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Quick question, is using a capacitor rated for high voltage (lets say 35v) in a dystem that lets say supplies 5V (like for LEDs or what have you) dangerous?

Since it can store up to 35V, will it like somehow store a bunch then release it at once damaging the system, or it is OK to use a higher rated capacitor than the voltage being supplied?

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  • \$\begingroup\$ I do agree with @Leon Heller answer, read his answer. Every device has its operating point. There are no universal devices in electronics. \$\endgroup\$ – Standard Sandun Jan 12 '13 at 18:18
  • \$\begingroup\$ Depends on the capacitor. An electrolytic capacitor relies on an oxidisation layer as an insulator so capacitance will drift more over time if you run it at low voltage. Other capacitors don't have this limitation so there is no downside to use a higher rated part.eccept for size and cost. \$\endgroup\$ – Warren Hill Aug 28 at 11:20
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While not a perfect analogy, think of the voltage on the capacitor similar to the liter capacity of a tank. It will hold "35 V" but you needn't fill it completely. But like @JustJeff said, you'd be wise to ensure the container can hold more than necessary to prevent spills (and in an electrolytic capacitor's case, the electrolyte can expand and quite literally "spill" out).

Note that a better analogy to capacity would be the farad unit, since that's a measure of a capacitor's charge capacity, so don't get that confused with voltage, which is the potential to do work.

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    \$\begingroup\$ Voltage is analogous to pressure, so the voltage rating is the same thing as the pressure rating on a pressurized tank. If you apply too much pressure, both a capacitor and a tank will explode. For an unpressurized tank, the voltage rating would be the height of the tank before it overflows? So you can have two tanks of different heights but the same capacity. This analogy is more direct, I think: wisc-online.com/objects/ViewObject.aspx?ID=ACE4803 amasci.com/emotor/cap1.html If the pressure is too high, the membrane fails. \$\endgroup\$ – endolith May 13 '11 at 19:21
  • \$\begingroup\$ @endolith I hadn't thought to use pressure for voltage and volume for charge, but that's a much better idea. Thank you! \$\endgroup\$ – JYelton May 13 '11 at 19:55
  • \$\begingroup\$ The capacitance is equivalent to the cross-sectional area of a tank, while the voltage limit is proportional to the height. If capacitance doesn't vary with voltage, the amount of charge that can be held is proportional to the product of capacitance and the voltage limit. If capacitance does vary with voltage (a situation equivalent to a tank of non-uniform cross section) the charge is proportional to the integral of the capacitance over voltage. \$\endgroup\$ – supercat Aug 25 '11 at 23:54
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No, having a higher rated cap will not somehow store up more voltage than is available in the circuit. You actually want a cap with a slightly higher voltage rating than the highest voltage you expect to put across it. In fact, if you put more voltage on a cap than it is rated for, it is apt to catastrophically fail, i.e., pop or explode.

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    \$\begingroup\$ a cautionary tale: when I was in college, I put together a simple power supply, using a 12.6V transformer, a 1N4001 diode, and an electrolytic cap, probably 100uF. Very shortly after plugging it in, there was this loud hissing/whistling noise, and I was alarmed to see a large jet of smoke blowing out the top of a very swollen, half split open capacitor. The problem: I had put 12.6*sqrt(2) volts (about 18) onto a 16V rated capacitor. \$\endgroup\$ – JustJeff Apr 15 '11 at 12:26
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    \$\begingroup\$ probably way more than that. The transformer is only 12.6V if you were drawing the recommended amperage from it at the same time. If you were just smoke testing it (heh) you likely put 30V+ into that poor little cap. \$\endgroup\$ – Bryan Boettcher Mar 13 '12 at 23:09
  • \$\begingroup\$ @insta - yeah, unloaded the secondary would probably have been something like 14 to 16V, so it would be even worse than stated above. a learning experience nonetheless. \$\endgroup\$ – JustJeff Mar 14 '12 at 2:05
  • \$\begingroup\$ You're reminding me how the mechanical and agricultural engineers used to regularly manage to fry equipment in our electronic engineering labs. \$\endgroup\$ – smci Nov 6 '16 at 12:51
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If a high-voltage electrolytic is used at low-voltage, the actual capacitance might be a lot lower than the stated value.

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    \$\begingroup\$ Explain? Im not sure I understand what you mean or why. \$\endgroup\$ – user3073 Apr 15 '11 at 16:25
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    \$\begingroup\$ @Sauron - the capacitance of an ideal capacitor is a constant, no matter what voltage you have across it, but for real capacitors, this is not always the case. \$\endgroup\$ – JustJeff Apr 15 '11 at 22:56
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    \$\begingroup\$ Think of a capacitor as being a water tank. The capacitance represents the horizontal cross-sectional area, while the voltage represents the height of the water. If the cross sectional area is uniform, adding a fixed volume of water will increase the height by a fixed amount, regardless of how much water the tank holds. If the cross-sectional area varies with height, adding a fixed amount of water may increase the height by a variable amount. \$\endgroup\$ – supercat Aug 26 '11 at 0:02
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    \$\begingroup\$ I was going to say this, well worth pointing out. Rule of thumb round here is that caps rated for 2x the working voltage is a good (reliable) part. \$\endgroup\$ – John U Feb 4 '13 at 10:50
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    \$\begingroup\$ You tend to find more like the opposite. A high voltage capacitor will have it's capacitance rated at low voltage meaning when operated close to it's rated voltage the capacitance will be much lower. This is why the different MLCC capacitor dielectric types exist, they guarantee a certain capacitance vs voltage characteristic (amongst other things) \$\endgroup\$ – hooskworks Aug 28 at 10:04
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The voltage rating of a capacitor is a measure of how strong its insulation is. A 35V cap can withstand at least 35 volts applied across it (a higher voltage may cause bad things like a short through the cap and burnup). It has nothing to do with how much voltage the capacitor will store; it can store nothing higher than is input to it. The voltage rating is describing how high its barrier is; electricity shall not pass through it as long as it does not get that high.

