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IamSmooth

Tue Mar 03 2015, 12:36AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

Is there a standard formula I should use to make sure my capacitors can handle the output current of a voltage doubler? Even if this is just a simple bridge rectifier/smoothing capacitor, what should I use to make sure the capacitor can handle the current? I'm talking about 30-50A output.

The capacitor is not in series with the circuit, so I don't think I need to figure out Z (1/2pifC). Is it a heating issue?

My capacitor is 1900uf/450v. I can put two in parallel, but I would like to know if one is sufficient as long as my ripple is low enough.

Thanks.

BigBad

Tue Mar 03 2015, 01:40AM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
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Posts: 600

Have you tried reading the manufacturer's data sheet?

dexter

Tue Mar 03 2015, 08:54AM

Registered Member #42796 Joined: Mon Jan 13 2014, 06:34PM
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Posts: 195

for a simple full wave rectifier the smoothing cap can be calculated with this formula C = I x t / V where I is the load current V is the voltage ripple we chose and t is the AC half cycle or charge time

so if we need a rectified mains (230Vrms 50Hx) of 320Vdc at 1 A with a maximum voltage ripple of 10% the cap would be
C=1*0.01/(0.1*230*1.41)=312uF

in case of a voltage multiplier the situation is worse as the charging time for the final stage is a multiple of the multiplier stages but i'm not sure how to calculate....

use this calculator

DerAlbi

Tue Mar 03 2015, 09:32AM

Registered Member #2906 Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727

Smooth, your concern is not applicable. Even if you overload your capacitor extremely, it wont damage it instaneous. If its only for charging/discharging.... The ripple current ratings are for long term operation only.

I ask: where is the difference between charing and discharging a capacitor?

You wouldnt hessitate to nearly short circuit the capacitor - infact i suspect you only charge it for excactly that reason..
...but think about heating when charging the cap? Why?

If ypu really have 40A average current thorugh your voltage doubler output then youve got much worse heating problem than only the capacitor.

Edit: some spelling. seems like i was drunk.

Wolfram

Tue Mar 03 2015, 10:16AM

Registered Member #33 Joined: Sat Feb 04 2006, 01:31PM
Location: Norway
Posts: 971

Using an electronics simulator is usually the easiest way to figure out these sorts of problems. LTSpice is a good and free alternative.

BigBad

Tue Mar 03 2015, 04:14PM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
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Posts: 600

Yes, the issue is resistive heating. As a decent rough rule of thumb, you square the current (50A) and multiply by the ESR.

So if the ESR is 0.005 and 50A then the power dissipation would be:

0.005 * 50 * 50 = 12.5 watts.

It doesn't matter that it's not in series, the capacitor charges and discharges over the cycle, most of the current going to the output comes from the capacitor in fact and it has to be recharged again; these are going to be very substantial currents.

You can confirm, and should, with a simulator.

Obviously running two capacitors makes things much better; the current through each capacitor ESR is halved, so the power loss is a quarter in each cap and the ripple is smaller too. I would think this would be desirable here, unless it's only a momentary use thing, in which case it doesn't matter.

Electra

Tue Mar 03 2015, 04:17PM

Registered Member #816 Joined: Sun Jun 03 2007, 07:29PM
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Posts: 156

Couldn't you just use the output current as an estimate, since the Cap has to provide the full load current when the rectifier is not conducting. On the other half cycle the same amount of charge has to be put back in. The peek currents may be slightly different. But it is the ( I ) squared x ESR that generates heat I think.

Edit
Yeah Bigbad thought the same way as I did while I was typing this...

Conundrum

Tue Mar 03 2015, 07:33PM

Registered Member #96 Joined: Thu Feb 09 2006, 05:37PM
Location: CI, Earth
Posts: 4061

Yeah, some psus use small low ESR ie PP cap in parallel with large not-so-low ESR cap to save costs.

Bored Chemist

Tue Mar 03 2015, 08:21PM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

The big problem would be that small capacitors would give a big voltage drop and high ripple

Edrop = I1/ (f*C) * (2 /3*n^3 + n^2/2- n/6)

from

Sulaiman

Wed Mar 04 2015, 11:21AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Simulation is the easiest,
if you don't want to build your own simulation use this simulator

which is quick and easy to use.

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