If ESR for a capacitor is not given, can you use f, C, and Q to calculate it?

To calculate ESR when it is not given, can you use the $$\Q\$$ value as follows?

According to Wikipedia, dissipation factor relates $$\Q\$$, $$\\text{ESR}\$$, and $$\\tan(\delta)\$$ as follows:

$$\frac{1}{Q} = \tan(\delta) = \frac{\text{ESR}}{\left|X_c\right|} = \text{DF}$$

Thus, if you know Q and $$\f\$$,

$$\frac{1}{Q} = \frac{\text{ESR}}{\left|X_c\right|} = 2\pi fC\cdot \text{ESR}$$

where

$$|X_c|=\frac{1}{2\pi fC}$$

and therefore,

$$\text{ESR} = \frac{1}{2\pi fC\cdot Q}$$

Pluging in numbers from a datasheet where Q >= 1000, f = 1GHz, C=1nF, I get an ESR of 0.159 $$\m\Omega\$$ (0.000159 $$\\Omega\$$), which seems far too small.

Question: Is my math wrong, or are datasheet Q values just too inaccurate to be useful in this context?

N.B., this is what I plugged into my calculator: 1/(2 * 3.14 * 1e9 * (1000e-12) * 1000)

Edit: @BitLauncher points out that 1GHz is above the cap's resonant frequency and that Q is only measured at 1MHz. So, recalculating with different frequencies. (Also I had a typo that has been fixed, I was doing 1000nF not 1000pF, so fixed and updated the original value).

with a 1000pF cap, assuming Q=1000:

• 1 MHz: 159 $$\m\Omega\$$
• 100 MHz: 1.59 $$\m\Omega\$$

and with a 10pF cap, assuming Q=1000:

• 1 MHz: 15.9 $$\\Omega\$$
• 1000 MHz: 15.9 $$\m\Omega\$$
• Which capacitor in the datasheet did you use? Commented Jul 20 at 22:15
• I like the question, +1, but if ESR is not given and it's important to you, you shouldn't use that capacitor. Commented Jul 20 at 22:15
• @RussellH, C=1nF. Thanks, I missed that info, updated question. Commented Jul 20 at 22:19
• Oh, I didn't see the link. Kyocera likely has an online tool to parameterize their caps. Commented Jul 20 at 22:20
• @BlairFonville, I agree with that assessment. This question is mostly a curiosity to see if this is even reasonably possible. Commented Jul 20 at 22:21