First slide

Opto electronic juction devices

Question

A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA. When it operates with a 6 V battery through a limiting resistor R, the value of .R is

Moderate

A

400 $Ω$
B

4 k $Ω$
C

200 $Ω$
D

40 k $Ω$

Solution

The term LED is abbreviated as 'Light Emitting Diode'. It is forward biased p-n junction which emits spontaneous radiation. Current in the circuit= 10 mA = 10 x $10^{3}$ A and voltage in the circuit = 6 - 2= 4V
From Ohm's law, V = IR
$∴ R = \frac{V}{I} = \frac{4}{10 \times 10^{- 3}} = 400 Ω$

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