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Which equation can be used to calculate the time taken to charge the capacitor at the given amount of current and voltage at a constant capacitance?

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  • \$\begingroup\$ en.wikipedia.org/wiki/Capacitance \$\endgroup\$
    – kenny
    Nov 20 '11 at 12:06
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    \$\begingroup\$ The problem is overconstrained - the current and voltage can't both be given, as one of them will vary (whichever is less "enforced" ie less approximates an ideal source) during the charging of the cap. \$\endgroup\$ Nov 20 '11 at 18:35
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If you want a "simple" equation, and it seems that you do, you could start with definition of current.

First, let's start with the farad. It is usually expanded as \$F=\frac {As}{V}\$.

Now let's write that with symbols for capacitance, current, voltage and time:

\$C=\frac {It}{U}\$

Since we have constant current and voltage and we need time, we'll divide the equation with current and multiply with voltage so that we can get time.

That gives us \$\frac{UC}{I}=t\$.

If this is just a school problem, then we have a solution.

In real life things will work differently. As the capacitor charges, the voltage on the capacitor will drop resulting in drop of current and the time will therefore be longer.

Here's an example: [capacitor charger Let's assume that at the beginning, the capacitor is discharged. First we have the voltage on the resistor which is \$U_r=Ri\$. Then we have voltage on the capacitor which is \$U_c=\frac{1}{C} \int {i \mbox{ }dt} \$.

So we know that \$E=Ri+\frac{1}{C} \int {i \mbox{ }dt}\$. To solve this, we need to turn it into differential equation.

\$(E=Ri+\frac{1}{C} \int {i \mbox{ }dt}) / \frac{d}{dt}\$

Since \$E\$ is constant, it will turn into zero. The integration and differentiation will cancel each-other out and we'll get:

\$R\frac{di}{dt}+\frac{i}{C}=0\$ Next we divide everything with \$R\$ and get
\$\frac{di}{dt}+\frac{i}{RC}=0\$

After that we move the \$\frac{1}{RC}i\$ to the other side and multiply everything with \$dt\$ and divide everything with \$i\$ and we get:

\$\frac{di}{i}=-\frac{1}{RC}dt\$

Now we integrate everything and get \$\int {\frac{di}{i}} = -\int {\frac{1}{RC}dt}\$ As a result, we get:

\$\ln{i}=-\frac{t}{RC}+C_1\$

Now to get rid of the logarithm, we raise everything to \$e\$
\$i=C_1 e^{-\frac{t}{RC}}\$

Now we have the general solution and we need to determine the constants. So first we look at what's happening when the time is equal to zero:

\$i=C_1 e^{-\frac{0}{RC}} = C_1\$.

We also know that the initial current is \$i_{(0)}=\frac{E}{R}\$. From that we can determine that \$C_1=\frac{E}{R}\$.

The complete equation for the current is:
\$i_{(t)}=\frac{E}{R} e^{-\frac{t}{RC}}\$

This is a classical capacitor charging equation and it is available on many sources on the Internet.

The \$RC\$ is also called the time constant, so \$\tau=RC\$. It is usually considered that five time constants are enough to charge a capacitor.

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    \$\begingroup\$ A possibly more descriptive version of your last sentence: after five time constants, the capacitor is charged to 1-e^-5 = .993 of fully charged (at the given voltage), which is usually considered charged. \$\endgroup\$
    – Cascabel
    Nov 20 '11 at 18:18
  • \$\begingroup\$ And if anybody is wondering why the initial current is E/r it is because the capacitor initially being fully discharged has 0 volts accross it, which is like a short. \$\endgroup\$ Apr 19 '16 at 21:53
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For this circuit:

When the everything starts out at 0 V and then the input is changed to Vin at time t=0:

\$V_{out}(t) = V_{in}(1 - e^{-\frac{t}{RC}})\$

When R is in Ohms and C in Farads, then t is in seconds.

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There are TWO cases, as Chris indicated.

Case 1 is where you charge a capacitor from a constant voltage source with resistance and capacitance known. (Resistance is any circuit resistance plus capacitor internal resistance plus any added resistance. This is the case covered by eg Andreja Ko & Olin.
You get

\$V_{out}(t) = V_{in}(1 - e^{-\frac{t}{RC}})\$

{stolen from Olin}, which you can rearrange for t.


