How to read a diode I–V curve

What the curve shows

The horizontal axis is the voltage across the diode and the vertical axis is the current through it. Positive voltage means forward bias (anode more positive than cathode) and negative voltage means reverse bias.

Unlike a resistor, whose I–V plot is a straight line through the origin, a diode's curve is strongly nonlinear. It is easiest to read in three zones: the forward region on the right, the reverse region on the left, and the breakdown region far to the left. The point where the forward current starts to rise noticeably is called the knee.

To follow along, open the PN diode simulator. Its default device is the 1N4148 preset, with I_s = 4 nA, n = 1.9, a 0.2 A current rating and a 100 V reverse rating, and the applied voltage starts at 0.7 V.

The Shockley equation

The ideal diode equation, named after William Shockley, is I = I_s × (e^(V/(n × V_T)) − 1).

• I_s is the saturation current, the tiny reverse current of an ideal junction. In the 1N4148 preset it is 4 nA.
• n is the ideality factor, between about 1 and 2 for real junctions. The preset uses 1.9.
• V_T is the thermal voltage, kT/q, about 25.85 mV at room temperature, the value Curvo Studio uses.
• V is the voltage across the junction.

Two limits are worth remembering. For negative V of a few V_T or more, the exponential vanishes and I ≈ −I_s. For positive V above about 0.1 V, the −1 can be ignored and I ≈ I_s × e^(V/(n × V_T)), so the current grows exponentially with voltage.

The forward region and the knee voltage

In the forward region the current grows by a constant factor for each equal step in voltage. With the 1N4148 preset, the Shockley equation gives about 0.11 mA at 0.5 V, 0.81 mA at 0.6 V, 6.2 mA at 0.7 V and 47 mA at 0.8 V. Each extra 0.1 V multiplies the current by about 7.7.

The same fact can be stated per decade: the voltage needed for a tenfold increase in current is n × V_T × ln(10), which is about 113 mV for n = 1.9.

Because the exponential is smooth, there is no single exact knee. Textbooks define it by convention, usually as the voltage where the diode carries a noticeable current. For this preset the current reaches 1 mA at about 0.61 V and 10 mA at about 0.72 V. This is why silicon diodes are said to drop roughly 0.6–0.7 V when conducting. Schottky diodes and other types have a lower knee, and the exact value depends on I_s, n, the current and the temperature.

The practical consequence is that a diode in forward conduction holds its voltage almost constant while the current changes a great deal, so something else in the circuit, usually a resistor, must limit the current.

Reverse leakage

With reverse bias, the equation gives a current of about −I_s, which is independent of the voltage. For the preset this is −4 nA, many orders of magnitude below the forward currents above, so the diode behaves almost like an open circuit.

Real diodes leak more than the ideal equation predicts, and the leakage rises sharply with temperature. Curvo Studio keeps the simple version, a constant −I_s in the reverse region, so you can see the idea without those extra effects.

Breakdown

If the reverse voltage grows large enough, the diode enters breakdown and the reverse current rises steeply. Ordinary rectifier and signal diodes are not meant to operate there, and the rated reverse voltage is a limit to stay under.

In the simulator the breakdown voltage is a parameter, 100 V for the 1N4148 preset. At exactly that voltage the model gives −8 mA, 4% of the rated current, and past it the reverse current rises quickly, for example to about −108 mA at 104 V. The current display is clipped at 1.5 times the rated current, and the device is flagged as overloaded when the current exceeds the 0.2 A rating.

The shape is a lesson model: a piecewise rule rather than a physical avalanche or Zener-effect calculation. Zener diodes are designed to operate in this region, which is the subject of the Zener regulator guide.

Try it in the simulator

Open the PN diode simulator, choose the 1N4148 preset and sweep the applied voltage.

• Set 0.5, 0.6, 0.7 and 0.8 V and note the current each time. You should see roughly 0.11 mA, 0.81 mA, 6.2 mA and 47 mA.
• Set a negative voltage such as −10 V. The current stays at about −4 nA.
• Set −100 V and below to see the breakdown region.
• Switch to the 1N4001 preset (I_s = 5 nA, n = 2, 1 A, 50 V) and compare where its forward current becomes noticeable.

These values come from the same equations described above, so you can also calculate them by hand and check the result.