Q
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?
A
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.