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Electronic Control Systems

Output devices and LEDs

  • 3 things to know
  • AS 1 Paper 2

Overview

LED series resistors with worked values, LEDs against lamps, current calculations for heaters and solenoids, relays and latching relays, solenoid valves, buzzers and piezo sounders.

What you need to know

Answer each one before you reveal it.

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Key term

What do COM, NO and NC mean on a relay's contacts?

Key term

Why must an LED always have a series resistor?

Numbers

Calculate the series resistor for an LED on 9 V, with a forward voltage of 2 V and a maximum current of 20 mA.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why must an LED always have a series resistor?

    Model answer

    An LED has very little resistance once it conducts, so the current would rise until the LED failed. The series resistor limits the current to a safe value.

  2. 2

    Calculate the series resistor for an LED on 9 V, with a forward voltage of 2 V and a maximum current of 20 mA.

    Model answer

    The minimum resistance is 350 Ω, so the next preferred value up, 390 Ω, is chosen. R = (9 − 2) ÷ 0.02 = 350 Ω.

  3. 3

    Calculate the power dissipated in the series resistor of an LED on 9 V, with a 2 V forward voltage and 20 mA flowing.

    Model answer

    The resistor dissipates 0.14 W, which is within a 0.25 W rating. P = V × I = (9 − 2) × 0.02 = 0.14 W.

  4. 4

    Why is the next higher preferred value chosen for an LED resistor, not the next lower?

    Model answer

    A higher resistance keeps the current just below the LED's maximum. A lower value would push the current above the maximum rating.

  5. 5

    Compare an LED with a filament lamp as an indicator.

    Model answer

    An LED draws far less current than a filament lamp and lasts much longer. A filament lamp wastes most of its energy as heat and has a fragile filament.

  6. 6

    A 12 V heater is rated at 36 W. What current does it draw?

    Model answer

    The heater draws a current of 3 A. I = P ÷ V = 36 ÷ 12 = 3 A.

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The output voltage is V.
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AThe total resistance is 2 kΩ.
BThe parallel pair gives (2 × 2) ÷ (2 + 2) = 1 kΩ.
CThe series resistor adds another 1 kΩ.

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