# understanding capacitor ESR ratings

Discussion in 'General Electronics Chat' started by DaveH, Sep 30, 2009.

1. ### DaveH Thread Starter Active Member

Jan 1, 2009
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Correct me if I'm wrong but formally speaking ESR is the resistive component of a capacitor's impedance.

If I've got that right, then I don't see why it should be frequency dependent. How come in all datasheets where ESR is listed it's always quoted at specific frequencies (usually 100Khz)?

I can see why it may be temperature dependent.

It seems to me that a lot of the web resources I've looked at on capacitor ESR are confusing. In wikipedia it even said that ESR includes the resistance of the dielectric? That can't be right as ESR would then be measured in megaohms. The quoted figures are usually a few ohms or a fraction of an ohm.

Usually I wouldn't bother looking at ESR but I've designed a circuit with the ICL7660. I've actually built this thing but I'm seeing a big ripple voltage on the output. I looked at the datasheet and it said that ESR is one of the main factors governing that ripple (and even gave a formula to calculate the peak to peak). I'm now trying to source alternative caps with low ESR ratings. Unfortunately some of the stores I've been to the people don't know what ESR is.

After looking at the web may be I don't know either.

2. ### ELECTRONERD AAC Fanatic!

May 26, 2009
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Dave,

I always think of ESR as the total internal resistance a component has.

Oct 29, 2008
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4. ### studiot AAC Fanatic!

Nov 9, 2007
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First you need to understand what ESR is not.

It is not the real or resistive component of the capacitor impedance, although it contributes to it.

It is not the leakage resistance of the capacitor.

These are separate parameters, which exist alongside ESR and must also be considered.

Now what ESR is.

Electrolytic capacitors are formed by using ions in a liquid gel as a dielectric.
This ionic gel has a measurable resistivity and a given capacitor a given resistance.
This resistance is in series with the capacitance formed by the plates of the capacitor and so is called the ESR.

Over time, especially at elevated temperatures, the liquid component of the dielectric dries out, reducing the ionic mobility and increasing the resistivity of the resultant material. Thus the ESR rises. The capacitor still possesses capacitance and still blocks direct voltage.

Now electrolytic capacitors are usually in a circuit to have low or negligable impedance at a given frequency so you can see that if the series component increases, even a little bit, it can have serious consequences for the functioning of a circuit.

As all this is internal to the low resistance leads of a capacitor, a connected voltmeter will still show the impressed voltage and a capacitance meter will still show significant, if reduced capacitance.

From a service point of view a capacitors which fail short or open would be revealed by the above. However, increased ESR would not show up so special ESR meters are used.

Here is a comparison:

http://www.anatekcorp.com/testequipment/esrcompar.htm

5. ### DaveH Thread Starter Active Member

Jan 1, 2009
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Studiot are you sure ESR is not the 'a' in Z = a + bj ?

If so then I'm now so confused I wish I'd continued in ignorance I'm going to have to do some rethinking.

6. ### bountyhunter Well-Known Member

Sep 7, 2009
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All I know is that it is indeed frequency dependent. I believe it is because it has some dependence on the dielectric material's characteristics which do change with frequency.

It is speciffied at 100kHz because that is a typical operating frequency for switching converters.

7. ### bountyhunter Well-Known Member

Sep 7, 2009
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I thought it was as well. We always measured the ESR using a spec analyzer measuring impedance. Sweep it until you get to the frequency where the impedance is minimum and that is where it is purely resistive, hence the impedance is the R component alone (ESR) at that frequency. I think it is true multiple effects contribute to what the R value is, but I thought ESR was equal to the real component of the impedance phasor.

8. ### bountyhunter Well-Known Member

Sep 7, 2009
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Here's a reference on ESR:

What Is ESR?
Every capacitor has some ESR. ESR is the electrical resistances in series with the capacitor plates. This includes the resistance of the metal leads and plates and the connections between them. An aluminum electrolytic capacitor also has resistance in the wet electrolyte solution, and in the layer of aluminum oxide which contains high levels of water (called the “hydrated oxide”).

