This effect is due to the effects of parasitic characteristics of the device. A capacitor has four basic parasitics:

### Equivalent Series Resistance - ESR:

A capacitor is really a capacitor in series with the resistances of its leads, the foil in the dielectric, and other small resistances. This means that the capacitor cannot truly discharge instantly, and also that it will heat up when repeatedly charged and discharged. This is an important parameter when designing power systems.

### Leakage current:

The dielectric is not ideal, so you can add a resistance in parallel with your capacitor. This is important in backup systems, and the leakage current of an electrolytic can be much greater than the current required to maintain RAM on a microcontroller.

### Dielectric Absorption - CDA:

This is usually of less interest than the other parameters, especially for electrolytics, for which leakage current overwhelms the effect. For large ceramics, you can imagine that there is an RC circuit in parallel with the capacitor. When the capacitor is charged for a long period of time, the imagined capacitor acquires a charge. If the capacitor is rapidly discharged for a brief period and subsequently returned to an open circuit, the parasitic capacitor begins to recharge the main capacitor.

### Equivalent Series Inductance - ESL:

By now, you shouldn't be too surprised that, if everything has capacitance as well as nonzero and non-infinite resistance, everything also has parasitic inductance. Whether these are significant is a function of frequency, which leads us to the topic of impedance.

We represent impedance by the letter Z. Impedance can be thought of like resistance, just in the frequency domain. In the same way that a resistance resists the flow of DC current, so does an impedance impede the flow of AC current. Just as resistance is V/R, if we integrate into the time domain, impedance is V(t)/ I(t).

You'll either have to do some calculus, or buy the following assertions about the impedance of a component with an applied sinusoidal voltage with a frequency of w:

\$
\begin{align}
Z_{resistor} &= R\\
Z_{capacitor} &= \frac{1}{j \omega C} = \frac{1}{sC}\\
Z_{inductor} &= j\omega L = sL
\end{align}
\$

Yes, \$j\$ is the same as \$i\$ (the imaginary number, \$\sqrt{-1}\$), but in electronics, \$i\$ usually represents current, so we use \$j\$. Also, \$\omega\$ is traditionally the Greek letter omega (which looks like w.) The letter 's' refers to a complex frequency (not sinusoidal).

Yuck, right? But you get the idea - A resistor doesn't change its impedance when you apply an AC signal. A capacitor has reduced impedance with higher frequency, and it's nearly infinite at DC, which we expect. An inductor has increased impedance with higher frequency - think of an RF choke that's designed to remove spikes.

We can calculate the impedance of two components in series by adding the impedances. If we have a capacitor in series with an inductor, we have:

\$
\begin{align}
Z &= Z_C + Z_L\\
&= \frac{1}{j\omega C + j\omega L}
\end{align}
\$

What happens when we increase the frequency? A long time ago, our component was an electrolytic capacitor, so we'll assume that \$C\$ is very much greater than \$L\$. At first glance, we'd imagine that the ratios wouldn't change. But, some trivial (Note: This is a relative term) complex algebra shows a different outcome:

\$
\begin{align*}
Z &= \frac{1}{j \omega C} + j \omega L\\
&= \frac{1}{j \omega C} + \frac{j \omega L \times j \omega C}{j \omega C}\\
&= \frac{1 + j \omega L \times j \omega C)}{j \omega C}\\
&= \frac{1 - \omega^2 LC}{j \omega C}\\
&= \frac{-j \times (1 - \omega^2 LC)}{j \omega C}\\
&= \frac{(\omega^2 LC - 1) * j)}{\omega C}
\end{align*}
\$

Well, that was fun, right? This is the kind of thing you do once, remember the answer, and then don't worry about it. What do we know from the last equation? Consider first the case where \$\omega\$ is small, \$L\$ is small, and \$C\$ is large. We have, approximately,

\$
\begin{align*}
\frac{(small * small * large - 1) \times j}{small * large}
\end{align*}
\$

which is a negative number (assuming \$small * small * large < 1\$, which it is for practical components). This is familiar as \$Z_C = \frac{-j}{\omega C}\$ - It's a capacitor!

How about, second, your case (High-frequency electrolytic) where \$\omega\$ is large, \$L\$ is small, and \$C\$ is large. We have, approximately,

\$
\begin{align*}
\frac{(large * small * large - 1) \times j}{small * large}
\end{align*}
\$

which is a positive number (assuming \$large * small * large > 1\$). This is familiar as \$Z_L = j \omega L\$ - It's an inductor!

