This is hard to answer without knowing the what exactly is going on inside the control loop, which is probably not described with much detail in the datasheet. Saying it is "optimized" for 30 mΩ doesn't say what it really needs or what happens when it's not exactly 30 mΩ. Since they make a issue of the output cap ESR, there should be a min/max ESR spec elsewhere in the datasheet.

Hopefully 0 ESR is OK, else it gets inconvenient. Lower ESR is generally better, and certainly closer to ideal, for a capacitor. Capacitor spec sheets therefore often only spec *maximum* ESR. In that case you either have to get the guaranteed minimum from the capacitor manufacturer, or add deliberate resistance. A better answer is to stay away from chips that specify a minimum output cap ESR to be stable.

That chip is probably old, as ceramic capacitors have come a long way in the last few years. These have much lower ESR than tantalum, and can now approach such capacitances at reasonable cost. Newer chips now make use of the lower available ESR to get better performance instead of requiring the capacitor to have a high resistance. This is a all around better strategy.

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

It's quite useful for repair purposes. The voltages are low enough that semiconductors don't come into play. When there are parallel e-caps, you should be comparing a total ESR of the parallel combination, but you're really looking for gross differences not even the last 2:1. The typical situation is an older piece of electronics where the electrolyte dries out, causing the capacitance to change a bit, but mostly the ESR to increase, so you get mains ripple or SMPS ripple, which eventually causes the thing to stop working adequately. E-caps do eventually wear out, and replacing them can often restore an older item. Especially the caps in the power supply- because they usually have less margin and because they tend to be in a hotter area and have more self-heating (due to ripple current and ESR-- so the higher ESR gets, the more self-heating occurs).

The other factor affecting usability is that a lot of recent electronics uses low-Z caps, so what's excellent for a standard e-cap is a bulging mess on a low-Z cap. You may have to look up the part number or guess from the schematic or circuit configuration (something like 1000uF/6.3V on a motherboard is just about guaranteed to be low-Z).

I wish I had one of those, there's a few times it would come in handy, but I don't do enough repair work to justify it. Usually I just tack a known-good cap across the suspect one and see if it makes a difference!