PN junction and diodes
A 1000 V diode and a 3 V Zener diode are the same PN junction with different doping. What decides how much voltage it withstands is the less doped side, not the more doped one.
01The junction at equilibrium
When a P region is joined to an N region, holes from the P side diffuse into the N side and electrons from the N side into the P side, following the concentration gradient described in the previous unit. Each carrier that crosses leaves behind an uncompensated fixed ion: negative acceptors on the P side, positive donors on the N side. That fixed charge creates an electric field that opposes diffusion, and the process stops when the drift current produced by the field exactly equals the diffusion current.
What remains is a strip almost empty of carriers, the depletion region or space-charge region, and between its edges a potential difference, the built-in potential:
The Electronics Technician program page on semiconductors and diodes tells the same story without equations; here it is quantified.
02The width of the depletion region
Solving Poisson's equation with uniform charge on each side (the depletion approximation), the field is triangular, with its maximum at the junction, and the potential grows as two parabolas. The total charge on each side has to be the same, so the region extends further into the less doped side:
With the doping levels of the example and no applied voltage, \( W \approx 0.33 \) µm, of which 0.30 µm lie on the N side. With 10 V in reverse, \( W \) grows to 1.24 µm and the maximum field reaches \( 1.7 \times 10^5 \) V/cm. Almost all real junctions are asymmetric, with one side much more heavily doped than the other: the depletion region lies entirely in the lightly doped side, and its doping sets the breakdown voltage and the capacitance.
You choose the two doping levels and the applied voltage. The three plots show, on the same axis, the fixed charge, the electric field and the potential along the junction; the horizontal scale adjusts itself. With a large reverse bias you can see when the field approaches the avalanche breakdown value.
03The diode equation
A forward voltage lowers the barrier, and diffusion wins out over drift: minority carriers are injected on each side. At the edge of the depletion region the minority-carrier concentration rises by the factor \( e^{V/V_T} \), and from there it decays with the diffusion length, exactly the exponential profile of the continuity equation. The current is the diffusion of that excess, and adding the two sides gives the Shockley equation:
With the previous example, \( \tau = 1 \) µs on both sides and an area of 1 mm², \( I_S \approx 6 \times 10^{-14} \) A: the diode conducts 1 µA at 0.43 V, 1 mA at 0.61 V and 100 mA at 0.73 V. Each decade of current costs \( V_T \ln 10 \approx 60 \) mV, which is why the “0.6 V” of the Electronics Technician program is such a good approximation over such a wide range. 89 % of that current is carried by holes injected into the N side: the current is dominated by the less doped side, something the bipolar transistor exploits so that the emitter injects almost only electrons.
\( I_S \) is proportional to \( n_i^2 \), which grows very fast with temperature: at constant current, the diode voltage drops about 2 mV per degree. This is the basis of integrated temperature sensors and the cause of thermal drift in amplifiers.
04The two capacitances of the diode
The junction stores charge in two different ways, and each one dominates in one bias condition:
| Junction capacitance | Diffusion capacitance | |
|---|---|---|
| Origin | The fixed charge of the depletion region, which changes with its width | The excess of injected minority carriers, proportional to the current |
| Expression | \( C_j = \dfrac{\varepsilon_s A}{W} = \dfrac{C_{j0}}{\sqrt{1 - V/V_{bi}}} \) | \( C_d = \dfrac{\tau_T\, I}{V_T} \) |
| Dominates in | Reverse bias | Forward bias |
| Put to use in | Varicap diode: electronic tuning | Limits the switching speed |
The exponent \( 1/2 \) is that of an abrupt junction; in a graded junction it is \( 1/3 \), and hyperabrupt varactors use special doping profiles so that the capacitance changes more with voltage.
05Reverse breakdown
In reverse bias only about \( I_S \) flows, until the maximum field reaches a critical value, on the order of \( 3 \) to \( 5 \times 10^5 \) V/cm in silicon. There are two mechanisms:
- Avalanche: carriers gain enough energy from the field to knock loose new pairs when they collide, and the multiplication runs away. It dominates in lightly doped junctions, with breakdown above about 7 V, and its voltage rises with temperature.
- Zener effect: with very high doping the depletion region is so thin that electrons cross it by tunneling. It dominates below about 5 V and its voltage falls with temperature.
