ESR of electrolytic capacitors is too high?

I would like to see how any cap. can suffer from the apparent rise in ESR below 10 Hz or even at 1 Hz.

I always treated it as fixed because it is insignificant , always being <2% of the total impedance.

Unlike at high freq. or square wave voltages or currents , where ESR dissipation is significant.

I highly doubt any cap discharge design at 1hz rate for LEDs is affected by the rise of ESR at 1Hz when the reactance is >50x greater.

i guess for energy storage it means efficiency is limited to 98-99% in low ESR caps due to dielectric loss resistance.

@ Treez, given that in your application (1 amp discharge, 0.4A charge) the caps don't get even a few degrees above ambient (I'm assuming), shows that the ESR is negligible given your application, you can easily check on the ESR with a scope on the cap terminals (AC coupled) capture the moment the 1 amp discharge is applied, (a few uS or 10's of uS) there will be an ESR jump downwards, before the voltage decays according to the discharge curve (i/C = dV/dt) if this ESR jump is 1V, the ESR is 1 ohm, etc (ohms = Vjump / current step applied), 100mV = 100milli-ohm for 1 amp step.
I hope this helps.... p.s. warm caps have a lower ESR than cold ones...

thanks, interestingly, a sudden "step" has infinite frequencies, and so presumably there is a lot of high frequencies, and so the esr should be low for these...just a thought

I agree with Orson Cart that overtemperature should be taken as criterion for capacitor suitability in your application.

thanks, and of all the components, electrolytic capacitors are the hardest ones to tell what is their internal temperature......you can get the temperature of their case, but they never come with a junction to case thermal resistance figure.

SunnySkyguy said:

I would like to see how any cap. can suffer from the apparent rise in ESR below 10 Hz or even at 1 Hz.

I always treated it as fixed because it is insignificant , always being <2% of the total impedance.

Unlike at high freq. or square wave voltages or currents , where ESR dissipation is significant.

I highly doubt any cap discharge design at 1hz rate for LEDs is affected by the rise of ESR at 1Hz when the reactance is >50x greater.

i guess for energy storage it means efficiency is limited to 98-99% in low ESR caps due to dielectric loss resistance.

Click to expand...

How does the fact that a capacitor's impedance is 50 times its ESR prevent the capacitor from suffering the ill effects of ESR? If a large enough AC current flows in the ESR to cause undue heating of the capacitor, what difference does it make if the impedance is much larger than the ESR?

- - - Updated - - -

treez said:

thanks, and of all the components, electrolytic capacitors are the hardest ones to tell what is their internal temperature......you can get the temperature of their case, but they never come with a junction to case thermal resistance figure.

Click to expand...

You can solve this problem like this: Look up the maximum DF at 120 Hz; from this you can calculate the maximum ESR. With the maximum ESR and the maximum allowed ripple current, you can calculate the maximum allowed internal dissipation (at 85 or 105 degrees).

Use a good oscilloscope to measure the capacitor instantaneous current and voltage in your circuit. Multiply and average over a reasonably long time. This will give the average power dissipation. If it's less than the maximum allowed internal dissipation calculated in the previous paragraph, you're ok.

CDE had an app note that pointed out that if your ambient temperature is less than the allowed 85 or 105 degrees, the maximum allowed internal dissipation can be increased.

thanks, but I would only measure the current in the cap as V*I product would be the apparent power. I would measure current over one period, and do irms^2*R.

treez said:

thanks, but I would only measure the current in the cap as V*I product would be the apparent power. I would measure current over one period, and do irms^2*R.

Click to expand...

If you do as I described you will not get the apparent power.

If you were to measure I as an RMS value, and then measure V as an RMS value, the product of the two would indeed be apparent power. But that's not what I said to do.

If you measure the instantaneous I on one channel of a scope, instantaneous V on another channel. Then you use the math function of the scope to calculate the instantaneous product of the two, and then average. That gives the real power, not the apparent power.