What Causes ESR To Change?
First, we can ignore the resistance contributed by the metal leads and plates, because it’s so small. There are two common causes of high ESR: 1) Bad electrical connections, and 2) Drying of the electrolyte solution. Electrical connection problems can happen in old or new capacitors, while drying is usually only a problem in old ones. Connection problems happen because the leads coming into the capacitor cannot be made of aluminum, since aluminum cannot be soldered. The electrical connection between the aluminum plates and the copper leads calls for a weld or a mechanical crimp. Problems with either method produces high series resistance. Drying problems occur because of the importance of water in the electrolyte solution. The solution soaks the paper spacer between the two aluminum plates. The water carries the electrical charge from the negative aluminum plate to the surface of the insulating oxide on the positive plate. The oxide forms the capacitor’s dielectric and the negative charge on the surface of the oxide forms the negative capacitor plate. As the water evaporates, the electrical resistance increases.

Last edited: Sep 30, 2009
9. ### Ratch New Member

Mar 20, 2007
1,068
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DaveH,

Chew on this PDF authored by a maker of ESR meters. I am not vouching for the validity of this, but it should be somewhat accurate.

Ratch

10. ### studiot AAC Fanatic!

Nov 9, 2007
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How would a perfect resistor in series with a perfect capacitor generate a perfect resistance in parallel with the series combination?

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Every physical two terminal component such as a resistor, capacitor or inductor can be characterized at a particular frequency by its impedance, or by its admittance. Those two will generally have a real and an imaginary part.

Impedance and its components are conventionally given as:

Z = R + jX

where Z is impedance, R is resistance and X is reactance.

Admittance and its components are conventionally given as:

Y = G + jB

where Y is admittance, G is conductance and B is susceptance.

If you measure (or calculate) the impedance of some component, the relationship Y = 1/Z will give you the admittance; that is, admittance is the reciprocal of impedance.

But it's not true generally that the individual components are related as reciprocals.

In other words, G ≠ 1/R and X ≠ -1/B generally.

But, if X = 0, then G does equal 1/R and if R = 0, then X does equal -1/B.

The acronym ESR stands for Equivalent Series Resistance. The word "Equivalent" is significant here, as is the word "Series".

If you measure the impedance of a real capacitor at some frequency, you will obtain a value R + jX, with a real part R and an imaginary part X. That impedance is equivalent to the series connection of an ideal resistor whose value is R and a capacitor whose reactance is X (X will be a negative quantity for a capacitor if the measurement is within the design operating range).

When you make a measurement like this, the value of R is due to the combined effect of all the physical aspects of the capacitor that cause loss when a current I is carried by the capacitor. The actual power loss is easily calculated as I^2*R. You can see why it's called "Equivalent" series resistance.

Wikipedia shouldn't have said "...resistance of the dielectric"; they should have said "...losses in the dielectric."

Imagine that you have a capacitor of reactance 10^4 (at some frequency), with parallel resistance (leakage) due to the dielectric of 100 megohms. Its admittance is 1*10^-8 + j 10^-4. Calculate its impedance by taking the inverse of that complex number. The result is 1 - j 10000; this is the impedance of an ideal capacitor with reactance equal to 10^4 in series with a 1 ohm resistor. The 100 megohm parallel leakage resistance became a 1 ohm series resistance when we converted the admittance to an impedance.

Now, if we reduce the frequency by a factor of 10, the capacitor's reactance will increase to 10^5, and the admittance (with the same 100 megohm leakage) will be 1*10^-8 + j 10^-5. Taking the reciprocal again gives an impedance of 100 - j 100000; the 100 megohm leakage converted into a 100 ohm series resistance. This arithmetic phenomenon is the reason why the ESR varies with frequency at the low frequency end of things.

At the higher frequencies, skin effect and the changes in the dielectric loss at high frequency come into play.

Capacitors certainly do have a series resistance due to the resistance of the metals used to construct the capacitor, and due to the resistance of the electrolyte in electrolytics. Those actual series resistances are contributors to ESR, but the term ESR does not refer to just those quantities; it refers to all the losses rolled into one equivalent series resistance.

The Sencore publication:

is mistaken in section 9 where it suggests that ESR can't be determined from "D". Contrary to what they say, ESR does vary with frequency for the reasons I've explained. They say "D includes the effects of (a number of things)". Those very things are part of ESR. ESR and "D" are essentially the same thing.

They have adopted their own definition of ESR which doesn't conform to usage in the rest of industry.

A correct explanation and description is given by:

www.lowesr.com/QT_LowESR.pdf

The capacitor could also be measured as an admittance G + jB. If this were done, then the admittance could be considered to be equivalent to the parallel connection of a resistance of value 1/G and a reactance of value -1/B. Then we would speak of the capacitor's EPC (Equivalent Parallel Conductance).

