I'm probably re-inventing the wheel, but here are a few variations of a mains-powered night-light for 230-250V, adaptable for 110-120V.
The component values given are for 240V 50Hz operation with an LED current of ~ 7mA. I've found that quite sufficient for a night-light for a child's bedroom.
If you do need a brighter light the current can be increased in proportion to the value of the capacitor C (e.g. double the capacitance gives 14mA), or additional LEDs (or LED inverse-parallel pairs in the case of the second circuit) can be connected in series with the LED(s) shown. If C is increased then a 1/2W resistor should be used for R.
Circuit operation
The fuse protects the circuit and its value must not be increased.
The reactance of C behaves like a 'resistor' to drop the voltage. For 240V 50Hz mains the RMS current through it will be ~ 7mA (70mA per microfarad). When first switching on the light the current may be much higher but is limited by R to protect the LED(s) and capacitor.
In the first circuit, a bridge rectifier feeds the LED so that it conducts on both halves of the mains cycle. This doubles the light output compared to half-wave rectification and also reduces flicker.
In the second circuit, one LED conducts on one half-cycle and the other LED conducts on the other half cycle. Each LED protects the other from reverse voltage greater than ~3V.
The only 'wasted' power for the circuits shown is ~ 49mW in R.
Adaptation for 110/120V operation
Double the capacitance of C and use a 1/2W resistor for R. Assuming a 60Hz supply the current will be ~8mA
CAUTION
All components are potentially at mains voltage and could give a fatal shock unless care is taken to shield them from touch. They must be housed in a fire-resistant insulated enclosure. For added safety a 1 megohm bleed resistor should be connected across the capacitor to discharge it when the mains is switched off.
DO NOT BUILD EITHER CIRCUIT UNLESS YOU FULLY UNDERSTAND AND ACCEPT THE HIGH VOLTAGE RISKS INVOLVED
The component values given are for 240V 50Hz operation with an LED current of ~ 7mA. I've found that quite sufficient for a night-light for a child's bedroom.
If you do need a brighter light the current can be increased in proportion to the value of the capacitor C (e.g. double the capacitance gives 14mA), or additional LEDs (or LED inverse-parallel pairs in the case of the second circuit) can be connected in series with the LED(s) shown. If C is increased then a 1/2W resistor should be used for R.
Circuit operation
The fuse protects the circuit and its value must not be increased.
The reactance of C behaves like a 'resistor' to drop the voltage. For 240V 50Hz mains the RMS current through it will be ~ 7mA (70mA per microfarad). When first switching on the light the current may be much higher but is limited by R to protect the LED(s) and capacitor.
In the first circuit, a bridge rectifier feeds the LED so that it conducts on both halves of the mains cycle. This doubles the light output compared to half-wave rectification and also reduces flicker.
In the second circuit, one LED conducts on one half-cycle and the other LED conducts on the other half cycle. Each LED protects the other from reverse voltage greater than ~3V.
The only 'wasted' power for the circuits shown is ~ 49mW in R.
Adaptation for 110/120V operation
Double the capacitance of C and use a 1/2W resistor for R. Assuming a 60Hz supply the current will be ~8mA
CAUTION
All components are potentially at mains voltage and could give a fatal shock unless care is taken to shield them from touch. They must be housed in a fire-resistant insulated enclosure. For added safety a 1 megohm bleed resistor should be connected across the capacitor to discharge it when the mains is switched off.
DO NOT BUILD EITHER CIRCUIT UNLESS YOU FULLY UNDERSTAND AND ACCEPT THE HIGH VOLTAGE RISKS INVOLVED
