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Chemistry LibreTexts

Chronopotentiometry/ Chronoamperometry Answer

  1. Since potentials can be positive or negative, we need a bipolar ADC. Because we might need potentials greater than 2.5 V, probably choose an ADC with ±5 V range. One rarely needs finer than 1 mV resolution; to get 1 mV out of 5 V is 1 part in 5000. That's an overall range of 1 part in 10,000. 13 bits is 1 part in 8192, so amazingly we need a 14 bit converter, just as in problem 1.
  2. Since current can have either sign, again choose a bipolar converter.
  3. ±10 V is a common maximum range for a bipolar ADC, and this ensures that we're well away from the ±15 V limits of the power supplies likely to be used. At 1 V/μA, that's ±10 μA range.
  4. 16 bits resolution says that each polarity is digitized with resolution of 1 part in 32768. 10 μA /32768 = about 1/3300 μA = 1/3.3 nA = 300 pA. At 1V/μA each bit corresponds to 1/3300 V = 1/3.3 mV = 300 μV. It is by no means obvious that 10 nV Hz-1/2 will be noisier than the shot noise in 300 pA observed for a short time, but the problem says to assume so, so we will. We must choose a bandwidth such that the noise is expected to be 300 μV. 0.01 μV Hz-1/2 × Δf1/2 = 300 μV. Δf = (30000)2 = 900 MHz. This seems absurd -- that's one measurement every nanosecond! 300 pA is 3 × 10-10 A = approximately 1.8 × 109 electrons per second. Thus, in 1 ns, we'd see only 2 electrons, and shot noise alone would give S/N ~ 1.4. What this means is that shot noise in the current rather than noise in the amplifier is what limits precision. To have S/N = 3 in 1 ns, we need 9 e-/ns or 1 nA. Amazingly, to get S/N = 3 from a 300 pA current, we need observe for only 4.5 ns, allowing digitization at 200 MHz.
  5. Besides amplifier noise, there's shot noise in the signal, shot noise in the background current, power supply ripple, and any noise in stirrers or mass transfer. Temperature drift could also be a problem.
  6. We've seen that 16 bits is adequate to the task. See for example the ADS8472.