Junction Diode Notes FE Sem 2 Semiconductor Physics | MU

Join a p-type and an n-type region in one crystal and you get the p-n junction, the simplest semiconductor device and the building block of almost every other: rectifier diodes, LEDs, solar cells, photodiodes and transistors. This module of the Semiconductor Physics elective explains how a p-n junction forms, how to calculate its barrier potential, the diode equation, the behaviour of the junction in forward and reverse bias, the current-voltage curve of a junction diode, and the LED and its working. It maps to course outcome CO2 (calculate barrier potential and plot the I-V characteristics of a p-n junction diode).

Summary

Formation of a p-n junction: A p-n junction is formed when one part of a single semiconductor crystal is doped p-type and the adjoining part n-type, for example by diffusing donor atoms into a p-type wafer or by growing one layer on another (epitaxy). It is not made by simply pressing two pieces together, because the surface roughness would prevent a clean junction. At the moment of formation, the n side has a large concentration of free electrons and the p side a large concentration of holes. Electrons therefore diffuse from n to p and holes from p to n, and recombine near the junction. Each electron leaving the n side uncovers a fixed positive donor ion, and each hole leaving the p side uncovers a fixed negative acceptor ion. A thin region on both sides of the junction is thus emptied of mobile carriers and contains only immobile ions: this is the depletion region (space charge region), typically about 0.1 to 1 μm wide. The fixed charges create an electric field directed from the n side to the p side, which opposes further diffusion of majority carriers. Equilibrium is reached when the diffusion current of majority carriers is balanced by the drift current of minority carriers swept across by the field, so the net current is zero. The potential difference across the depletion region is the barrier potential (built-in potential) \(V_0\). In equilibrium the Fermi level is constant throughout the junction, so the bands on the p side are raised by \(eV_0\) relative to the n side.

Calculation of barrier potential: In equilibrium the Fermi level is flat. On the n side, \(E_F - E_{in} = kT\ln(N_D/n_i)\), and on the p side, \(E_{ip} - E_F = kT\ln(N_A/n_i)\). The bands bend by \(eV_0 = E_{ip} - E_{in} = kT\ln(N_D/n_i) + kT\ln(N_A/n_i)\). Hence \(V_0 = \dfrac{kT}{e}\ln\dfrac{N_AN_D}{n_i^2}\). An equivalent derivation uses the electron concentrations: on the n side \(n_n \approx N_D\), on the p side \(n_p = n_i^2/N_A\), and Boltzmann statistics gives \(n_n/n_p = e^{eV_0/kT}\). The barrier potential increases with doping and decreases as temperature rises (mainly because \(n_i\) rises steeply). Typical values: about 0.6 to 0.7 V for Si and 0.2 to 0.3 V for Ge at room temperature. It cannot be measured with a voltmeter because contact potentials at the metal-semiconductor contacts cancel it. The width of the depletion region is \(W = \sqrt{\dfrac{2\varepsilon V_0}{e}\left(\dfrac{1}{N_A} + \dfrac{1}{N_D}\right)}\) in equilibrium; it extends more into the lightly doped side.

Diode equation: The current through an ideal p-n junction diode is given by the Shockley diode equation \(I = I_0\left(e^{eV/\eta kT} - 1\right) = I_0\left(e^{V/\eta V_T} - 1\right)\), where V is the applied voltage (positive for forward bias), \(I_0\) is the reverse saturation current, \(V_T = kT/e\) is the thermal voltage (about 25.9 mV at 300 K) and η is the ideality factor (1 for an ideal diode and for Ge; between 1 and 2 for Si, often taken as 2 in textbooks for low currents). For forward voltages larger than a few \(V_T\), the exponential dominates and \(I \approx I_0e^{V/\eta V_T}\): current rises exponentially, about tenfold for every \(\eta V_T\ln10 \approx 60\) mV (η = 1). For reverse voltages larger than a few \(V_T\), \(I \approx -I_0\): a small constant reverse saturation current. \(I_0\) is due to thermally generated minority carriers; it is proportional to \(n_i^2\) and roughly doubles for every 10 °C rise in temperature. Typical \(I_0\) is in nA for Si and μA for Ge.

p-n junction in forward bias: The p side is connected to the positive terminal and the n side to the negative terminal of the battery. The applied voltage opposes the built-in field, so the barrier is lowered from \(V_0\) to \(V_0 - V\) and the depletion region becomes narrower. Holes from the p side and electrons from the n side can now cross the junction in large numbers (minority carrier injection); they diffuse and recombine on the other side, and the battery replaces them, so a large forward current flows. Little current flows until the applied voltage approaches the barrier, called the cut-in (knee or threshold) voltage, about 0.7 V for Si and 0.3 V for Ge; above it the current rises very steeply. In practice a series resistor must limit the current, or the diode will overheat.

