Special-purpose diodes: Zener, LED, photodiode, solar cell
The Zener diode
This is an extra lesson. The current NCERT book has dropped special-purpose diodes, but they still come up in competitive exams like JEE and NEET, so they are well worth knowing.
An ordinary diode in reverse bias lets almost no current through, until, at the breakdown voltage, the current suddenly shoots up. A Zener diode is designed to work right there.
Both its p side and its n side are heavily doped. Heavy doping makes the depletion region very thin, much less than a micrometre.
So even a small reverse voltage, about 5 volt, sets up a huge field across the junction: about 5 × 10⁶ volt per metre.
Such a field can pull valence electrons right out of their bonds on the p side and send them into the n side. This is called internal field emission, and it gives the sudden rise in current at the Zener voltage, V Z.
Look at the reverse characteristic. Below V Z, almost no current. At V Z the current rises steeply, while the voltage hardly changes. That nearly constant voltage is what makes the Zener useful.
Here is a Zener voltage regulator. The unsteady d.c. input passes through a series resistor R s, to a reverse biased Zener, with the load R L connected across the Zener.
If the input rises, more current flows through R s. The extra current goes through the Zener, the extra voltage drops across R s, and the load still gets V Z. If the input falls, the Zener current falls instead.
An example. A 6.0 volt Zener must supply a load current of 4.0 milliampere from a 10.0 volt input. Choose the Zener current as five times the load current: 20 milliampere. So R s carries 24 milliampere.
R s equals 10.0 minus 6.0, divided by 24 × 10⁻³ ampere: about 167 ohm. The nearest standard value, 150 ohm, will do.
The photodiode
In optoelectronic devices, light and charge carriers exchange energy. The first one is the photodiode, which turns light into an electric current.
A photon can break a bond only if its energy, h ν, is more than the band gap E g. Then it makes one electron–hole pair.
A handy rule: photon energy in electron volt is 1240 divided by the wavelength in nanometre. Red light of 620 nanometre carries 2.0 electron volt.
In a photodiode, light reaches the junction through a thin transparent window. Pairs made in or near the depletion region are pulled apart by its field before they can recombine.
Electrons are swept to the n side and holes to the p side. Inside the diode this is a current from n to p: the reverse direction.
So a photodiode is used in reverse bias. Here are its characteristics, one curve for each intensity. More light makes more pairs, and a larger reverse current.
Why reverse bias, when forward current is so much bigger? In forward bias, the light adds a few microampere to many milliampere of majority-carrier current. That change is hard to notice.
In reverse bias, the dark current comes from only a few minority carriers. The photocurrent can be many times larger, so the change is easy to measure.
Because the reverse current follows the light intensity, photodiodes are used as light detectors, for example in optical fibre receivers and remote-control receivers.
Light with photons below E g does nothing. A silicon photodiode, with E g equal to 1.1 electron volt, cannot detect light longer than 1240 divided by 1.1: about 1130 nanometre.
LEDs and solar cells
A light-emitting diode, or LED, works the other way round: current goes in, light comes out. It is a heavily doped p-n junction under forward bias.
Forward bias pushes electrons from n into p, and holes from p into n. Near the junction they recombine with the majority carriers there.
In each recombination, an electron drops from the conduction band to the valence band. The energy it loses, about E g, comes out as a photon.
So the band gap sets the colour: λ equals h c over E g. For visible light, E g must be at least about 1.8 electron volt. 1240 divided by 1.8 is 689 nanometre, deep red.
Gallium arsenide, with 1.4 electron volt, gives infrared, used in TV remotes. Gallium arsenide phosphide, about 1.9 electron volt, gives red. Wider gaps give green and blue.
Compared with filament bulbs, LEDs need low voltage and little power, switch on and off very fast, last much longer, and give light of nearly one colour.
A solar cell is a p-n junction that turns sunlight into electrical power, with no battery at all. A thin n layer sits on a p-type wafer, with metal fingers on top to let light in.
Three steps. Generation: light makes electron–hole pairs near the junction. Separation: the junction field sends electrons to n and holes to p. Collection: the p side becomes positive, the n side negative, and current flows through the load.
Its I–V curve lies in the fourth quadrant, because the cell gives out power. It cuts the voltage axis at the open-circuit voltage, V oc, and the current axis at the short-circuit current, I sc.
Sunlight is most intense for photons of about 1.5 electron volt, so band gaps near 1.5 electron volt are ideal. Silicon, 1.1 electron volt, and gallium arsenide, 1.43 electron volt, are common choices.
Try it: special-diode bench
- Zener: this is the worked example (6 V Zener, R_s = 167 Ω, I_L = 4 mA). Slide V_in between 8 V and 14 V and watch I_Z change while V_out stays near 6 V.
- Zener: make R_L small or V_in low. When I_Z drops to zero, the Zener stops regulating.
- Photodiode: raise the intensity and the reverse current rises. Push λ past about 1130 nm and the photocurrent vanishes (hν < E_g of silicon).
- LED: click GaAs, GaAsP, GaP and InGaN. A wider band gap gives a shorter wavelength and needs a larger forward voltage.
- Solar cell: change the load R and find the maximum-power point. Compare V_oc for silicon and GaAs.
The input goes up
In the worked-example regulator (V_Z = 6.0 V, R_s = 167 Ω, load current 4.0 mA), the input rises from 10.0 V to 12.0 V. What happens?
- V_out rises to 7.2 V; I_Z stays 20 mA
- V_out stays 6.0 V; I_Z rises to about 32 mA
- V_out stays 6.0 V; the load current rises to about 8 mA
- V_out stays 6.0 V; I_Z falls to about 8 mA
Answer
B. The Zener holds the output at 6.0 V, so the load current stays 6.0 V / 1.5 kΩ = 4.0 mA. The current through R_s is (12.0 − 6.0)/167 = 35.9 mA, so I_Z = 35.9 − 4.0 ≈ 32 mA. The extra 2 V drops across R_s. Scaling the output (6 × 12/10 = 7.2 V) is exactly what a regulator prevents.
What can it detect?
A photodiode is made from a semiconductor with a band gap of 2.8 eV. Which of these wavelengths can it detect?
- 6000 nm
- 600 nm
- 500 nm
- 400 nm
Answer
D. A photon is detected only if hν ≥ E_g. E (eV) = 1240/λ (nm): 6000 nm → 0.21 eV, 600 nm → 2.07 eV, 500 nm → 2.48 eV, 400 nm → 3.1 eV. Only 400 nm is above 2.8 eV. The longest wavelength it can detect is 1240/2.8 ≈ 443 nm. A longer wavelength means less energy per photon, not more.
Summary
- Zener regulator series resistor:
R_s = (V_in − V_Z)/(I_Z + I_L) - Photon energy E = hν = hc/λ:
E (eV) ≈ 1240/λ (nm) - Absorption or emission across the gap:
hν ≥ E_g ⇔ λ ≤ hc/E_g
- A Zener diode (both sides heavily doped, thin depletion layer) breaks down by internal field emission at V_Z. Used in reverse bias, it keeps the voltage across a load nearly constant.
- A photodiode is used in reverse bias: light with hν > E_g makes electron–hole pairs, and the reverse current grows with intensity.
- An LED is a heavily doped junction in forward bias; recombination emits photons of energy ≈ E_g. Visible LEDs need E_g ≳ 1.8 eV.
- A solar cell needs no bias: generation, separation and collection of pairs give a photovoltage. Its I–V curve is in the fourth quadrant, with V_oc and I_sc; E_g ≈ 1.5 eV is ideal.
- This topic is beyond the current NCERT syllabus but is useful for competitive exams.