Author Topic: Capacitor question  (Read 1673 times)

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JWilliams354

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Capacitor question
« on: July 26, 2005, 01:27:32 PM »
I bought a capacitor for alittle under 10$.  It is bigger than I expected.  But, I was wanting to know what the 33000MFD means.  It is 200VDC, and I know that means, 200 volts direct current.  But I'm not 100% on the 33K MFD.
« Last Edit: July 26, 2005, 01:27:32 PM by (unknown) »

pyrocasto

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Re: Capacitor question
« Reply #1 on: July 26, 2005, 07:50:26 AM »
Someone correct me if I'm wrong:


MFD stands for MicroFarad I think, though it's usually the u sysmbol instead of an M. It's the so called "amp hour" rating for it. A farad is a 1 amp for 1 second. So you should have 1 amp for 33 seconds.

« Last Edit: July 26, 2005, 07:50:26 AM by pyrocasto »

pyrocasto

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Re: Capacitor question
« Reply #2 on: July 26, 2005, 07:51:45 AM »
Wait, make that .033 Farads giving you an amp for .033 seconds. ;)
« Last Edit: July 26, 2005, 07:51:45 AM by pyrocasto »

thunderhead

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Re: Capacitor question
« Reply #3 on: July 26, 2005, 07:53:10 AM »
33k mfd means 33,000 microfarads - or 0.033 farads.


A farad is the capacitance that stores one amp-second per volt applied.  So if you apply the maximum 200V to your capacitor, it will store 6.6 amp-seconds, or 1.8333.. milliamp-hours.


In battery terms, that's not much, but for a capacitor it's quite a hefty beast - which is why it's large.


I hope this explanation helps. :-)

« Last Edit: July 26, 2005, 07:53:10 AM by thunderhead »

JWilliams354

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Re: Capacitor question
« Reply #4 on: July 26, 2005, 07:58:05 AM »
thank you very much

« Last Edit: July 26, 2005, 07:58:05 AM by JWilliams354 »

Ungrounded Lightning Rod

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Re: Capacitor question
« Reply #5 on: July 26, 2005, 09:53:32 AM »
Or 33 miliamps for one second.


But that's to change its voltage by ONE VOLT.  (The voltage is directly proportional to the state of charge - no funny curves like batteries.)  Since this can charge to 200 volts you can store 200 times as many miliamp seconds between fully charged and discharged.


Note that the amount of energy stored per amp-second goes with the voltage, so the total energy stored goes with the square of the state of charge.


Not that that matters a whole lot:  A capacitor is still puny compared to a battery.  They're normally used for the very short-term storage that's involved in things like smoothing out voltages from fluctuations or transferring energy from one part of a cycle to another in an AC application.  Filtering after rectification, keeping signals in one part of a circuit from getting to another by way of the power supply, stabilizing the voltage on a circuit element, transferring AC signals across voltage differences and thus isolating the DC offset, shifting the phase of AC power to drive a second phase to drive a winding in a motor (to create a spinning field to get it started or improve its running), correcting power factor in a transmission system (so the current is in more phase with the voltge, reducing the resistive losses and increasing the fraction of energy deliverable), or forming part of a tuned circuit to select certain frequencies while suppressing others.


They ARE sometimes used to keep electronics running for a short time after power is removed, or even to store energy for significant periods, and recent "supercapacitors" are starting to get into the low end of the power-to-weight ratio characteristic of batteries.  But they're more for efficiently delivering or absorbing high currents for short times than for storing energy for long periods.  (They have VERY low losses compared to batteries.)


This is a pretty big non-super cap, and with a DC rating.  So I suspect it was intended for power supply filter service.

« Last Edit: July 26, 2005, 09:53:32 AM by Ungrounded Lightning Rod »