# Simple power Supply for LEDS

Posted on Jul 18, 2012

LED's are great display tools. Their prices have decreased to a point where they are replacing more conventional light sources. In one sense their characteristic need for low voltages is an advantage e.g compatibility with I.C drives, but this voltage requirement can also be a disadvantage. I wanted to light some LED's in a simple sign application, nothing fancy, but I needed to decrease the utility line voltage to a LED-compatible value for this application.

The line voltage can be reduced in various ways, via dropping resistor, a transformer, a "brick" power supply, a capacitor etc. A dropping resistor wastes considerable power and generates heat. A "brick" power supply must be adapted to the particular design needs, i.e a specific voltage and current and also may have an expense associated with it that is undesirable for a simple circuit. A transformer can be bulky, weigh a lot and may be expensive. capacitor can act as a voltage dropping component and in this instance, was chosen to light the static LED display shown below. The considerable drop in voltage from the line (117 v.a.c) down to approximately 2-1/2 volts a.c results in an almost negligible power loss in the capacitor(s). It is simple, low cost and small. The capacitor value may be computed via the following empirical formula. This formula is valid for the circuit shown in Fig.2: Where "C" is in uFd and "I" is in miliamperes. A design current of 10 mA per LED was selected. Thus for ten LED's, 100mA is the total current. Inserting this current value into the formula, yields a capacitor value of 3.1uFd. To achieve this value, two capacitors connected in parallel were used. Recall that the total capacitance of capacitors connected in parallel is simply the arithmetic sum of the individual capacitors:

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