## Beginner question: capacitor & 555

 Hello, I'm making my way through an electronics intro book. I've just started on ICs, specifically the 555. (i couldnt find a 555 symbol so i used two switches in the diagram). it's going to be part of an alarm system. I'm told to measure the voltage on my meter, from the left of R1, to ground. 9V, slowly reduces to 1/3 and the LED turns on. it all works fine and well, but I'm wondering where the Volts are coming from in the first place? I thought the capacitor blocks the DC,, and the trigger pin(2) of the 555timer doesn't receive volts? sorry if the answer is right there, but i've been trying to figure it out for a while Thanks!! by BeefJeff October 27, 2021 That is true that a capacitor untimely blocks DC, but that is about charging and discharging the capacitor first. In a "DC source-resistor-capacitor" loop, if you start with an uncharged capacitor, consider that the capacitor is like two plates in parallel. One side of the battery (the - pole) will emit electrons toward its connected plate of the capacitor. The electrons will be stopped there. But they repulse themselves and repulse the electrons of the other plate too, and these one, the electrons of the second plate, will start moving on toward the + side of the battery. So, on total, even if no single electron did make the whole trip, there is a current, nonetheless. (Like a relay race). At one point though, there will be so many electrons accumulated on the (-) plate that the DC source won't be able to push more electrons on it. Same, for the (+) plate of the capacitor, so many electrons will have left, that any additional electron trying to leave toward the + side of the battery will simply be attracted back by the atoms missing an electron (and thus, being locally charged positively). That is why the initial current will gradually decrease and finally, essentially stop (statistically, on average, even if noise is generally observable). Under constant voltage for the DC source, it is considered that the capacitor is fully charged after a time, in second, equals to 5RC, R in Ohm and C in Farad. So, for R of 1 MegaOhm and F of 1 microFarad, that results into 5 seconds. To reach one third of the source's DC voltage, at the plates of the capacitor, the time required is around 1.1 R C. by vanderghast October 28, 2021 thank you for this explanation, it's very helpful. Do all Capacitors work in this fashion? where they allow this 'Fake' current though.. or only ones in an RC loop? when the empty capacitor is connected to the power in my circuit, does it (almost) instantaneously reach 9V? because that's what my meter tells me. I would have thought it needs time to charge up to 9V to then slowly release it. sorry for all the questions thanks again! by BeefJeff October 28, 2021 Without resistor, R is almost zero ohm (the source and the capacitor have a small resistance valeur, negligeable most of the time). So 5RC equals 0, which is almost instantaneous. You can see it is not exactly zero with an oscilloscope set into the nano seconds. An ideal capacitor works like that, in a completed loop. A disconnected capacitor or connected to just one leg does not, ideally, change its stored voltage (the voltage of the charge at its plates). At a very high frequency, a real capacitor changes its behavior. by vanderghast October 28, 2021 awesome, thanks vanderghast! i think i have a much better understanding of what is going on :D by BeefJeff October 28, 2021

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