up diffusion and drift. Diffusion is driven by the concentration difference (majority carriers); drift is driven by the junction field (minority carriers).
Writing the forward-bias barrier as $V_{0} + V$ . Forward bias lowers it to $V_{0} - V$ and narrows the depletion layer; reverse bias raises and widens.
Reading the reverse part of the V-I graph in mA. The two halves use different scales: forward mA, reverse $μ A$ .
Taking the full-wave output frequency as $f$ , or using $I_{S} = \frac{V _{in}}{R _{S}}$ in a Zener regulator. It is $2 f$ , and $I_{S} = \frac{V _{in} - V _{Z}}{R _{S}}$ (after checking the Zener is in breakdown).
Biasing optoelectronic devices wrongly. Only the LED is forward biased; the photodiode is reverse biased and the solar cell has no bias.

Frequently Asked Questions

What is the difference between intrinsic and extrinsic semiconductors?

An intrinsic semiconductor is pure silicon or germanium, where every free electron leaves a hole, so $n_{e} = n_{h} = n_{i}$ . An extrinsic semiconductor is doped: pentavalent donors make n-type (electrons majority), trivalent acceptors make p-type (holes majority). Doping raises conductivity enormously while $n_{e} n_{h} = n_{i}^{2}$ still holds.

What are the depletion region and the barrier potential of a p-n junction?

When the junction forms, electrons and holes diffuse across and recombine, leaving fixed negative ions on the p side and positive ions on the n side. This carrier-free layer, about $0.1 μ m$ wide, is the depletion region. Its ions set up a potential difference, the barrier potential, with the n side higher, which stops further diffusion.

Why does a p-n junction diode conduct in only one direction?

In forward bias the external voltage lowers the barrier to $V_{0} - V$ , so majority carriers cross and a current of milliamperes flows. In reverse bias the barrier rises to $V_{0} + V$ , so only a few minority carriers drift, giving microamperes. This huge difference in resistance lets the diode rectify alternating voltage.

Why is the reverse current of a diode almost independent of voltage?

The reverse current is carried by minority carriers, whose number is fixed by temperature, not by the applied voltage. Even a small reverse voltage sweeps all of them across the junction, so raising the voltage further cannot increase the current, until breakdown. That is why it is called the reverse saturation current.

How does a Zener diode regulate voltage?

A Zener diode is heavily doped and is used in reverse breakdown, where its voltage stays at $V_{Z}$ over a wide range of current. Placed in reverse bias across the load, with a series resistor, it absorbs changes: extra input voltage drops across the series resistor, while the load voltage stays at $V_{Z}$ .

Why is a photodiode operated in reverse bias?

Light creates equal numbers of extra electrons and holes. In reverse bias the current is carried by minority carriers, which are few, so the fractional change in current caused by light is large and easy to measure. In forward bias the same extra carriers would be lost in the much larger majority carrier current.

Which semiconductor questions are common in NEET?

NEET often asks which dopant makes n-type or p-type material, band gap order of carbon, silicon and germanium, forward and reverse bias of a diode from given potentials, output frequency of rectifiers, identifying Zener, LED, photodiode and solar cell characteristics, and simple numericals using the mass action law.

How is the p-n junction diode tested in JEE Main?

JEE Main asks circuit problems where each diode must be judged on or off, Zener regulator currents and power, rectifier output frequency and average voltage, mass action law and conductivity numericals, LED or photodiode wavelength from the band gap, and reading of V-I characteristics, sometimes with dynamic resistance.

Previous year questions on Semiconductors, p-n Junction Diode and Applications

34 questions from past papers, each with a step-by-step solution.

Show all 34 questions

Ready to master Electronic Devices?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test