Figure 1.  A circuit for driving a chain of LED's directly from the AC line with a half-wave rectifier

Figure 1 shows a circuit for driving a chain of white LED's directly from the AC line without the need for a transformer-based DC power supply.  This is the half-wave version of a similar circuit with full-wave rectification.

This circuit simply connects enough LED's in series with a current-limiting resistor so that the high AC voltage is evenly distributed across the LED's and resistor at safe levels.  Note that this must be done only if the cut-in voltage of the LED's to be used is known.

The peak voltage of 120 VAC is around 170 V.  Assuming that the cut-in voltage of each white LED is 3 V, the 25 LED's in series will only light up when the voltage across them is 75 V.  This means that the voltage drop across the resistor at that point is about 95 V.

If the maximum current rating for each LED is 25 mA, then the resistor must have a value of R = V / I = 95 V divided by 25 mA =  3.8 kilo-ohms. The power rating of the resistor should be P = VI = 95 V x 0.025 A = 2.375 W.  A regular 5-W ceramic resistor is recommended for this purpose.

Note that the values above assume that the LED's are conducting at peak current 100% of the time, which is not the case. With a duty cycle that's half of what it is for the similar full-wave circuit, the actual resistor value to be used may be reduced to 1/2 this value, or about 1.9 kilo-ohms.  In fact, an even lower value than this may be used, since the LED's are actually conducting only when the voltage across them is 75 V, which only happens 71% of the time during each cycle.

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