LED circuit

Let’s say we want to run a light-emitting diode (LED) using a 9 V battery as the power supply. To do that a current limiting resistor is necessary unless the immediate destruction of the LED is the main aim. OK, a resistor. But with what amount of resistance?

First, we need to know the characteristics of the LED, which can be looked up in the datasheet: Forward voltage = 1.8 V, maximum current = 10 mA

Now two principles are at work here: The notorious law of Ohm and something known as KVL: Kirchhoff’s voltage law (aka Kirchhoff’s second law). It states that the sum of all potential differences (i.e. voltages) in a circuit equals zero. Voltages are to be observed across individual components within a circuit. In the example above these components are the battery (9 V), the LED (with its forward voltage or voltage drop of 1.8 V) and the resistor. (For the purpose at hand we can safely neglect the real but in fact comparatively very low resistances of the wires.) Therefore, the resistor has to deal with the rest of the voltage: 9–1.8 = 7.2 V. Kirchhoff at work.

The rest is easy. Just plug the now known values into Ohm’s law:

R = V/I = 7.2/0.01 = 720 Ω

This means that a resistor of 720 Ω will be able to make sure that the LED gets exactly what it wants (and what it is capable of handling, LEDs like to operate right at their limits): 1.8 V and 0.01 A (or 10 mA).

As in the diagram above, the resistor is commonly placed before the LED. But this is not an unalterable necessity. Swapping the two components has absolutely no effect on the functionality of the circuit.

Would it be possible to dispense with the resistor altogether by running the LED with a voltage equal to its forward voltage drop?

The answer is yes. Try it!

But the margins of error are quite small. Only a modest increase in the voltage of the power supply will simply blow out the LED. It is interesting to note that the LED might have a certain resistance but unlike a true resistor, the LED’s resistance exhibits a peculiar behavior as it decreases with rising voltage.

To make things clearer an experiment is warranted: Measure the current at increasing voltage settings in a circuit with only a power source and our little friend, the LED, and without its guardian resistor.

1.8 V: R = V/I = 1.8/0.01 =180 Ω

1.82 V: R = 1.82/0.0123 = 148 Ω

1.84 V: R = 1.84/0.015 = 122.7 Ω

1.86 V: R = 1.86/0.0185 = 100.5 Ω

1.88 V: Now the LED is positively dead!

This is experimental proof that an LED is a nonohmic device.

On the other hand, with a power supply that definitely does not exceed its prescribed voltage — a battery would fulfil this criterion quite well — the danger of burning the LED to cinders is practically non-existent.