Case 2 is charging at constant current.
More current charges quicker.
More capacitance takes longer.
Charging to a higher voltage takes longer

So:

\$t=\frac{VC}{I}\$

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  • \$\begingroup\$ would be really helpful if you did rearrange the case 1 for t. \$\endgroup\$
    – MightyPork
    Apr 29 '16 at 19:45
  • \$\begingroup\$ @MightyPork Assignment or ??? What is your experience? | (1-Vout/Vin) = e^-it/.... | Take ln (log base e) of both sides: ln(1-Vo/Vi) = -t/rc | t = -ln(1-Vo/Vi)/(rc) -> make SURE you follow that !!! [[E&OE]] \$\endgroup\$
    – Russell McMahon
    Apr 30 '16 at 12:34
  • \$\begingroup\$ No need to be rude, I'm just not that good with e^x. I found the question on Google but was disappointed the answer didn't really answer it.. \$\endgroup\$
    – MightyPork
    Apr 30 '16 at 12:47
  • \$\begingroup\$ @MightyPork No need to be rude ! :-) | I assure you ABSOLUTELY NO RUDENESS INTENDED. Really. The questions were serious. You will note that I wholly answered your question. Yes? If you don't think I did pse read through it again and ask again if needs be. When people ask a question like that it is good to know if we are doing their homework for them - which does not help them. We are happy to help but the approach would be different. And knowing what your experience is is part of knowing what level to answer at. Your question amounts to algebra and knowing why you need to ask helps with ... \$\endgroup\$
    – Russell McMahon
    Apr 30 '16 at 12:58
  • \$\begingroup\$ ... further input. [You may note that I tend to be fairly helpful n:-) ]. | PS: I think you may have accidentally missed a "thank you" off the end of your response ? :-) \$\endgroup\$
    – Russell McMahon
    Apr 30 '16 at 13:00
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The basic equations are

time constant = R C = 1/(2 Pi fc)
i = dq/dt = C du/dt

Links: wikipedia Capacitor Charge and Discharge

Hope it helps!

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  • \$\begingroup\$ Please format your equations using MathML instead of blockquote. \$\endgroup\$ Apr 19 '16 at 22:09
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While all the above formulae are useful, I prefer to see it as a bucket with current flowing into it to charge it to a certain voltage level. What determines the amount of time taken to fill the bucket? Its capacitance or C of the bucket.

Now lets say the current flowing in is I. C acts as resistance to prevent the water level to rise. How fast can the water level rise? If I is more, it will be faster. If C is more, it will be slower. Hence, water level rise wrt time is just I/C.

Now we know how fast water level can rise. How do we know the time to get a certain value? Slope= (Final Voltage-Initial Voltage)/Time= I/C Hence, Time= Delta(V)* C/I

Again to quickly picturise this, time is lesser if C is less, current flowing in more or Vfinal lesser.

Hope this helps.

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  • \$\begingroup\$ Please format your equations using MathML. \$\endgroup\$ Apr 19 '16 at 22:10
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The possibility that the capacitor has a non-zero initial voltage also has to be considered. Following from AndrejaKo's calculations, the current through the circuit if the initial voltage across the capacitor \$V_{C0} = 0\$ is:

\$i_{(t)}=\frac{E}{R} e^{-\frac{t}{RC}}\$

But if \$V_{C0}\$ is non-zero then,

\$i_{(t)}=\frac{E - V_{C0}}{R} e^{-\frac{t}{RC}}\$

So that the voltage across R is

\$V_R(t) = i_{(t)}R = (E - V_{C0}) e^{-\frac{t}{RC}}\$

And the voltage across C at any time t is:

\$V_C(t) = E - V_R(t) = E - (E - V_{C0}) e^{-\frac{t}{RC}}\$

\$V_C(t) = E + (V_{C0} - E)e^{-\frac{t}{RC}}\$

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The capacitor (C) in the circuit diagram is being charged from a supply voltage (Vs) with the current passing through a resistor (R).
enter image description here
The voltage across the capacitor (Vc) is initially zero but it increases as the capacitor charges.
The capacitor is fully charged when Vc = Vs.
The charging current (I) is determined by the voltage across the resistor (Vs - Vc):

 Charging current,  I = (Vs - Vc) / R   (note that Vc is increasing)  

At first Vc = 0V so the initial current,

 Io = Vs / R 

Vc increases as soon as charge (Q) starts to build up (Vc = Q/C),
This reduces the voltage across the resistor and therefore reduces the charging current.
This means that the rate of charging becomes progressively slower.

links,links use these link that may be helpful for you.

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  • \$\begingroup\$ Please format your equations using MathML instead of blockquote. \$\endgroup\$ Apr 19 '16 at 22:10

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