Simple LED Circuit

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For this exercise we want to create one of the simplest circuits possible - a simple red LED driven by a battery:

Led1 p1.svg

Since the voltage drop over the LED is more than the standard battery voltage of 1.5 V we will be using 2 batteries in series, providing a total voltage of Vtot = 1.5 V + 1.5 V = 3 V. Unfortunately 3 V is higher voltage than the LED will tolerate. If we examine the LED datasheet (a typical one is linked at the bottom of this document),

Led data.png

we will learn that the voltage drop over a red LED (of this particular type) is {{math|2.1 V} typically. In other words, we will need a voltage drop over the resistor R1 of UR1 = 3 V - 2.1 V = 0.9 V.

We also learn that the LED will operate optimally at a current of 20 mA. Using Ohm's law, we can now calculate the value of the resistor R1:

R1 = {{0.9 V \over 20 mA}} = {{0.9 V \over 0.02 A}} = 45 \Omega.

While 45 Ω is available, picking a 47 Ω resistor from the E12 series is more readily available.


The following is a breadboard setup of the above circuit. Two 1.5 V AA-cells are used to supply a 3 V voltage, a 47 Ω resistor limit the current in the circuit, a red LED and 3 multimeters are added to show what goes on.

As can be seen, the voltage drop over the resistor is slightly higher than our calculation, but that is because the resistance is slightly above our calculated "ideal" value.

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