Consider the following circuit, and the recorded current curve (amp-meter current_in). enter image description here

enter image description here

How do I correctly determine the capacitor's (C_1) capacity, without looking at the time constant? My idea was to calculate the integral of the capacitive part of the current and divide it by the voltage VG1. This does not yield the correct result. What is wrong about my approach?

//Edit I tried to derive the current at the amp-meter: \$ I = V_G1/(R_1+R_2) + (R_2*VG1*\exp(-t(R_1+R_2)/(C_1*R_1*R_2)) +1/(R_1(R_1+R_2)) \$

Is that correct? Is the RC-circuit a high-pass?

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    integrating the current over time would give you the capacitance times the voltage change. Assuming the initial voltage on the cap is zero, you'll need to figure out the final voltage. From that you'll be able to find C. – dirac16 Jan 30 '17 at 14:44
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    Due to high unbalanced R values for +/- inputs , the input current will cause large Vio offset voltage and contribute excessive errors. If you have a known precision C value as a reference, C_1 can be used in feedback instead and generate large AC gain ( e.g. 1nF = x100) from C ratio from a small signal AC coupled with 100M bias to Vin+. Measuring current with unknown R leakage contributes DC errors. Normally LCR meters inject current pulse or sine wave at 1kHz or 100kHz such as a Howland Constant current source and then voltage can indicate impedance and C value. – Tony EE rocketscientist Jan 30 '17 at 16:39
  • RC values are unreliable with 2 unknowns unless you have 2 other known reference R and C values in a bridge to match RC1=RC2 or some ratio. – Tony EE rocketscientist Jan 30 '17 at 16:41

If you integrate the current over the time, you get \$[As]\$. This is the charge \$Q\$. The charge of a cap is given by \$Q_C=C\cdot U\$.

Have a look at the energy of a cap, it defined by \$E_{C}=\frac{1}{2}\cdot C\cdot U^2\$ and the relation to the current is \$i_C(t) = C\cdot\frac{du_C(t)}{dt} \$.

With this equations you can see, that only looking at time and current can not load to the capacitance, you also have to take the voltage curve in consideration. VG1 might not be correct, because there is voltage drop above R1.

  • In this case the voltage across the capacitor is also an exponential function. According to the equation C = Q/U I should just divide by U. But how do I determine U if it changes over time? – luis Jan 30 '17 at 15:12

You can easily find the final voltage when the capacitor is fully charged, this is when the current through the cap is negligible and from voltage divider you simply get $$V_{final}=\frac{R2}{R1+R2}*VG1.$$

Now from $$ i_c=C\frac{dv_c}{dt},$$ capacitance simply becomes $$ \frac{\text{The area below the I-t curve}}{R_2VG_1}.(R_2+R_1),$$ assuming the initial voltage across the cap was zero. Now what you should do is find the area below the I-t curve.

  • I implemented that in matlab, there is still a significant error. The error gets larger when R1 and R2 are in the same range. If R2 is way larger than R1 the error is quite small. Are you sure the voltage divider is sufficient? I mean you only consider the final voltage but the change from the initial voltage (0V) to the final value is non-linear. Does that not play a role? – luis Jan 30 '17 at 17:18

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