Question 5.15: The circuit of Fig. 5-24 shows an LED indicator for the ac p...
The circuit of Fig. 5-24 shows an LED indicator for the ac power line. The idea is basically the same as in Fig. 5-23, except that we use a capacitor instead of a resistor. If the capacitance is 0.68 μF, what is the average LED current?
Calculate the capacitive reactance:
X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi (60 \ Hz)(0.68 \ \mu F)}=3.9 \ k\OmegaIgnoring the LED voltage drop, the approximate peak LED current is:
I_S=\frac{170 \ V}{3.9 \ k\Omega}=43.6 \ mAThe average LED current is:
I_S=\frac{43.6 \ mA}{\pi }=13.9 \ mAWhat advantage does a series capacitor have over a series resistor? Since
the voltage and current in a capacitor are 90° out of phase, there is no power dissipation in the capacitor. If a 3.9-kΩ resistor were used instead of a capacitor, it would have a power dissipation of approximately 3.69 W. Most designers would prefer to use a capacitor, since it’s smaller and ideally produces no heat.