Capacitor question. - Page 1
Forum section: electronics / beginners

author: Jimbz
author_rank: Contributor
timestamp: January 17, 2014, 01:11:49 am
content: Greetings everyone, I have a small question regarding capacitor.I was reading the section about capacitors in allaboutcircuits.com and I'd like to ask a question.Say we fully charge a capacitor at some 15 volts and remove it from the circuit. Now let us discharge that same capacitor on some random active load in a circuit without a power supply. My question is: what determines the peak voltage of a fully charged cap when discharging it ? Voltage is highest on the first moment of discharge I understand that, but how do we calculate that ? Does a capacitor that is not in a circuit keep storing its energy in the form of an electric field and we can perhaps continue from there ?Thanks for reading.

author: c4757p
author_rank: Super Contributor
timestamp: January 17, 2014, 01:13:54 am
content: The voltage will in theory not change; it will still be 15V at peak. In practice it will have discharged a bit through internal leakage.Q = C VCharge = Capacitance * VoltageIt keeps the charge, and the capacitance doesn't change, so the voltage doesn't change either.

author: Rudane
author_rank: Regular Contributor
timestamp: January 17, 2014, 01:49:19 am
content: The TOTAL charge will not change. If the new circuit introduces some capacitance, then yes the voltage will change. This is a typical text book example of charging a capacitor and moving it to a new circuit with other capacitors.

author: IanB
author_rank: Super Contributor
timestamp: January 17, 2014, 02:04:56 am
content: Quote from: Jimbz on January 17, 2014, 01:11:49 amIt is as c4757p says. The voltage between the plates of a capacitor depends on the charge difference between those plates and the geometry of a capacitor. In an ideal world it doesn't matter whether the capacitor is in a circuit or not, or whether it is being discharged or being charged or if it is at rest. If we had some way of measuring the voltage without disturbing the capacitor we would see that the voltage is present even when the capacitor is not in a circuit.The formulaQ = C Vsays that the voltage is proportional to the charge. The constant "C" is the capacitance, and this depends on the construction of the capacitor (the plate area, the distance between the plates, the dielectric constant of the material separating the plates, etc.).

author: Psi
author_rank: Super Contributor
timestamp: January 17, 2014, 02:16:06 am
content: At the instant of contact (between the charged cap and a load) the caps terminal voltage will drop based on the caps ESR and load current.This leads to a momentary dip in measured cap voltage while high current flows out in order to charge up other caps in the load/circuit.

author: c4757p
author_rank: Super Contributor
timestamp: January 17, 2014, 02:35:11 am
content: Quote from: Rudane on January 17, 2014, 01:49:19 amTrue. The problem is solved, however, simply by considering that multiple capacitors together are equivalent to asinglecapacitor. Thus,same Q = (C + more C) VThink in terms of capacitance, not capacitors.

author: Rufus
author_rank: Super Contributor
timestamp: January 17, 2014, 03:19:07 am
content: Quote from: Jimbz on January 17, 2014, 01:11:49 amThe voltage on a capacitor is proportional to the charge (number of electrons) it holds. Q = C * V where Q is in coulombs. 1 coulomb is about 6.2 x 10^18 electrons. A current of 1 amp flowing for 1 second is also 1 coulomb of electrons.The change in charge of a capacitor with current flowing in or out is I * T where I is the current in amps and T the time in seconds.

author: Jimbz
author_rank: Contributor
timestamp: January 21, 2014, 06:29:22 am
content: Thanks for the replies, it helped clear things up a bit.One more question. Say we take a real life capacitor rated at 25V breakdown voltage and we apply 15V to it. Am I right to assume this capacitor does not hold its maximum charge Q, as the charge is dependant on the voltage applied to the capacitor ? At say 20 volts we have more charge (more ?current between the plates? since charge is amperage ?).If thats the case then I think I get the whole concept of this black magic.PS. Sry for long reply, totally forgot this thread :O

author: Rerouter
author_rank: Super Contributor
timestamp: January 21, 2014, 06:48:05 am
content: Quote from: Jimbz on January 21, 2014, 06:29:22 amSeems the waters are still a bit muddy,Charge is electrons, just sitting there packed onto the plates with one side having more than the other (because a voltage source pushed / pulled them there) creating a difference in voltage,amperage is the rate at which x amount of electrons flow per second (made pretty with the word coulomb)as far as the voltage question, technically your correct that the capacitor could hold more charge by increasing the voltage before it breaks down (heck depending on brand you could get to 1.5 times the rating before you see the breakdown) but you can sum it up by saying this is the most charge this size capacitor can hold for this voltage, (Q = CV)

author: Jimbz
author_rank: Contributor
timestamp: January 21, 2014, 07:51:40 am
content: Quote from: Rerouter on January 21, 2014, 06:48:05 amWhat is the relationship between the charge and voltage ?I assumed that the capacitance C is constant.1.When we apply voltage to a capacitor and charge it full, when the amperage reaches 0 we hold in it a certain amount of charge Q.2.When we apply even more voltage than previously to the same capacitor, when the amperage reaches 0, do we not hold in a bigger amount of charge Q ?If we happen to discharge them, do we not have a longer flow of electrons on the second case (assuming capacitors are the "constant voltage" type of batteries, are they not?)Now yes, I understand the concept of amperage-electron flow in a cross-section in 1 second.

author: Rudane
author_rank: Regular Contributor
timestamp: January 21, 2014, 10:00:24 pm
content: I might be barking up the wrong tree, but I think there is a discrepancy here. What does Voltage mean? Charges are all around us, all the time. There is no way to do anything useful (in electrical terms) unless we can separate charges. Why? Stop reading right now and think about this. Why do you have to separate positive and negative charges to do anything?Electrons just floating around are met with an equal amount of positive charge, and there will be no NET current flow. That is, if you consider all the charges floating around they balance one another. Now, assume we can separate those charges, that is, gather up a whole bunch of negative charge and dump them somewhere and do the same with the positive charges. Let's call it an electron cloud and a positive charge cloud. So, now I have potential to do something. If I connect a wire between these two clouds, a current will flow.How do we define how much potential for current flow we have? Let's make up some term and quantify the potential here. This should start to sound familiar .... this made up term is called Voltage. That's right, voltage doesn't exist, we made it up to make our lives easier. Why we made it up and defined it the way we did gets complicated, but essentially we did it to make our lives easier.

author: Nerull
author_rank: Frequent Contributor
timestamp: January 22, 2014, 12:19:19 am
content: The charge and energy stored in a capacitor is as this series of equations:A 1000uF capacitor charged to 10V will store 0.05J of energy and a charge of .01C. The same capacitor charged to 20V will store 0.2J of energy with a charge of .02C.So yes, charge - and energy - are voltage dependent. Capacitance is fixed, assuming an ideal capacitor. Which isn't always a good assumption.A product I work on contains 2 2000V 1000uF capacitors. That's 400J.