- Capacitor i-v equations
- A capacitor integrates current
- Capacitor i-v equation in action
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- Inductor kickback (2 of 2)
- Inductor i-v equation in action
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- RC step response solve (2 of 3)
- RC step response example (3 of 3)
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Capacitor i-v equations
The capacitor current-voltage equation in derivative form and integral form. Created by Willy McAllister.
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- Really enjoying this subject matter. I have been stopping periodically to get other source material on how capacitors work and why they are important in electronics. I appreciate the need to break things down to the math, but, it seems that a more practical explanation of what is actually happening would enhance the lesson plan. For example, I assume a capacitor is effectively a charge container. I gather the equations tell us things like, how much charge a given capacitor can store, perhaps how it can receive a charge, how it can be discharged, etc. For my purposes, jumping into the equations before i have some conceptual view of what the component is supposed to be doing and how, matters greatly.(20 votes)
- the youtube channel great scott in his electronic basics series explains them very well.(5 votes)
- why is the integral from -infinity to T rather than 0 to T??(13 votes)
- Good question. The right side of the integral equation is the term "cv". This product is equivalent to the amount of charge stored in the capacitor, q = cv. The integral's lower bound is t = -infinity to make sure we account for all the charge that has ever flowed into or out of the capacitor.
We often assign a moment where t=0, but that is always specific to the problem at hand. At t=0 we don't necessarily know how much charge is on the capacitor. So the most general way to express the equation is to go all the way back in time as far as possible to keep track of all the charge the capacitor has ever seen.
Now, that said, for many problems we often pick a time to assign as t=0. We pick a time when we know (or define) the charge on the capacitor to be q = cv = 0. When we do this, we can change the lower bound on the integral from t = -inf to t=0. If the capacitor is empty at t=0 ("empty" means zero charge stored), then everything that ever happened in the past to this capacitor has integrated to zero, and we can leave the past out of the bounds on the integral. This allows us to concentrate on the time interval of interest, from t=0 to t=T.(19 votes)
- I'm sorry, but do you have time to re-address the "t vs. tau" question? Your lessons are excellent btw, and I am a khan contributor, but I am heartily confused by t and tau. As to when t=0, wasn't the only time when t was actually zero was at minus infinity? When does t become zero AFTER minus infinity? Is charge gained and then lost over the period from minus inf to zero? How do we know the answer to any of this? And finally, is the "trick" actually parametrization? it seems similar, but not quite. I don't understand the description of t as "continuously running time", and"Big-T" as " the length of the current pulse. " Is big-T the "tau"? Please forgive my ignorance, but I am utterly perplexed.(4 votes)
- Beyond7:30, everything is confusing, when he transforms from t to T. Can anybody explain to me that?(4 votes)
- From the author:I can understand how this switch from t to T happens quickly without a lot of explanation. Introducing the greek tau variable is a calculus trick.
There are two time variables in this video, little-t is continuously running time. Big-T is the length of the current pulse.
At7:30the equation is written as v(big-T) where big-T is the argument to v and it is also the upper limit on the integral symbol.
At7:37I change my mind about how I want to use the variable names. From now on I want to use little-t as the length of the pulse. This is strictly a matter of style and taste on my part. I want the answer to come out v(little-t) just because I thing it looks prettier. So the equation at7:37now starts out v(little-t) = ...
In this version of the equation, little-t becomes the upper limit on the integral. Now I've substituted little-t for Big-T.
But there's one more thing I have to do before it's ok to proceed. Before, I was using little-t inside the integral, as the dt term. Since little-t now has another job, I change i(t) dt to i(tau) d-tau.
All of this is just a symbol change. There is no math happening.(3 votes)
- at2:59dq/dt is the change of q of capacitor over a infinitesimal change of time.but how are we sure that it is exactly the same as the current that passes through it.i mean it is reasonable that if the current is high then the change of the charge of capacitor over a infinitesimal change of time(dq/dt of capacitor)is high.but why are we sure that they are the same.(2 votes)
- Current is exactly dq/dt because that's how we chose to define the meaning of current.(2 votes)
- Hi, I'm just stuck on part of the math
How come when you do the integral, on the one side you use a definite integral between negative infinity and T, but then on the other side of the equation you do an indefinite integral? Shouldn't they both be the same type of integral?
- 3:40begins "find v in terms of i"4:20shows an equation with an integral on each side. There are no limits shown.
By5:00we are down to one integral. The other integral has vanished and become just 'v'.5:16the bounds are added to the integral, making it clear it's a definite integral. It has been all along.
(admission: I'm not always a purist when it comes to integral notation indefinite/definite.)
There is some informality in how this is presented. The point is to see how the voltage now depends on the whole history of the current. It is similar to how the water level in a bucket (voltage) depends on the entire history of how water has been poured in and poured out.(2 votes)
- At5:51, the author integrates the right hand side without adding an arbitrary constant. Is this ever taken care of? In particular, it should be "Cv + d" instead of merely "Cv," where d is some arbitrary constant, right?(2 votes)
- In the integral form of the capacitor i-v equation the arbitrary constant turns out to be equivalent to the starting voltage on the capacitor.
The initial version of the equation has a lower limit of t = -infinity. When you don't see an arbitrary constant at the end of the equation, that means we are assuming the capacitor was manufactured and first shipped with no charge on it (the constant is 0).
The i-v equation tells us that the voltage on the capacitor right now depends on the entire past life of the capacitor, all the way back to its birth. That's not so convenient to work with, so we start the time clock ticking at some arbitrary time we call t = 0. We can pick any t = 0 we want as long as it is a moment when we happen to know the voltage on the capacitor. The i-v equation changes its lower limit to t = 0 and that known voltage becomes Vo, the constant at the end of the equation.
