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How would you find the mathematical function of the voltage discharge curve for a capacitor considering the following circuit and conditions:

  • When time t=0, the capacitor is fully charged with the voltage V.

  • The current through resistor R is not negligible.

  • The constant current source is "powerful" enough to bring down the voltage of that node.

enter image description here

If the exact function cannot be determined, maybe at least some individual points could be found.

I tried to apply Kirchhoff's law:

I_R (t)+I_C  (t)=I_cst

which means:

enter image description here

and after integrating

enter image description here

enter image description here

At this point I don't know what to do since I cannot integrate an unknown function: Uc(t) (which is also the function I want to find).

Any ideas or different approaches? Thanks!

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After you get to this point:

enter image description here

you have to solve a first order differential equation:

https://www.mathsisfun.com/calculus/differential-equations-first-order-linear.html

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  • \$\begingroup\$ Thank you! from the link you provided I was able to solve the differential equation which turned out as: RC*((Icst*R-V)+Constant) Unfortunately, after I plotted this function into a graph drawer and swapped the R, C, Icst, V and varied the constant I only got a straight horizontal line. I'll have to check everything again. \$\endgroup\$ – CharvelFan Jun 15 at 19:58
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    \$\begingroup\$ It should have an exponential in it... here is [WolframAlpha's answer]( wolframalpha.com/input/…) \$\endgroup\$ – ErikR Jun 15 at 20:04
  • \$\begingroup\$ Finally got the link in my previous comment correct! \$\endgroup\$ – ErikR Jun 15 at 20:09
  • \$\begingroup\$ Ohh, it makes a lot of sense. Turned out that I messed up an integration. Now it is very clear. Thank you so much for the help and for the fast responses! \$\endgroup\$ – CharvelFan Jun 15 at 20:16
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Replace the \$V\$, \$R\$ combination with its Norton equivalent: a current source \$V/R\$ in parallel with \$R\$. Combine the two current sources (add) then convert back to the Thevenin equivalent: voltage source \$(V/R-I_{cst})R\$ in series with \$R\$. Then, instead of connecting the current source at \$t=0\$, solve for a capacitor charged to \$V\$ discharging through \$R\$ to \$(V/R-I_{cst})R\$ starting at \$t=0\$. This is a familiar exponential with time constant \$RC\$.

After you're introduced to the concept of a time constant, probably in second semester college physics, you'll know that the answer will have an exponential. \$V_C\$ starts at \$V\$ and ends at \$V - I_{cst}R\$, so you write $$ V_C(t) = V - I_{cst}R(1-e^{-t/RC}). $$ Done.

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  • \$\begingroup\$ nice approach, thank you! \$\endgroup\$ – CharvelFan Jun 15 at 22:57
  • \$\begingroup\$ Thanks for the later edit also. Eventually I figured out that I should see how the capacitor charging trough a resistor formula is determined, and used the the same principle to find the formula for my particular case. Thanks again for your input! \$\endgroup\$ – CharvelFan Jun 19 at 17:05

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