Capacitor Voltage-Current Relations

Capacitor Voltage-Current Relations

In this lab, we wire the circuit in the diagram above and predict the voltage though the capacitor V_c with regards to the input voltage V_in. First, we will apply a sinusoidal wave for V_in, predict the capacitor voltage V_c and then compare it with the observed capacitor voltage. Next we repeat the experiment using a different frequency wave, and finally will do it using a triangle wave.

Here we see the circuit wired up with wires connected for ground and measuring channels for the Analog Discovery.

Above is the equation for current in the circuit. We can integrate it and multiply it by the inverse of the capacitance to get the voltage across the capacitor. Plugging in the two frequencies we will be using, we can find the equations for the two sinusoids.

Sinusoidal - 2V input @ 1 kHz

We use WaveForms software to apply a sinusoidal signal for V_in with amplitude of 2 V at 1 kHz. In the oscilloscope window, C1 (orange) shows the signal for the input voltage V_in, C2 (blue) shows the voltage across the capacitor V_c, and M1 (red) shows the current through the resistor i_r. We see that the capacitor voltage has an integral relationship with the current as predicted earlier. However, the amplitude of the voltage is about 1 V. Compared to our predicted voltage of 2.8 V, our voltage has a tremendous error of 64%!

Sinusoidal - 2V input @ 2 kHz

The experiment is now repeated for a frequency of 2 kHz. Again we see that the capacitor voltage is a multiple of the integral of the current. We obtain a voltage amplitude of 1.5 V. Compared to our predicted value of 1.4, we have a 7.1% error, which is still a bit higher than we would like, but still a lot better than our previous try.

Triangle - 4V input @ 100 Hz

A triangle wave of 4 V amplitude at 100 Hz frequency is now used. Qualitatively speaking, we see the capacitor voltage looks more like the derivative of the current than the integral. The current takes on the form of a piecewise function that remains linear and constant in slope magnitude but reverses its sign. The integral of a linear function is a quadratic function. If we look at the peaks, where the current switches directions, we see that the voltage has a slightly curved shape. 

The current is given by the piecewise function above, and likewise, the voltage is also a piecewise function and the integral of the current multiplied by the inverse of the capacitance. The period of the function is 10 ms and the ranges of each function are given in terms of the period T.

We believe that the flat regions in the voltage graph are indicative of the time periods when the capacitor is full and cannot gain any more voltage.