modified 8 years ago

Not much

02:01:34

Just cool I guess

published 8 years ago

Robert_Kidd

8 years ago

Move the ground from the secondary winding to the junction of the two diode anodes. Open the left hand switch once the capacitor is charged sufficiently. Close the other switch to bring LED into circuit. Capacitor will now discharge but notice LED is burning since current is far too high. You need a resistor in series with the LED to limit the current flowing to around 20mA. Note, this will slow the discharge.

Robert_Kidd

8 years ago

You could omit the resistor and replace the LED with a light bulb. You will need to adjust the bulb working voltage to match the fully charged capacitor voltage. Of course, in reality such a bulb may or may not be available so you may have to add a suitable resistor if you built this circuit.

torrid

8 years ago

Put a Voltmeter parallel to the Capacitor, and you see, that it's only charged to a few hundered Millivolt. That is too low for the LED. Make the Capacitor much smaller, and the frequency much higher, then you get a decent amount of charge within seconds.

Robert_Kidd

8 years ago

With ground moved as I describe there is more than 60V DC across the capacitor.

EagleSnipez

8 years ago

Alright, thank you Robert and torrid for the help!

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