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I've been running into problems calculating the DC behaviour the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When I try to calculate the voltage of capacitor C1 I get different results based on the way I try to find it. Here are the two ways I tried to calculate it:

  1. Solve Kirchhoffs current law

$$ I_1 = i_{c1}(t)+i_{c2}(t)= C_1 \cdot \frac{\mathrm{d}u_{c1}(t)}{\mathrm{d}t} + C_{p} \cdot \frac{\mathrm{d}(u_{c1}(t)+u_{r1}(t))}{\mathrm{d}t} \\ I_1 = (C_1 + C_2)\cdot \frac{\mathrm{d}u_{c1}(t)}{\mathrm{d}t} + C_{2} \cdot R_1 \frac{\mathrm{d}i_{c1}(t)}{\mathrm{d}t} \\ I_1 = (C_1 + C_2)\cdot \frac{\mathrm{d}u_{c1}(t)}{\mathrm{d}t} + C_1 \cdot C_{2} \cdot R_1 \frac{\mathrm{d}^2 u_{c1}(t)}{\mathrm{d}^2 t} \\ \int I_1 \mathrm{d}t= \int ( (C_1 + C_2)\cdot \frac{\mathrm{d}u_{c1}(t)}{\mathrm{d}t} + C_1 \cdot C_{2} \cdot R_1 \frac{\mathrm{d}^2 u_{c1}(t)}{\mathrm{d}^2 t} ) \mathrm{d}t\\ \frac{I_1\cdot t}{ C_1 \cdot C_{2} \cdot R_1} + K_1= \frac{(C_1 + C_2)}{C_1 \cdot C_{2} \cdot R_1}\cdot u_{c1}(t) + \frac{\mathrm{d} u_{c1}(t)}{\mathrm{d} t} $$

When solving this first order differential equation I get:

$$ u_{c1}(t)= \frac{I_1}{C_1+C_2}(t - \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) +\frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} \cdot K_1 + K_2 \cdot \mathrm{e}^{-t\frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}}\\ $$

The initial value of the capcitor C1 is assumed as uc1_0. This leads to:

$$ u_{c1}(t)= \frac{I_1}{C_1+C_2}(t - \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) +\frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} \cdot K_1 + (u_{c1_0}+ \frac{I_1}{C_1 + C_2}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} - \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} \cdot K_1) \cdot \mathrm{e}^{-t\frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}}\\ $$

I don't know how i can actually finde K_1 since I only have one initial value which I use for finding K_2. Can it be omitted or combined with K2?

  1. Solution with capacitator current

So far so good. I calculated the current ic1 through the capacitor C1 and the resistor R1 as:

$$ i_{c1}(t)= I_1 \cdot \frac{C_1}{C_1+C2}-( I_1 \cdot \frac{C_1}{C_1+C2} - i_{c1_0})\cdot \mathrm{e}^{-t \frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}}\\ $$

When I try to calculate the voltage uc1 of the capacitator c1 by multipying the current ic1(t) with 1/C1 and integrating it over dt I dont receive the same result:

$$ u_{c1}(t)= \frac{1}{C_1} \int i_{c1}(t)\mathrm{d}t = \int ( \frac{I_1}{C_1 + C_2} - ( \frac{I_1}{C_1+C2} - \frac { i_{c1_0}}{C_1})\cdot \mathrm{e}^{-t \frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}})\mathrm {d}t \\ u_{c1}(t)= K_1 + \frac{I_1}{C_1 + C_2} \cdot t - ( \frac{I_1}{C_1+C2}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} - \frac { i_{c1_0}}{C_1}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) \mathrm{e}^{-t \frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}} \\ u_{c1}(0)= u_{c1_0} = K_1 - ( \frac{I_1}{C_1+C2}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} - \frac { i_{c1_0}}{C_1}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) \\ K_1 = u_{c1_0} +( \frac{I_1}{C_1+C2}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} - \frac { i_{c1_0}}{C_1}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) \\ u_{c1}(t) = u_{c1_0} + \frac{I_1}{C_1+C2}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} - \frac { i_{c1_0}}{C_1}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} + \frac{I_1}{C_1 + C_2} \cdot t - ( \frac{I_1}{C_1+C2}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} - \frac { i_{c1_0}}{C_1}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) \mathrm{e}^{-t \frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}} \\ u_{c1}(t) = u_{c1_0} + \frac{I_1}{C_1 + C_2} \cdot t -\frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} ( \frac{I_1}{C_1+C2} - \frac { i_{c1_0}}{C_1} ) \cdot( 1- \mathrm{e}^{-t \frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}}) $$

Which one of the solution is correct?

