3
\$\begingroup\$

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.

Full wave bridge points of measurement

Using my oscilloscope (grounded) and an isolation transformer to "float" the circuit I am measuring, I made the following observations:

Input waveforms A and B

On points A (CH1-blue) and B (CH2-yellow).

Please note the "width" of the waveforms at their base.

Channel one of the scope on point C

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.)

Filtered DC pulsating wave - 100uF on 1k load

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:

Point A (CH1-blue) and point C (CH2-yellow)

That above is point A on channel 1 (blue) and point C on channel 2 (yellow).

Point B (CH1-blue) and point C (CH2-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?

\$\endgroup\$
1
\$\begingroup\$

You may consider this when measuring any voltage before the rectifier and after it, at the same time. When you add the capacitor, the rectifier bridge is open during most of the time:

enter image description here

I've used different colors because at each cycle a different pair of diodes will conduct. There will be a short burst of current to recharge what the capacitor has lost during the last cycle. As soon as the AC voltage drops below the capacitor voltage plus the diode drops, they open.

So, during most of the time it is the oscilloscope that closes the circuit between the transformer and the DC circuit. Your probes (probably 1Meg or 10Meg, depending on the 1x or 10x selection) are in parallel with the diodes, which are not directly polarized and also allow only extremely low currents.

Take the following moment as an example. The instant voltage at the secondary (which is floating) is very low, the diodes are opened, and the voltages you measure at A or B are determined by the current flowing through the probe and the reversed polarized diodes.

This may also explain why your last two pictures are not identical. In one it seems that one diode started conducting a few ms before its pair. In the other, one opened a few ms before the other. It is hard to know for sure (minor differences between the probes, if you switched them, of between the diodes could make a difference).

enter image description here

Sugestion: do you have a transformer with a central tap at the secondary? You could build a full wave rectifier with only two diodes and perform good measurements, since the central tap will always be the "0 V" reference.

Update: disregarding the first page of the datasheet and looking into the detailed curves, the operating conditions of the circuit would put the reverse current of the diodes well below 100nA.

enter image description here

\$\endgroup\$
7
  • \$\begingroup\$ So, what I thought it was the negative lead of the capacitor pushing the point A up by actually pulling GND down is incorrect. What is really happening is I am measuring the capacitor through the unpolarized (open) diode via the reverse current? That's something I didn't know it was possible. I didn't know such small residual current could be measured so trivially. \$\endgroup\$
    – Sieg Novak
    Mar 17 at 13:49
  • \$\begingroup\$ I did not switched probes and I was using them on 10x. The diodes I used were all 1N007 and it's data sheets indicate a reverse current "Ir" of 5uA. So, if the "advancement" of the rise of potential - as I called it - was up to about 14V before the coil further elevated it (and recharged the capacitor) and my probe was on 10M, there was a current of about 1,4uA flowing through my probe? \$\endgroup\$
    – Sieg Novak
    Mar 17 at 14:09
  • \$\begingroup\$ I'm not sure I understood since this is about voltage measurement on the floating secondary. Thanks for the extra info. I'll try to clarify with a simulation. Have you tried the other topology I suggested? \$\endgroup\$
    – devnull
    Mar 17 at 14:11
  • \$\begingroup\$ Actually I have not. But I will soon, alongside some other suggestions from the course I'm following. The transformer I own has indeed a center tap, so I'm looking forward to it. \$\endgroup\$
    – Sieg Novak
    Mar 17 at 14:33
  • \$\begingroup\$ I've just added a graph from the datasheet to show that the expected leakage current for diodes in the circuit would be dozens of times inferior the current through the 10M probe. \$\endgroup\$
    – devnull
    Mar 17 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.