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I am starting to learn transient analysis series RC which requires our own equations to be derived, but I don't understand how the voltage across the capacitor is given in the equation below from a KVL loop around the circuit (underlined in yellow):

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I'm assuming it comes from the basic capacitor voltage/current relationship formula below, but would appreciate if someone would explain how it is given in this form mathematically.

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Thanks

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  • \$\begingroup\$ Do you want us to derive it? because this looks like a homework problem and homework problems need an attempt at a solution. \$\endgroup\$ – laptop2d Jan 29 at 20:30
  • \$\begingroup\$ It's from my theory notes and I'm just unsure where it came from, as I've never seen it before. I am happy deriving it myself, if someone could just confirm if it comes from the basic voltage/current capacitor relationship I posted above? \$\endgroup\$ – David777 Jan 29 at 20:47
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    \$\begingroup\$ Yeah, it comes from integrating the capacitor current, this will yield a voltage function. You plug that in KVL equation for the capacitor voltage and you get what you show in your question. \$\endgroup\$ – Big6 Jan 29 at 20:59
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    \$\begingroup\$ Integrate both sides of \$i = C \frac{dv}{dt}\$ over time and you get \$\int i \, dt + i_0 = C\, v\$. Can you take it from there? \$\endgroup\$ – TimWescott Jan 29 at 21:21
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    \$\begingroup\$ @David777 As I'm sure you already know, there are initial conditions needed to resolve constants of integration. So, you do need to know some condition at some moment in time to pin down the position of the resulting curve. Usually, if unstated and no other hints are present, the only assumption you can easily defend is that at the start (which is, again, probably assumed to be at \$t=0\$), the voltage on the capacitor (and it's stored charge) are assumed zero. \$\endgroup\$ – jonk Jan 30 at 0:15
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Your questions aren't entirely clear to me, so please accept my apologies if I wander just a bit. I'll avoid using integrating factors below, as well, since it's not necessary for this problem.

For ideal capacitors (few are), the basic idea is \$Q=C\:V\$. Taking the full derivative, this is \$\text{d} Q = C\:\text{d} V+V\:\text{d}C\$. But it's usually assumed that \$\text{d}C=0\$ -- that there is no change in capacitance, since it is a fixed value in an ideal capacitor. So \$\text{d} Q = C\:\text{d} V\$. You can now introduce time by simply dividing both sides by \$\text{d} t\$, giving \$\frac{\text{d} Q}{\text{d} t} = C\:\frac{\text{d} V}{\text{d} t}=I\$. (The ad-hoc introduction of the infinitesimal of time can be done where it pleases.)

Your KVL loop equation is:

$$\begin{align*} V &= R\:I_{\left(t\right)}+\frac{1}{C}\int I_{\left(t\right)}\:\text{d}t\tag{1}\\\\ \text{d} V&=R\:\text{d}I_{\left(t\right)}+\frac{I_{\left(t\right)}}{C}\text{d} t=0\tag{2}\\\\ \text{d}I_{\left(t\right)}&=-\frac{I_{\left(t\right)}}{R\,C}\text{d} t\tag{3}\\\\ \frac{\text{d}I_{\left(t\right)}}{I_{\left(t\right)}}&=\frac{-1}{R\,C}\text{d} t\tag{4}\\\\ \int\frac{\text{d}I_{\left(t\right)}}{I_{\left(t\right)}}&=\frac{-1}{R\,C}\int\text{d} t\tag{5}\\\\ \operatorname{ln}\left(I_{\left(t\right)}\right)&=\frac{-t}{R\,C} + A_0\tag{6}\\\\ I_{\left(t\right)}&=e^{\left[\frac{-t}{R\,C} + A_0\right]}=e^{A_0}e^{\frac{-t}{R\,C}}=A\,e^{\frac{-t}{R\,C}}\tag{7} \end{align*}$$

Where \$A\$ is the constant of integration you need to worry about, now. At \$t=0\$, it must be that \$A=I_{t=0}\$. If the voltage across the capacitor at \$t=0\$ is \$V_C=0\:\text{V}\$, then it follows that \$A=\frac{V}{R}\$. So the result must be:

$$I_{\left(t\right)}=\frac{V}R\,e^{\frac{-t}{R\,C}}\tag{Series Current}\label{E1}$$

And,

$$\begin{align*} V_{C\left(t\right)} &= \frac{1}{C}\int I_{\left(t\right)}\:\text{d}t\tag{8}\\\\ &=\frac{1}{C}\int \frac{V}R\,e^{\frac{-t}{R\,C}}\:\text{d}t\tag{9}\\\\ &=\frac{V}{R\,C}\int e^{\frac{-t}{R\,C}}\:\text{d}t\tag{10}\\\\ &=\frac{V}{R\,C}\left[-R\,C\,e^{\frac{-t}{R\,C}} + A_0\right]\tag{11}\\\\ &=V\left[\frac{A_0}{R\,C}-e^{\frac{-t}{R\,C}}\right]\tag{12} \end{align*}$$

Again, at \$t=0\$ and assuming that \$V_{C\left(t=0\right)}=0\:\text{V}\$, it follows that \$A_0=R\,C\$. So:

$$V_{C\left(t\right)}=V\left[1-e^{\frac{-t}{R\,C}}\right]\label{E2}\tag{Capacitor Voltage}$$

That's it. The \$\ref{E1}\$ equation provides the current in both \$R\$ and \$C\$, since both must be the same. The \$\ref{E2}\$ equation provides the voltage that accumulates onto the capacitor over time.

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    \$\begingroup\$ @David777 If you select one or two that are troublesome, let me try and help explain them better. I'll tag them to make it a little easier for you to name them. \$\endgroup\$ – jonk Feb 2 at 19:52
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    \$\begingroup\$ @David777 Whenever you do an indefinite integral, there is a constant of integration and needs to be solved, somehow, with known initial conditions. In integrating both sides of (4) to get (5), there must be a constant of integration for each integral. No need for two of them, though. So I just put it on the right side as \$A_0\$. I could have put it on the left side. But I chose the right. Either way it is the same thing. I assumed you already knew that \$\int \text{d}t=t\$ and since there was already a -1 there, it should be clear where the -t comes from. Isn't it? \$\endgroup\$ – jonk Feb 3 at 21:32
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    \$\begingroup\$ @David777 Yes. I just chose to use \$A_0\$ instead of \$C\$ because of the possible confusion with the capacitance. In doing so, I see I caused confusion. Sorry about that. \$\endgroup\$ – jonk Feb 3 at 21:48
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    \$\begingroup\$ @David777 The integral on the left side of (5) is solved as ln(). I then raised both sides of (6) as the power of e in order to get rid of the ln() function to get (7). \$\endgroup\$ – jonk Feb 3 at 22:40
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    \$\begingroup\$ @David777 Well, getting from (11) to (12) should be simple algebra. The RC cancels in the first term of the second factor because of the denominator of the first factor. And I just moved \$A_0\$ to the front and, of course, again applied the first factor's denominator to it, as I should. Just basic algebra there. The final equation had to apply initial conditions to figure out \$A_0\$. But that just turns out to be \$R\:C\$ because to force the equation to produce 0 volts at time 0, that term had to become a "1" somehow. That "somehow" is by setting \$A_0=R\:C\$. \$\endgroup\$ – jonk Feb 6 at 1:13
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Simple answer is after switch closes, Ip=V/R then decays with a slope dV/dt=V/RC = V/T then exponential decay. So Vc rises with this slope with same exponential rise toward Vc with a linear asymptote at the t=T=RC at V=~64%Vcc then slower as current reduces.

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  • \$\begingroup\$ Thank you for your answer, it was very helpful. \$\endgroup\$ – David777 Feb 1 at 20:01

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