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.
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.
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.
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.
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.
You are confusing ESR, that stands for Equivalent Series Resistance, and the leakage. The first is modeled as a series resistor, and take account of leads resistance, leads-internal plates resistance and so on, and is ideally zero. The second is modeled as a resistor in parallel with the capacitor and takes account of small leakage currents in the dielectric, and is ideally infinity.
The formula you use is correct, but the value you come out with is NOT the ESR, is the leakage resistance. Once the capacitor is charged, if you leave it it slowly discharges trough the leakage resistor with a time constant \$R_{leak}\cdot C\$, so \$R_{leak}\$ is what you calculated, approximately \$50M\Omega\$, that is plausible.
To calculate the ESR you need to measure how long does it take the capacitor to discharge through a much smaller resistor, let's call it \$R_{dis}\$. When you discharghe the capacitor through \$R_{dis}\$ the total resistance through which it discharges is actually \$R_{dis}+R_{ESR}\$, so using the very same formula you used for the leakage resistance you can calculate the ESR.
But is it really that easy? Of course not.
The ESR is hopefully quite small, tenths of milliohms if you have a very good capacitor up to a few ohms. Since in the formula you have \$R_{dis}+R_{ESR}\$ you don't want an eccessive \$R_{dis}\$ to mask \$R_{ESR}\$. Ok then! Why don't we choose \$R_{dis}=0\Omega\$? Easy question:
- \$0\Omega\$ resistance does not exist. But i can make it small!
- Time. You need to be capable to measure how long does it take to the capacitor to discharge.
If you charge the capacitor to a certain voltage it will take \$\tau\ln{2}\approx0.7\cdot\tau\$ where \$\tau=RC\$. If \$R=R_{ESR}+R_{dis}=1\Omega+1\Omega=2\Omega\$ and \$C=680\mu F\$ that's less than 1ms. Without proper equipment, that is a properly set oscilloscope, you can't easily measure the ESR.
Last but not least, keep in mind that electrolytic capacitors values have a tolerance of \$\pm10\%\$, that leads to:
$$
R_{ESR}=\frac{t_{dis}}{\left( C\pm C/10\right)\ln{2}} - R_{dis}
$$
with the above numbers, t=1ms, C=\$680\mu F\$, \$R_{dis}=1\Omega\$, this translates to:
$$
R_{ESR}\in\left[0.91,1.33\right]\Omega
$$
That's 10% down and over 30% up.
Best Answer
The short answer is: insulator thickness. But that probably will not suffice as an actual answer.
It is helpful to understand how capacitors are made. All capacitors we frequently use in electronics are essentially flat plates with insulation in between them. The plates are some kind of metal and the insulation is often plastic or ceramic material. We usually call capacitors by one of these materials: aluminum electrolytic capacitors are literally aluminum foil with a liquid electrolyte in between. Ceramic multilayer capacitors have lots of layers of conductor and ceramic slurry in between them, and tantalum capacitors have the element Ta as its conductor (plate).
The key to making a capacitor is getting two plates with the largest possible area as close as possible to each other without them conducting to each other. You can imagine that this is pretty hard; mechanically it is very hard to get two surfaces exactly parallel and have only in the order of microns or hundreds of nanometers in between them, which is what you need for the thousands or tens of thousands of microfarads, somtimes even tens to thousands of farads we use in electronics today. So an insulator is used in between two conductive plates, but in order to guarantee that the insulator is mechanically strong enough to survive manufacturing and useful life, it usually needs to be fairly thick - tens to hundreds of microns is typical for for instance MKP capacitors. This increases the distance between the plates to far beyond what is necessary for electrical insulation, only to serve a mechanical purpose.
So, in aluminum electrolytic capacitors (and all electrolytic capacitors in fact), a trick is used to reduce the plate distance without actually putting the plates in very close proximity to each other. What they do: - One plate is oxidized to form a very thin, well-controlled insulating layer on its surface which can be as little as tens of nanometers thick - The other plate is left with a conductive surface - A sponge-like, fairly thick mechanical divider is placed in between the plates, but it is drenched in an electrically conductive liquid called the electrolyte.
This electrolyte effectively moves the plates much closer together, as if only the oxide layer on the first plate is the actual distance between them. This allows for a much, much higher capacitance in electrolytic capacitors than other methods of constructing capacitors.
Alright, with this kind of information under our belt we can see why leakage is more pronounced in electrolytic capacitors. Leakage is caused by four major mechanisms:
The first one is simple: even the best insulator still conducts a little bit of electricity. The fact is that inside a large-value capacitor, there is a ton of surface area of the oxide layer, and even with a very low conductivity, if you have a lot of surface area the amount of leakage will be significant. This is especially significant for special super high-capacity electrolytics, where the aluminum foil inside is not just a coiled flat piece of foil, but the manufacturer has etched patterns into the foil to dramatically increase the amount of surface area of the aluminum.
The second one is also simple. Electrolytics are made by winding up coil and electrolyte 'sponge' together, and during this winding process mechanical stress and material imperfections cause tiny microfractures in the oxide surface. This allows electrons to directly flow between the plates, i.e. leakage current.
Then there is electrolyte dielectric effects. The electric field between the plates does cause some dielectric effects in the electrolyte, which change with the voltage (strength of electric field) over the plates. This means that usually at higher voltages, there is disproportionally more leakage current but at the same time higher ESR because the electrolyte ions are less free to move and are pulled in one direction by the electric field.
Lastly I should mention electron tunneling over very thin oxide layers. This is not an issue for run-of-the-mill electrolytics, but especially ultracapacitors suffer from this phenomenon a lot. Quantum-mechanically, electrons can be regarded not as a single point of charge, but a probability density of the electron being somewhere (wave-like behaviour of elementary particles). The probability of an electron being on the conductor is very high, and as you move further away from the conductor the probability of the electron being there decreases until it becomes essentially zero a few nanometers away. When an oxide layer is extremely thin, the probability of an electron that sits on the conductor to be on the other side of the oxide layer is nonzero, i.e. there is a probability that the electron 'jumps' over the oxide layer. This is called quantum tunneling and starts playing a role in leakage current when you approach nanometer-scale insulators.
So, to get back to your question: even though it may seem like it, by far the most important reason that electrolytic capacitors seem to leak more is that they simply have more capacitance and, by association, larger surface areas and thinner insulators which both contribute to higher leakage. If you would make any other construction of capacitor with the same specifications, they will exhibit fairly similar leakage characteristics. Yes, there are some unique reasons why electrolytics fare a little bit worse on the leakage side, but this is not something that puts the technology in a completely different order of magnitude of leakage.