12. ### DaveH Thread Starter Active Member

Jan 1, 2009
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I'd like to thank all my superiors here for trying to help. I think I'm starting to understand it a little bit.

Also Studiot, thanks for pointing out my flawed understanding. The Electrician, you must have written books I'm sure, but you've pointed out some very good stuff. It will take me a while to process all these links and figure it out.

13. ### studiot AAC Fanatic!

Nov 9, 2007
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Z = R + jX

Let us consider this model of a capacitor at zero frequency, by connecting a capacitor across a battery.

According to the model, Z = ∞ so I = 0

But we all know that a real capacitor will exhibit a (measurable) leakage current and eventually drain the battery.

14. ### bountyhunter Well-Known Member

Sep 7, 2009
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All real capacitors have a finite resistance (leakage path) across their terminals which can be modeled as a resistor. That is separate from the ESR which is modeled as a resistor in series with the "ideal" capacitor. The leakage resistor is sometimes omitted from the model because in some cap types (ceramic, film) it is so high that it is negligible in AC effect.

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
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You can't use this model at zero frequency. The leakage resistance of a capacitor is in parallel with the capacitance. If you invert G + jB, you will get zero in the denominator at zero frequency, leading to an undefined impedance.

The ESR concept is only defined and useful at non-zero frequencies.

The EPC (equivalent parallel conductance) model will work at zero frequency.

16. ### studiot AAC Fanatic!

Nov 9, 2007
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But you can use a capacitor at zero frequency.

This simply proves what I said earlier, that the model is imperfect or incomplete. A perfect model could be used in all situations.

I agree with BH, mostly (ie in a good quality cap) some terms are insignificant and can be ignored.

But, in any situation - not just with capacitors, we should always bear these terms in mind and check that we are not operating outside a situation where they can be ignored.
Further we should not use an imperfect model to explain all possible mechanisms, effects and modes of operation.

17. ### alim Senior Member

Dec 27, 2005
113
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Hi not a response to the question about ESR, and hope I am not hijacking the post. M y question is do any one has a chart of ESR values of different capacitance and voltage, or is there a method of calculating them. I have a DICK SMITH ESR meter and it operates at 100khz. Thanks for any assistance.

18. ### The Electrician AAC Fanatic!

Oct 9, 2007
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You are leading the discussion in a direction that should be in the off-topic forum. This is a matter of philosophy.
Philosophy switch on.

There's no such thing as "zero frequency" or "DC" in the real world; it's just a conceptual description. If a constant voltage is applied to a circuit for a time which is less than infinite, then it's not "zero frequency". In other words, for a voltage to be properly described as "zero frequency" it would have to be constant for all time. If it's ever turned off during the life of the universe, then it's not "DC", without even considering the question of whether it's been on since the beginning of time.

It proves no such thing. You just can't use simple arithmetic to calculate the relevant quantities when some of the parameters of the calculation become "infinite".

I suppose I should have said that you can't make calculations with the impedance model at zero frequency with simple arithmetic. I didn't bother because it seems to me that the case of zero frequency is not something one normally cares about in connection with the ESR of a capacitor.

But if you insist on a "perfect" model, then you just need to realize that when frequency becomes "zero" or "infinite", you may have to use concepts from calculus. In this case, if you calculate the loss in a combination of a capacitor in parallel with a resistance, using the impedance description of the combination, you just need to take limits as frequency approaches zero.

Consider what happens if you apply a current source of 1 amp to the combination as the frequency approaches zero. If you use the impedance description of the combination, the voltage rises to "infinity", and in order to calculate the power loss you need to take the ratio of "infinite" quantities, the ratio of voltage squared to resistance. You apply l'Hospital's rule and you get the right answer.

If you're going to insist on a perfect model, then no terms are insignificant.

The ESR model is not imperfect, as I explained above.

19. ### studiot AAC Fanatic!

Nov 9, 2007
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I don't go in for complaining that reality is wrong because my model can't handle it.
That would be in the realms of philosophy or something.

We construct models by first defining some idealised (perfect if you like) components in terms of (mathematical) rules etc.

Then we create arrangements of these perfect components to model real world components and to explain the imperfect effects the real world components exhibit.

I did not propose a current source, real or perfect. I proposed a voltage source.

I do not have mathematical difficulty with the concept that ∞/∞ or 0 * ∞ equals a finite value.
So when I apply Ohm's law I find that I have no difficulty with the concept that

a finite voltage can equal the product of a zero current and infinite impedance.

