Understanding Vout Waveform and Voltage Ripple in Voltage Multiplier Circuits

AI Thread Summary
The discussion focuses on simulating a Cockcroft-Walton voltage multiplier circuit using Pspice, specifically regarding the output voltage waveform and voltage ripple. Participants clarify that the circuit converts AC input to DC output and emphasize the importance of using AC coupling to measure ripple accurately. The ripple amount is influenced by various factors, and resources are suggested for further understanding. Additionally, it's confirmed that while voltage multipliers increase voltage, they decrease current, maintaining the principle that power input cannot exceed power output. Understanding these dynamics is crucial for effective circuit design and simulation.
sean23
Messages
4
Reaction score
0
I am trying to simulate a voltage multiplier circuit on Pspice simulation package, which will produce an output voltage Vout.

What is the waveform of Vout supposed to look like?
And also how will i work out the voltage ripple of the waveform?
 
Engineering news on Phys.org
Info ??

Sean you have not given enough info here -- I assume you mean a dc to dc multiplier but there is more than one way of doing this and the results differ.
Ray.
 
If your talking about a Cockroft Walton voltage multiplier then you put AC in and get DC out. I simulated this in Electronic workbench and used the o-scope to measure ripple, use AC coupling. The amount of ripple depends on a bunch of things, check out this website http://home.earthlink.net/~jimlux/hv/cw1.htm
 
cockroft walton voltage multiplier

yes it is the cockroft walton multiplier circuit, putting in an ac source and producing dc.

i have an output waveform but i cannot work out the voltage ripple

cheers guys
 
hey random question about voltage multipliers (im only posting here because ppl have responded to this as is...lol) although they increase voltage they also decrease current, correct?
 
Yes. It's sort like a transformer. The power in can't be less than the power out (on a continuous basis). Since P = V*I, the output current must decrease in proportion with the voltage increase.
 
k, just makin sure. lol
 
Back
Top