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I have a staircase with 35 steps. I am standing on the fifth step. What if I fall down? Is it dangerous? Falling down 35 steps can hurt!

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    \$\begingroup\$ As funny as this is, i'd rather keep away from Sarcastic answers. What's obvious to you may not be obvious to another person. \$\endgroup\$ – user3073 Apr 15 '11 at 11:03
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    \$\begingroup\$ It was not intended as sarcasm. Sometimes an analogy helps understand a new concept. \$\endgroup\$ – markrages Apr 15 '11 at 14:52
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    \$\begingroup\$ then you need to say it in such a way, instead of just saying the analogy, which might not be obvious to others. \$\endgroup\$ – akaltar Sep 9 '16 at 15:00
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Everyone else has explained well, the "pressurised water tank" analogy is very good.

Just to add;

  • If you look (wikipedia etc.) at how capacitors are constructed, and the factors that determine capacity and voltage-tolerance, that may help explain why the different ratings exist and why using a 1000v capacitor in your 5v circuit may be just as poor an idea as using a 3v one.

  • Rule of thumb: Always add a bit / round up to the next preferred value for "safety" specs like capacitor voltage, wire current carrying capacity, component power dissipation, etc.

  • Bear in mind that your 5v circuit is not a perfect 5v, there can be spikes, drops, surges, etc., and powering something from 5v does not guarantee that any part of the circuit will not exceed 5v due to oscillations or whatnot.

We generally spec ~2x the working voltage (so a 12v circuit would have 24v caps, and generally the available rating is 25v so that's what we use), the closer you get to the working voltage the harder the thing is working and the less reliable it will be.

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Yes, the voltage is the high end rating of the capacitor but the capacitor is for storing electrons measured in farads or microfarads.

If you forget about the technical jargon, think of it like a battery. Not quite the same but if you have a 24 volt battery supplying a circuit that has a cut off of 19 volts and you only charge it to 12 volts, you have a lot less electrons to supply your circuit than what is needed and chances are the circuit won't work.

A 25 \$\mu\$F capacitor that is rated at 16 volts will have a 25 \$\mu\$F capacitance when operated near the 16 volts but if you substitute a 25 \$\mu\$F capacitor rated at 35 volts you will not have 25 \$\mu\$F capacitance if you only apply 16 volts.

These capacitors have many functions in circuits. One main function is to supply electrons to a circuit when the normal plug in supply has dropped lower than needed such as with alternating current. As the voltage and current reverse, 60 times a second, the level goes from around 170 volts peak down to zero volts and on down to -170 volts and then it repeats. The capacitors filter this drop by supplying the appropriate voltage to keep the circuit smooth. As the voltage rises back up again, it recharges the capacitor.

A leaky capacitor has the effect of a large rated capacitor that leaks and keeps the circuit from working properly. In most cases, you can over rate a capacitor and get away with it. If you double the voltage value of the capacitor but keep the supply voltage low you might want to also double the Farad value. Ex: 25 \$\mu\$F at 16 volts to become 50 \$\mu\$F at 35 volts running on 16 volt supply.

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    \$\begingroup\$ Welcome to the site. You will find people appreciate answers more, and they will therefore get upvoted more, if they are easily readable. Your answer would be lot more readable if it were broken up into paragraphs of related thoughts. Also be careful with units and how they are written. I know what you were trying to say, but "25 uf of charge" is the wrong way to express it. Farads is not a unit of charge (that's what Coulombs are), and the abbreviation for farad is "F", not "f". These "little" things matter in engineering, and people here care about that. \$\endgroup\$ – Olin Lathrop Jan 12 '13 at 19:11
  • \$\begingroup\$ Your second paragraph is too technical for me to comprehend....!! \$\endgroup\$ – soosai steven Apr 3 '16 at 14:19
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Basic capacitor construction is two plates with leads separated by insulating material...

However, electrolytic capacitor is much more than this. If you open up an electrolytic capacitor, you will find two long strips of plated aluminum separated by paper, all rolled in the cylindrical form. The cylinder is soaked in electrolyte and packaged in an aluminum can.

One most important thing to take note is the real insulation between the strips is not the paper. The porous paper is merely to avoid the strips from directly contacting each other. The real insulation is the chemical layer that forms on the aluminum strips when the capacitor is connected to the DC source in the correct polarity. When connected in wrong polarity the conductive layer forms causing continuous current flow. Temperature quickly rise because the DC resistance of the conductive layer that forms is not very low, causing ohmic power loss.

So, answering the question, the optimum amount of chemical insulating layer forms when the capacitor is operated almost near the rated voltage in correct polarity. Operating a high voltage capacitor at lower dc voltage cause some low continuous current to flow through the capacitor, thus rendering the capacitor not behaving ideally as a capacitor.

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