Question

5. A light emitting diode (LED) is a diode that lights up when a current flows across it. The symbol for an LED is shown below (sometimes wiggly arrows are used to represent the light instead.) For this question we will assume we are working with LEDs that have a turn-on voltage of VoN-1V and a reverse breakdown voltage of VBR-8V a. You hook up a single LED to a power supply that can put out vol tage from -10 V to +10 V and the forward direction of the diode points from positive to negative voltage. Make a sketch of this crcuit. Draw an IV curve of the diode over the entire voltage range and label when the light goes on b. Now you reverse the direction of the LED from part a). Make another circuit diagram and draw an IV curve of the diode, again labeing when the light goes on c. You now use two diodes and hook them up as shown below. One end of the configuration is hooked up to the positive output of the power supply and the other is hooked up to the negative end of the power supply. Draw an IV curve of the circuit and label when a light will be on For all of your circuits /graphs, assume the same polarity of the power supply. d. Do you ever have to worry about either diode breaking down due to too large of a reverse voltage? Explain your answer

Answer 1

Before solving this problem you have the following knowledge about LED:

An LED is not able to withstand a reverse bias of more than a few volts. Most can survive a reverse bias of -8 V, but no more. This is quite different from most diodes, which can withstand reverse bias of hundreds of volts.

CURRENT THROUGH A DIODE

When forward biased:

1. No current flows when the voltage is less than 1 Volts
2. When the voltage is a little over 1 V, a small current flows. With higher voltage, a large current flows.

When reverse biased:

1. No current flows.
2. No diode is perfect, so there is always a small leakage current. This is only a few nanoamps, which is too small to measure easily.

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