accapacitorrectifier
I disassembled a Chinese-made flashlight and found they use just capacitors to bring down the voltage instead of a transformer. They are using it to charge small lead-acid battery.
My question is how a capacitor alone can be used to bring 220 V down to 6 or 12 V?
R30 limits the charging peak current to the capacitor somewhat, but at 1 ohm it still allows for 12 A, so of little use there. Also would limit the current through the zener if there are peaks above 20 V.
The larger capacitors work less well at higher frequencies, and that's where the smaller ones take over.
470 µF in a non-polarized version would be expensive, but there would be nothing against it. All large capacitors are polarized.
On the output it would give an extra load for the 7805 to charge it. On the input too for the battery, but that can deliver more than enough current.
No, except that they're larger.
Keep in mind that the input-output difference for the 7805 is about 5 V (12 V - 1 V for the diode and 1 V for R30 - 5 V out) and that at 1 A out the regulator will dissipate 5 W, so it will need considerable cooling (sizable heatsink) if you want to draw that much current.
Firstly, I'd use this guide to the BQ2031 - it has a fuller description and includes a circuit diagram. I'm saying this because the diagram in the question is probably not going to work.
As for the output capacitor, \$C_O\$, this should be a decent low ESR type and placed close to where the inductor and \$R_S\$ is on the circuit board: -
I've circled the diode that is needed (and missing from the diagram in the question). Also circled is \$R_S\$ which is in the milli-ohm range and is required when the battery leads are very short.
It should also be noted that the diode (preventing battery discharge when the charger is not powered) should not be used where the question diagram proposes. See DB1 on diagram above.
Best Answer
A capacitor is an impedance and at (say) 50 Hz, a 1 uF capacitor will have an impedance of 3183 ohms - if you put this capacitor directly across the AC (at 220 V), the current that would flow is about 70 mA.
OK, so it's not directly across the AC supply because it feeds a bridge rectifier and battery but it will provide a decent enough source of current to charge the battery. The voltage is limited by the battery itself.
Be aware that this type of circuit is not isolated from the incoming AC and stripping it down is OK providing you don't switch it on while still stripped down—the voltages are lethal.
Here's something like the circuit it uses: