Thanks for the explanation. I read through the sensor's patent again. The s= ensor is powered just as you had explained it to me. Sincerely, Sai On Friday, January 3, 2014 6:16 PM, IVP wrote: =20 > Did the protection circuitry of the power supply cause the problem > I was having? Hi Sai, No, your power supply has a very low output impedance, compared with the sensor. As Bob said, it is too "stiff" for the sensor to overcome this low impedance and cause a significant dip in the PSU's o/p voltage What the resistor does is limit the current, or raise the output impedance of the PSU (as the sensor sees it) high enough for the sensor's drive circuitry to be able to pull the 'cold' side of the 100R down to 0V. The supply voltage for the sensor will be fairly stable as it presumably will have a reservoir capacitor and isolating diode on its Vcc pin that keeps it powered during the pulse It's very much like a pull-up on a micro output pin like RA4. You wouldn't expect RA4 to be able to overpower a battery or PSU Your overall circuit is probably something like the attached Joe --=20 http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist --=20 http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .