2
\$\begingroup\$

How is the instantaneous (inrush) current calculated for the capacitor in this circuit?

enter image description here

Both Falstad Circuit Simulator and LTSpice give the same answer for inrush current (500 uA).

LTSpice enter image description here

Falstad Circuit Simulator enter image description here

How does this get calculated?

\$\endgroup\$
8
  • 1
    \$\begingroup\$ Looks like the method is to calculate the step response for the circuit. From what I've seen, this is done via Laplace Transform and/or second order equations. Is this the right track? \$\endgroup\$ – mrbean Aug 18 '20 at 4:23
  • 1
    \$\begingroup\$ It's not a straightforward algebraic equation if that's what you're hoping. There are ways to calculate it. But honestly, you've already performed the easiest one ;) (Simulate it) \$\endgroup\$ – Kyle B Aug 18 '20 at 4:36
  • 1
    \$\begingroup\$ To calculate this by hand, you do KVL but you will end up with a differential equation due to the derivatives and integrals in the capacitor and inductor terms. There are multiple methods to solve the differential equation, but Laplace transforms are the easiest. \$\endgroup\$ – DKNguyen Aug 18 '20 at 4:41
  • 1
    \$\begingroup\$ Search this website for "RLC differential equation" and you will come up with at least a few examples. electronics.stackexchange.com/questions/480405/… \$\endgroup\$ – DKNguyen Aug 18 '20 at 4:46
  • 1
    \$\begingroup\$ @mrbean That should help out. One thing to take note of is that the series resistor (in series with the inductance) obviously doesn't have much impact. Your peak current only implies at most 2.5 mV drop across it. So you could simplify your analysis by just removing it. This still leaves the other resistor (in parallel with the capacitor.) And that one will definitely impact the moment when the peak is reached -- a LOT because it is so conductive. Even a few mV across it will be an issue. \$\endgroup\$ – jonk Aug 18 '20 at 6:14
4
\$\begingroup\$

Well, I am trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1=\text{I}_3+\text{I}_4\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_2}{\text{R}_3}+\frac{\text{V}_2}{\text{R}_4}\\ \\ \frac{\text{V}_1-\text{V}_2}{\text{R}_2}=\frac{\text{V}_2}{\text{R}_3}+\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag3 $$

Now, we can solve for \$\text{V}_2\$:

$$\text{V}_2=\frac{\text{R}_3\text{R}_4\text{V}_\text{i}}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)+\text{R}_4\left(\text{R}_1+\text{R}_2+\text{R}_3\right)}\tag4$$

Using \$(2)\$, we can see that:

$$\text{I}_3=\frac{\text{V}_2}{\text{R}_3}=\frac{\text{R}_4\text{V}_\text{i}}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)+\text{R}_4\left(\text{R}_1+\text{R}_2+\text{R}_3\right)}\tag5$$


Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

  • $$\text{R}_2=\text{sL}\tag6$$
  • $$\text{R}_3=\frac{1}{\text{sC}}\tag7$$
  • The input voltage is a stable DC voltage equal to \$\hat{\text{u}}\$, so: $$\text{v}_\text{i}\left(\text{s}\right)=\frac{\hat{\text{u}}}{\text{s}}\tag8$$

So, we get:

$$\text{i}_3\left(\text{s}\right)=\frac{\hat{\text{u}}}{\text{s}}\cdot\frac{\text{R}_4}{\frac{1}{\text{sC}}\left(\text{R}_1+\text{sL}\right)+\text{R}_4\left(\text{R}_1+\text{sL}+\frac{1}{\text{sC}}\right)}\tag9$$

Now, when we use inverse Laplace transform (using your values) we can see:

$$\text{I}_3\left(t\right)=\frac{2 e^{-\frac{5}{2} \left(\sqrt{15976001}+4001\right) t} \left(e^{5 \sqrt{15976001} t}-1\right)}{\sqrt{15976001}}\tag{10}$$

