Recently one of my LED Flash Lights (Chinese made) stopped working. Being the curious person I am I opened it only to be surprised to see the circuit that was doing the charging. The Battery is a Sealed Lead Acid type and it has 4 Bright White LEDS.
But the charging circuit just had:
2 Resistors (One was burnt and the other was 560KΩ)
1 Capacitor (1µF probably 250v or above)
4 1N4007 diodes connected as a Bridge
1 Indicator LED
THATS IT ... I have seen charger circuits before but something this simple was :roll: ... I am amazed it worked for like 6 months using just a bunch of components.
I was wondering if anyone has come across this circuit. It would be great if someone could explain this circuit for me.
Yes...its very simple. The battery is charged through the dropped down AC by the capacitor and then rectified by the 4 diodes. Probably the 560K is across the capacitor to discharge the same when plugged off and other may be for the indicator led. We often get these chineese products for dirt cheap rates (This torch with 4-5 white LED's cost just 40-50?- rupees (1$).in every corner of India.
Yup they are sure cheap... I got a dozen of them for 1/4th the price of a Maglite
So from your explanation the wiring is like:
Resistor Parallel to the Capacitor connected to AC I/P of the Bridge and then the O/P of the Bridge has the Battery attached and the LED with the resistor.
But how low will the voltage get by using this method? Or is there just current reduction?
Is there some calculation involved to figure out the Capacitor value?
Is this a safe way to charge a sealed Lead Acid type battery? I have heard they are notorious for exploding under wrong charging conditions :O
The said technique of charging only suits for the lead acid SMF batteries. The current is reduced by the capacitor and the volatge without the battrey may be somewhere arround 23-24 volts. When the battery is connected, it drops to 4-5 volts and thus charges the battery. Exploding is normally experienced when batteries are connected with more volatge with high current and thus over charged...Cheers