What happens if \$\omega^2 LC = 1\$? Then the impedance is zero!?!? Yes! This is called the resonant frequency - It's the point at the bottom of the curve you showed in your question. Why isn't it *actually* zero? Because of ESR.

TL,DR: Weird stuff happens when you increase the frequency a lot. Always follow the manufacturers' datasheets for decoupling your ICs, and get a good textbook or take a class if you need to do high speed stuff.

This is a YMMV / Caveat Emptor / DTTAH / ACNR type question and answer*:

ie the following may help but you MAY blow things up.

Is the buzzing because of the capacitor is bad or that a different component could be failing and having a secondary, audible effect on the capacitor?

Could be either. Capacitor is possible but I've not heard one do this. You may get a better idea by using a 'sound transfer unit' eg a wooden ruler or length of solid plastic bar or similar. **ENSURE** it is not electrically conductive to the extent that applying 600 VDC on one end and your ear on the other will not cause "problems" [tm]. You may wish to use a piece of rubber glove on one end as well BUT in practice even touching the capacitor outer SHOULD be safe.

Place one end of STU (sound transfer unit) against object to be tested and other end against ear or bone of head near ear. Sound can be localised and heard much better this way., Try on surrounding objects as well. Applying EHT to far end is to be frowned on - you should not have any EHT in a UPS.

I thought the AC capacitors didn't have electrolytes in them and hence can't understand why they would be buzzing loudly when they did not previously

AC capacitors will probably contain two x electrolytic capacitors of twice the target capacitance each, connected in opposed polarity.

Should not both these caps be interchangeable in this application?

There is a "reasonable chance" that the cheap fan cap WILL work OK here. Also, used or dead fans may be cheaper again.

I am thinking of plugging out the 20uF 250V pedestal fan capacitor and putting this capacitor into such a fan instead, to confirm if the cap itself is causing the noise - sounds like a good idea?

Yes. Sounds good. Always a risk that the fan will stress it worse than the UPS does but if it is good it should work OK.

I am not sure why the fans use 250V caps but my assumption is that these Chinese makers just stock up on one variety of the caps since most of the world are at or below 250V.

Many countries are 230 VAC. Some nominally 220 VAC. Voltages can be and are sometimes higher than nominal. 250 VAC is saying it will work anywhere.

Hence, for this experiment, using a 150V cap instead of the 250V one for the fan (in the US) should not damage either fan or cap?

Probably. Cap MUST be AC rated and suitable for whatever class of service it sees. X rates is phase-phase or phase-neutral or line-line. Y rated is phase-ground. Such ratings allow for surges, spikes etc. 250 VAC cap in 110 VAC system should be fine.

If it's confirmed that the cap's at fault, could I put in the 20uF 250V pedestal fan capacitor on the APC UPS board and expect it to have no detrimental effect on the UPS?

You can always **EXPECT** :-) ... .

But, yes, it will probably be OK BUT no guarantee.

My understanding is that this is a power cap and as long as the capacitance and voltage rating is good, other cap chars like ESR don't matter?

See above re X and Y rated. Also, some designs pass high currents at all times and some don't and a cheap cap **may** get sadder quicker.

*YMMV / Caveat Emptor / DTTAH / ACNR =
Your milage may vary

Let the buyer beware.

Don't try this at home (you can but ...)

All care, no responsibility

## Best Answer

Yellow axial caps like that are (almost) invariably polyester film (for which Mylar is a trade name). They have very low leakage and fairly mediocre dielectric absorption performance.

The part number seems more like a 4uF/100V than a 15uF. Unless the OPs fingers are huge the size would better fit as well. The other types of film cap that are relatively common (PP and PC and PS) would be much bigger again. Stick in on a capacitance meter and check it. Put 100V across it and verify it doesn't fail. There's not much that can go wrong with a cap like that.

There are more modern films such as PPS but they are uncommon in that style of capacitor.

As to where you would use such a part- it might be used for an analog filter or an analog controller. Much of this kind of thing is done in the digital domain these days, but there are still applications.

Your bypass application would probably be better served by a ceramic capacitor (perhaps in parallel with a low-ESR aluminum electrolytic)- lower inductance, smaller and cheaper most likely (depending on voltage). The film caps do have longer lifetime than electrolytics.