Between 5 and 6 V the two coefficients cancel: that is why the reference diodes with the best stability sit in that range. In an asymmetric junction, the avalanche breakdown voltage is set by the doping of the lightly doped side:
06Switching and reverse recovery
A forward-biased diode has the charge of the injected minority carriers stored in it. If the voltage is suddenly reversed, the diode keeps conducting in reverse until that charge is removed: first with a reverse current limited by the circuit during the storage time, and then decaying as the depletion region re-forms:
| Diode | Recovery | Use |
|---|---|---|
| Ordinary rectifier (1N400x) | Microseconds | 50 or 60 Hz |
| Signal (1N4148) | 4 ns maximum | Logic, detectors, protection |
| Fast and ultrafast | Tens of nanoseconds | Switching power supplies |
| Schottky (metal-semiconductor) | Almost none: they conduct with majority carriers, so there are no stored minority carriers | Low-voltage switching power supplies, with lower forward drop |
07Models for circuit analysis
| Model | What it keeps | When it is enough |
|---|---|---|
| Ideal | Only the direction of conduction | Circuit voltages much larger than 0.7 V |
| Constant drop | A 0.6 to 0.7 V source in forward bias | Everyday bias analysis |
| Piecewise linear | The drop plus a series resistance | Large currents, where the resistance of the material matters |
| Exponential | The Shockley equation | Precision design, simulation |
| Small signal | The slope at the operating point: \( r_d = n V_T / I_D \) | Small variations around the bias point |
The small-signal resistance is remarkable: at 1 mA it is 25.85 Ω, whatever the diode, because it depends only on the current and the temperature. The same idea gives the transconductance of the bipolar transistor in the next unit.
08In the lab
Measure the voltage of a 1N4148 at currents of 1 µA, 10 µA, 100 µA, 1 mA and 10 mA. Plot \( V \) against \( \log I \), obtain the slope and deduce the ideality factor \( n \).
Use an LCR meter to measure the capacitance of a varicap diode (or of a reverse-biased 1N4007) between 0 and 20 V. Plot \( 1/C^2 \) against \( V \): the line crosses the axis at \( V_{bi} \) and its slope gives the doping of the lightly doped side.
Drive a 1N4007 and a 1N4148 with a ±5 V square wave at 100 kHz through a 1 kΩ resistor, and use the oscilloscope to measure the reverse-current pulse across the resistor in each one.
09Common mistakes
- Believing that the built-in potential can be measured. The metal contacts compensate it.
- Assuming that the depletion region is split equally. It extends into the less doped side.
- Treating the 0.7 V drop as a constant of silicon. It depends on the current, the area and the temperature.
- Forgetting reverse recovery in a switching power supply. The slow diode conducts in reverse on every cycle and heats the transistor.
- Confusing the two capacitances. The junction capacitance is the reverse-bias one; the diffusion capacitance, the forward-bias one.
10Self-assessment
What equilibrium is reached in a junction with no applied voltage?
The diffusion current of each carrier equals the drift current produced by the field of the depletion region, and the net current is zero.
In a P⁺N junction, on which side does almost the entire depletion region lie?
On the N side, the less doped one, because the charge on each side has to be equal.
How much does the voltage of a diode increase if the current is multiplied by one hundred?
\( V_T \ln 100 \approx 119 \) mV, with \( n = 1 \).
Why does a Schottky diode switch faster than a junction diode?
Because it conducts with majority carriers and does not accumulate minority-carrier charge that has to be removed when the voltage is reversed.
How does the capacitance of an abrupt junction change when \( V_{bi} - V \) is multiplied by four?
It is halved, because the width \( W \) grows with the square root.
11Further reading
- Robert F. Pierret. Semiconductor Device Fundamentals. Addison-Wesley, 1996. The electrostatics of the junction and the derivation of the diode equation, step by step.
- Ben G. Streetman and Sanjay Kumar Banerjee. Solid State Electronic Devices. 7th ed., Pearson, 2015. Very clear on switching, reverse recovery and the breakdown mechanisms.
- Richard S. Muller, Theodore I. Kamins and Mansun Chan. Device Electronics for Integrated Circuits. 3rd ed., Wiley, 2003. The junction seen from the integrated circuit, with the models that simulators use.