GATE EC

P-N junctions and Zener diodes

How a junction between p-type and n-type silicon builds a built-in potential and a depletion region, and how the depletion width, peak field and depletion capacitance depend on doping and bias. Covers how the ideal (Shockley) diode equation, the saturation current and the diffusion capacitance come from minority-carrier injection and storage, and how reverse breakdown by avalanche multiplication or Zener tunnelling is used in the Zener diode as a voltage reference and regulator. A short section compares the p-n junction with metal-semiconductor (Schottky and ohmic) contacts. Every formula states its conditions and sign convention.

6 previous-year questions · 5 solved · 4 practice questions

P-N Junction Formation and Electrostatics

When p-type and n-type regions of the same semiconductor are joined, mobile carriers diffuse across the junction. The space charge they leave behind (ionized acceptors on the p-side, ionized donors on the n-side) builds an electric field, and diffusion stops when that field is strong enough to balance it. At equilibrium the Fermi level is flat across the structure, and this alignment is what fixes the built-in potential. Conventions used throughout:

electron charge $q = 1.6 \times 1 0^{- 19} C$ , taken positive;
thermal voltage $V_{T} = k T / q \approx 25.9 mV$ at 300 K, the value used unless another temperature is stated;
$n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ for silicon at 300 K when a number is needed;
doping concentrations in $cm^{- 3}$ , with all dopants ionized;
energies in eV;
applied bias $V_{A}$ positive for forward bias (p-side at the higher potential), and reverse bias written $V_{A} = - V_{R}$ with $V_{R} > 0$ ;
diode current $I$ positive when it flows from p to n through the diode.

The depletion approximation (no mobile carriers inside the depletion region, abrupt edges) is assumed.

BUILT-IN POTENTIAL

V_{bi} = V_{T} ln (\frac{N _{A} N _{D}}{n _{i}^{2}})

Holds at thermal equilibrium with complete ionization and non-degenerate doping ( $N_{A}, N_{D}$ well below the effective densities of states). $N_{A}$ and $N_{D}$ are the net acceptor and donor concentrations on the two sides. $V_{bi}$ is positive: the n-side sits $V_{bi}$ above the p-side. It cannot be read with a voltmeter across the diode, because the contact potentials at the metal contacts cancel it.

BUILT-IN POTENTIAL FROM CARRIER RATIOS

V_{bi} = V_{T} ln (\frac{n _{n 0}}{n _{p 0}}) = V_{T} ln (\frac{p _{p 0}}{p _{n 0}})

Equilibrium majority and minority concentrations on the two sides, with $n_{n 0} \approx N_{D}$ and $p_{p 0} \approx N_{A}$ . This is the same result as above, written with the mass-action law.

How the depletion region forms (teaching order)

Before contact: separate flat-band p and n regions with different Fermi levels.
On contact: electrons diffuse from n to p and holes from p to n, leaving fixed ionized dopant charge near the junction.
This space charge creates a built-in field pointing from n to p, which opposes further diffusion.
At equilibrium the drift and diffusion currents of each carrier cancel exactly: zero net current and a flat Fermi level.
The result is a depletion (space-charge) region with almost no mobile carriers, across which the full built-in potential $V_{bi}$ is dropped.

DEPLETION WIDTH (STEP/ABRUPT JUNCTION)

W = \frac{2 ϵ _{s}}{q} (\frac{1}{N _{A}} + \frac{1}{N _{D}}) (V_{bi} - V_{A})

For an abrupt (step) junction under the depletion approximation. It holds in reverse bias and in forward bias only while $V_{A}$ stays well below $V_{bi}$ ; the approximation fails as $V_{A} \to V_{bi}$ . In reverse bias $V_{A} = - V_{R}$ , so the bracket becomes $V_{bi} + V_{R}$ and $W$ grows. $ϵ_{s} = ϵ_{r} ϵ_{0}$ ; for Si, $ϵ_{r} \approx 11.7$ , giving $ϵ_{s} \approx 1.04 \times 1 0^{- 12} F/cm$ .

