I am looking at SLAU208P Revised October 2016. I am in the ADC12_A section 28.2.5.3 Sample Time Considerations Page 737. It says:
The resistance of the source RS and RI affect tsample. The following equation can be used to calculate the
minimum sampling time tsample for a n-bit conversion, where n equals the bits of resolution:
t
sample > (RS + RI) × ln(2n+1) × CI + 800 ns
Substituting the values for RI and CI given above, the equation becomes:
t
sample > (RS + 1.8 kΩ) × ln(2n+1) × 25 pF + 800 ns
For example, for 12-bit resolution, if RS is 10 kΩ, tsample must be greater than 3.46 µs.
How do you get from
(10 kOhm + 1.8kOhm) * ln(2^(13)) * 25 pF + 800 ns to being greater than 3.46 us?
If I do not convert units I get 11.8 kOhms * 2.564949 * 25 pF = 756.66 which I can not see any way I can add 800 ns to indicate it must be greater than 3.46 us
Normally I would think you go to base units so 11800 Ohms * 2.564949 * 0.000000025 F = .000757 seconds and then add 800 ns you get 0.0007578 sec or 757.8 us.
I have so many problems with equations in TI documentation. One more intermediate step would clear up a lot. But you never do it.
What am I missing? How does this work?
Since I am asking, in my case I have about 50 volts through a resister divider of 237.4K on top, and 10K on bottom (to ground) with a .1 MFD in parallel with the 10K on the bottom. I would guess the 0.1 MFD would swamp the 25 pF on the other side of the 1.8K resistor, so I should model it as just the 1.8KOhm resistor. Is that correct? But I still need to know how the equation works.
Kip