p-n junction in reverse bias: The p side is connected to the negative terminal and the n side to the positive terminal. The applied voltage adds to the built-in field, so the barrier becomes \(V_0 + V\) and the depletion region widens as majority carriers are pulled away from the junction. Majority carriers cannot cross, so only a very small reverse current flows, carried by minority carriers (electrons from p and holes from n) that are swept across by the field. This reverse saturation current \(I_0\) is almost independent of the reverse voltage but increases with temperature. The widened depletion region acts like the dielectric of a capacitor, giving a junction capacitance that decreases as reverse voltage increases (used in varactor diodes). If the reverse voltage is increased to the breakdown voltage, the reverse current rises suddenly due to avalanche multiplication (carriers gain enough energy to break bonds by collision, in lightly doped junctions, usually above about 6 V) or Zener breakdown (direct tunnelling by the very strong field in heavily doped, thin junctions, usually below about 5 V). Ordinary diodes are damaged if operated in breakdown without current limiting.

Current-voltage curve of a p-n junction diode: The I-V curve is drawn with forward voltage and current in the first quadrant and reverse voltage and current in the third quadrant (with different scales: mA for forward current and μA or nA for reverse current). Forward region: almost zero current up to the cut-in voltage, then an exponential rise; the curve becomes nearly linear at high current because of the bulk resistance. Reverse region: a small, nearly constant reverse saturation current up to breakdown, then a sharp increase at the breakdown voltage. From the curve we define the static (DC) resistance \(R = V/I\) and the dynamic (AC) resistance \(r_d = \Delta V/\Delta I = \eta V_T/I\) in forward bias, about 2.6 Ω at 10 mA for η = 1. The curve shifts left with rising temperature (the forward voltage at constant current falls by about 2 mV/°C) and reverse current increases. In the lab the forward characteristic is obtained with the diode, a series resistor, a variable DC supply, a voltmeter across the diode and a milliammeter in series; for reverse characteristics the supply is reversed and a microammeter is used.

LED and its working: A light emitting diode (LED) is a forward-biased p-n junction that emits light when electrons and holes recombine. In forward bias, electrons injected from the n side into the p side and holes injected into the n side recombine near the junction. In a direct band gap semiconductor, an electron falling from the conduction band into a hole in the valence band releases its energy as a photon of energy approximately equal to the band gap: \(h\nu \approx E_g\), \(\lambda = hc/E_g\) (λ in μm ≈ 1.24/Eg in eV). This process is called electroluminescence (injection luminescence). Silicon and germanium have indirect gaps, so recombination there releases energy mostly as heat; LEDs use compound semiconductors. Colour depends on the band gap: GaAs (1.42 eV) emits infrared at about 870 nm (remote controls), GaAsP gives red, orange or yellow depending on the phosphorus content, AlGaInP gives bright red to yellow, GaP gives green (with nitrogen doping), and InGaN gives green, blue and violet. The blue InGaN LED, developed by Isamu Akasaki, Hiroshi Amano and Shuji Nakamura (Nobel Prize 2014), made white LEDs possible: a blue LED coated with a yellow phosphor gives white light. Construction: a thin p layer on top of an n layer (or a multiple quantum well active layer between them) so that light generated near the junction escapes from the top; a small top contact, reflective back contact and a domed epoxy lens to direct light. LEDs need forward voltage of about 1.8 to 2 V (red) up to about 3 V (blue, white) and a series resistor to limit current. Advantages: high efficiency, long life (tens of thousands of hours), low voltage, small size, fast switching, robustness, no mercury. Uses: indicators, displays, traffic lights, general lighting, backlights, vehicle lamps, optical communication and remote controls.

Mind map

Deriving the barrier potential

Important definitions

Formulas

Key points to remember

• The junction is made in one crystal, not by pressing two pieces together.
• The depletion region contains immobile ions: positive on the n side, negative on the p side.
• The built-in field points from n to p and stops further diffusion.
• V0 depends on doping and temperature; about 0.7 V for Si and 0.3 V for Ge.
• Forward bias narrows the depletion region; reverse bias widens it.
• Forward current is due to majority carriers crossing the lowered barrier; reverse current is due to minority carriers.
• I0 is in nA for Si and μA for Ge; it doubles every 10 °C.
• Forward voltage at fixed current falls by about 2 mV/°C.
• Avalanche breakdown in lightly doped junctions; Zener breakdown in heavily doped junctions.
• LEDs need direct gap materials; photon energy ≈ Eg; colour is set by the band gap.
• Always use a series resistor with a diode or LED.

