Looks like a good idea at first glance. No immediately obvious "gotchas.
Won't compete with the best test instruments but should be viable.
I'd guess there would be some learning required along the way for a practical device.
These two Burr Brown application notes were cited in your reference. Don't seem to specifically address the test issue BUT cover the high impedance fron end issues well enough that they will provide useful material.
SBOA035 -
PHOTODIODE MONITORING WITH OP AMP
SBOA060 -
NOISE ANALYSIS OF FET TRANSIMPEDANCE AMPLIFIER
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 sounds like you know the current (800uA when 8kV is applied).
It's possible (with a great deal of care) to simply put a milliammeter in series with the capacitor. You should put it in series with the grounded side of the supply so that the meter leads are at approximately ground potential and make sure that connection is reliable. While 800uA is probably not enough to cause injury, it would not be nice. Contact with your supply (the ungrounded side) could result in much higher (possibly fatal) current.
Note that at 8kV you may be getting corona discharge (especially if you have pointy bits facing a strong electric field gradient). This behaves similarly to leakage through the dielectric at a fixed voltage, but will disappear at a somewhat lower voltage (say 1kV).
Edit:
simulate this circuit – Schematic created using CircuitLab
The 10W resistor is physically large and intended to not break over if the capacitor shorts. If your supply is capable of a lot of current, more precautions may be necessary so this is just a preliminary though. Safety is paramount, so get this checked out by someone local and competent before flipping the power on.