CHARGE BALANCE AND INDIVIDUAL DEPLETION WIDTHS

N_{A} x_{p} = N_{D} x_{n}, x_{n} = W \frac{N _{A}}{N _{A} + N _{D}}, x_{p} = W \frac{N _{D}}{N _{A} + N _{D}}

Charge neutrality: the ionized charge on the two sides is equal and opposite. The more lightly doped side always holds most of $W$ .

PEAK ELECTRIC FIELD (STEP JUNCTION)

E_{ma x} = \frac{q N _{D} x _{n}}{ϵ _{s}} = \frac{q N _{A} x _{p}}{ϵ _{s}} = \frac{2 ( V _{bi} - V _{A} )}{W}

The peak is at the metallurgical junction ( $x = 0$ ). The field falls linearly to zero at each depletion edge, so the profile is a triangle whose area equals $V_{bi} - V_{A}$ .

DEPLETION WIDTH (LINEARLY GRADED JUNCTION)

W = [\frac{12 ϵ _{s} ( V _{bi} - V _{A} )}{q a}]^{1/3}

For a doping profile $N_{D} - N_{A} = a x$ that changes linearly through the junction, where $a$ is the grading constant in $cm^{- 4}$ . Here $W \propto (V_{bi} - V_{A})^{1/3}$ , compared with the square-root dependence of the step junction.

Depletion (Junction) Capacitance

The depletion region acts like a parallel-plate capacitor whose plate separation is the bias-dependent width $W$ . It is a small-signal (incremental) capacitance $C_{j} = d Q / d V$ : the change in depletion charge for a small change in bias. Since $W$ grows with reverse bias, $C_{j}$ falls as reverse bias increases. Varactor diodes use this voltage-tunable capacitance.

DEPLETION CAPACITANCE (GENERAL)

C_{j} = \frac{ϵ _{s} A}{W}

$A$ is the junction cross-sectional area and $W$ is the depletion width at the bias point. This holds for any doping profile as a small-signal capacitance.

DEPLETION CAPACITANCE VS BIAS

C_{j} (V_{A}) = \frac{C _{j 0}}{( 1 - \frac{V _{A}}{V _{bi}} ) ^{m}}

$C_{j 0}$ is the capacitance at zero bias. The grading exponent is $m = 1/2$ for an abrupt junction and $m = 1/3$ for a linearly graded one. In reverse bias $V_{A} = - V_{R}$ , so the bracket is $1 + V_{R} / V_{bi}$ and $C_{j} < C_{j 0}$ . The formula applies in reverse bias and at small forward bias only; it wrongly diverges as $V_{A} \to V_{bi}$ .

1/C_J^2 VS REVERSE BIAS (ONE-SIDED ABRUPT JUNCTION)

\frac{1}{C _{j}^{2}} = \frac{2 ( V _{bi} + V _{R} )}{q ϵ _{s} N _{B} A ^{2}}

$N_{B}$ is the doping of the lightly doped side. A plot of $1/ C_{j}^{2}$ against $V_{R}$ is a straight line: its slope gives $N_{B}$ , and its intercept on the voltage axis is at $V_{R} = - V_{bi}$ .

EXAM TIP

For a strongly one-sided junction (for example $N_{A} ≫ N_{D}$ ), $1/ N_{A} + 1/ N_{D} \approx 1/ N_{light}$ , so the lighter doping sets both $W$ and $C_{j}$ . The relative error in the bracket is $N_{light} / N_{heavy}$ , which is 1% for a 100:1 ratio, so you can use the shortcut directly for a p $^{+}$ n or n $^{+}$ p diode.

Diode Current: Shockley Equation and Saturation Current

The applied bias injects minority carriers across the depletion region. They diffuse into the neutral regions and recombine over a diffusion length. The minority-carrier concentration at each depletion edge follows the law of the junction. Integrating the resulting diffusion currents at the depletion edges gives the ideal (Shockley) diode equation.

LAW OF THE JUNCTION

p_{n} (x_{n}) = p_{n 0} e^{V_{A} / V_{T}}, n_{p} (- x_{p}) = n_{p 0} e^{V_{A} / V_{T}}

Minority-carrier concentrations at the n-side and p-side depletion edges. Valid for low-level injection, meaning the injected minority density stays well below the majority density.

