Diodes & LEDs
A diode is a one-way valve for current. It conducts from anode to cathode and blocks the reverse direction. Conduction isn’t free: a silicon diode drops about 0.7 V while conducting, and that drop stays roughly constant over a wide range of current. Schottky diodes drop less, around 0.3 V, with some trade-offs, which makes them common in power paths.
Two jobs dominate. Rectification turns AC into DC; a bridge of four diodes flips the negative half of the waveform upward and a capacitor smooths the result. Protection uses the blocking direction. Put a diode in series with a power input and a reversed battery simply doesn’t conduct, instead of frying the board.
An LED is a diode that emits light when it conducts. Its forward drop depends on color, roughly 2 V for red and 3 V or more for blue and white. An LED does nothing to limit its own current, so it always needs a series resistor: R = (Vsupply − Vf) / I. From 5 V, a red LED at 10 mA wants about 330 Ω. Wire an LED straight across a supply and it dies in a flash.
Key points
- Diodes conduct one way and block the other. Silicon drops about 0.7 V while conducting.
- The two headline uses are rectification (AC to DC) and reverse-polarity protection.
- LED forward voltage varies by color: about 2 V for red, 3 V and up for blue and white.
- LEDs never limit their own current. Size the series resistor with R = (Vsupply − Vf) / I.
Practice
0 of 3 answered · Not startedWrong answers just let you try again, and hints are there if you want them. Answering every question first time, without hints, is what earns mastery.
- 1
Why does a Schottky diode often replace a standard silicon diode in a power path?
- 2
What is the resistor in series with an LED actually doing?
- 3
An LED with a 2 V forward voltage should run at 15 mA from a 5 V supply. What series resistor do you need?
Ω
