Quote:
Originally Posted by MazdaMatt
jyanof, can you make a quick description of how the circuit operates? ... (I'm no power electronics engineer!)
|
Me neither!
But, I think the way you described the circuit matches my understanding of it.
The 47k resistor was left over from a sample circuit in the BQ2031 datasheet. Admittedly, I'm not sure of it's purpose, but I suspect it could help reduce initial voltage differences, or allow the cap to charge slowly when it's plugged in (though that would take forever). I can't see anything wrong with leaving it in, so I haven't taken it out...
But otherwise, the FET closes and the battery is connected to the DC powersource (rectified AC). A rapid rise in current is limited by the inductor. The FET then closes and the inductor releases it's energy through the diode and batteries. I don't know what to expect for voltage across the FET at this point. If my mental exercise is right, the voltage at the bottom of the diode must be greater than the top if current is going to flow through it, so that's something greater than about 180V near the end of charge. I suspect the 500V FET can handle whatever it is, but might be something to keep note of.