Question
I plan to use the AD7709 to read two temperature signals with two PT100. One
application on data sheet shows a solution based on internal current source.
However, in this way I can reach only a 20mV input range and, a 13bit
resolution. So, I plan to use an external current source of about 1mA in order
to have a better resolution. How accurate must be the external source in order
to reach my goal? i.e. is ok the same grade of the internal source (200ppm) or
not?
Answer
If the analog input range is +/-20mV and the reference is 2.5V, the customer
will get 13 bits of resolution at the default update rate (19.79Hz). By using
a 1 mA current source and using a 5k precision resistor to generate the
reference voltage, you will have a reference voltage of 2.5V and the analog
input range will be 0.1V max. So, you need to set the gain to 16 so the
allowable span is +/-160mV. The rms noise is 0.65 uV under these conditions.
So, the p-p resolution will be 14 bits for an analog input signal of 0 - 100mV
or 15 bits for an analog input range of +/-100mV. The external current source
should have similar performance to the internal current source.
The AD7792 is a 16-bit part which has programmable excitation currents. The
current sources can be programmed to 1 mA. The rrms noise is 0.18uV when the
gain equals 16 and an external 2.5V reference is selected. So, for an analog
input signal of 0 - 100mV, the p-p resolution will be 16 bits. This may be a
better choice if you are only beginning the design since it saves on
purchasing an external current source.