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I have this HW problem: enter image description here

It has me completely stumped because the usual equation for the transient response of a capacitor assumes the circuit has a voltage source, not a current source, and requires a value for the time constant, which is normally defined as \$ RC \$. If there were a resistor in parallel with the current source, I could use Thevenin's theorem to get the equivalent circuit with a voltage source and resistor in parallel. But this circuit has no resistor.

What's the correct way to approach modeling the voltage over the capacitor as a function of time? My thought was that I could just assume the current source isn't ideal and add a resistor in series with it, to represent its internal resistance. But even then, I still wouldn't have a value for the initial voltage to plug into the usual capacitor voltage equation, since I wouldn't have a specific value for the resistance of the current source, so I don't see how I could get a numerical time value for part b.

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3 Answers 3

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A capacitor that is fed with a constant current will experience a constant voltage rise.

$$\frac{dV}{dt}=\frac{I}{C}$$

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  • \$\begingroup\$ Why I/C instead of I*C? \$\endgroup\$ Sep 7, 2023 at 2:21
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    \$\begingroup\$ The larger the C, the more charge that is needed to raise the voltage by a given amount. \$\endgroup\$ Sep 7, 2023 at 2:24
  • \$\begingroup\$ Wow, OK, I was COMPLETELY overthinking this. It didn't even occur to me to start out with the diff eq and solve for V assuming I is a constant. \$\endgroup\$ Sep 7, 2023 at 2:43
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Well, the voltage- and current relation in a capacitor is given by:

$$\text{I}_\text{C}\left(t\right)=\text{V}_\text{C}'\left(t\right)\text{C}\tag1$$

So, when \$\displaystyle\text{I}_\text{C}\left(t\right)\$ is a constant and stable DC-current equal to \$\displaystyle\hat{\text{I}}\$, we can solve:

$$\hat{\text{I}}=\text{V}_\text{C}'\left(t\right)\text{C}\space\Longleftrightarrow\space\text{V}_\text{C}\left(t\right)=\int\frac{\hat{\text{I}}}{\text{C}}\space\text{d}t=\frac{\hat{\text{I}}}{\text{C}}\int1\space\text{d}t=\frac{\hat{\text{I}}}{\text{C}}\cdot t+\text{K}\tag2$$

Looking at your circuit, we can see that \$\text{V}_\text{C}\left(0\right)=0\space\text{V}\$, so we find \$\text{K}=0\$ and end up with:

$$\text{V}_\text{C}\left(t\right)=\frac{\hat{\text{I}}}{\text{C}}\cdot t\tag3$$

Finding the time when \$\displaystyle\text{V}_\text{C}\left(t\right)=1\space\text{V}\$, we see:

$$1=\frac{\hat{\text{I}}}{\text{C}}\cdot t\space\Longleftrightarrow\space t=\frac{\text{C}}{\hat{\text{I}}}=\frac{10\cdot10^{-12}}{1\cdot10^{-3}}=10^{-8}\space\text{s}=10\space\text{ns}\tag4$$

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Start with the integral equation. The voltage across a capacitor is equal to the integral of the current as a function of time. If the current is a constant, it falls out of the integral. That leaves the integral of dt, which is just t. The integral equation reduces to grade school arithmetic:

EC = IT

E (voltage across capacitor) times C (capacitance) = I (the constant current) times T (time).

This is a handy little puppy when doing an R-C charge or discharge calculation. If the time period is much less than one time constant, you are working on an almost linear portion of the exponential curve. For example, if you are trying to find the correct capacitor size to reduce power supply ripple to a certain value, and the ripple value is relatively small, such as 2 V of ripple with a peak value of 10 or 15 V, the discharge current is essentially constant. You can get to a first order approximation in your head. It's even easier in Europe, where t = 10 ms.

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