When comparing the two solutions the most obvious difference (and the most baffling to me) is uc1_0. Once it is a constant value in the equation and then it is multiplied with e ^-x which causes its influence to subside.

I'm at my wits end. I can't find my error but there has to be one.

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  • \$\begingroup\$ It rises to negative infinity. \$\endgroup\$
    – Andy aka
    Jul 28, 2017 at 11:30
  • \$\begingroup\$ Thank you for your input but that doesn't answer my question or help me find my error. \$\endgroup\$
    – M. Lavery
    Jul 31, 2017 at 12:34

3 Answers 3

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First order differential equation analysis gives:

Current flowing \$\small down\$ through \$\small C_1\$: $$\small I_1=-\frac{C_1}{C_1+C_2}\left(1-e^{-\frac{(C_1+C_2)}{RC_1C_2}t}\right)=-\frac{1}{11}\left(1-e^{-t/9.1\times 10^{-5}}\right)$$

Current flowing \$\small down\$ through \$\small C_2\$: $$\small I_2=-\left(\frac{C_2}{C_1+C_2}+\frac{C_1}{C_1+C_2}e^{-\frac{(C_1+C_2)}{RC_1C_2}t}\right)=-\frac{1}{11}\left(10+e^{-t/9.1\times10^{-5}}\right)$$

Voltage across \$\small C_2\$:

$$\small V_{C2}=\frac{1}{C_2}\int_0^t I_2\:dt= -\frac{1}{C_2}\int_0^t\left(\frac{C_2}{C_1+C_2}\right)+\left(\frac{C_1}{C_1+C_2}\right)e^{-\frac{(C_1+C_2)}{RC_1C_2}t}\:dt$$

thus: $$\small V_{C2}=-\left(\small\frac{1}{C_1+C_2}\right)t+\frac{RC_1^2}{(C_1+C_2)^2}\left(1-e^{-\frac{(C_1+C_2)}{RC_1C_2}t}\right) $$

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So using KVL,KCL equations, you would get something like below, considering the voltages across C1 and C2 as v1 and v2 $$C_1\frac{dv_1}{dt}+C_2\frac{dv_2}{dt}+I_1 = 0$$ $$v_2=v_1+i_1R=v_1+RC_1\frac{dv_1}{dt}$$ Substituting for v2 in the above equation, $$C_1\frac{dv_1}{dt}+C_2\frac{d}{dt}(v_1+RC_1\frac{dv_1}{dt})+I_1=0$$ $$(C_1+C_2)\frac{dv_1}{dt}+RC_1C_2\frac{d^2}{dt^2}v_1+I_1=0$$ Now applying laplace transform to the above time domain expression we get, $$(C_1+C_2)[sV_1(s)-v_1(0^-)]+RC_1C_2[s^2V_1(s)-sv_1(0^-)-\frac{dv_1}{dt}|_{t=0^-}]+\frac{I_1}{s}=0$$ Assuming the experiment starts at t=0, and the capacitor voltages v1 and v2 be 0V, $$V_1(s) = \frac{-I_1}{s^2(C_1+C_2+RC_1C_2s)}$$ The time response would be, (after partial fractions) $$v_1(t)=\frac{-I_1}{C_1+C_2}[t-\frac{RC_1C_2}{C_1+C_2}(1-e^{-t\frac{C_1+C_2}{RC_1C_2}})]u(t)$$

