Figure 1.  A circuit for driving a chain of LED's directly from the AC line with a full-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. The full-wave rectifier of this circuit allows the LED's to light up in both AC directions. See also: a similar circuit with half-wave rectification.

There is no rocket science in this circuit - it simply involves connecting 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, so the bridge rectifier to be used to turn the AC current into DC must be able to handle this large voltage, hence the use of a 200-V bridge in the circuit.  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 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 may be used for this purpose.

Note that a higher peak current can be tolerated by the LED's because they are not conducting during periods when the AC voltage level across them is less than 75 V.  As such, correspondingly lower resistor values (~2.5 kilo-ohms) may be used in this circuit.

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