Estimating self-discharge rate of ceramic capacitors from datasheet values DF and ESR

Question

I'm trying to make an estimate for self-discharge rate of a typical small capacitor, such as 0.1 uF ceramic capacitor. Horowitz and Hill says "A better measure of loss at low frequencies is Dissipation Factor [...] the low-frequency loss mechanism is dielectric loss, not metallic resistivity." (Art of Electronics, the X Chapters, p36).

Kemet's datasheet for one of its most common parts C0805C104K5RACTU (0.1 μF, 10%, 50 VDC, X7R) says:

• Dissipation factor: 2.5% 1 kHz 1.0 Vrms
• ESR at 100 Hz: ~150 Ω (The ESR graph only goes down to 100 Hz.)

Can we use these values to estimate the decay time to say 1% of original? Following the capacitor charging rule, V = e^-t/RC, I'm hoping for something like t ≈ 5RC, but don't see what to use for R.

Am I chasing rainbows or is there a simple way to estimate this?

A test circuit would be this: after C1 is completely charged, we disconnect from the charging circuit and measure the V_output.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

I'm trying to make an estimate for self-discharge rate of a typical small capacitor

You should read the extended data sheet and look at the insulation resistance on page 14: -

So, I calculate the insulation resistance to be 10 GΩ using their guidance. This will determine the typical CR decay time (1000 seconds for a 100 nF capacitor).

What you linked to is what Kemet call their spec sheet.

after C1 is completely charged, we disconnect from the charging circuit and measure the V_output.

Most typical multimeters will have a 10 MΩ voltage measuring input resistance and, this would be a thousand times lower than the insulation resistance = ineffective test.

Estimating self-discharge rate of ceramic capacitors from datasheet values DF and ESR

Dissipation factor is for AC scenarios and at DC, this would imply an infinite resistance because reactance would be infinite (\$D.F. = \frac{R_S}{X_C}\$). Insulation resistance takes over at very low frequencies because it becomes dominant.