Solved examples

Example 1. A silicon p-n junction has NA = 10^22 m^-3 and ND = 10^21 m^-3. Find the barrier potential at 300 K (ni = 1.5 × 10^16 m^-3).

\(kT/e = 0.02585\) V. \(\dfrac{N_AN_D}{n_i^2} = \dfrac{10^{43}}{2.25\times10^{32}} = 4.44\times10^{10}\). \(\ln(4.44\times10^{10}) = 24.52\). \(V_0 = 0.02585\times24.52 = 0.634\) V. (With the newer value \(n_i = 1.0\times10^{16}\) m⁻³, \(V_0 = 0.02585\ln(10^{11}) = 0.655\) V.)

Example 2. A germanium junction has NA = ND = 10^22 m^-3 and ni = 2.4 × 10^19 m^-3. Find V0 at 300 K.

\(\dfrac{N_AN_D}{n_i^2} = \dfrac{10^{44}}{5.76\times10^{38}} = 1.736\times10^{5}\). \(\ln(1.736\times10^{5}) = 12.06\). \(V_0 = 0.02585\times12.06 = 0.312\) V, much lower than Si because \(n_i\) of Ge is larger.

Example 3. A diode has I0 = 10 nA and η = 1 at 300 K. Find the current at a forward bias of 0.3 V and at a reverse bias of 0.3 V.

\(V/V_T = 0.3/0.02585 = 11.61\); \(e^{11.61} = 1.10\times10^{5}\). Forward: \(I = 10^{-8}(1.10\times10^{5} - 1) = 1.10\times10^{-3}\) A \(= 1.10\) mA. Reverse: \(I = 10^{-8}(e^{-11.61} - 1) \approx -10^{-8}\) A \(= -10\) nA, essentially \(-I_0\).

Example 4. By how much must the forward voltage of a diode (η = 1) be increased to increase the current tenfold? Find its dynamic resistance at 10 mA.

\(\Delta V = V_T\ln10 = 0.02585\times2.303 = 0.0595\) V \(\approx 60\) mV. \(r_d = V_T/I = 0.02585/0.01 = 2.59\) Ω.

Example 5. The reverse saturation current of a diode is 5 nA at 25 °C. Estimate it at 75 °C.

Rise = 50 °C = five 10-degree steps. \(I_0 = 5\times2^5 = 160\) nA.

Example 6. Find the wavelength emitted by LEDs made of GaAsP (Eg = 1.9 eV) and GaN (Eg = 3.4 eV). What band gap gives blue light of 450 nm?

GaAsP: \(\lambda = hc/E_g = 6.63\times10^{-34}\times3\times10^{8}/(1.9\times1.6\times10^{-19}) = 6.54\times10^{-7}\) m \(= 654\) nm (red). GaN: \(\lambda = 1.989\times10^{-25}/(3.4\times1.6\times10^{-19}) = 3.66\times10^{-7}\) m \(= 366\) nm (ultraviolet). For 450 nm: \(E_g = hc/\lambda = 1.989\times10^{-25}/(450\times10^{-9}\times1.6\times10^{-19}) = 2.76\) eV, as in InGaN.

Example 7. A red LED drops 2 V and needs 20 mA. Find the series resistor for a 9 V supply.

\(R = (9 - 2)/0.02 = 350\) Ω; the nearest standard value above, 360 Ω or 390 Ω, is used.

Important questions and answers

Q1. Explain the formation of a p-n junction and the depletion region. (5 marks)

When p and n regions are formed in one crystal, electrons diffuse from n to p and holes from p to n because of the concentration gradient. They recombine near the junction, leaving uncompensated positive donor ions on the n side and negative acceptor ions on the p side. This layer without mobile carriers is the depletion region. The ions set up an electric field from n to p, and a potential barrier V₀, which opposes further diffusion of majority carriers but helps minority carriers drift across. Equilibrium is reached when diffusion and drift currents are equal, so the net current is zero and the Fermi level is uniform. Diagrams: (a) p and n regions with mobile carriers and ions; (b) depletion region with fixed charges; (c) potential versus distance showing V₀; (d) energy band diagram showing bent bands and flat E_F.

Q2. Derive the expression for the barrier potential of a p-n junction. (5 marks)

See the deep-dive above: equate the Fermi levels, use \(E_F - E_{in} = kT\ln(N_D/n_i)\) and \(E_{ip} - E_F = kT\ln(N_A/n_i)\), add to get \(eV_0 = kT\ln(N_AN_D/n_i^2)\). Discuss: V₀ rises with doping logarithmically; falls with temperature as n_i increases; larger for wider-gap materials (Si > Ge). Include the band diagram showing \(eV_0\) as the difference in band edges.

