Rate of change of voltage across a capacitor

Thread Starter

WaltB

Joined Jan 17, 2018
5
Hello. I'd like to get some confirmation on a question. It's about the formula that describes current through capacitors. i = C.dV/dt .So let's start with C. As you increase the capacitance, it takes more and more time for the capacitor to charge to a specific voltage and thereby by increasing C you effectively decrease the potential difference between the Source voltage(i'll call this one Vs) and the capacitor voltage Vc and this leads to a bigger instantenious current through the capacitor. Now the rate of change of the voltage dV/dt.

The picture i tried to upload is a potentiometer with a capacitor connected to it. The potentiometer can move up to deliver more Voltage to the capacitor. Now if we assume that Vc at first ist 0v and we begin to move the wiper up, increasing the voltage, the faster we move it from p.A to p.B the less time would the current have had to flow, thus less charge would have exited the capacitor and less voltage would the capacitor have accumulated for that short period of time dt. Thus when the final Vs voltage across the capacitor is established, the potential difference Vs - Vc would be greater than if the voltage was changed with a slower rate, and the current that will flow through the capacitor will be greater too.
Is that right?
 

dl324

Joined Mar 30, 2015
11,897
Welcome to AAC!

We need the schematic you're referencing so we can better understand what you're describing.
 

MrChips

Joined Oct 2, 2009
22,100
Here is a link to the AAC eTutorials:

https://www.allaboutcircuits.com/textbook/direct-current/chpt-16/capacitor-transient-response/


upload_2018-1-17_12-48-36.png

Fundamentally, if you change the value of the source voltage, the charging time does not change, relatively speaking. Only the final voltage on the capacitor will change.

The charge time is a fundamental characteristic known as the time constant.

time constant = resistance x capacitance

in this RC example circuit.

In this example, time constant is the time it takes for the capacitor to reach 63% of the final voltage.
If you wait 5 times the time constant, the capacitor will reach 99% of the final voltage.

In the example shown above,
time constant = resistance x capacitance = 10kΩ x 100μF = 1 second
If you wait 5 seconds the capacitor will reach 99% of the source voltage, regardless of the value of the source voltage.
 

Thread Starter

WaltB

Joined Jan 17, 2018
5
Yes, i know. It takes 5 time constants to charge up to about 99.3% of the source voltage, but increasing Vs wont change that. It will only increase the charging current. You give the capacitor more voltage to charge up to but that increased Vs is also proportional to the current so it's like it has more voltage to accumulate but on the other hand the current has now proportionaly increased too and it will deliver the required charge in time. But was i right about i = C.dV/dt ? I mean C influences the time constant, and dV/dt doesn't influence it, but depending on it it will give bigger or smaller current based on how fast the potentiometer is moved.
 

crutschow

Joined Mar 14, 2008
25,666
The rate of voltage change on (and current into) the capacitor will be a function of both the pot resistance and how fast the pot wiper is changed.
 

WBahn

Joined Mar 31, 2012
26,301
The relationship i(t) = C·dv(t)/dt is fundamental for a capacitor. But the v(t) and i(t) refer to the voltage across the capacitor and the current through the capacitor, respectively.

If you want to talk about time constants and particular relationships between voltage or current and time (such as the classic first-order exponential response), you are then referring to particular circuit topologies that are described by those relationships and it depends on the entire circuit, not just the capacitor.

If you are using a changing potentiometer, then that is not one of the circuit topologies that is described by that exponential equation. The response for that has to be determined by analyzing the specific circuit and the specific way that the potentiometer is changed. For that, you need to show the specific circuit you are talking about and carefully define how the potentiometer is to be changed.
 

#12

Joined Nov 30, 2010
18,210
I have a feeling this guy is trying to understand concepts, but the questions are not well formed in his mind. That leaves several people out because there is no circuit, no numbers, and a vague kind of question. I can provide an equation:
Vcap = deltaV times e to the (-time/RC)
Not awfully helpful today, but something to put in the tool box. It will be handy later.
 

Thread Starter

WaltB

Joined Jan 17, 2018
5
Thanks. I was away for about a week and so I couldn't helped you help me so to speak heh. I have only one question, dV/dt is that the rate of change of voltage OF the capacitor or of the source ACROSS the capacitor. And I guess I wanted also to ask why I is direct proportional to C and I asked if it is because more capacitance requires more charge and therefore more current. The main problem in my head is if dV/dt is the rate of change of voltage OF the capacitor or of the source voltage ACROSS the capacitor.
Thank you!
 

kubeek

Joined Sep 20, 2005
5,734
dV/dt is the voltage across the capacitor. Other voltages in the circuit will be related to it, but dV/dt an I in the capacitor equation is the voltage on the capacitor and current through the capacitor, irrespective of any other pat of the circuit.
 

Thread Starter

WaltB

Joined Jan 17, 2018
5
dV/dt is the voltage across the capacitor. Other voltages in the circuit will be related to it, but dV/dt an I in the capacitor equation is the voltage on the capacitor and current through the capacitor, irrespective of any other pat of the circuit.
So dV/dt is the voltage that is being build on the plates, and the more voltage over a short period of time there is the more current it will require(the capacitor) to build this voltage, and it is not the source voltage changing across the capacitor, forcing it to adapt?
 

kubeek

Joined Sep 20, 2005
5,734
So dV/dt is the voltage that is being build on the plates, and the more voltage over a short period of time there is the more current it will require(the capacitor) to build this voltage, and it is not the source voltage changing across the capacitor, forcing it to adapt?
Exactly, current causes more charge difference between the plates (charges arriving at one plate and leaving the other), which shows as higher voltage.
 

WBahn

Joined Mar 31, 2012
26,301
Thanks. I was away for about a week and so I couldn't helped you help me so to speak heh. I have only one question, dV/dt is that the rate of change of voltage OF the capacitor or of the source ACROSS the capacitor. And I guess I wanted also to ask why I is direct proportional to C and I asked if it is because more capacitance requires more charge and therefore more current. The main problem in my head is if dV/dt is the rate of change of voltage OF the capacitor or of the source voltage ACROSS the capacitor.
Thank you!
The proper terminology is that the voltage appears ACROSS something and the current flows THROUGH something. But saying something like the voltage ON the capacitor or the current IN the capacitor are also widely used, particularly in casual speech. I don't hear the use of the voltage OF a component too often, but you often hear things like the voltage OF a power supply or a battery.

But in any event, the rate of change of the voltage ON or ACROSS or OF a capacitor are all the same thing.
 

WBahn

Joined Mar 31, 2012
26,301
And I guess I wanted also to ask why I is direct proportional to C and I asked if it is because more capacitance requires more charge and therefore more current.
It's due to the definition of capacitance.

Q = CV

If you have some component that can store charge proportional to the voltage you put across it, the capacitance is the proportionality constant. The higher the capacitance, the move charge is stored for the same voltage. In order to change the voltage by a certain amount, you will need to move more charge onto or off of the capacitor. If you do that in the same amount of time, then the current is greater because current is, by definition, the time rate of change of charge.
 

noweare

Joined Jun 30, 2017
115
Yes, moving the potentiometer faster or slower will cause the the magnitude of I to be more or less, respectively.
So if the voltage is changing at 1 volt per second and C is 1 farad the current will be 1 amp. If the voltage change stays at that rate forever the current will always be 1 amp through the capacitor.

So think of the voltage as a ramp going up at some slope with respect to time lets say
Volt=1v @1second 2v@2 second etc then the current will be be a constant (level line) 1 amp
thru a 1 farad capacitor
 
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