14-02-2014, 05:54 PM
Remember that a diode dropper halves the power, not the voltage. Hence for 240V mains the effective voltage is about 170V, not 120V. The dropper resistor can then be calculated using ohm's law in the usual way. If the heater chain is 112V and the current is 200mA this will be 240R and will dissipate 9.6W.
Modern recitifiers are probably more reliable than a BY126 etc. The crowbar method is simple and effective providing the fusing is right and you put the diode before the dropper resistor. 2 diodes in series is another cheap and simple belt and braces approach.
Calculating a capacitor dropper is a little more complicated but not difficult once you have realised that the reactance and resistance add pythagorus style. If I'm doing it without a spreadsheet here's a simple way:
Total mains voltage/total current = required impedance
Chain voltage/chain current = chain resistance
Cap reactance = SQRT (Impedance ^2 - Resistance ^2)
Cap value = 1/(2 * PI * f * reactance) = 1/(314 * reactance)
Modern recitifiers are probably more reliable than a BY126 etc. The crowbar method is simple and effective providing the fusing is right and you put the diode before the dropper resistor. 2 diodes in series is another cheap and simple belt and braces approach.
Calculating a capacitor dropper is a little more complicated but not difficult once you have realised that the reactance and resistance add pythagorus style. If I'm doing it without a spreadsheet here's a simple way:
Total mains voltage/total current = required impedance
Chain voltage/chain current = chain resistance
Cap reactance = SQRT (Impedance ^2 - Resistance ^2)
Cap value = 1/(2 * PI * f * reactance) = 1/(314 * reactance)
www.borinsky.co.uk Jeffrey Borinsky www.becg.tv







