Introduction to Electronics
Senior 6 · Physics
Introduction to Electronics
Lesson 2 of 5
25 min
A small solar lantern has three jobs: collect sunlight by day, store the energy in a battery, and give light at night. It does all three with the same structure, a junction between p-type and n-type semiconductor. In the solar panel, light falling on junctions drives a current. A diode between the panel and the battery lets charge flow into the battery by day but stops the battery from discharging back through the panel at night. And the light itself comes from LEDs, junctions that turn current back into light. This lesson explains how one junction can do all of these.
Suppose one part of a silicon crystal is doped p-type and the neighbouring part n-type. Near the boundary, electrons from the n side diffuse across into the p side, and holes diffuse the other way; where they meet they recombine (an electron fills a hole) and both carriers disappear. Left behind is a thin region, about a micrometre wide, with no mobile carriers: the depletion layer. It is not neutral. It contains fixed positive donor ions on the n side and fixed negative acceptor ions on the p side.
Figure 2.1: The p–n junction in equilibrium.
These fixed charges create an electric field from the n side to the p side, which pushes back any more electrons trying to cross into the p region (and holes trying to cross into the n region). Diffusion stops when this field is strong enough. The potential difference across the depletion layer, the barrier potential, is about 0.7 V for silicon and 0.3 V for germanium. It cannot be measured with a voltmeter, because contact potentials at the metal connections cancel it.
Definition
A p–n junction diode is a single crystal with a p-type region (the anode) and an n-type region (the cathode). The depletion layer is the region near the junction emptied of mobile carriers, containing only fixed ions; the p.d. across it is the barrier potential (about 0.7 V for silicon).
Figure 2.2: Diode symbols (IEC 60617). On a real diode a band marks the cathode.
Figure 2.3: The current–voltage characteristic of a silicon diode (not to scale: the reverse current is far smaller than shown).
The forward current of an ideal junction follows the Shockley equation, , where is the tiny reverse saturation current. Because at room temperature, the current is multiplied by about ten for every extra 60 mV, which is why the curve rises so suddenly and why a diode in a circuit is often treated as "a fixed 0.7 V drop when conducting, an open switch when not".
Worked example 1: the steepness of the forward curve
A silicon diode at 300 K carries 2.0 mA at a forward p.d. of 0.600 V. Using the Shockley equation (with ), find (a) the current at 0.660 V, (b) the forward p.d. needed for 20 mA. Take .
Solution
(a) , so .
(b) From (a), 20 mA needs about 0.66 V. Exactly: .
Conclusion: a tenfold change in current needs only 60 mV. Over a wide range of currents a conducting silicon diode drops roughly 0.6 V to 0.7 V.
Figure 2.4: Circuit for measuring a diode characteristic. The resistor limits the current if the supply is turned up too far.
Increase the supply in small steps from zero and record the diode p.d. and the current, taking closer readings once the current starts to rise. Then reverse the diode and repeat (a microammeter is needed for the reverse current). Plot against . Replacing the diode with LEDs of different colours gives a second experiment: note the p.d. at which each LED just begins to glow (viewed through a tube in a dim room). An electron crossing the junction gains energy , which is roughly the energy of the photon it produces.
In a forward-biased junction, electrons crossing into the p side fall into holes. The energy released is about the band gap . In silicon it becomes heat (vibrations of the lattice), but in compound semiconductors such as gallium arsenide phosphide (red) or indium gallium nitride (blue and green) it is emitted as a photon of energy . That is a light-emitting diode (LED). The colour is set by the material, and a larger band gap gives shorter wavelengths and a larger forward voltage: about 1.8 V to 2.0 V for red, 3 V or more for blue. White LEDs use a blue LED coated with a phosphor that turns part of the blue into yellow. An LED needs far less power than a filament lamp for the same light, and lasts much longer.
An LED must always have a current-limiting resistor in series: because of the steep characteristic, connecting it straight across a supply even slightly above its forward voltage would let a destructive current flow.
Worked example 2: choosing the resistor for an LED
A red LED has a forward p.d. of 2.0 V and should carry 15 mA. It is to be run from a 5.0 V supply. Find (a) the series resistance, (b) the power dissipated in the resistor and in the LED.
Solution
(a) The resistor takes the rest of the supply: . (The nearest standard value, 220 Ω, would give a slightly smaller, safe current: .)
(b) ; . A 0.25 W resistor is ample.
Worked example 3: Planck's constant from LEDs
In Experiment 2.1 the following were found (illustrative readings).
| LED colour | red | yellow | green | blue |
|---|---|---|---|---|
| wavelength λ / nm | 630 | 590 | 525 | 470 |
| / V | 1.88 | 2.02 | 2.28 | 2.55 |
| / | 1.587 | 1.695 | 1.905 | 2.128 |
Since , a graph of against should be a straight line of gradient . Estimate using the red and blue points.
Solution
Gradient .
, close to the accepted .
Limitations: "just glowing" is a judgement, each LED emits a spread of wavelengths, and the junction p.d. is not exactly . Using the gradient of several points, rather than one LED, cancels a constant offset.

A Zener diode is heavily doped so that its depletion layer is very thin and its reverse breakdown happens sharply at a chosen, stable p.d., the Zener voltage (from about 2 V to over 100 V). In breakdown, the p.d. across it stays almost exactly over a wide range of currents. It is used reverse-biased, with a series resistor, to hold a voltage steady: a voltage regulator (Figure 2.5). If the input rises, the extra p.d. appears across the series resistor and the Zener simply takes more current; if the load draws more current, the Zener takes less. Either way .
Figure 2.5: A Zener voltage regulator.
Definition
A Zener diode is a heavily doped p–n diode designed to operate in reverse breakdown, where its p.d. stays close to a fixed value over a wide range of current. Example: a 5.1 V Zener holds the supply to a circuit at 5.1 V while the battery feeding it falls from 12 V to 8 V.
Worked example 4: designing the regulator
In Figure 2.5, find (a) the current in , the load current and the Zener current, (b) the power dissipated in the Zener, (c) the Zener current when the input falls to 9.0 V, and (d) the lowest input p.d. at which the output stays at 5.1 V.
Solution
(a) ; ; by Kirchhoff's first law .
(b) (81 mW).
(c) ; : still positive, so the output is still 5.1 V.
(d) Regulation fails when falls to zero, i.e. when : .
Warning
Common mistakes. (1) Connecting a Zener regulator the wrong way round: forward biased, it just drops about 0.7 V. (2) Forgetting the series resistor with an LED or a Zener. (3) Saying the depletion layer contains no charge: it has no mobile charge, but it holds fixed ions, which create the barrier. (4) Saying a diode has "no resistance" in forward bias: it drops about 0.7 V, and its p.d. depends (slightly) on the current.
Tip
The arrow in the diode symbol is the way the current is allowed to go. Read it as "anode to cathode: Allowed to Keep flowing". In any diode circuit, first decide which diodes are forward biased; replace each by a 0.7 V drop, each reverse-biased one by a gap (and each Zener in breakdown by ), then use Kirchhoff's laws.
How does a diode work: the PN junction (with animation), by CircuitBread. Watch the depletion layer form, then widen in reverse bias and narrow in forward bias.
Practice
Key points
Sources and credits
The five diagrams, the experiment readings (illustrative), the worked examples and the practice exercises are written by Ganzaa. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.