# Analig circuit, explanation

#### wantedSpidy

##### Senior member
Can anyone explain to me, how the author came up with the times of 4.5-10 mins by adjusting the 1M potentiometer?

Does it have to do with the time it takes to charge the 220uF capacitor?

#### PsYcHoCoW

##### Member
Indeed, the potentiometer lowers or increases time required to charge the 220uF capacitor, which determines the output timing...

http://cache.national.com/ds/LM/LM555.pdf

#### bobsmith1492

##### Diamond Member
Capacitors charge exponentially; on chargeup, their voltage is Vin(1-e^(-t/(RC))). 555s trigger at about 1/3Vin or 2/3 Vin (depending on the setup) assuming the same supply as is charging the capacitor, so the equation relating your time to your resistance and capacitance would be:

2/3Vin = Vin(1-e^(-t/(RC))) -> Vin cancels, good - the time is independent (roughly) of the supply).

2/3-1 = -e^(-t/(RC));
1/3 = e^(-t/(RC));
ln(1/3) = -t/(RC);
t = -ln(1/3)(RC)

Done! That's your turn-on trigger time given R and C.

For 1M and 220uF, t = -ln(1/3)((10^6)(220*10^-6)) = -ln(1/3)220 ~= 241.7 seconds, or 3 minutes.

Plug in your numbers and have at it.

Thanks!!