# help understanding capacitors

Discussion in 'General Electronics Chat' started by bodisha, May 29, 2013.

1. ### bodisha Thread Starter New Member

May 29, 2013
2
0
Hello and thanks for any help anyone can offer me... I apologize if this isn't the correct forum to post this in

I'm trying to understand capacitors... And so far I understand the math behind charging & discharging, so that's not the issue. What I'm having a hard time understanding is what event triggers the discharge of a capacitor in an AC circuit.

Everything I've read simply says as the voltage starts to decrease, a capacitor will discharge. I have yet to read anything that states exactly when a capacitor will discharge as the voltage decreases. Is it as soon as it crosses the peak voltage and begins to decrease? When it crosses the threshold from positive to negative? When it reaches negative peak voltage?

I apologize for the noob question... and once again thank you to anyone that can help me understand this.

2. ### LDC3 Active Member

Apr 27, 2013
920
161
A capacitor will start to discharge whenever the potential across the capacitor is greater then the voltage on the wires. For example, if you charge a capacitor with 5 volts, the electrons pushed into the capacitor is in equilibrium with the potential. When the voltage decreases (even by a small amount), the electrons push their way out of the capacitor until equilibrium is again reached.

3. ### #12 Expert

Nov 30, 2010
18,078
9,619
Capacitors aren't very smart. They don't actually decide when to charge or discharge. They are merely pushed around by the voltage. At the very moment the voltage applied is different from the voltage on the capacitor plates, current flows in or out of the capacitor in the direction to make the capacitor voltage become equal to the applied voltage.

It's always difficult to find the right words to make it click for you. That's a major advantage of a forum. Different people say the same thing in different ways until one of them makes sense to you.

PackratKing and Metalmann like this.
4. ### Mike33 AAC Fanatic!

Feb 4, 2005
349
25
Right, #12 - sometimes it just takes a while to get the concept. It sure too me a long time!

Capacitors will begin to discharge just as soon as the applied voltage goes below the voltage they are charged at; and if the voltage comes back up, they stop discharging and re-charge as that voltage goes above whatever level the capacitor is now at.

Now, if the voltage continued dropping, the cap would discharge completely, of course. Without resistance in series with it, it would lose its charge very quickly (as the OP stated, via that math). Or, if the applied voltage came back up to a new level (say, didn't go back to 5V, but only 3V), the cap would charge that that NEW level and stay there as long as the applied voltage is at that new level! Capacitors just follow along, lagging a little bit behind the power supply they're connected to, like a buffer, in a way.

That's why some think of them sort of as small 'batteries', sometimes. This is how they are able to filter varying voltages....by discharging during the 'gaps' when a voltage decreases, and charging when the voltage 'comes back'. If the drop and return to proper voltage is fast enough, the cap fills in that gap. If it's longer, or the capacitance is too small, then the charge on the cap also drops.

5. ### bodisha Thread Starter New Member

May 29, 2013
2
0
I'd like to thank everyone for their answers. It helped me understand capacitors better. To reiterate what everyone was saying.. Just to make sure I understand. After a capacitor is 100% charged, the voltage of the capacitor will discharge to a voltage level to match the applied voltage.

6. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
523
Just a small point.

There is no such thing as 100% charge on a capacitor.

The voltage across the capacitor is never greater than the charging voltage, so it doe not 'discharge to match the charging voltage'

Start from zero across the capacitor and apply some charging voltage V.

Current (charge) flows into the capacitor from V, at a rate deopending upon the resistance between the two (even connecting wires have some resistance).

Initially all the voltage V is across the resistor and the voltage across the capacitor is zero.

As the charge flows in the voltage across the resistor (Vr) diminishes and the voltage across the capacitor (Vc) increases.

The sum of these voltages is always equal to the charging voltage.

Vr + Vc = V at all times.

Eventually when enough charge has entered the capacitor Vc = V for all practical purposes and we say the capacitor is fully charged (to V).

However for a different applied voltage say V' the time and quantity of charge would be different.

Does this help?