modified 7 years ago

Make Electronics-Exp 4e

01:19:10

Here we have a LED with specs 2V&20mA. Our power supply is 9V and we want to operate at the optimum 20mA in the circuit so that the LED drops the expected 2V at this level of current. We want to add a resistor and calculate its value so that we operate at 20mA in the circuit with the 2V&20mA LED in the circuit. Since the LED will drop 2V, the potential difference of the resistor will have to be 9-2=7V. So, what resistance will operate at 7V and generate 20mA? Simple using Ohm's law. R=E/I=7/0.02=350 Ohm. As mentioned in previous experiments, if we do not operate at 20mA as per the LED's datasheet, then the LED's voltage drop will be slightly different than 2V. You can play around with the Resistor's value and observe the impact.

published 7 years ago

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