I am learning some basic electronics, and I'm trying to wrap my mind around some measurements I made on a full wave bridge rectifier.
Using my oscilloscope (grounded) and an isolation transformer to "float" the circuit I am measuring, I made the following observations:
On points A (CH1-blue) and B (CH2-yellow).
Please note the "width" of the waveforms at their base.
On point C (CH1-blue). Same base width.
Without the filtering capacitor on point C, that's the expected waveform. Once I connected an electrolytic 100uF 25V capacitor as a filter, I got another expected waveform as well (DC level with ripple.)
Point C on channel 1 (blue) now with parallel capacitor.
What I'm not sure I understood is the waveform of points A and B again with the capacitor now filtering the output wave:
That above is point A on channel 1 (blue) and point C on channel 2 (yellow).
And now, point B on channel 1 (blue) and point C again on channel 2 (yellow).
Please note the wider base of the waveforms of point A and B with the capacitor when compared with the same points without it.
To explain this change, I came to the conclusion that since the ground alligator clamp of my oscilloscope is conected to the negative of the now charged capacitor, during the positive semi-cicle on point A (the rising of the waveform) point B is now 14V "beneath" point A or in other words, point A is 14V above ground - i.e. the leftover stored charge on the capacitor (first smaller peak). And then, the EMF of the transformer takes this waveform to its nominal value in volts shortly after (second taller peak). Point B is symmetrically opposite.
So, the negative lead of the electrolytic capacitor is in some sense advancing the rise time of the waveform on point A and in the other hand, delaying the fall time of the waveform in point B.
Is my conclusion accurate?