It is amusing (or cosmically deep) that the entire life of the capacitor all the way from t = -infinity up until t = 0 is completely captured in the value of Vo.(2 votes)
- What exactly is an integral?(1 vote)
- Check out this article on Preparing to Study EE on KA... https://www.khanacademy.org/science/electrical-engineering/introduction-to-ee/intro-to-ee/a/ee-preparing-to-study-electrical-engineering?modal=1
Scroll down to the section on Calculus and check out the links to other parts of KA that answer your question.(2 votes)
Can I have the the link for the explanation on what are integrals?
- Q=CV --> Where we got it from ? please help ?(1 vote)
- [Voiceover] We're gonna talk about the equations that describe how a capacitor works, and then I'll give you an example of how these equations work. The basic equation of a capacitor, says that the charge, Q, on a capacitor, is equal to the capacitance value, times the voltage across the capacitor. Here's our capacitor over here. Let's say we have a voltage on it, of plus or minus V. We say it has a capacitance value of C. That's a property of this device here. C is equal to, just looking at the equation over there, C is equal to the ratio of the charge, stored in the capacitor, divided by the voltage of the capacitor. What we mean by stored charge is, if a current flows into this capacitor, it can leave some excess charge on the top. I'll just mark that with plus signs. There will be a corresponding set of minus charges, on the other plate of the capacitor. This collection of excess charge will be Q , and this down here will be Q-, and they're gonna be the same value. What we say here, is when the capacitor's in this state, we say it's storing this much charge. We'll just name one of these numbers here. They're gonna be the same, with opposite signs. That's what it means for a capacitor to store charge. What I want to do now, is develop some sort of expression that relates the current through a capacitor, to the voltage. We want to develop an IV characteristic, so this will correspond, sort of like, Ohm's Law for a capacitor. What relates the current to the voltage. The way I'm gonna do that, is to exercise this equation, by causing some changes. In particular, we'll change the voltage on this capacitor, and we'll see what happens over here. When we say we're gonna change a voltage, that means we're gonna create something, a condition of DV, DT. A change in voltage per change in time. I can do that by taking the derivative of both sides of this equation here. I've already done it for this side. Over here, what I'll have is DQ, DT. I took the derivative of both sides, just to be sure I treated both sides of the equation, the same. Let's look at this little expression right here. This is kind of interesting. This is change of charge, with change of time. That's equal to, that's what we mean by current. That is current. The symbol for current is I. DQ, DT is current, essentially, by definition, we give it the symbol I, and that's gonna be equal to C DV, DT. This is an important equation. That's, basically, the IV relationship, between current and voltage, in a capacitor. What it tells us, that the current is actually proportional to, and the proportionality constant is C, the current's proportional to the rate of change of voltage. Not the voltage itself, but to the rate of change of voltage. Now, what I want to do is find a expression that expresses V, in terms of I. Here we have I, in terms of DV, DT. Let's figure out if we can express V, in terms of some expression containing I. The way I do that is, I need to eliminate this derivative here. I'm gonna do that by taking the integral of this side of the equation, and at the same time, I'll take the integral of the other side of the equation, to keep everything equal. What that looks like is, the integral of I... With respect to time, is equal to the integral of C DV, DT, with respect to time, DT. On this side, I have basically, I do something like this. I have the integral of DV. This looks like an anti-derivative. This is an integral, acting like an anti-derivative. What function has a derivative of DV? That would be just plain V. I can rewrite this side of the equation, constant C comes out of the expression, and we end up with V, on this side. Just plain V. That equals the integral of I DT. We're part way through, we're developing what's gonna be called an integral form of the capacitor IV equation. What I need to look at next is, what are the bounds, on this integral? The bounds on this integral are basically minus time, equals minus infinity, to time equals sub-time T, which is sort of like the time now. That equals capacitance times voltage. Let me take this C, over on the other side, and actually, I'm gonna move V over here, onto the left. Then, I can write this, one over C. This is the normal looking version of this equation. I DT, minus infinity to time, T. Time, big T, is time right now. What this says, it says that the voltage on a capacitor has something to do with the summation, or the integral, of the current, over its entire life, all the way back to T equals minus infinity. This is not so convenient. What we're gonna do instead, is we're gonna pick a time. We'll pick a time called T equals zero, and we'll say that the voltage on the capacitor was equal to, let's say, V not, with some value. Then, what we'll do, is we're gonna change the limit on our integral here, from minus infinity, to time, T, equals zero. Then, we'll use the integral from, instead, zero to the time, we're interested in. That equation looks like this. We're just gonna change the limits on the integral. We have the integral now, but we have to actually account for all the time, before T equals zero. What we do there, is we just basically add V not. Whatever V not is, that's the starting point, at time equals zero, and then the integral takes us, from time zero until time now. This is the integral form of the capacitor equation. I want to actually make one more little change. This is the current at V, as a function of T. What we really want to write here, is we wanna write V of a little T. This is just stylistically, this is what we like this equation to look like. I want the limits on my integral, to be zero to t. Now, I need to sort of make a new replacement for this T that's inside here. I can call it something else. I can call it I of, I'll call it tau. This is basically just a little fake variable. D tau plus V not. This is, now, we finally have it, this is the integral form of the capacitor equation. We have the other form of the equation that goes with this, which was I equals C DV, DT. There's the two forms of the capacitor equation. Now, I want to do an example with this one here, just to see how it works, when we have a capacitor circuit.