In many instances, the actual stray capacitance is closer to 10 pF than 1 pF. If the amplifier is driving a coaxial cable, the cable has a capacitance of 4 pF/foot (12 pF/meter). That slows things down even more. Why not use smaller resistance? Then the current goes up together with power dissipation. Why not use smaller capacitance? Any pair of conductors has mutual capacitance, so there's a floor underneath the capacitance of the circuit.

The above argues for the RELATIVE error in jumping from one potential to another. What is the ABSOLUTE error? Let's consider two scenarios, both with a 16 bit straight binary DAC and with the same 10 kΩ resistor and 1 pF capacitance we've used in the previous exercise.

1) Big Jump. In the first scenario, a jump is made from 0 to 2^{16}-1 in a single step. We already know that in 110 ns, the DAC has settled within 1 least-significant bit (LSB) of the final value. But what voltage is that? Assume that the DAC has a range from 0 to 65535/65536 of 10.00000V. 1 LSB is then 1/65536 of 10.00000 V = 153 μV.

2) Little Jump. Suppose the setting is changed by only 1 count. Then the increment or decrement of the output is only 153 μV altogether. It will still take 110 ns to get within 1/65536 of the amount of the change, but can we detect that small a voltage? 1/65536 * 153 μV = 2.3 nV. In this case, how long it takes for the output to settle depends on the resolution of the measurement. If we can only measure 1 μV, then we, in effect, only have to wait to settle to 1 part in 153. But from above, that takes only 5 time constants or 50 ns. The bigger the jump, the longer it takes for the output to settle within our ability to detect change.