IDEAL DIODE EQUATION

I = I_{0} (e^{V_{A} / V_{T}} - 1)

$I$ flows from p to n through the diode. $V_{A} > 0$ is forward bias and $V_{A} < 0$ is reverse bias. When $∣ V_{A} ∣ ≫ V_{T}$ in reverse bias, $I \to - I_{0}$ . Assumes low-level injection and no generation or recombination inside the depletion region.

DIODE EQUATION WITH IDEALITY FACTOR

I = I_{0} (e^{V_{A} / (η V_{T})} - 1), 1 \leq η \leq 2

$η \approx 1$ when diffusion current dominates. $η$ approaches 2 when recombination in the depletion region dominates, which is typical of Si at low forward current. Use $η = 1$ unless the question gives another value.

REVERSE SATURATION CURRENT (LONG-BASE DIODE)

I_{0} = q A (\frac{D _{p} p _{n 0}}{L _{p}} + \frac{D _{n} n _{p 0}}{L _{n}})

$D_{p}$ and $D_{n}$ are the minority-carrier diffusion coefficients, with $D = μ V_{T}$ (Einstein relation). $L_{p} = D_{p} τ_{p}$ and $L_{n} = D_{n} τ_{n}$ are the diffusion lengths. This form holds when each neutral region is much longer than its diffusion length. In a short-base diode, replace $L_{p}$ and $L_{n}$ by the neutral-region widths.

EQUILIBRIUM MINORITY-CARRIER CONCENTRATIONS

p_{n 0} = \frac{n _{i}^{2}}{N _{D}}, n_{p 0} = \frac{n _{i}^{2}}{N _{A}}

$p_{n 0}$ is the equilibrium hole concentration on the n-side and $n_{p 0}$ the equilibrium electron concentration on the p-side. From the mass-action law, for non-degenerate doping.

SMALL-SIGNAL (DYNAMIC) RESISTANCE IN FORWARD BIAS

r_{d} = \frac{d V _{A}}{d I} = \frac{η V _{T}}{I}

$I$ is the DC forward current at the bias point, with $I ≫ I_{0}$ . For example, $r_{d} \approx 25.9 Ω$ at 1 mA with $η = 1$ at 300 K.

Temperature. $I_{0} \propto n_{i}^{2}$ and $n_{i}^{2} \propto T^{3} e^{- E_{g} / k T}$ , so $\frac{1}{I _{0}} \frac{d I _{0}}{d T} = \frac{3}{T} + \frac{E _{g}}{k T ^{2}}$ . For Si near 300 K ( $E_{g} \approx 1.12$ eV) this is about $0.15$ per kelvin, so the ideal $I_{0}$ roughly doubles every 5 K.

In a practical Si diode the reverse current is usually dominated by generation current, which goes as $n_{i}$ rather than $n_{i}^{2}$ . Half the rate gives a doubling roughly every 9–10 K; this is where the often-quoted "doubles every 10°C" comes from.

At a fixed forward current, the forward voltage falls as temperature rises. The temperature coefficient is negative, typically about $- 2 mV/°C$ for Si; the exact value depends on the current.

The Shockley equation predicts a reverse current that saturates at $- I_{0}$ . In real Si diodes at room temperature, carriers generated thermally inside the depletion region add a current proportional to $n_{i} W$ . This generation current is usually larger than the ideal $I_{0}$ , and since $W$ grows with $V_{R}$ , it keeps rising slowly with reverse bias instead of saturating cleanly.

EXAM TIP

When a question gives $I_{0}$ at one temperature and asks for it at another, use whatever rule the question states: $I_{0} \propto n_{i}^{2}$ with the given $E_{g}$ , or a stated doubling temperature. Don't assume a fixed doubling interval unless the question gives one.

Diffusion Capacitance and One-Sided Junctions

Under forward bias, excess minority carriers are stored in the neutral regions while they diffuse and recombine. The stored charge changes with $V_{A}$ , and this defines a diffusion capacitance. It matters only in forward bias and grows in proportion to the forward current. Unlike $C_{j}$ , it is negligible in reverse bias.

