How to measure a capacitor with an oscilloscope. 

Andre Sant'Anna Thanks very much for your input Andre.

Even the capacitance of your bum on the chair....

@@muppetpaster The capacitance of the bum on the chair is inversley perportional to how interesting the video is and like a capacitor it will short to ground as soon as the interest peeks and it will need to dump its load..

I think that is why he starts with explaining that he is saving up for a good and precise LCR-meter.....

And how would you measure the parasitic capacitance? Connect the function generator through a resistor directly to the scope?

"You should measure the parasitic capacitance of the system first (without holding the wires with your hands), then connect your capacitor, measure the new value and subtract the parasitic value from the compound value."

With what frequency should I measure ESR in circuit then?

@@thuglifescorpion What do you mean?

Roman Kyselý My pleasure Roman, thanks for the post.

These are great suggestions. The exponential fit will give the best results, but obviously requires a bit more work. Using the method in this video, you were relying on the capacitor being nearly saturated by the time it hit the end of the waveform. In particular, you took y2-y1 to be the completely charged voltage. In the process you effectively measured two voltages (bottom to top of the waveform, and bottom to 63%) and one time (beginning of waveform to 63%V). There is an in-between option that would offer improvement without much more work. By adding one measurement, you can ignore the top voltage, and get better results. The big advantage is that you can use bigger resisters, since you no longer care about saturating the capacitor during a pulse, which makes for a slower rise time and better measurements. The procedure is as follows: 1) As before, pick a time that puts the voltage about halfway up the curve. This doesn't need to be precise. Forty percent or sixty percent up the curve is fine. Write down the time (measured from the beginning of the waveform) as t1. 2) Write down the corresponding voltage (measured from the bottom of the waveform) as V1. 3) Now choose t2 equal to twice t1. In other words, if t1 was 1.00 milliseconds, choose t2 to be 2.00 milliseconds. This does need to be precise in order to take advantage of an algebraic trick, and to avoid solving a nonlinear algebraic equation for the final answer. Just double t1, and measure V2 at t2. 4) The time constant, RC, is equal to (V1*t1)/(V2-V1). Note that t2 does not appear in the equation, because it is assumed to be double t1. The equation is derived from the assumption that the voltage history has the form V(t) = Vfull*(1-e^(t/RC)) and plugging in (t1,V1) to give a first equation a and (2*t1,V2) to give a second equation. Solving the pair of equations for RC gives the simple ratio above. 5) Fix my algebra and repeat step (4). I did the algebra by hand, which means I screwed it up. One more note for the future: you did a nice job of eyeballing the measurement locations, except that you measured from the top of the noise band for the top voltage, and from the bottom of the noise band at the low voltage. In this procedure, the errors roughly cancelled each other out, but in most cases your results will be better if you measure from the middle of a noise band. Thanks for the video! I'm coming back to electronics after many years, and it's great to see this kind of tinkering in a lab. Lots of fun.

Thanks a lot to you both for your comments, they helped me a lot! :-)