This measurement may be difficult to make on a high quality capacitor, but it shouldn't be too difficult on a typically lossy electrolytic. I'll try it myself tomorrow or the next day.

Of course, as I said in post #17, you could just operate your circuit for a while and see how hot the capacitors become. If the ambient is approximately normal room temperature, and the capacitors are only warm, not hot, you should be ok.

I have seen manufacturer's app notes where they give typical values for the thermal resistance from capacitor internal hot spot to case as inferred from given ratings.

With most digital oscilloscopes, loss power/energy measurement can be made cascading multiply and integrator math functions. The accuracy respectively the chance to get a correct real power value when accompanied with a large reactant power is limited by digital resolution and channel offsets.

I believe that the calorimetric measurement is easier and directly giving the intended information as you already pointed out.

FvM said:

To be serious I don't get your distinction between ESR (effective series resistance) and "actual" series resistance. It makes no sense in theoretical or practical design regard, I think.

The frequency dependent ESR is the number that gives the actual capacitor losses when multiplied with squared capacitor current.

Click to expand...

I admit that frequency dependent actual series resistance does not make any sense. I wanted to mean the ESR (Resr of non-simplified series equivalent circuit model) which is independent of leakage current loss . Generally used simplified series equivalent circuit model of capacitor includes the leakage current loss in the ESR, but non-simplified model excludes the leakage current loss from the ESR. Below the pictures of the two models.



Reference link : https://en.wikipedia.org/wiki/Electrolytic_capacitor

The ESR in the simplified series equivalent circuit model rises too much when frequency becomes very low (1Hz) and it tends to infinity when frequency tends to zero. It happens due to the inclusion of leakage current loss in ESR. At very low frequency leakage current loss dominates over all losses so ESR derived from this loss mostly represents leakage current loss. It can be seen in the figure 5 of the document link: **broken link removed** (also referenced on post #20) that leakage current loss dominates over all losses at very low frequency.

The Resr in the non-simplified series equivalent circuit model does not include leakage current loss so it should not vary too much like the ESR of simplified series equivalent circuit model at very low frequency range.

This capacitor bank is charged up to 333V, and then, every second, they are discharged with a 1 Amp discharge current for 0.3 seconds, then during the proceeding 0.7 seconds they are charged back up to 333V with a 429mA charge current. This happens repeatedly.

Click to expand...

One approach to calculate the dissipated power in the capacitor bank is to use simplified series equivalent circuit model then calculate I2R loss for all the harmonic components of the capacitor current with ESR at each harmonic frequency, then add all the losses. This approach would give accurate result but due to lack of ESR information at very low frequency and being a lengthy process, this approach seems harder to me. I would use non-simplified series equivalent circuit model then get the value of Resr(at 120Hz) which is nearly equal to ESR(at 120Hz) of the datasheet. As Resr does not vary too much with frequency at very low frequency, so assuming it relatively constant, I would not go for fourier analysis, instead I would do the calculation like Irms^2*Resr or for constant charging/discharging cycle, [charging time*(charging current)^2*Resr+discharging time*(discharging current)^2*Resr]/time period. Then I would add the leakage current loss calculated separately.

So the estimated worst case power dissipated in the capacitor bank is
[0.7s*(429mA)^2*0.3ohm + 0.3s*(1A)^2*0.3ohm]/1s + leakage current loss
=0.128W + leakage current loss.

The difference between a physical plausible equivalent circuit and the simplified circuit that concentrates the real impedance part in a single frequency dependent ESR component is generally interesting, e.g. to make a LCR model, but it's irrelevant for describing the capacitor losses. If you have determined the frequency dependent (total) ESR, you know the losses for different current waveforms. In a behavioral description, you don't need to know if the losses are caused by an actual series resistance or leakage. They are just losses.

