Electronic – n’t electrolytic capacitors be short-circuited

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.

Summary:

• Yes "polarised" aluminum "wet electrolytic" capacitors can legitimately be connected "back-to-back" (ie in series with opposing polarities) to form a non-polar capacitor.

• C1 + C2 are always equal in capacitance and voltage rating
  Ceffective = = C1/2 = C2/2

• Veffective = vrating of C1 & C2.

• See "Mechanism" at end for how this (probably) works.

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It is universally assumed that the two capacitors have identical capacitance when this is done.
The resulting capacitor with half the capacitance of each individual capacitor.
eg if two x 10 uF capacitors are placed in series the resulting capacitance will be 5 uF.

I conclude that the resulting capacitor will have the same voltage rating as the individual capacitors. (I may be wrong).

I have seen this method used on many occasions over many years and, more importanttly have seen the method described in application notes from a number of capacitor manufacturers. See at end for one such reference.

Understanding how the individual capacitors become correctly charged requires either faith in the capacitor manufacturers statements ("act as if they had been bypassed by diodes" or additional complexity BUT understanding how the arrangement works once initiated is easier.
Imagine two back-to-back caps with Cl fully charged and Cr fully discharged.
If a current is now passed though the series arrangement such that Cl then discharges to zero charge then the reversed polarity of Cr will cause it to be charged to full voltage. Attempts to apply additional current and to further discharge Cl so it assumes incorrect polarity would lead to Cr being charge above its rated voltage. ie it could be attempted BUT would be outside spec for both devices.

Given the above, the specific questions can be answered:

What are some reasons to connect capacitors in series?

Can create a bipolar cap from 2 x polar caps.
OR can double rated voltage as long as care is taken to balance voltage distribution. Paralleld resistors are sometimes used to help achieve balance.

"turns out that what might LOOK like two ordinary electrolytics are not, in fact, two ordinary electrolytics."

This can be done with oridinary electrolytics.

"No, do not do this. It will act as a capacitor also, but once you pass a few volts it will blow out the insulator."

Works OK if ratings are not exceeded.

'Kind of like "you can't make a BJT from two diodes"'

Reason for comparison is noted but is not a valid one. Each half capacitor is still subject to same rules and demands as when standing alone.

"it is a process that a tinkerer cannot do"

Tinkerer can - entirely legitimate.

So is a non-polar (NP) electrolytic cap electrically identical to two electrolytic caps in reverse series, or not?

It coild be but the manufacturers usually make a manufacturing change so that there are two Anode foils BUT the result is the same.

Does it not survive the same voltages?

Voltage rating is that of a single cap.

What happens to the reverse-biased cap when a large voltage is placed across the combination?

Under normal operation there is NO reverse biased cap. Each cap handles a full cycle of AC whole effectively seeing half a cycle. See my explanation above.

Are there practical limitations other than physical size?

No obvious limitation that i can think of.

Does it matter which polarity is on the outside?

No. Draw a picture of what each cap sees in isolation without reference to what is "outside it. Now change their order in the circuit. What they see is identical.

I don't see what the difference is, but a lot of people seem to think there is one.

You are correct. Functionally from a "black box" point of view they are the same.

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MANUFACTURER'S EXAMPLE:

In this document Application Guide, Aluminum Electrolytic Capacitors bY Cornell Dubilier, a competent and respected capacitor manufacturer it says (on age 2.183 & 2.184)

• If two, same-value, aluminum electrolytic capacitors are connected in series, back-to-back with the positive terminals or the negative terminals connected, the resulting single capacitor is a non-polar capacitor with half the capacitance.

  The two capacitors rectify the applied voltage and act as if they had been bypassed by diodes.

  When voltage is applied, the correct-polarity capacitor gets the full voltage.

  In non-polar aluminum electrolytic capacitors and motor-start aluminum electrolytic capacitors a second anode foil substitutes for the cathode foil to achieve a non-polar capacitor in a single case.

Of relevance to understanding the overall action is this comment from page 2.183.

• While it may appear that the capacitance is between the two foils, actually the capacitance is between the anode foil and the electrolyte.

  The positive plate is the anode foil;

  the dielectric is the insulating aluminum oxide on the anode foil;

  the true negative plate is the conductive, liquid electrolyte, and the cathode foil merely connects to the electrolyte.

  This construction delivers colossal capacitance because etching the foils can increase surface area more than 100 times and the aluminum-oxide dielectric is less than a micrometer thick. Thus the resulting capacitor has very large plate area and the plates are awfully close together.

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ADDED:

I intuitively feel as Olin does that it should be necessary to provide a means of maintaining correct polarity. In practice it seems that the capacitors do a good job of accommodating the startup "boundary condition". Cornell Dubiliers "acts like a diode" needs better understanding.

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MECHANISM:

I think the following describes how the system works.

As I described above, once one capacitor is fully charged at one extreme of the AC waveform and the other fully discharged then the system will operate correctly, with charge being passed into the outside "plate" of one cap, across from inside plate of that cap to the other cap and "out the other end". ie a body of charge transfers to and from between the two capacitors and allows net charge flow to and from through the dual cap. No problem so far.

A correctly biased capacitor has very low leakage.
A reverse biased capacitor has higher leakage and possibly much higher.
At startup one cap is reverse biased on each half cycle and leakage current flows.
The charge flow is such as to drive the capacitors towards the properly balanced condition.
This is the "diode action" referred to - not formal rectification per say but leakage under incorrect operating bias.
After a number of cycles balance will be achieved. The "leakier" the cap is in the reverse direction the quicker balance will be achieved.
Any imperfections or inequalities will be compensated for by this self adjusting mechanism. Very neat.