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.

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1 Answer 1


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 %.

enter image description here Source: https://de.wikipedia.org/wiki/RC-Glied

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.

  • \$\begingroup\$ What would be impact of not maintaining this contstant? \$\endgroup\$ Commented Feb 19, 2020 at 7:42
  • 3
    \$\begingroup\$ 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. \$\endgroup\$
    – jusaca
    Commented Feb 19, 2020 at 7:48

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