Regarding plausible models, I can tell you that the simplified two resistor model is wrong as well. To reproduce ESR versus frequency realistically, you need a complex RC ladder circuit. It can be fitted to the empirical observed impedance and gives a useful RLC model.

ahsan_i_h said:

The ESR in the simplified series equivalent circuit model rises too much when frequency becomes very low (1Hz) and it tends to infinity when frequency tends to zero. It happens due to the inclusion of leakage current loss in ESR. At very low frequency leakage current loss dominates over all losses so ESR derived from this loss mostly represents leakage current loss. It can be seen in the figure 5 of the document link: **broken link removed** (also referenced on post #20) that leakage current loss dominates over all losses at very low frequency.

Click to expand...

The image you included from Wikipedia showing Rleakage separate from Resr is contradicted by later text. Look under the heading "ESR and dissipation factor tan δ" where it says:

"ESR summarizes all resistive losses. These are the terminal resistances, the contact resistance of the electrode contact, the line resistance of the electrodes, the electrolyte resistance and the dielectric losses."

Also have a look at this application information from a large capacitor manufacturer:

https://www.cde.com/resources/catalogs/AEappGUIDE.pdf

At top of page 8 you will find this:

"The equivalent series resistance (ESR) is a single resistance representing all of the ohmic losses of the capacitor and connected in series with the capacitance"

This usage is universally used by the manufacturers of capacitors, AFAIK; they do not exclude leakage and dielectric losses from ESR. I am unaware of any exceptions to this usage.

In your posts you are ignoring the effect of dielectric loss; you haven't even mentioned dielectric loss. Why haven't you included it?

Figure 5 in this link: **broken link removed** has a mistake. The image shows the dielectric loss (blue curve) decreasing with decreasing frequency at very low frequencies, but the expression for the loss is given as D3 = 1/(ω Rd C). This expression must continue to increase as ω continues to decrease. So the continuing increase in D (and ESR) as frequency decreases is due to both leakage and dielectric loss (Rd), not leakage alone.

A person might wonder why the equivalent series resistance of the leakage should rise with decreasing ω; why shouldn't it be constant with frequency?

The reason is because leakage is a shunt loss; it appears as a resistance in parallel with the capacitor, not in series. The same is true of the dielectric loss; it is effectively a parallel loss resistance. So why does its series equivalent increase with decreasing frequency?

Both the leakage resistance and the dielectric loss resistance are effectively parallel losses, but the industry standard loss parameter ESR (which includes Rleakage and Rd) is a series equivalent resistance.

When the parallel losses are converted to a series equivalent resistance, that leads to a (series equivalent) resistance which varies with frequency. Even though the parallel resistance does not vary with frequency, its series equivalent does. Here is the reason: Figure 2 in the reference **broken link removed** shows how this occurs. The parallel resistance Rp in the figure becomes an equivalent series resistance Rs = Rp/(1+ω² Cp² Rp²). The ω factor in the denominator causes the equivalent Rs to continue to increase as the frequency decreases even if Rp itself does not vary with frequency.

Even in capacitors with completely negligible leakage, when the ESR is plotted versus frequency, there is a continuing rise in ESR at low frequencies because the parallel loss factor due to dielectric loss has been converted to a series equivalent resistance.

Here is a sweep of a good quality mylar film capacitor on the impedance analyzer. The leakage of this capacitor is difficult to measure. Using the conductance measurement of a Fluke DMM, I find that the leakage resistance is greater than 20 gigaohms. This makes a completely negligible contribution to ESR compared to all the other losses; nevertheless, you can see the continuing large rise in ESR at lower and lower frequencies:



This rise in ESR at low frequencies is due to the parallel resistance of dielectric losses converted to a series equivalent in the ESR; it is not caused by leakage.

In a good quality electrolytic with low leakage, the rise in ESR at low frequencies is similarly due to dielectric losses. This is not to say that leakage cannot contribute to ESR rise at low frequencies, but it won't be the dominant factor when the leakage is small as it should be in a new, good quality electrolytic. Dielectric losses cannot be ignored.

The Electrician, what analyzer are you using.

