The datasheet says IOut max for analog devices is 50 µA.
Do I have limit this current myself? I plan on connecting the output of the PIR to an ADC input on a PIC microcontroller. The PIC ADC input impedance is 10K. If I connected the PIR output directly to that then 5V / 10K = 500µA which would exceed the 50µA max of the PIR.
Also,
On page 20 there is a wiring diagram (attached pic).
What is RL for? (connected from the sensor's output to ground).
I'm guessing it means "R-Low" and it's a pull-down resistor?
What value should it be? Anything under 100K would exceed the 50µA max.
No that is the analog diagram (digital is attached to this post).
Oh, and sorry about my mistake on ADC input impedance (10K is the recommeded _source_ impedance for the ADC). The input impedance on my current chip (PIC 12F1822) is actually about 7K (@5V Vdd).
Still -- that's even lower. Can I damage the PIR by connecting it's output directly to the PIC's ADC input???
No, it's the typical sampling switch series resistance which determines the < 1 µs current peak during acquisition. Can be ignored related to sensor output current rating.
No, it's the typical sampling switch series resistance which determines the < 1 µs current peak during acquisition. Can be ignored related to sensor output current rating.