# Calculate capacitance from current / voltage (dV, dI & dT) pulse

How to calculate the capacitance value based on a current and voltage pulse ?

From some data, I would like to calculate the capacitance value of a specific point based on a current and voltage pulse. Unfortunately, all the resources I could find online always refers to calculating the capacitance from a static voltage or current or a instant current / voltage from a capacitator, but not from dI/dV.

My input are two curves, I & V with known time [us]:

I've calculated the dI/(dV/dT) for each step and integrated the values, but I doubt that is correct.

• Show the circuit that is being tested and indicate where the measurements are being made. It doesn't look like a capacitor to me; it looks more complicated than Ic vs Vc. You also need to indicate what units are in the x-axis. Commented Sep 24, 2022 at 11:08
• The circuit is a high hysteresis solar panel, which capacitance varies with the applied voltage, here a variance in the load V is applied at a specific voltage (A "step" per say). dI, dV and dT are known input, the ultimate goal is to repeat these steps across the whole voltage range and extract the capacitance for each "step" from the dI/dV(& dT) to extract a C/V curve. @Andyaka Commented Sep 24, 2022 at 11:11
• I and V must be the capacitor’s current and voltage respectively. Commented Sep 24, 2022 at 13:27
• How do you a step response is purely capacitive? That is, what is the model (physical/circuital) of your panel? What about an LCR meter, instead?
– edmz
Commented Sep 24, 2022 at 13:57

In lieu of a schematic as yet, I will take the most obvious interpretation I suspect given the comments:

simulate this circuit – Schematic created using CircuitLab

A Thevenin voltage source, where the impedance is, not just a series resistance, but an (R || C) + R, with the values shown derived from the plots given.

This was determined by the following procedure:

## (1) Extract Current Waveform

In lieu of a data series, a curve was fitted. The nearly flat initial segment suggests a time delay of about 3µs, and the fairly smooth curve suggests an RLC equivalent response. An exponential polynominal was therefore used:

$$i(t) = \begin{cases} i_0 & t < 0 \\ i_0 + A \left( e^{\frac{-t}{\tau_a}} - 1 \right) + B t e^{\frac{-t}{\tau_b}} + C t^2 e^{\frac{-t}{\tau_c}} & t \ge 0 \end{cases}$$

Where,

$$\begin{array} \\ i_0 &= 3.852 \, A\\ i_s & = -0.275 \, A\\ A &= 0.275\, A\\ \tau_a &= 1 \, \mu s \\ B &= 0.15\, A\\ \tau_b &= 1 \, \mu s \\ C & = -0.0181 \, A\\ \tau_c & = 2.82 \, \mu s \\ \end{array}$$

Finally, 2.7µs was subtracted off to get the correct time offset.

This is assumed to be I1 in the above schematic (not a fixed current as shown, but a dependent signal current as above).

## (2) Fit Voltage Waveform

The circuit elements were solved by fitting to the voltage curve. Whereas the above method uses an analytic form suggestive of an RLC network transient response, this function is assumed to have a filtering action upon that waveform. Which requires calculating the convolution of impulse responses, in the time domain: not something I'm going to do in Excel, and not trivial in the analytic form. (At least, I don't feel like writing it out to solve it based on circuit values.) Therefore, an incremental method was chosen.

First, a parallel RC was suspected. This was found to fit a bit poorly during the rising edge, with error seemingly proportional to current. Therefore, series resistance was added, and this seems to reduce error to acceptable levels. Remaining error appears fairly random, or perhaps a higher frequency interfering signal (ca. 50 kHz).

The method chosen was calculation by forward difference, i.e., evaluating the differential system as a fixed-timestep difference equation, also known as Newton integration. First, a finer timestep was chosen, to help avoid numerical error. Then the recurrence relation was entered, solving for the state variable, C1's voltage:

$$v_c[n] = v_c[n-1] + \left( i[n] - \frac{v[n-1]}{R} \right) \frac{t_s}{C}$$

Where $$\v_c[n]\$$ corresponds to $$\v_c(n t_s )\$$, the voltage on C1, and $$\t_s\$$ is in µs, R in Ω, and C in µF. Finally, these voltages (the capacitor voltage, and the voltage drop on R2 which is simply proportional to current, not shown here) were subtracted from the source, to give the output plot. These values were found suitable:

$$\begin{array} \\ t_s & = 0.05 \, \mu s \\ V_1 & = 6.154 \, V \\ R_1 & = 0.1223 \, \Omega \\ R_2 & = 0.04091 \, \Omega \\ C_1 & = 88 \, \mu F \\ \end{array}$$

The result is shown below:

Entering these formulas into Excel and using SOLVER to reduce the (peak or RMS) differences between measured values and fitting curves, is one method to automate this process. Presumably, the circuit will not change with bias, so a range of values can be calculated fairly straightforwardly in this way. (Further fitting could be applied to solve for R and C's as a function of V and I.)