the discharge time of 10uF capacitor which put in the VDD pin of 51822

RE: the discharge time of 10uF capacitor which put in the VDD pin of 51822

bjorn-spockeli — Thu, 17 Dec 2015 09:37:38 GMT

If this answered your question I would appreciate if you could mark it as correct and accept the answer.

jessie — Wed, 16 Dec 2015 11:46:29 GMT

Clearly. Thank you very much.

bjorn-spockeli — Wed, 16 Dec 2015 10:40:43 GMT

The voltage across a capacitor with constant current is given by

U(t) = U(t=0) + (I*t)/C )

Solving the equation for t to obtain the discharge time

t = C*( U- U(t=0) )/I

The power management system of the chip will put the chip in a reset state if the supply voltage is below 1.7 V, i.e. Brown-out reset.

Thus with U = 1.7V, U(t=0) = 3V, I = -0.33uA, C = 10uF the discharge time is

t = 10uF * (1.7V -3.0V)/-0.33uA = 39s

and the voltage after 10 seconds will be

U(t=10) = 3V + (-0.33uA*10s)/10uF = 2.67 V