Oscilloscope Waveform

[staticshock]

Can someone help explain the waveform trigger screenshot?

The signals being looked at are currents from a 3 phase 400Hz. Current clamps (40mV/A) was used on each phase wire and directly connected to an oscilloscope. The smaller section of the waveforms was measured to be 2A and then the bigger section of the waveforms was measured to be 18A. I know their values because prior we used a Fluke 434 to measure it. However, looking at this graph I don't understand how to explain that the small section is 2A and the bigger section is 18A. Is there some scaling that I'm missing. The current clamps are 1M ohm input, does that make a difference than 50 ohm, which I'm more familiar with?

[donlisms]

The current clamp doesn't have a 1M ohm input; rather, it's output is calibrated to expect a 1M input on the measuring device.  So your scope input impedance should be 1M ohm.  If it is 50 ohms, you may be burdening the output of the meter excessively, and all bets are off.  In other words, it might not just be calibrated into a 1M ohm load, it might be incapable of driving a lower-impedance load.  The readings would be lower than expected.

The other related error is a device that produces an output expecting to go into 50 ohms.  Here's the thing: if you have an output impedance of 50 ohms, and an input of 50 ohms, and you connect them together, you have a voltage divider.  The output produced by the (whatever) is living half within output impedance of itself, and half within the output of the receiving device.  So the output that's produced is twice what you expect the reading to be, because you're only reading half of it.

On the other hand, a device that expects to see a 1M ohm input impedance will typically have a very low output impedance, relatively speaking.  So we have, more or less, a nearly-zero output impedance connected to a nearly-infinite input impedance.  This, again, forms a voltage divider, and almost all the output voltage the instrument is producing will live in the input of the measuring device.  In such a case, the output of the producer should expect to produce the voltage that should actually be read, since none of it's getting lost in its own output impedance.

You are measuring three lines; each appears to be at about 280mV peak or so, which means each one is carrying about 7 amps peak, assuming peak is what you want to know.  I have a suspicion it should really be RMS, and that puts it at about 5 amps each, but I do not know.

And then I also suspect there is something you'll have to do with combining the three legs.  They are 120 degrees out of phase with each other, and so they are not simply adding up; there are vectors involved.  But they might be adding up if you're using RMS values... these are things I do not know.  If I were in your position, I'd start studying three phase power and how to measure it.  There seems to be some references on the interweb; poke around a little and see if any of them help you understand what's going on.  They speak of the sqrt(3) / 2 and that sort of thing.  I would guess it applies here somehow.  Maybe.

[vk6zgo]

Can someone help explain the waveform trigger screenshot?

The signals being looked at are currents from a 3 phase 400Hz. Current clamps (40mV/A) was used on each phase wire and directly connected to an oscilloscope. The smaller section of the waveforms was measured to be 2A and then the bigger section of the waveforms was measured to be 18A. I know their values because prior we used a Fluke 434 to measure it. However, looking at this graph I don't understand how to explain that the small section is 2A and the bigger section is 18A. Is there some scaling that I'm missing. The current clamps are 1M ohm input, does that make a difference than 50 ohm, which I'm more familiar with?

I don't possess a DS4014, or, indeed, any DSO, so I had to do a bit of research to translate your screenshot into "old school" analog speak, but after finding a User Guide for your 'scope online, & trawling through piles of extraneous stuff, I can definitely say your Rigol is set for 1M input impedance, so if the clamps were designed to look into such an impedance, that is not your problem.
 
Several possibilities come to mind:

(1) Are you sure that 400Hz is within the frequency response of your clamps, as many are designed for 50/60 Hz?

(2) Your time/div(H) is very long, maybe you should look at reducing it to display just one or two cycles of 400Hz on the screen.

(3) The resultant waveform of all three phases simply added together is going to be quite complex, & difficult to analyse.
If you have separate clamps for each phase, perhaps you should display them separately on different channels to prove whether the clamps or other parts of your setup are appropriate for your test.

The traces for the three phases will then appear displaced in time w.r.t each other, proportionally to the 120 degree phase displacements between phases.
The current contributions of each phase can then be seen, & the required mathematics applied to get the result you require.
The DS4014 is quite a capable device, so possibly it may have provision to do this latter internally, if you can find the right page in the "Guide".

[Damianos]

Can someone help explain the waveform trigger screenshot?

The signals being looked at are currents from a 3 phase 400Hz. Current clamps (40mV/A) was used on each phase wire and directly connected to an oscilloscope. The smaller section of the waveforms was measured to be 2A and then the bigger section of the waveforms was measured to be 18A. I know their values because prior we used a Fluke 434 to measure it. However, looking at this graph I don't understand how to explain that the small section is 2A and the bigger section is 18A. Is there some scaling that I'm missing. The current clamps are 1M ohm input, does that make a difference than 50 ohm, which I'm more familiar with?

I assume what you mean is that the current clamps are for instruments with a 1 MΩ  or higher input. Set the input of the oscilloscope to accommodate that.
Select the probe as 25X and the unit as Ampere (1V at the current clamp is 25A).
Then search in the manual and the menus on how to do measurements in selected parts of the waveform.