# What is an RC time constant and why do I have to wait for the capacitor to carge up before taking samples with ADC?

I am using LTC6403 in my project. It is used for single-ended to differential conversion, which will be driving LTC2185 ADC.

In the application information of "Interfacing the LTC6403-1 to A/D Converters" on page 19 of the datasheet, it is mentioned to have a discrete RC filter between Differential outputs of the amplifier and ADC input signals.

In the second paragraph, it is mentioned that " 16-bit applications require a minimum of 11 R-C time constants to settle". What is the meaning of this? I know that 16 bit refers to the number of bits of ADC.

The time constant of an RC element is calculated $$\ \tau = R \cdot C \$$.

Example: With $$\R = 10 k\Omega \$$ and $$\C=10nF\$$ this gives you $$\\tau = 10k\Omega \cdot 10nF = 100µs\$$.
If you have to wait for 11 RC time constants you have to wait for $$\11 \cdot \tau\$$, in the example above this would be 1.1ms.

Details: $$\\tau\$$ corresponds to the time it would take a capacitor to charge up to the applied voltage over a resistor R, if the initial charging speed (at t=0) would charge the whole capacitor. As you probably know, the charging speed decreased while charging, because the voltage drop over the resistor is decreasing. So after 1 $$\\tau \$$ the capacitor is only charged to 63 %.

In typical applications it is assumed that a capacitor is charged up to full voltage after $$\5\tau\$$ (> 99%), but 99% is not enough if you want to measure with 16 Bit resolution:

$$1 - \frac{1}{2^{16}} = 1 - \frac{1}{65536} = 1 - 0.00001526... = 0.99998474...$$

So the input capacitance has to be charged up to 99.99847%. With a charging of the capacitor with

$$U_{(t)} = U_{max} \cdot (1-e^{-\frac{t}{\tau}})$$

this is reached after

$$\frac{U_{(t)}}{U_{max}} = 0.99998474 = (1-e^{-\frac{t}{\tau}})$$ $$e^{-\frac{t}{\tau}} = 0.00001256$$ $${\frac{t}{\tau}} = 11.09$$ $$t = 11.09 \tau = 11.09 \cdot R\cdot C$$

As you can see, it takes the capacitor little over 11 RC time constants to charge to a high enough voltage to measure with 16 bit accuracy.

• What would be impact of not maintaining this contstant? Commented Feb 19, 2020 at 7:42
• I added a detailed explanation. As you can see you will get smaller conversion results when not waiting that long, because the capacitor is not charged up far enough. Commented Feb 19, 2020 at 7:48