Q3. Explain the working of a p-n junction in forward and reverse bias. (5 marks)

Forward bias: positive terminal to p, negative to n. External field opposes built-in field; barrier reduces to \(V_0 - V\); depletion region narrows; majority carriers cross in large numbers; current rises exponentially beyond the cut-in voltage; current is in mA. Reverse bias: positive terminal to n, negative to p. Fields add; barrier rises to \(V_0 + V\); depletion region widens; majority carriers blocked; only minority carriers drift across, giving the small saturation current \(I_0\) (nA to μA), nearly independent of voltage until breakdown. Diagrams: circuit and band diagram for each case.

Q4. Draw and explain the I-V characteristics of a p-n junction diode. (5 marks)

Circuit: DC supply, rheostat or series resistor, diode, voltmeter across diode, milliammeter (forward) or microammeter (reverse). Graph: forward quadrant shows near-zero current up to the knee (0.7 V Si, 0.3 V Ge) and then a steep exponential rise; reverse quadrant shows a tiny constant I₀ until breakdown voltage V_BR, where current rises sharply. Explain using the diode equation \(I = I_0(e^{V/\eta V_T} - 1)\): for V ≫ V_T current is exponential, for negative V current is −I₀. Parameters: cut-in voltage, forward resistance (static and dynamic), reverse saturation current, breakdown voltage, maximum forward current and peak inverse voltage. Effect of temperature: curve shifts left, I₀ rises.

Q5. Write the diode equation and explain each term. (3 marks)

\(I = I_0(e^{eV/\eta kT} - 1)\). I: diode current; I₀: reverse saturation current, set by minority carriers, proportional to \(n_i^2\); e: electronic charge; V: applied voltage, positive for forward bias; η: ideality factor, 1 for Ge and ideal diodes, about 2 for Si at low current; k: Boltzmann constant; T: absolute temperature. At room temperature \(kT/e = 25.9\) mV.

Q6. Explain the construction and working of an LED. Why are Si and Ge not used for LEDs? (5 marks)

Construction: a direct gap compound semiconductor (GaAs, GaAsP, AlGaInP, InGaN) p-n junction, with a thin top p layer, a small top contact, a reflective base, and an epoxy dome lens; leads called anode (longer) and cathode. Working: forward bias injects electrons into p and holes into n; they recombine in and near the junction; each radiative recombination emits a photon of energy about E_g; light escapes through the top. Colour is decided by the band gap: λ = 1.24/E_g μm. Si and Ge have indirect band gaps: the conduction band minimum and valence band maximum are at different crystal momenta, so recombination needs a phonon, is slow, and releases energy mostly as heat; hence they emit very little light. Applications: indicators, displays, lighting, signals, IR remotes, optical communication.

Q7. Distinguish between avalanche and Zener breakdown. (3 marks)

Avalanche: occurs in lightly doped junctions with wide depletion regions; minority carriers accelerated by the field gain enough energy to ionise atoms by collision, producing more carriers in a chain; breakdown voltage usually above about 6 V; increases with temperature. Zener: occurs in heavily doped junctions with very thin depletion regions; the field (about \(10^8\) V/m) is strong enough to pull electrons directly out of covalent bonds by quantum tunnelling; breakdown voltage usually below about 5 V; decreases with temperature.

Exam tips

• Draw the depletion region with ions (not free carriers) and show the direction of the built-in field from n to p.
• In barrier potential numericals compute N_AN_D/n_i² carefully with powers of ten before taking the logarithm.
• Use different scales for forward (mA) and reverse (μA) currents on the I-V graph and label the knee and breakdown voltages.
• Remember V_T = 25.9 mV at 300 K and the 60 mV per decade rule.
• For LED questions, always state that the material must have a direct band gap.

FAQs

Why can't we measure the barrier potential with a voltmeter?

When metal leads are attached, contact potentials appear at the metal-semiconductor junctions that exactly cancel the barrier potential in equilibrium, so the voltmeter reads zero.

Why does the depletion region extend more into the lightly doped side?

The total positive and negative charges must be equal. The lightly doped side has fewer ions per unit volume, so a wider region is needed to hold the same charge.

Why is an LED's forward voltage higher than a silicon diode's?

LED materials have larger band gaps (about 1.4 to 3.4 eV), so the barrier and the voltage needed for significant current are higher, roughly 1.8 V for red and 3 V for blue.

Does a diode conduct at all below the cut-in voltage?

Yes, but the current is extremely small. The current rises exponentially, and the cut-in voltage is only the point where it becomes noticeable on a milliampere scale.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Semiconductor Physics, Semester II) and were prepared with AI help and checked.