Changing current through a capacitor

I have been reading a bit lately on the basics, but I can't find an answer this question: why does the current to/from (through) a capacitor change?

Here's a little background, first, to help clarify my question. I understand that the instantaneous current through a capacitor is its capacitance multiplied by the rate of change in voltage at that point in time.

I = C dv/dt

But, in a basic circuit I've simulated, it appears that the current itself also changes. For the current to change, and with C being constant, then the rate of change of the voltage must also change, akin to acceleration in physical motion.

This makes me think of "voltage acceleration" but searching for that doesn't turn up anything except for "acceleration voltage" which is related to particle acceleration. I also get no results if I search for something like "capacitor current changes over time"... at least, I get no results going any further than the initial formula above.

Any help?

Thanks!
-Phillip

Are you talking about a RC circuit (a resistor in series with a capacitor)?
In a RC circuit, when charging, the voltage across the capacitor increases thus the voltage across the resistor decreases. And with Ohm's law we know that if the voltage across the resistor decreases, the current decreases as well.

A capacitor stores charge and the basic relation is Q (total charge stored) = CV, where V is the voltage across the capacitor.

If the voltage changes, the charge must change, so current I = dQ/dt must flow into or out of the capacitor.

phillipt:
I have been reading a bit lately on the basics, but I can't find an answer this question: why does the current to/from (through) a capacitor change?

Here's a little background, first, to help clarify my question. I understand that the instantaneous current through a capacitor is its capacitance multiplied by the rate of change in velocity at that point in time.

I = C dv/dt

Rearranging that equation might help alleviate your confusion, I think:

dv/dt = I/C

which says that the rate of change of voltage across a capacitor is proportional to the current flowing through the capacitor and inversely proportional to its capacitance; and when the above is expressed in units of volts, seconds, amperes and farads, the relationship is exact.

In a sense, I think your confusion comes from reversing cause and effect: usually, the current through the capacitor is largely determined by other things going on in the circuit (voltages, resistances, etc.), and the effect of the current is to change the voltage across the capacitor over time.

Of course, as the capacitor voltage changes it, in turn, may cause some change in the current flowing through the capacitor (as in a simple RC filter); but for the most part, current can be thought of as the "cause", and dv/dt the "effect."

There are some simpler ways to think about this stuff if you haven't taken calculus.

In DC circuits, no current flows through once the capacitor is charged-up. So in a series circuit, capacitors "block" DC while passing AC. Or similarly, they pass high frequencies better than low frequencies.

Capacitors tend to "resist changes in voltage". The capacitors in power supplies and the bypass capacitors on PC boards are there to "smooth" out the power supply and to prevent power supply "glitches" and noise.

For example, if you put a capacitor in series with an LED, the led will flash-on for a brief period while the capacitor charges, then the LED will begin to dim and then go dark.

If you put a capacitor in parallel with an LED, the LED will turn-on slowly, as the current flows through the capacitor charging it instead of lighting the LED. Then if you remove power, the capacitor will hold the LED on for a time while it discharges through the LED.

You'd need about 1000uF or more for a long-enough time constant to see these effects with an LED.

A high-pass RC filter puts the capacitor in series with the signal flow where it passes high frequencies while blocking/reducing low frequencies (including DC, which is zero Hz).

A low-pass RC filter puts the capacitor in parallel with the signal where high frequencies tend to be "shorted out" by the capacitor.

Another formula for calculating capacitive reactance (impedance in Ohms) in AC circuits is: Xc = 1/(2Pi x f x C). Where f is frequency in Hz, and C is capacitance in Farads.

Or, the RC [u]RC time constant[/u] can be calculated as R x C. (With resistance in Ohms, and capacitance in Farads).

The dV/dt term is delta voltage with respect to time not delta velocity as your enquiry suggested

As the capacitor charges the dv/dt term reduces as the difference between supply voltage and capacitor voltage decreases, hence the charge current reduces.

The charge rate is inverse exponential (maximum at t=0 and minimum at t= infinity) and hence the charge current decreases exponentially

Thank you all for the answers. I understand the relation of I = C dv/dt... what I didn't (don't) understand is how or why dv/dt changes. I posited that it must because I observed the current changing. If dv/dt did not change and C does not change, then I must not change.

Even if we invert the relation to get dv/dt = IC... still, if the current changes then dv/dt must change. I'm not talking about voltage changing. I'm talking about a change in the rate at which the voltage changes. After 1ms, at t=1, say the voltage has dropped 1 volt. But after 2ms, at t=2, perhaps it dropped only 0.5 volts more. The rate (dv/dt) from t=0 to t=1 is different than the rate from t=1 to t=2.

Dave_D_in_PA:
usually, the current through the capacitor is largely determined by other things going on in the circuit (voltages, resistances, etc.), and the effect of the current is to change the voltage across the capacitor over time.

jackrae:
As the capacitor charges the dv/dt term reduces as the difference between supply voltage and capacitor voltage decreases, hence the charge current reduces.

Dave and Jack - That's what I was looking for!!
I figure the same is true in the opposite direction as the capacitor discharges... yes?

Also, yes, I meant "voltage" and not "velocity". That was just a typo as I guess I was thinking about the acceleration analogy as I typed.

