# Required Capacitance in Voltage Multiplier?

Discussion in 'General Electronics Chat' started by Hugs Are Drugs, Mar 11, 2013.

1. ### Hugs Are Drugs Thread Starter New Member

Mar 8, 2013
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Hello, this is my first thread so I'm not sure if this is the right forum to post.

Here's a pretty basic question I'm curious about, I know in a voltage multiplier circuit the capacitors and diodes need to withstand the current and voltages, however I'm wondering what are required capacitances of the capacitors for it to provide power long enough and how I can calculate them and are less required capacitances required with higher frequencies?

I appreciate any help given, thanks.

2. ### #12 Expert

Nov 30, 2010
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As a complete guess, I guess you are using some frequency attached to a string of diodes and capacitors. The capacitors can be modeled as having an impedance based on Xc=1/(2PiFC). rinse and repeat.

That's all the math you need, and yes, the impedance of a capacitor becomes lower when the frequency becomes higher.

3. ### Hugs Are Drugs Thread Starter New Member

Mar 8, 2013
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I don't understand how impedance is relevant to this, I'm quite inexperienced compared to anyone here so you're all experts relative to me. I'm talking about the actual capacitance rating of the. capacitors in farads and such, can you explain to me what impedance is used for?

4. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Impedance is the "AC Resistance" of a capacitor at various frequencies.

That depends on the circuit (usage) and the frequency (resulting in an impedance).

5. ### Hugs Are Drugs Thread Starter New Member

Mar 8, 2013
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I understand the concept of impedance however I'm quite unfamiliar with it, can you try and explain how impedance affects the behaviour of capacitors?

Also, I'm just putting this information into the thread if it can help, I am wanting to use ~80KHz into the voltage multiplier.

6. ### JMac3108 Active Member

Aug 16, 2010
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Impedance is not the point. The OP is asking how to calculate the required capacitance value in a voltage multiplier, to provide some specified load current.

I've done the calculation before but don't have time to figure it out right now. Basically you're looking at the energy stored in the capacitor 1/2CV^2, during the charging time, compared to the energy taken by the capacitor by the load during the discharge time.

Most people use voltage multipliers for very small loads such as a bias voltage for an LCD. In these cases you can usually just use a 0.1uf ceramic and it will work fine. But its always good to perform the calculations!

Last edited: Mar 12, 2013
7. ### Hugs Are Drugs Thread Starter New Member

Mar 8, 2013
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Well, my output is going to be very low current, so I guess there's really no need to do calculations other than just for practice? Thanks for telling me this, I really appreciate the help.

However I still have one more question, do you requre less capacitance with higher frequencies since the capacitors need to supply energy for shorter periods of time at higher frequencies?

Feb 19, 2009
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718