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Diode and LED

Curricula: KS3, GCSE, Sek I, cycle 4

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Aim

Confirm that a diode passes current in one direction only, and connect an LED with a current-limiting resistor.

The bench

At the top are two 5 V circuits with 4.5 V / 1 W lamps: on the left the diode is connected in the forward direction, on the right in the reverse direction. At the bottom are a 5 V source, a 220 Ω resistor and an LED.

Procedure

  1. Compare the lamps in the two upper circuits.
  2. Look at the LED.
  3. In the settings of the LED's resistor change 220 Ω to 1 kΩ, then to 100 Ω.
Expected result

With the diode connected forward the lamp lights; with it reversed it does not: a diode passes current only from anode to cathode. The LED lights; the current through it is about 14 mA: about 1.8 V across the LED and the remaining 3.2 V across the resistor. With 1 kΩ the LED dims; with 100 Ω it becomes brighter — the current is about 31 mA. For an ordinary 5 mm LED the working current is 20 mA and the maximum is 25–30 mA: 31 mA is already at the limit. The resistor is what limits the current: without it only the source and the LED itself would limit the current, and a real LED would burn out at once.

Questions

  1. How can you tell from the circuit diagram which direction the diode passes current?
  2. What resistor does a 2 V, 20 mA LED need with a 9 V supply?
Answers
  1. The diode symbol is a triangle pointing at a bar. Current passes in the direction the triangle points, from the anode on the triangle side to the cathode at the bar.
  2. The resistor must take the remaining 9 − 2 = 7 V at 20 mA: \(R = (U - U_{LED}) / I\) = 7 V / 0.02 A = 350 Ω. The nearest standard value above it, 390 Ω, gives about 18 mA.