The p-n junction and the diode

A one-way gate for current

A tap lets water through in both directions. A valve lets it through in one. A p-n junction is a valve for electric current. It is made from a single crystal that is p-type on one side and n-type on the other. The two sides are not glued. The junction is made inside the crystal during the manufacture.

What happens at the junction

On the n side there are many electrons and few holes. On the p side there are many holes and few electrons. At the moment the two sides meet:

1. Diffusion. Electrons move from the n side into the p side, because there are many more of them on the n side. Holes move the other way.
2. A layer without carriers. An electron that leaves the n side leaves a fixed positive donor ion. A hole that leaves the p side leaves a fixed negative acceptor ion. Near the junction, the diffused carriers recombine. A region with only fixed ions remains. This is the depletion layer. Its width is about $10^{-6}$ m.
3. A built-in field. The positive ions on the n side and the negative ions on the p side make an electric field that points from n to p. It pushes electrons back to the n side and holes back to the p side. So it opposes the diffusion.
4. A barrier. The field gives a barrier potential $V_0$ across the layer. For silicon it is about $0.7$ V and for germanium about $0.3$ V. A field of $10^6$ V/m across a layer of $0.5$ micrometre gives $0.5$ V.
5. Balance. The field also sweeps the few minority carriers across, as a drift current. In balance, the diffusion current of the majority carriers equals the drift current of the minority carriers, and the net current is zero.

Forward bias

Connect the battery with its positive end to the p side and its negative end to the n side. This is forward bias.

• The battery field opposes the built-in field. The barrier falls to $V_0 - V$.
• The depletion layer becomes narrower.
• Many more majority carriers can climb the lower barrier. The diffusion current grows very fast.
• Below about $0.7$ V (silicon) the current is small. Above it, the current rises steeply, to milliamperes and beyond, and the diode conducts.

Reverse bias

Connect the positive end to the n side. This is reverse bias.

• The battery field adds to the built-in field. The barrier rises to $V_0 + V$.
• The depletion layer becomes wider.
• The majority carriers cannot climb the barrier. Only the minority carriers drift across.
• The reverse saturation current $I_s$ is tiny, microamperes or less, and it hardly changes with the voltage. It grows fast with temperature, because minority carriers come from heat.

The current-voltage curve

For an ideal junction the current follows the diode equation:

Here $V_T = \dfrac{kT}{e}$ is the thermal voltage, $25.85$ mV at $300$ K. The ideality factor $\eta$ is between $1$ and $2$. For forward voltages above a few $V_T$, the term $-1$ is negligible and the current grows by a factor of $e^{V/\eta V_T}$. For a large negative $V$, the exponential goes to zero and $I \to -I_s$.

The curve is not a straight line, so a diode does not obey Ohm's law. Two resistances describe it:

• The static resistance is $V/I$ at a point.
• The dynamic resistance is the slope, $r_d = \dfrac{\Delta V}{\Delta I} = \dfrac{\eta V_T}{I}$. At $1$ mA it is about $26\ \Omega$, and it falls as the current rises.

Set the voltage to $0.6$ V, then to $0.7$ V. By what factor does the current change? Compare with the value of $e^{0.1/0.02585}$.

A diode in a circuit

In most circuits we do not use the full equation. We use a simple model.

• Ideal diode: forward, a closed switch with $0$ V across it. Reverse, an open switch.
• Constant drop model: forward, a switch with a fixed drop, $0.7$ V for silicon and $0.3$ V for germanium. Reverse, an open switch.

A worked example. A silicon diode is forward biased by a $5.0$ V battery through a resistor of $430\ \Omega$. Find the current. Then reverse the diode.

1. Forward. The diode takes $0.7$ V. The resistor gets $5.0\ \text{V} - 0.7\ \text{V} = 4.3$ V. So $I = \dfrac{4.3\ \text{V}}{430\ \Omega} = 10$ mA.
2. Reverse. The diode is an open switch. The current is close to zero, and the whole $5.0$ V appears across the diode.

What if it is germanium? The drop is $0.3$ V. Then $I = \dfrac{4.7\ \text{V}}{430\ \Omega} = 10.9$ mA. The current changes only a little, because the battery is much larger than the drop.

Practise with new numbers

Try it in Python

The program uses the diode equation for a silicon diode with $I_s = 1$ pA. It prints the current from $-1$ V to $0.7$ V. Then it finds the dynamic resistance near $0.6$ V from the slope.

Common mistakes

• Saying no current flows in the reverse direction. A tiny leakage current, $I_s$, flows, and it comes from minority carriers.
• Saying the depletion layer holds free electrons. It is a layer with only fixed ions, because the free carriers have left or recombined.
• Using Ohm's law for a diode. The curve is not a straight line. Use the diode equation, or the constant drop model.
• Leaving out the diode drop. In a silicon circuit, subtract $0.7$ V before you divide by the resistor.
• Mixing up the bias direction. Forward means the p side is joined to the positive end. Think of the arrow of the symbol: current goes with the arrow.
• Believing the junction is two pieces joined. It is one crystal with two doping types.

Remember these

• A p-n junction has a depletion layer of fixed ions and a barrier potential: about $0.7$ V (Si) and $0.3$ V (Ge).
• Forward bias lowers the barrier and narrows the layer. The current grows exponentially. Reverse bias raises the barrier and widens the layer. Only a tiny $I_s$ flows.
• $I = I_s\left(e^{V/\eta V_T} - 1\right)$, with $V_T = 25.85$ mV at $300$ K. The dynamic resistance is $r_d = \dfrac{\eta V_T}{I}$.
• In a circuit, a conducting silicon diode drops about $0.7$ V.

References

• NCERT Physics Class 12, Semiconductor Electronics, the sections on the p-n junction and the semiconductor diode.
• OpenStax University Physics Volume 3, the chapter Condensed Matter Physics, on semiconductor devices.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In which direction does the built-in field of an unbiased p-n junction point, and what does it do to the majority carriers?

2. A silicon diode in series with a resistor of $1.0\ \text{k}\Omega$ is forward biased by $9.0$ V. What is the current?

3. What happens to the width of the depletion layer when the reverse voltage is raised?

4. The reverse current of a germanium diode is much larger than that of a silicon diode. Which has the smaller gap, and how does this fit?