pull down resistor in PIR circuit
Well, to calculate it the mathematical way, set it up as a ratio and solve for X:

47k / 5v = x / 8v

Get rid of "/8" by doing the opposite to the left:

(47 * 8) / 5 = x

376 / 5 = x

x = 73k. That will provide a scaled pull-down resistor value for 8v instead of 5v. We can check the actual current by using Ohm's Law. Assuming the full supply voltage was being imposed across the pulldown resistor,

I = E / R       I = E / R
I = 5v / 47k    I = 8v / 73k
I = 0.000106A   I = 0.000109A
I = 106uA       I = 109uA


Close enough. Give it a try, should work just fine. Test a +/-20% variance and see what effect it has on the circuit. I doubt the value is super-critical, but one particular value may work best.

73k * 0.20 + 73k = 87.6k

73k - 73k * 0.20 = 58.4k

If a particular pull-down current is specified as best, then calculate the resistor value for the given operating voltage by:

E = I * R

R = E / I

R = 12v / 100uA

R = 120k