DIFFUSION CAPACITANCE (ONE-SIDED DIODE)

C_{d} = \frac{d Q}{d V _{A}} = \frac{τ I}{η V _{T}}

Charge-control result for a one-sided long-base diode in forward bias ( $I ≫ I_{0}$ ). The stored charge is $Q = τ I$ , where $τ$ is the minority-carrier lifetime on the lightly doped (injected) side. In a short-base diode, replace $τ$ by the transit time $W_{n}^{2} / (2 D_{p})$ , where $W_{n}$ is the neutral n-region width. A full small-signal analysis of a long-base diode at low frequency gives half this value, $τ I / (2 η V_{T})$ ; use whichever form the question specifies.

One-sided junction (e.g. p$^+$n): what simplifies

Injection is essentially one-way: holes injected into the lightly doped n-side dominate, and electron injection into the heavily doped p $^{+}$ -side is negligible.
$I_{0} \approx q A D_{p} p_{n 0} / L_{p}$ (drop the term for the heavily doped side).
$W$ and $C_{j}$ are set almost entirely by the doping of the lightly doped side.
$C_{d} = τ_{p} I / (η V_{T})$ , using the hole lifetime on the n-side.

EXAM TIP

In reverse bias and at low forward bias, $C_{j}$ dominates the diode's small-signal capacitance. At the forward currents used in switching, $C_{d}$ dominates. The stored charge $τ I$ has to be removed before the diode can turn off, so the reverse-recovery (storage) time scales with $τ$ . A fast-switching diode needs a short $τ$ .

Breakdown Mechanisms: Avalanche and Zener Tunnelling; the Zener Diode

Under enough reverse bias every p–n junction breaks down and conducts heavily in the reverse direction. The two physical mechanisms are avalanche multiplication and Zener (band-to-band) tunnelling. Doping decides which one dominates, because it sets both the peak field and the depletion width at a given reverse voltage.

Avalanche vs Zener tunnelling

Property	Avalanche breakdown	Zener tunnelling
Mechanism	Carriers accelerated by the field gain enough energy to impact-ionize lattice atoms; the new carriers ionize further, so the current multiplies	Heavy doping on both sides makes $W$ very narrow and the field very high (of order $1 0^{6}$ V/cm). Under reverse bias, filled valence-band states on the p-side line up with empty conduction-band states on the n-side, and electrons tunnel directly across
Doping level	Lightly to moderately doped (at least one side)	Heavily doped on both sides
Typical breakdown voltage (Si)	Higher, roughly above 6–7 V	Lower, roughly below 5 V; the two mechanisms coexist in between
Knee of the I–V curve	Sharp	Softer, more gradual
Temperature coefficient of $V_{Z}$	Positive: breakdown voltage rises with temperature	Negative: breakdown voltage falls with temperature, since $E_{g}$ decreases

AVALANCHE BREAKDOWN VOLTAGE (ONE-SIDED ABRUPT JUNCTION)

V_{B R} \approx \frac{ϵ _{s} E _{cr i t}^{2}}{2 q N _{B}}

$N_{B}$ is the doping of the lightly doped side. $E_{cr i t}$ is the critical field for avalanche, a few $\times 1 0^{5}$ V/cm in Si, and it changes only weakly with doping. $V_{bi}$ is neglected next to $V_{B R}$ . So $V_{B R} \propto 1/ N_{B}$ : lighter doping gives a higher breakdown voltage.

The two temperature coefficients have opposite signs. Diodes that break down near the crossover (roughly 5–6 V in silicon) therefore have a net temperature coefficient close to zero, which makes them a good choice for stable voltage references.

The Zener diode as a device

A diode designed to operate continuously in reverse breakdown at a specified voltage $V_{Z}$ . The name covers both mechanisms: low-voltage devices (below about 5 V in Si) break down by tunnelling and need heavy doping on both sides, while higher-voltage devices break down by avalanche.
In breakdown the current can change by orders of magnitude for a small change in voltage. This near-vertical part of the $I$ – $V$ curve is what makes it useful as a voltage reference or regulator.
It must be used with an external series resistor to limit the current; once breakdown starts, the device itself does not limit current.
It is rated by $V_{Z}$ at a specified test current, by its dynamic resistance $r_{z}$ in breakdown (ideally small), by the minimum (knee) current $I_{Z, min}$ needed for regulation, and by its maximum power $P_{Z, ma x} = V_{Z} I_{Z, ma x}$ .