Plotting that, gives:

enter image description here

The maximum occurs when:

$$\hat{t}=\frac{2 \text{arccoth}\left(\frac{4001}{\sqrt{15976001}}\right)}{5 \sqrt{15976001}}\approx0.000380305\space\text{s}\tag{11}$$

And at that time \$\$ we have a current of:

$$\text{I}_3\left(\hat{t}\right)\approx0.000498226\space\text{A}\tag{12}$$


Using Mathematica I found a general expersion for the inverse Laplace transform:

In[1]:=R2 = s*L;
R3 = 1/(s*c);
Vi = u/s; FullSimplify[
 Solve[{I1 == I3 + I4, I1 == (Vi - V1)/R1, I1 == (V1 - V2)/R2, 
   I3 == V2/R3, I4 == V2/R4}, {I1, I3, I4, V1, V2}]]

Out[1]={{I1 -> ((1 + c R4 s) u)/(s (R1 + R4 + L s + c R1 R4 s + c L R4 s^2)),
   I3 -> (c R4 u)/(R1 + R4 + L s + c R1 R4 s + c L R4 s^2), 
  I4 -> u/(s (R1 + R4 + L s + c R1 R4 s + c L R4 s^2)), 
  V1 -> ((R4 + L s + c L R4 s^2) u)/(
   s (R1 + R4 + L s + c R1 R4 s + c L R4 s^2)), 
  V2 -> (R4 u)/(s (R1 + R4 + L s + c R1 R4 s + c L R4 s^2))}}

In[2]:=FullSimplify[
 InverseLaplaceTransform[((R4 u)/(
    s (R1 + R4 + L s + c R1 R4 s + c L R4 s^2)))/R3, s, t]]

Out[2]=(2 c E^(-(((L + c R1 R4) t)/(2 c L R4))) R4 u Sinh[(
  Sqrt[-4 c L R4 (R1 + R4) + (L + c R1 R4)^2] t)/(
  2 c L R4)])/Sqrt[L^2 + c^2 R1^2 R4^2 - 2 c L R4 (R1 + 2 R4)]
\$\endgroup\$
9
  • 2
    \$\begingroup\$ Very nicely arranged, Jan! It's approachable for anyone. +1 \$\endgroup\$ – jonk Aug 18 '20 at 23:46
  • 2
    \$\begingroup\$ It was really a wonderful approach!! I loved the simple and accessible way you started out, and then the way to segued into transitioning into Laplace replacements, and the use of the inverse transform. It was a beautiful thing to see. And it teaches how useful and yet also simple, once you get comfortable with the ideas, that Laplace approaches are towards differential equations. You really brought it. I loved it! I hope you stick around and continue to share your perspective and vision. \$\endgroup\$ – jonk Aug 19 '20 at 5:10
  • 2
    \$\begingroup\$ I've no doubt of it, Jan!! I'm so glad to see you here. I can't express that enough. It's the kind of vision that you can share with others that will make differences. I very much hope to see more of you here, helping others see as you do. I think your vision is one of crystal clarity and beauty. It just "sings" in my mind. Some others, those who can hear, will hear that melody well. Best wishes!! \$\endgroup\$ – jonk Aug 19 '20 at 5:31
  • 2
    \$\begingroup\$ It's sometimes hard to pass along visions. When I saw your approach I couldn't think of a better way. It was just ... really good. I don't think I could have done better. Likely, I'd have done a little worse to be honest. And that matters. We have our visions. But we cannot always present it as best we might. And having you here adds yet another who may provide those insights, in a novel and still better way, to help others grasp and accumulate the ways we see the world. We learn from our betters, not for their conclusions so much, but for how they "see the world." You offer something here. \$\endgroup\$ – jonk Aug 19 '20 at 5:53
  • 2
    \$\begingroup\$ I'm just glad to see you, to recognize your mind, to hear your thoughts sing in my mind. Keep it up and offer yourself to others. That's enough. It's more than enough. \$\endgroup\$ – jonk Aug 19 '20 at 6:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.