Also, is there any combination of capacitor types you can parallel up to get low ESR at low frequency's (1 HZ).

FlapJack said:

The Electrician, what analyzer are you using.

Also, is there any combination of capacitor types you can parallel up to get low ESR at low frequency's (1 HZ).

Click to expand...

The analyzer is a Hioki IM3570:
**broken link removed**

Since the ESR at very low frequency is highly dependent on the dielectric losses, having a low ESR there requires a capacitor with very low dielectric losses. Mylar (polyester) is not good as you see in the sweep above. Polypropylene is good. The best film capacitor I've seen is a GE motor run capacitor.

With respect to electrolytics, I've seen some that don't have much rise in ESR at very low frequencies, but since the manufacturers don't generally spec for that, a person might have to investigate for himself among various electrolytics. And, if you do find a good one, the manufacturer might change their formula in the future.

As far as paralleling caps, it seems the best one could do would be to find a really good one, and parallel a bunch of the same ones.

@ The Electrician, I agreed that rapid rise of ESR at 4Hz on the pictures of your posts are due to dielectric loss. Because when I convert any reasonable parallel leakage resistance to series resistance to contribute in ESR at 4Hz, the converted series resistance value is well below than the value of the ESR at 4Hz supplied by the graph. I guess at the frequency when leakage loss starts to dominate over all losses is far below than 4Hz, it must be very very low frequency.
In my opinion, temperature measurement is easier way to get an idea about the loss as you already pointed out in post #17.

Electrolytics are commonly spiral wound and very inductive (as well as
having the long / high aspect ratio sheet resistance). Point being that,
at some standard test condition like 1MHz, your ESR datasheet value
may be large and not too relevant to a 1Hz application where only ESR
(real R, not inductive effective R) applies.

However just because the cycle time is 1Hz does not mean that you
do not care about ESL. You have to look at your discharge event
and call the discharge time a half cycle, and then what is your
care-about frequency on that basis?

So how much are the losses with your design?

given ;

- tan delta at 120 Hz (=0.163 ohms) per five caps = 220uF @333V or CV= 73260 [uFV]
- the datasheet specifies leakage current as ;

After 1 minute's application of rated voltage, CV > 1000 :I= 0.04CV+100 (µA) or less

- thus 0.04*73260 = 2.93 mA per cap @ 333V = 976mW worst case.

Lets call it charging power loss Pc= 1W

- Endurance spec indicates Less than or equal to the initial specified value ( ! )
- Caps are poor heat conductors but no thermal resistance is given.

- The leakage is equivalent to 333V/2.93mA = 113 kΩ
- for an RC decay rate of 113 kΩ* 220uF ~ 25 seconds or a breakpoint in rising ESR of near ~ 0.01 Hz ( ! )

The impulse discharge current will be regulated by the LED array ESR.

- Since the leakage time constant is much longer than the discharge periodic rate, the tan delta at 120Hz or Rs=0.163 Ω is valid a
nd added to the ESR of the LED string to determine the impulse discharge time constant.
- ESL can be a factor including wiring inductance , but is unknown and needs more input.

- The Discharge for cap. from total current is Ic= I/5 ( assuming matched ESR)
- The power dissipated in the cap during discharge is Pd = I²Rs*D for duty cycle, D = discharge time (RC) over discharge repetition interval.

Treez, you have the more accurate test results to calculate total discharge heat loss.

As the "Electrician" said you cannot ignore dielectric loss, nor ignore discharge loss.

The heat loss is Pc +Pd
Pc from the parallel leakage resistance during charging
AND Pd from the series resistance during discharge. as I_discharge >> I_leak

Both values are combined into ESR as function of frequency but must be converted to parallel when using predominantly average DC during charging and impulses during discharging.
This is why they do not specify ESR. However for rapid pulse applications, when ultralow ESR caps are required, they do specify ESR as voltage tends to be low and leakage power as well.

I presume the discharge voltage < 25% of 333V.