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I'm a newbie with the oscilloscope. I have a digital RIGOL with two probes.

I'm testing a simple 12V AC to 5V DC converter.

I'm trying to measure the effect of the 7805 on the DC signal in output and comparing this signal to the AC signal before the rectifier bridge.

If I connect the probes as you can see in the pictures I see sparkles (short circuit) on the GND probe of the AC line. This make me think that the the two probes of the oscilloscope have common GND.

enter image description here

So now I have two question:

  • Have I to connect only one GND of the probe to the GND of the circuit to study the signals properly?

  • What can I look to understand if the output signal is "clean" (no ripple, no big oscillation)? Have I to select AC coupling in the oscilloscope and see Vrms, frequency, ecc...?

Sorry to bother you with this very simple question but I see that using the oscilloscope properly is difficult.

Thanks.

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    \$\begingroup\$ Besides your GNDs being connected inside the scope, before you start, please search for a youtube video "how not to blow up your osciloscope". \$\endgroup\$ – PlasmaHH Jun 9 '17 at 11:40
  • \$\begingroup\$ Why measuring both at the same time? Just measure the output and use AC trigger to sync on it. The ripple will be in sync with the mains. \$\endgroup\$ – Sredni Vashtar Jun 9 '17 at 12:05
  • \$\begingroup\$ This is the video mentioned by @PlasmaHH: youtu.be/xaELqAo4kkQ \$\endgroup\$ – Christian Sep 25 at 20:27
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First of all:

Yes. All the ground probes of your different channels are connected internally. This means you can't connect grounds a different voltage levels because they're shorting on the inside of the oscilloscope.

Second:

Your ground probe from each channel is loosely connected to the earth wire from the oscilloscope's plug. This means that you should never use the oscilloscope to measure the voltage from the wall unless you're using an isolation transformer. Otherwise you risk frying it and/or activating the differential protection.

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For what you have it seems that it would be more interesting to see the waveforms at VI and VO. This share a common ground so you can see them both at the same time. Don't use AC coupling for this, since it removes all the DC component which is what we're trying to see at the output.

Start with that and then see what happens when you increase/decrease the input/output capacitance. Learning by doing will work wonders.

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You can't connect scope probe ground connections simultaneously to two different points on the target circuit without problems (as others have pointed out) however...

Given the likely light loading of the 7805, you could try and get away with a half wave rectifier circuit and this means your AC input would naturally connect to the 0 volts on the 7805: -

enter image description here

I say this because you have a 12 volt AC input and, after rectification and filtering the DC voltage presented to the 7805's input will be about 17 volts minus two diode drops (bridge) and 1 diode drop (half wave).

So with a bridge you get about 15.6 volts DC and with a single diode rectifier you get about 16.4 volts. Under load, these will produce ripple voltages of about 20% and 40% respectively (hand waving approximation) and so, you might be feeding an average voltage into the 7805 of possibly 14 volts or 13 volts (not much difference).

This means that the 7805 cannot deliver too much current to its output load without over-heating. Over-heating occurs because the 7805 drops (loses) 8 or 9 volts in producing a 5 volt output and at (say) 300 mA this is a power dissipation of around 2.5 watts. Without a heatsink the regulator will be stinking hot hence....

That is why I said "likely light loading" earlier on.

So, your circuit doesn't necessarily need a bridge rectifer and therefore you can possibly get away with a single rectifier and "scope" your AC input voltage and DC output voltage with impunity and without fear of damaging anything.

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