Forward and Reverse Bias: the Diode Curve

Lowering the barrier

Connect a battery across a junction with the p side to its positive terminal and the n side to its negative terminal. This is forward bias. The battery's voltage \(V\) across the junction opposes the built-in step \(V_0\). The barrier falls to \(V_0 - V\), and the depletion layer becomes thinner.

Recall the balance of lesson 3. The diffusion current depends very strongly on the height of the barrier, because only carriers with enough energy can climb it. Lower the hill a little, and many more carriers get over. The drift current is unchanged. So the diffusion current now wins, and a net current flows from p to n. It grows very rapidly as \(V\) approaches \(V_0\).

Try \(V = 0.4\), \(0.6\) and \(0.7\,\text{V}\) in the sim. The current is about 0.01 mA, then 0.55 mA, then 4 mA. Below about 0.6 V the current is too small to notice on the graph. Above about 0.7 V it shoots up. The barrier is nearly gone, and a large current of electrons and holes streams across the junction.

Raising the barrier

Now turn the battery round: p side to the negative terminal, n side to the positive. This is reverse bias. The applied voltage adds to the built-in step, so the barrier rises to \(V_0 + V\). The positive terminal pulls electrons away from the junction, and the negative terminal pulls holes away. More ions are uncovered on each side, so the depletion layer widens.

Almost no majority carrier can climb the higher barrier, so the diffusion current stops. What remains is the drift of minority carriers, swept across by the field. There are very few of them, so the current is tiny, about a microampere or less. It is called the reverse saturation current. It hardly changes with the reverse voltage, because the field already sweeps across every minority carrier that arrives.

The minority carriers are made by heat, so the reverse current grows quickly with temperature. It roughly doubles for every 10 K rise. Heating also lowers the barrier a little, by about 2 mV for each kelvin. With more carriers about, a smaller barrier is enough to balance diffusion. At a very large reverse voltage the junction finally breaks down, and the reverse current rises sharply. The next chapter takes up this breakdown and the Zener diode that uses it.

The diode and its curve

A p-n junction with a lead on each side is a junction diode. The lead to the p side is the anode and the lead to the n side is the cathode. Its symbol is a triangle pointing at a bar. The triangle points the way conventional current passes easily, from anode to cathode. The bar marks the cathode, the n side.

The graph of current against voltage in the sim is the diode's characteristic. In forward bias the current stays small until the knee voltage, and then climbs almost straight up. The knee is about 0.7 V for silicon and 0.3 V for germanium, the same as the barrier. In reverse bias the current is only microamperes, so the two halves of the graph need different scales. The forward half is in milliamperes and the reverse half in microamperes.

Beyond the knee a small change of voltage gives a large change of current. The ratio of the two is the dynamic resistance \(r_d\). It is the inverse of the slope of the curve at that point, not \(V/I\). Read two points from the sim: 11.0 mA at 0.75 V and 30.0 mA at 0.80 V. Then \(r_d = 0.05/0.019 = 2.6\,\Omega\). In reverse bias a change of many volts barely changes the current, so \(r_d\) there is millions of ohms.

Two simple models

A real characteristic is a curve, which is awkward in a circuit. Two simple models stand in for it. The ideal diode is a perfect switch. In forward bias it conducts with no voltage across it, like a closed switch. In reverse bias it passes no current at all, like an open switch.

The 0.7 V model is closer to a real silicon diode. It passes no current until the forward voltage reaches 0.7 V. Once it conducts, it keeps 0.7 V across itself, whatever the current. For germanium the drop is 0.3 V. In reverse bias it is an open switch, like the ideal diode.

A way to solve any diode circuit

Guess the state of each diode, on or off. Replace an "on" diode by its drop and an "off" diode by a gap. Solve the circuit with Ohm's law, \(V = IR\) for each resistor. Then check the guess. An "on" diode must carry current from anode to cathode. An "off" diode must have less than its drop across it in the forward direction. If a check fails, change that guess and solve again.

Now the torch of the opening makes sense. With the cells the right way, the LED is forward biased, and a current of some milliamperes lights it. Turned round, the LED is reverse biased. Only the reverse saturation current flows, a microampere or less, far too little to light it. The cells are not being used up either: they push almost no current round the loop.