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  • \$\begingroup\$ Thank you very much for your response. I get the same results for v1 using a laplace transform. But now I have three different answers and I cant tell which one is correct. \$\endgroup\$
    – M. Lavery
    Jul 31, 2017 at 12:12
  • \$\begingroup\$ The answers you and I get using the laplace transform don't make sense unitwise. The unit of the numerator is Current times t which equals Asec. The unit of the denominator as Resistance [V/A] times capacitance [Asec / V] squared equals [A*sec^2/V]. Numerator divided by denominator equals V/s. Yet we are trying to calculate a voltage [V]. So there's an error in the laplace transform calculation as well. \$\endgroup\$
    – M. Lavery
    Jul 31, 2017 at 12:26
  • \$\begingroup\$ Thank you Lavery for your comments, I have made some edits, please look at it. I missed the step input current source as I1/s. Please correct me if there are any mistakes. Looks like it matches your second approach with minor sign corrections. \$\endgroup\$ Jul 31, 2017 at 15:06
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The second solution for \$u_{c1}(t)\$ is correct. I had calculated \$i_{c1}(t)\$ correctly and also calculated \$u_{c1}(t)\$correctly via integration.

To calculated \$u_{c1}(t)\$ correctly via the first method \$K1\$ must not be omitted but kept throughout the calculation. We need more information though to determine \$K1\$. My error was forgettint the second initial value.

We have two initial values
1. \$u_{c1_0}\$ for the voltage of C1: We used this value to calculate \$K_2\$

  1. \$i_{c1_0}\$ for the current through C1: We can use this value to determin \$K_1\$

In order to use the initial value for the current \$i_{c1_0}\$ we have to calculate the current through

$$ i_{c1}(t)= C_1 \cdot \frac{\mathrm{d} u_{c1}(t) }{\mathrm{d}t}$$ $$ i_{c1}(t)= \frac {I_q \cdot C_1} {C_1 + C_2} -( C_1\cdot \frac{C_1 + C_2 }{C_2 \cdot R_1 \cdot C_1} \cdot u_{c1_0} + \frac{I_q \cdot C_1}{C_1 + C_2 } + C_1\cdot K_1 ) \cdot \mathrm{e}^ { - \frac{C_1 + C_2 }{C_2 \cdot R_2 \cdot C_1} \cdot t} $$

If we assume \$t=0\$ we receive :

$$ K_1 = \frac{i_{c1_0}}{C_1}+ \frac{C_1 + C_2 }{C_2 \cdot R_2 \cdot C_1}\cdot u_{c1_0} $$

We can use \$K_1\$ in the originally calculated \$u_{c1}(t)\$:

$$ u_{c1}(t)= \frac{I_1}{C_1+C_2}(t - \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) +\frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} \cdot K_1 + (u_{c1_0}+ \frac{I_1}{C_1 + C_2}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} - \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} \cdot K_1) \cdot \mathrm{e}^{-t\frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}}\\ $$

$$ u_{c1}(t)= \frac{I_1}{C_1+C_2}(t - \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) +\frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} \cdot ( \frac{i_{c1_0}}{C_1}+ \frac{C_1 + C_2 }{C_2 \cdot R_2 \cdot C_1}\cdot u_{c1_0}) + (u_{c1_0}+ \frac{I_1}{C_1 + C_2}\cdot \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} - \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2} \cdot (\frac{i_{c1_0}}{C_1}+ \frac{C_1 + C_2 }{C_2 \cdot R_2 \cdot C_1}\cdot u_{c1_0})) \cdot \mathrm{e}^{-t\frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}}\\ $$

The \$u_{c1_0}\$ inside the parenthesis cancel each other out and we receive:

$$ u_{c1}(t)= \frac{I_1}{C_1+C_2}(t - \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) +\frac{i_{c1_0}\cdot C_2\cdot R_1}{C_1 + C_2}+ u_{c1_0} + ( \frac{I_1 \cdot C_1\cdot C_2\cdot R_1}{(C_1 + C_2)^2} - \frac{i_{c1_0}\cdot C_2\cdot R_1}{C_1 + C_2} )) \cdot \mathrm{e}^{-t\frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}} \\ u_{c1}(t)= u_{c1_0} +\frac{I_1}{C_1+C_2} \cdot t - ( \frac{C_1\cdot C_2\cdot R_1}{C_1 + C_2}) +\frac{i_{c1_0}\cdot C_2\cdot R_1}{C_1 + C_2})\cdot ( 1- \mathrm{e}^{-t\frac{C_1+C_2}{C_1\cdot C_2 \cdot R_1}})\\ $$

Which is what I had calculated before and therefore assume to be right.

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