A friend of mine brought a little device he made its a bridge rectifier fed straight off the mains then fed into a 900 uF capacitor the purpose of this little device is to try to make a CFL stay on for 10 seconds I don't know why it's his project. We tried this device out and after a few switch ons it popped a 5 amp fuse in the feed to my test bench so the question is what would be the charge current for a 900 uF capacitor? It made the ammeter in the feed to the test board kick up. So is there a way to work it out?
also be aware that even after the lamp goes out, the capacitor will hold charge, possibly for hours after if the electronics switches off totally at low voltage.
A resistor across the C to bleed it down such that R * C is a sensible time - normally 5 secs in commercial products, but you actually want than that storage as run-on time, but perhaps a few mins.
900uF even if the lamp runs it downto 100V is 4.5 joules
energy is half C V squared - so 4.5 joules - which may not sound like a lot but it can still hurt.
To bleed down to 1/e of the voltage in say 50 secs , R*C=50 would be ~ 50 k ohms.
also be aware that even after the lamp goes out, the capacitor will hold charge, possibly for hours after if the electronics switches off totally at low voltage.
A resistor across the C to bleed it down such that R * C is a sensible time - normally 5 secs in commercial products, but you actually want than that storage as run-on time, but perhaps a few mins.
900uF even if the lamp runs it downto 100V is 4.5 joules
energy is half C V squared - so 4.5 joules - which may not sound like a lot but it can still hurt.
To bleed down to 1/e of the voltage in say 50 secs , R*C=50 would be ~ 50 k ohms.