ZENER PIECEWISE-LINEAR MODEL (IN BREAKDOWN)

V_{Z} = V_{Z 0} + r_{z} I_{Z}

$I_{Z}$ is the reverse current, taken positive, with $I_{Z} \geq I_{Z, min}$ . $V_{Z 0}$ is the knee voltage and $r_{z}$ the dynamic resistance. An ideal Zener has $r_{z} = 0$ .

SHUNT REGULATOR: SERIES-RESISTOR LIMITS (IDEAL ZENER)

R_{ma x} = \frac{V _{S, min} - V _{Z}}{I _{L, ma x} + I _{Z, min}}, R_{min} = \frac{V _{S, ma x} - V _{Z}}{I _{L, min} + I _{Z, ma x}}

Supply $V_{S}$ feeds the Zener in parallel with the load through a series resistor $R$ . $R_{ma x}$ keeps the Zener in regulation in the worst case (lowest supply, highest load current). $R_{min}$ keeps the Zener current within its rating in the worst case (highest supply, lowest load current). Assumes $r_{z} = 0$ .

EXAM TIP

First check that the Zener is actually in breakdown. With the Zener removed, the load voltage is $V_{S} R_{L} / (R + R_{L})$ . If that is below $V_{Z}$ , the Zener is off and the output is just this divider voltage, not $V_{Z}$ .

Metal–Semiconductor Contacts: Schottky and Ohmic

A metal on a semiconductor forms either a rectifying (Schottky) junction or a non-rectifying (ohmic) contact. Which one you get depends on the work-function difference and the doping. The band-bending logic is the same as for a p–n junction, with the metal taking the place of one doped region. The relations below are the ideal Schottky–Mott model, which ignores interface states; real barrier heights are often pinned by those states. Energies and work functions are in eV.

IDEAL SCHOTTKY BARRIER HEIGHT

ϕ_{B n} = ϕ_{M} - χ, ϕ_{B p} = E_{g} - (ϕ_{M} - χ)

$ϕ_{M}$ is the metal work function and $χ$ the semiconductor's electron affinity. $ϕ_{B n}$ is the barrier to electrons on n-type material and $ϕ_{B p}$ the barrier to holes on p-type material. On n-type, the built-in potential of the contact is $ϕ_{B n} - (E_{C} - E_{F})$ , in volts.

SCHOTTKY DIODE THERMIONIC-EMISSION CURRENT

I = A A^{**} T^{2} e^{- ϕ_{B} / V_{T}} (e^{V_{A} / V_{T}} - 1)

$A$ is the contact area, $A^{**}$ the effective Richardson constant, and $ϕ_{B}$ the barrier height in volts (numerically equal to its value in eV). The prefactor $A A^{**} T^{2} e^{- ϕ_{B} / V_{T}}$ is the saturation current, which replaces the diffusion-based $I_{0}$ of a p–n junction. For typical barrier heights it is several orders of magnitude larger than a comparable p–n junction's $I_{0}$ .

Rectifying or ohmic (ideal model)

Rectifying (Schottky), n-type: $ϕ_{M} > ϕ_{S}$ , the semiconductor work function. Electrons flow into the metal and the bands bend upward, leaving a barrier to electrons. For p-type the condition is reversed: $ϕ_{M} < ϕ_{S}$ .
Ohmic, n-type: $ϕ_{M} < ϕ_{S}$ . The bands bend down into electron accumulation, so electrons meet no barrier in either direction. For p-type the condition is $ϕ_{M} > ϕ_{S}$ .
Ohmic by tunnelling: heavily dope the semiconductor under the contact. The barrier becomes thin enough for carriers to tunnel through it freely whatever its height. This is the usual practical way ohmic contacts are made, since interface states often prevent the ideal work-function condition.

EXAM TIP

At the same moderate current, a silicon Schottky diode typically drops about 0.2–0.4 V, against about 0.6–0.7 V for a p–n diode. The reason is its much larger saturation current, which comes from the relatively small barrier. It is also a majority-carrier device, so it stores almost no minority charge and switches off quickly, with negligible reverse-recovery time.

Worked Examples

WORKED EXAMPLE · EXAMPLE 1: BUILT-IN POTENTIAL AND DEPLETION WIDTH

A silicon step junction has $N_{A} = 5 \times 1 0^{17} cm^{- 3}$ and $N_{D} = 5 \times 1 0^{14} cm^{- 3}$ . Take $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ , $V_{T} = 25.9 mV$ and $ϵ_{s} = 1.04 \times 1 0^{- 12} F/cm$ , with zero applied bias. Find $V_{bi}$ and the depletion width $W$ .

Solution

$V_{bi} = V_{T} ln (N_{A} N_{D} / n_{i}^{2})$ . $N_{A} N_{D} = 2.5 \times 1 0^{32}$ and $n_{i}^{2} = 2.25 \times 1 0^{20}$ , so the ratio is $1.11 \times 1 0^{12}$ .
$ln (1.11 \times 1 0^{12}) = 27.74$ , so $V_{bi} = 0.0259 \times 27.74 \approx 0.718 V$ .
Since $N_{A} = 1000 N_{D}$ , the junction is strongly one-sided: $1/ N_{A} + 1/ N_{D} \approx 1/ N_{D} = 2 \times 1 0^{- 15} cm^{3}$ (error 0.1%).
$W = (2 \times 1.04 \times 1 0^{- 12} /1.6 \times 1 0^{- 19}) \times 2 \times 1 0^{- 15} \times 0.718$ .
$2 ϵ_{s} / q = 1.3 \times 1 0^{7}$ . Multiplying by $2 \times 1 0^{- 15}$ gives $2.6 \times 1 0^{- 8}$ , and multiplying by $0.718$ gives $1.867 \times 1 0^{- 8} cm^{2}$ .
$W = 1.867 \times 1 0^{- 8} \approx 1.37 \times 1 0^{- 4} cm = 1.37 μ m$ , almost all of it on the lightly doped n-side.

Answer: $V_{bi} \approx 0.718 V$ and $W \approx 1.37 μ m$ , essentially all on the n-side.

WORKED EXAMPLE · EXAMPLE 2: ZENER REGULATOR SERIES RESISTOR

A silicon Zener diode with $V_{Z} = 6.2 V$ (treat it as ideal, $r_{z} = 0$ ) is fed from a 12 V source through a series resistor $R$ . The load draws a fixed 20 mA, and the Zener must carry at least 5 mA at all times to stay in regulation. Find the largest allowed $R$ , and the Zener's dissipation if the load is disconnected.

Solution

The current through $R$ must supply both the load and the minimum Zener current: $I_{R} \geq I_{l o a d} + I_{Z, min} = 20 + 5 = 25 mA$ .
The source and the regulated output fix the voltage across $R$ : $V_{R} = 12 - 6.2 = 5.8 V$ .
$R$ is largest when the current is at its minimum: $R_{ma x} = V_{R} / I_{R} = 5.8 V /25 mA = 232 Ω$ .
With $R = 232 Ω$ and the load disconnected, all 25 mA flows through the Zener: $P_{Z} = 6.2 \times 0.025 = 0.155 W$ . The Zener's power rating must be at least this.

Answer: $R_{ma x} = 232 Ω$ . A larger resistor would starve the Zener of its minimum regulating current at full load. With no load, the Zener dissipates 155 mW.

WORKED EXAMPLE · EXAMPLE 3: DEPLETION CAPACITANCE UNDER REVERSE BIAS

An abrupt-junction diode has zero-bias capacitance $C_{j 0} = 10 pF$ and $V_{bi} = 0.8 V$ . Find $C_{j}$ at a reverse bias $V_{R} = 7.2 V$ .

Solution

Abrupt junction, so $m = 1/2$ , and reverse bias gives $V_{A} = - 7.2 V$ .
$C_{j} = C_{j 0} / 1 + V_{R} / V_{bi} = 10/ 1 + 7.2/0.8 = 10/ 10$ .
$C_{j} \approx 3.16 pF$ .

Answer: $C_{j} \approx 3.16 pF$ .

PREVIOUS-YEAR QUESTIONS

6 questions from the EC papers

Try each question first, then open the answer. One shows the official answer only. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE ECE 2025 · GATE ECE 2020 · GATE ECE 2018

GATE 2026 EC Q.14

GATE 2026 · ECQ.14 · 1 mark · MCQ

Consider carrier transport in a Zener diode in the breakdown region.

Which is the dominant transport mechanism for current flow in this case?

(A)
Drift
(B)
Diffusion
(C)
Tunneling
(D)
Ballistic transport

Answer and solution

GATE 2026 EC Q.14: answer and solution

See the question in the paper

Tap to open full size

GATE 2026 EC Q.14: answer and solution

GATE 2026 EC Q.25

GATE 2026 · ECQ.25 · 1 mark · MSQ

Consider a p-n junction diode when it is forward biased with 2 V. Which of the following is/are the correct magnitude(s) of the energy difference between quasi Fermi-levels, $E_{f n}$ in the n-side and $E_{f p}$ in the p-side?

(A)
2 eV
(B)
1 eV
(C)
2 V
(D)
1 V

Answer and solution

GATE 2026 EC Q.25: answer and solution

See the question in the paper

Tap to open full size

GATE 2026 EC Q.25: answer and solution

GATE 2025 EC Q.63

GATE 2025 · ECQ.63 · 2 marks · NAT

An ideal p-n junction germanium diode has a reverse saturation current of 10 μA at 300 K.

The voltage (in Volts, rounded off to two decimal places) to be applied across the junction to get a forward bias current of 100 mA at 300 K is ______.

(Consider the Boltzmann constant $k_{B} = 1.38 \times 1 0^{- 23} J/K$ and the charge of an electron $e = 1.6 \times 1 0^{- 19} C$ .)

Answer and solution

GATE 2025 EC Q.63: answer and solution

See the question in the paper

Tap to open full size

GATE 2025 EC Q.63: answer and solution

GATE 2020 EC Q.31

GATE 2020 · ECQ.31 · 2 marks · MCQ

A one-sided abrupt pn junction diode has a depletion capacitance $C_{D}$ of 50 pF at a reverse bias of 0.2 V. The plot of $1/ C_{D}^{2}$ versus the applied voltage $V$ for this diode is a straight line as shown in the figure below. The slope of the plot is ____ $\times 1 0^{20} F^{- 2} V^{- 1}$ .

GATE 2020 EC question 31: Axes: vertical axis labeled 1/C_D^2, horizontal axis labeled V, origin marked 0. — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
-5.7
(B)
-3.8
(C)
-1.2
(D)
-0.4

Official answer

Official answer: marks to all candidatesSource: Official answer key, GATE 2020 EC (IIT Delhi)

GATE 2020 EC Q.31: official answer

GATE 2018 EC Q.7

GATE 2018 · ECQ.7 · 1 mark · MCQ

In a p-n junction diode at equilibrium, which one of the following statements is NOT TRUE?

(A)
The hole and electron diffusion current components are in the same direction.
(B)
The hole and electron drift current components are in the same direction.
(C)
On an average, holes and electrons drift in opposite direction.
(D)
On an average, electrons drift and diffuse in the same direction.

Answer and solution

GATE 2018 EC Q.7: answer and solution

See the question in the paper

Tap to open full size

GATE 2018 EC Q.7: answer and solution

GATE 2018 EC Q.18

GATE 2018 · ECQ.18 · 1 mark · NAT

A p-n step junction diode with a contact potential of $0.65 V$ has a depletion width of $1 μ m$ at equilibrium. The forward voltage (in volts, correct to two decimal places) at which this width reduces to $0.6 μ m$ is ______.

