The diode’s I-V curve
A diode is a P/N junction with two leads: the anode (P) and the cathode (N). Its current follows the Shockley equation. In reverse it passes almost nothing; in forward it passes almost nothing until the knee, then the current climbs steeply, roughly tenfold for every 60 mV more.
I = Is · (exp(V / (n·VT)) − 1) ΔV ≈ n·VT·ln(10) ≈ 60 mV per decade of current (n = 1)
Load-line analysis
In a circuit, the diode’s voltage and current are tied to the rest of the loop by Kirchhoff’s law: VS = I·R + VD. Plotted on the same axes, that is a straight line from (VS, 0) to (0, VS/R): the load line. The operating point Q is where it crosses the diode’s curve; both the diode and the circuit are satisfied there.
I = (VS − VD) / R
Three models, three levels of detail
Hand analysis rarely needs the full exponential. Pick the simplest model that answers the question:
- Ideal: a switch. On in forward with 0 V drop, off in reverse. Good for seeing which way current flows.
- Constant drop: 0.7 V once on (0.3 V for germanium). Good for most circuit calculations.
- Exponential (Shockley): the real curve. Needed for small signals, temperature effects and the exact operating point.
LEDs: the voltage is the colour
In an LED each electron crossing the junction gives up its energy as a photon. The photon’s energy is about the band gap, so the forward voltage tracks the colour: about 1.9 V for red, 2.2 V for green, 2.9 V for blue. Shorter wavelength, more energy per photon, higher voltage. An LED always needs a series resistor to set its current: R = (VS − Vf)/I.
E_photon = hc/λ ≈ 1240 / λ(nm) eV ≈ q·Vf R = (VS − Vf) / I_LED
Worked example (the simulator’s default values)
- Silicon diode, VS = 5 V, R = 430 Ω.
- Constant-drop model: I = (5 − 0.7)/430 = 10.0 mA.
- Exponential model: Q at VD = 0.700 V, I = 10.0 mA. The 0.7 V model is spot on here.
- Ideal model: I = 5/430 = 11.6 mA, 16% too high because it ignores the drop.
- Swap in a red LED: VD = 1.88 V, I = 7.25 mA. For exactly 10 mA you would need R = (5 − 1.9)/0.01 = 310 Ω.
Common mistakes
- Connecting an LED straight across a supply with no resistor. The exponential curve means the current, and the heat, run away.
- Using the ideal model when the supply is only a few volts. A 0.7 V drop out of 5 V is a 14% error.
- Reading reverse current as zero forever. It is tiny, but every diode breaks down at a high enough reverse voltage.
Common questions
How do you find a diode’s operating point?
Draw the load line I = (VS − V)/R on the diode’s I-V curve. The point where they cross gives the diode voltage and current, satisfying both the diode and the circuit.
Why is the diode voltage drop 0.7 V?
For a silicon diode at a few milliamps, the Shockley equation gives about 0.6–0.7 V. The voltage changes only about 60 mV for each tenfold change in current, so 0.7 V is a good fixed estimate.
How do you calculate the resistor for an LED?
R = (VS − Vf)/I. A red LED (Vf ≈ 1.9 V) at 10 mA from 5 V needs about 310 Ω; use the next standard value, 330 Ω.