Question

(6) A blue LED with a forward voltage drop of 3.5V is to be energized from a 12 volt battery. To light the blue LED to ful brightness requires a current of 20 mA through the LED. The blue LED has a peak inverse voltage rating of 5 V. i) Draw the circuit dingram of the arrangement you will use to light the LED to full brightness. (ii What is the value of the resistor you will use in your circuit? (ii) What power is dissipated in the blue LED when 20 mA passes through it iv) The blue LED is connected to a transformer that has a penk output voltage of 12 volts. The LED has a resistor of the same value that you calculated in (i) connected in series with it. However even though the transformer only has a peak voltage of 12 V (the same voltage as the 12 volt battery) the LED keeps blowing. Explain why. v) Explain, using a sketch of the current-vs-voltage plot for a pn junction diode, why you need a resistor in your circuit i.e. why can't you just connect the LED directly across the 12 volt battery?

Answer 1

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