A diode conducts mainly in one direction. In forward bias the junction allows current once the voltage exceeds a device-specific threshold; in reverse bias it blocks, up to its rated reverse voltage. That asymmetry makes diodes useful for rectification, reverse-polarity protection and freewheeling.
Forward voltage and current limiting
The forward voltage of a diode is a property of the device and its current, not a universal constant: a silicon rectifier, a Schottky diode and an LED of a given colour each have their own value. An LED must always have its current limited by a resistor or a current source.
LED series resistor
Problem. A 12 V supply drives an LED with a forward voltage of 2.0 V at a target current of 15 mA. Calculate the resistor and check the current with the nearest common value, 680 Ω.
Solution. R = (12 − 2.0) ÷ 0.015 = 10 ÷ 0.015 = 667 Ω. With 680 Ω, I = 10 ÷ 680 = 0.0147 A = 14.7 mA, close to the target.
The transistor as a switch
A bipolar junction transistor is controlled by base current; a MOSFET is controlled by gate-source voltage. In switching service the control signal drives the device fully on or fully off, so it dissipates little power. The load current flows through the output path and must be within the device ratings.
- A diode drops a device-specific forward voltage when conducting.
- Reverse voltage rating must exceed the worst-case reverse voltage in the circuit.
- An LED needs a current-limiting element; never connect it directly across a voltage source.
- A switching transistor needs a base or gate drive that guarantees full enhancement.
- Inductive loads need a flyback path to protect the switching device.
- A diode conducts in forward bias and blocks in reverse bias.
- Forward voltage is device-specific, not a universal number.
- An LED always needs current limiting.
- A transistor switches a load using a small control signal.
Worked example
For a 5 V supply, an LED with 2.0 V forward voltage at 10 mA needs R = (5 − 2.0) ÷ 0.010 = 300 Ω. A 330 Ω resistor gives I = 3.0 ÷ 330 = 9.1 mA.