Thanks again!
-Phillip

phillipt:
But, in a basic circuit I've simulated, it appears that the current itself also changes. For the current to change, and with C being constant, then the rate of change of the voltage must also change, akin to acceleration in physical motion.

...or decceleration - that's what makes the charge or discharge characteristic in a simple RC circuit follow an exponential curve.

The fundamental equation is Q = CV

If you differentiate that you get your equation dQ/dt = I = CdV/dt but only if C is constant. It never is but the change is usually so small it can be ignored.

Russell

what I didn't (don't) understand is how or why dv/dt changes.

It changes because of the signal YOU apply to the capacitor. Saying dv/dt is just another way of saying you apply a signal to it. All a signal consists of is some voltage that changes over time.

Grumpy_Mike:
It changes because of the signal YOU apply to the capacitor. Saying dv/dt is just another way of saying you apply a signal to it. All a signal consists of is some voltage that changes over time.

It’s easier to think of it the other way round. You apply a current to it and that makes the voltage change with time.

You can’t apply a voltage to a capacitor from a perfect voltage source. If you did the current would become infinite for a zero time and that is clearly impossible. All real voltage sources have some series resistance built in. You have to consider the complete circuit.

So if you change the voltage of a source having a resistance R by V volts a current of V/R amps will flow. That current will decay exponentially with a time constant of RC.

Russell.

If you did the current would become infinite for a zero time and that is clearly impossible.

Grandmother, egg sucking lessons here.

You can't apply a voltage to a capacitor from a perfect voltage source.

I never said you could. I said a signal, all signal sources have impedance. I was very careful not to use the word voltage for just such pedantry.

It's easier to think of it the other way round.

Well you might be for you but you never actually apply a current as such, you apply a signal Which as I say is a voltage derived from some impedance source.

The problem here is that the OP never actually said what sort of circuit he was envisioning which makes simple easy to understander answers impossible. Therefore you have to resort mathematical general soloutions, something which the OP can not cope with, hence him not being able to envisage dv/dt.

You can consider two extreme cases first, the voltage source and the current source.

If you connect the capacitor to a perfect voltage source (zero impedance) which applies
a particular waveform, then the capacitor equation will mean that the current is determined
by C.dV/dt, simple to understand.

If you connect the capacitor to a perfect current source (infinite impedance), which
controls the current waveform, the the capacitor equation means the voltage across the
capacitor will be the integral of I/C dt.

If you connect the capacitor to some circuit that is neither a perfect voltage or current source
then the equation acts both ways - the voltage and current are related by the equation and
each affects the other such that the equation holds. Solving the equation in this context
must involve the mathematics of the rest of the circuit.

In reality there are no perfect voltage or current sources or perfect capacitors (for instance they
breakdown if the voltage is too high and melt if the current is too high, and
have resistance and inductance)

Therefore you have to resort mathematical general soloutions, something which the OP can not cope with, hence him not being able to envisage dv/dt.

That's quite presumptuous and insulting.

I put forth that I do understand the equation I = C dv/dt and asked, essentially, about dI/dt. Where do you get off figuring I'm wrong about my level of understanding?

Everyone thus far, save for a few, have continued to discuss dv/dt.

If there is a dI/dt and C is constant, then dv/dt must also change. Not V, voltage; we've already established that change in dv/dt. I'm talking about dv/dt istelf changing, as well.

Thanks to Dave_D and jackrae for rather clearly answering that question.

And russelz added this bit, too, which helps clarify more about changing current.

So if you change the voltage of a source having a resistance R by V volts a current of V/R amps will flow. That current will decay exponentially with a time constant of RC

Where do you get off figuring I'm wrong about my level of understanding?

I am telling it like I see it, if you do understand it you are doing a fine job of covering it up.

Considering you said:-

what I didn't (don't) understand is how or why dv/dt changes.

Grumpy_Mike:
Grandmother, egg sucking lessons here.

Sorry, I wasn't trying to explain it to you Mike but to the OP who seemed to me to have better knowledge of maths than of electronics.

Russell

Considering I said: what I didn't (don't) understand is how or why dv/dt changes..

Exactly, Grumpy_Mike - I understand dv/dt, that voltage changes, but what I was curious about is how that quantity also changes. The rate of change of dv/dt, which yes is itself a rate of change.

Hence my analogy to physical motion. Velocity is a rate of change in position. But acceleration is a rate of change in velocity... i.e. a rate of change of a quantity that is itself a rate of change of yet another quantity.

I = C dv/dt

That I changes in time means we have dI/dt and since we take C to be constant, it must be that dv/dt changes as well. That is, the rate at which voltage changes is also a changing quantity.

Voltage changes.
The rate at which voltage changes also changes.
These are two different things, and my question centered around the latter.

Thank you.

I think perhaps you are failing to comprehend that the equation i=Cdv/dt relates to the instantaneous current, which is a constant based upon the rate at which the voltage changes (the dv/dt term)

This article might explain things better http://www.allaboutcircuits.com/vol_1/chpt_13/2.html

... but the rate at which voltage changes can be different between two measurements. And directly related to that, the currents at the respective moments will be different.

That's what I asked about, and you even gave an answer. So thanks!