The potential which we sample comes from some source with non-zero internal resistance. We thus recognize that, when the switch is closed, we are looking at an RC circuit. The capacitance is obvious -- it's the C in the circuit. What about the R? The output resistance of typical operational amplifiers is between 1 Ω and 1000 Ω. If we always put a buffer amplifer before the switch, then the RC time constant for this circuit is between C seconds and 1000C seconds, with C in Farads. Review time!

Fill in the following table. What is the RC time constant for each combination of R and C?

1 pF

1 nF

1 μF

1 Ω

10 Ω

100 Ω

1000 Ω

From this, we see that the rate of response can vary over many orders of magnitude. The lower the capacitance, the less charge is needed to represent a given potential, but the greater the influence of shot noise on the measurement. 1 V stored on 1 pF is about 16 million electrons, which has an uncertainty of 1 part in 4000. Such a small capacitor is inappropriate for even a 12 bit converter, much less something with greater resolution. 1 V stored on 1 μF is about 1.6 × 10^{12} electrons, with an uncertainty of 1 part in 4 × 10^{6}, about the resolution of a 24 bit converter.

To think about: even without any knowledge of the inside workings of an analog-to-digital converter, why will high resolution conversions always be slower then low-resolution conversions?