Preparing Signal for Sampling

Amplifier U1 helps to make the circuit as close to "ideal" as possible. In the pH cell the impedances involved re very high and any variations from ideality are reflected in the results.

The "challenge" is given in the following section from the application note:

• The output of amplifier U1, which is set up in a unity-gain configuration, biases the reference electrode of the pH electrode with the same voltage, 512 mV, at low impedance.

  The pH-measuring electrode will produce a voltage which rides on top of this 512 mV bias voltage. In effect, the circuit shifts the bipolar pH-electrode signal to a unipolar signal for use in a single-supply system.

ie any error in this voltage is directly reflected in the output voltage as an error in pH reading.

The source impedance of the 2 x 10 k resistors in series is 5 K ( Reffective = R1 x R2 / (R1+ R2)). If the cell were to load this with a 1 megohm impedance the change in actual voltage would be 5k // 1 M = 0.005 = 0.5%. Loading with 10 Megohm would give 0.05% error etc. This does not sound much (& isn't much) but the sensitivity of the cell is 1 mV per pH. So 5/1000 x 512 mV ~= 2.5 mV or 2.5 pH error. And 10 megohm loading = 0.25 pH error. Even 100 megohm loading = 0.025 pH error.

If pH is read to even 0.1 pH units an error of 0.025 pH is 1/4 of a "bit". If pH was read to 0.01 units then 0.025 pH = 2.5 "bits " - and that's with 100 only megohm load!

Reducing R1 & R2 to 1 k or 100 ohms would help, at the expense of increased current drain U1 provides a better solution at acceptable cost.

If you have a negative power rail then use a shunt voltage reference - basically it's a precision zener diode so it can work on either supply rail. On the negative rail the cathode connects to ground and a resistor feeds the anode from the rail.

Here is one from LT - it's the LT1389 and the first one I found. It's a 1.25V reference and you can use two resistors to divide this down to 300mV. You need to feed it with a minimum of 0.6 uA but because you'll have a potential divider in parallel you'll need to feed the reference and potential divider with a slightly bigger current. You might as well use 1 mA (the device is rated up to 20 mA anyway).