Cantor proved all this at the end of the 19th cent.

So in the 'perfect world' the maths works out.

But in the real world I find I cannot produce such a component. The charge always leaks away from capacitors.

So I am forced to amend my model to include a parallel leakage resistor as an additional ideal component.

This path, parallel to my idealised capacitor component, is capable of dissipating a ( measurable) quantity of energy. I can propose various physical phenomenon associated with real world capacitors and identify them with this parallel resistance.

I find that, in use in the real world, capacitors dissapate further energy that cannot be accounted for by the (ideal) resistance in this parallel path.
So I add a series perfect resistance to my model.
I can also propose various physical phenomenon associated with real world capacitors and identify them with this series resistance.

I now am in proud possession of a model which accounts for all known effects and is in accord with all known mathematical processes.

This model has a perfect resistance in series with a perfect capacitance with another perfect resistance in parallel with the combination.

If I use this model with frequency dependent equations these work fine and allow X+jR type calculations, and reduce satisfactorily to handleable arithmetic if I input the value of zero for frequency.

Are you, by the way, suggesting that a battery produces an output at other than zero frequency?

20. ### The Electrician AAC Fanatic!

Oct 9, 2007
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What is your point for saying this?

One can determine the impedance of a two terminal component by applying a current source and finding the ratio of the resultant voltage to the applied current. Or, one can apply a voltage source and find the ratio of the applied voltage to the resultant current. It matters not how you do it.

You are only forced to do this if you care about the loss at DC, and don't want to use limits in calculating the loss.

But, if the capacitor isn't defective, a designer will hardly ever care what the leakage at DC is; it's just not a significant loss factor.

In some special case where the DC leakage matters, it will have to be measured, not calculated.

But, the fact is that if the capacitor is carrying ripple current, and it's of good quality, the DC leakage will be totally negligible.

For example, I have a WIMA polypropylene 5uF capacitor. It has a measured leakage resistance of 20 gigohms. It's measured resistance, using a series model, at 10 kHz is 3 mΩ. This means that the actual metallic series resistance can't be greater than that. If I make the measurement using a parallel model at 100 Hz, the measured parallel resistance is 3 megohms. But, the series resistance is transformed into a component of the parallel resistance when using the parallel model for the measurement.

If we calculate what the parallel component should be due to the 3 mΩ series resistance, it comes out to about 30 megohms. But, the measured parallel resisistance is 3 megohms; why is it so low? The reason is that the dielectric losses, even at 100 Hz, look like a parallel AC resistance of 3.33 megohms; these losses are much greater than the DC leakage losses.

Even in a (good quality) capacitor carrying ripple current at frequencies as low as power line frequencies, the DC leakage resistance is not worth worrying about. This is also true for good quality electrolytics; the dielectric losses at low audio frequencies dominate.

There's no need to have a model where the parallel loss component is explicit, with its slight advantage of allowing simpler arithmetic at DC. If the capacitor is carrying an AC current, the DC loss component is completely swamped by the parallel AC loss component.

The losses in the parallel component of resistance are included in the ESR when using the simple impedance characterization Z = ESR + jX, even at "zero frequency". In fact, all the losses at the particular frequency are included in the ESR.

The ESR figures that manufacturers give are just the real part of the impedance that you would measure with an LCR meter at some particular frequency. They include the loss in the parallel resistance (leakage, dielectric loss) component, and it's the AC loss in this parallel component that is responsible for the rise in ESR at low audio frequencies.

This is all well and good, but the resistance in parallel with the capacitor is not going to be the DC leakage resistance. It's not necessary to use such a model in situations where one cares about ESR (switching power supplies, usually); situations where the capacitor is carrying an AC current.

You said "mostly (ie in a good quality cap) some terms are insignificant and can be ignored." The leakage in a good quality cap is one of these things that can be ignored in situations where ripple current is the dominant loss factor.

Characterizing the impedance of a capacitor with a parallel resistance as simply ESR + jX gives the correct result with simple arithmetic at any frequency other than zero, and by using limits it gives the correct result at "zero frequency", using a frequency dependent ESR.

This doesn't mean that one can't use a model that explicitly uses a parallel resistance. But it does show that the R + jX model isn't imperfect.

What's an X+jR calculation? Do you mean R+jX?

In hyper-philosophical mode, yes. There is undoubtedly noise on the battery's output, and as I said, unless the voltage is absolutely constant for all time, then it's output isn't "zero frequency".