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Legend333
modified 5 years ago

Voltage Multiplier

3
2
249
04:08:44
Multiplies the voltage such that the output voltage is higher than the input voltage...current however reduces as power must remain constant throughout. The input voltage is AC but the output voltage is a much higher DC. The first two capacitors and two diodes make up stage one where the voltage output on capacitor 2 is two times the input voltage... 250VAC to 498VDC The third and fourth capacitors and diodes make up stage two where the output voltage between capacitor 4 terminals is still double the input voltage but the voltage between capacitor 4 positive terminal and ground (negative terminal of capacitor 1) is four times the input voltage...250VAC to 996VDC The fifth and sixth capacitors and diodes make up stage three where the output voltage between capacitor 6 terminals is still double the input voltage but the voltage between the positive terminal of capacitor 6 and ground (that's the negative terminal of capacitor 1) is now six times the input voltage...250VAC to 1.49KVDC The seventh and eighth w capacitors and diodes make up stage four where the output voltage between capacitor 8 terminals is still double the input voltage but the voltage between the positive terminal of capacitor 8 and ground (that's the negative terminal of capacitor 1) is now eight times the input voltage....250VAC to 1.99KVDC Such that from 250VAC we get 1.99KVDC from the entire circuit (taking into account the diodes saturation voltage) ....just as the voltage is output is now eight times the input, the current output is now 1/8 times the input current
published 5 years ago
birdlaw
5 years ago
Why?
PrathikP
5 years ago
Set the amplitude of your voltage source to 340V (240*1.4142) as the parameter amplitude is the peak voltage and not the RMS voltage. Also, reduce your diode's resistance to something like 100mohm or 50mohm for better performance.

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