Question 5.14: LEDs are often used to indicate the existence of ac voltages...
LEDs are often used to indicate the existence of ac voltages. Figure 5-23 shows an ac voltage source driving an LED indicator. When there is ac voltage, there is LED current on the positive half-cycles. On the negative half-cycles, the rectifier diode turns on and protects the LED from too much reverse voltage. If the ac source voltage is 20 V_{rms} and the series resistance is 680 Ω, what is the average LED current? Also, calculate the approximate power dissipation in the series resistor.
The LED current is a rectified half-wave signal. The peak source voltage is 1.414 × 20 V, which is approximately 28 V. Ignoring the LED voltage drop, the approximate peak current is:
I_S=\frac{28 \ V}{680 \ \Omega }=41.2 \ mA
The average of the half-wave current through the LED is:
I_S=\frac{41.2 \ mA}{\pi }=13.1 \ mA
Ignore the diode drops in Fig. 5-23; this is equivalent to saying that there is a short to ground on the right end of the series resistor. Then the power dissipation in the series resistor equals the square of the source voltage divided by the resistance:
P=\frac{(20 \ V)^2}{680 \ \Omega }=0.588 \ W
As the source voltage in Fig. 5-23 increases, the power dissipation in the series resistor may increase to several watts. This is a disadvantage because a highwattage resistor is too bulky and wasteful for most applications.