The following information is for the EMANT300, EMANT380

Measure a medical grade 1.2M @ 25c thermistor

Question: I would like to use the EMANT300 USB DAQ to measure a medical grade thermistor which has a resistance @ 25c of around 1.2M. I know that I will need to reduce the current excitation level to below 39uA, what would be the best way to do this?

Buffered Analog Input Schematic

Answer: Connect the thermistor to the 2.5V voltage reference through a 100K 1% resistor, the voltage (V) across the 100K resistor is

V = 2.5*100000/ (R_th+100000)

where R_th is the resistance of the thermistor in ohms.

Moving the equation around gives

R_th = (250000 - 100000V)/V

Enable the input buffer so that the analog input impedance does not load your circuit. Requires the latest development language driver. Download driver and example program from info.emant.com (for EMANT customers only)

LabVIEW Diagram

LabVIEW Buffered Analog Input

C# code

Emant300 DAQ = new Emant300();
DAQ.Open();
DAQ.ConfigAnalog(2.5, Emant300.POLARITY.Unipolar, 100, Emant300.VREF.V2_5, true);
volt = DAQ.ReadAnalog(Emant300.AIN.AIN3, Emant300.AIN.AIN2);
R = (250000 - 100000 * volt) / volt;
Console.WriteLine(R);
DAQ.Close();