I opened an emergency rechargeable light and saw the charger circuit which is quite simple: a bridge rectifier and then a big capacitor (I don't know the rating). There is no transformer, but the battery terminal voltage is appx 4.5V, but how does it work? When I measured the output of the rectifier I got nearly 270VDC. In India domestic voltage level is 220-240VAC. How is it possible to get more voltage? How does the battery charge with high voltage?
Electronic – How does this battery charger from an emergency rechargeable light work
battery-chargingchargerintegrated-circuit
Related Solutions
It's fairly common to see a lead-acid battery charged using rectified AC. As long as the charging current isn't beyond the capability of the battery, it will 'work'. If there isn't a series resistor somewhere, or some primary-side limiter, the winding resistance of the transformer could be what's limiting the charging current.
A handheld multimeter is sensitive to 60Hz AC, so yes, your DC reading was likely skewed by the low-frequency ripple.
The circuit you have drawn is a full-wave rectifier.
If there isn't any explicit current-limiting protection in the charger, it is possible that the charger can become damaged if subjected to long-term overload. Wall-wart adapters have similar failure modes (usually the transformer goes high-impedance).
If you are confident that the transformer output is appropriate for charging your battery, one solution would be to simply increase the current-carrying capacity of your bridge rectifier.
You mention that you can't obtain higher current diode bridges. Can you obtain higher current discrete diodes? If not, and your application can handle the cost of 2 diode bridges, you can make an effective 20A-rated full bridge rectifier out of two 10A full diode bridge modules.
If you look at the schematic for a full bridge rectifier, you'll notice that it is possible to connect the two AC input terminals together to obtain two diodes in parallel between the AC input and each of + and -. Since the diodes in a bridge rectifier are usually on the same die, they are likely to be reasonably well matched and likely to current share reasonably well (though never perfectly, so some derating is warranted). Simply use two full bridges connected in this way (one bridge for each AC input with the rectifier + and - terminals each connected in parallel)
Best Answer
Firstly, the AC mains voltage is rated with RMS value. So in India, it is 220 Vrms. When you rectify it and smooth it with a capacitor, the DC voltage becomes 312 V because of the following calculations.
Vdc = Vpk = Vrms x sqrt(2) = 220 x sqrt(2) = 312 V
After inefficient conversion and some losses, you measured it at 270 V.
Secondly, the circuit you saw for charging the battery was a very clever application of voltage divider rule and capacitor impedance.
It is similar to the following.
The capacitor C1 is the large capacitor you saw. It's value is about 0.68 uF or somewhere near it. At the mains frequency (50 Hz in India) C1 has a reactive impedance Xc1 = 1/(2*pifC1). This Xc1 and the resistance of R1 come in series to form a voltage divider. The diode bridge simultaneously rectifies the AC wave.
The value of Xc1 and R1 is chosen such that with 220Vrms at the input, the voltage after the rectification is a little more than the battery voltage, to keep it charging.
C2 smooths out the ripples, but it is omitted sometimes, because the battery does not mind any ripples while charging.