Answer and solution

GATE 2018 EC Q.18: answer and solution

See the question in the paper

Tap to open full size

GATE 2018 EC Q.18: answer and solution

PRACTICE

4 questions to test yourself

PRACTICE 11 mark · MCQ

A silicon step p–n junction at 300 K has $N_{A} = 1 0^{16} cm^{- 3}$ , $N_{D} = 1 0^{15} cm^{- 3}$ and $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . Taking $V_{T} = 25.9 mV$ , the built-in potential $V_{bi}$ is closest to:

(A)
0.35 V
(B)
0.64 V
(C)
0.90 V
(D)
1.10 V

Answer and solution

Answer: (B)

$V_{bi} = V_{T} ln (\frac{N _{A} N _{D}}{n _{i}^{2}})$
$N_{A} N_{D} = 1 0^{31} cm^{- 6}$ and $n_{i}^{2} = 2.25 \times 1 0^{20} cm^{- 6}$
The ratio is $4.44 \times 1 0^{10}$ , and $ln (4.44 \times 1 0^{10}) = 24.52$
$V_{bi} = 0.0259 \times 24.52 \approx 0.635 V$

PRACTICE 21 mark · MCQ

In the junction above ( $N_{A} = 1 0^{16} ≫ N_{D} = 1 0^{15} cm^{- 3}$ ), charge balance $N_{A} x_{p} = N_{D} x_{n}$ shows that the depletion region extends mainly into the:

(A)
p-side
(B)
n-side
(C)
Equally on both sides
(D)
Depends only on the applied bias sign

Answer and solution

Answer: (B)

Charge balance gives $\frac{x _{n}}{x _{p}} = \frac{N _{A}}{N _{D}} = 10$
So $x_{n} = 10 x_{p}$ : most of $W$ lies on the lightly doped side
Here the lightly doped side is the n-side

PRACTICE 32 marks · MSQ

For a silicon p–n junction diode at 300 K, which of the following statements are correct?

(A)
$I_{0}$ increases with temperature because $n_{i}^{2}$ increases with temperature
(B)
Diffusion capacitance dominates over depletion capacitance under strong forward bias
(C)
Depletion capacitance decreases as reverse-bias magnitude increases
(D)
The peak electric field occurs at the neutral-region boundary, away from the junction

Answer and solution

Answer: (A), (B), (C)

A: $I_{0} \propto n_{i}^{2}$ , and $n_{i}^{2}$ (proportional to $T^{3} e^{- E_{g} / k T}$ ) rises steeply with $T$ , so $I_{0}$ increases with temperature. Correct.
B: under strong forward bias, the stored excess minority charge makes the diffusion capacitance ( $C_{d} = τ I / V_{T}$ , growing with current) far larger than the depletion capacitance $C_{j}$ . Correct.
C: $C_{j} = ϵ_{s} A / W$ , and $W$ grows with reverse-bias magnitude, so $C_{j}$ falls as reverse bias increases. Correct.
D: false. In a step junction the peak field is at the metallurgical junction ( $x = 0$ ), where the space-charge density changes sign. At the depletion edges, next to the neutral regions, the field is zero.

PRACTICE 42 marks · NAT

A one-sided (p $^{+}$ n) silicon diode has reverse saturation current $I_{0} = 2 \times 1 0^{- 14} A$ at 300 K ( $V_{T} = 25.9 mV$ , ideality factor 1). Using $I \approx I_{0} e^{V_{A} / V_{T}}$ , find the forward voltage $V_{A}$ (in volts) needed to reach $I = 1 mA$ .

Answer and solution

Answer: $0.638$ (accepted range $\pm 0.02$ )

$V_{A} \approx V_{T} ln (\frac{I}{I _{0}})$ (the $- 1$ in the Shockley equation is negligible in forward bias)
$\frac{I}{I _{0}} = \frac{1 0 ^{- 3}}{2 \times 1 0 ^{- 14}} = 5 \times 1 0^{10}$
$ln (5 \times 1 0^{10}) = 24.64$
$V_{A} = 0.0259 \times 24.64 \approx 0.638 V$

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

P-N junction, Zener diode

Preparing in Kolkata? IAE Academy's course for GATE ECE.