36 LED String on 240Volt AC
36*3.2= 115 Volts
240*√2= 336 volts (peak value)
240 input voltage produce 336 volt peak value but we can use RMS value to select correct resistor.
with this approximation:
use one diode bridge rectifier to 240 volt and use one resistor about (240-115)/20mA= 6K = 6.2 Kohm in input path.
resistor power dissipation will be 125*125/6.2K=2.5Watts
(use minimum 5W for resistor)
also you can use one resistor and one capacitor (in series) in place of resistor.
in this case power dissipation in resistor is lower.
Important note:
this formulas are not precise for this application,
because voltage on resistor isnt sinosoidal, but
for first try we can use these.
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