Measuring DC current??

I have a 3d printer with a 12vdc-30A power supply (PS). The PS has a common +/- rails with three screw connectors for positive and three for negative. The printer has two sets (+/-) of wires coming from the PS to a junction b lock. Then there are two sets of wires (+/-) going to two screw terminals o n the controller card.

I recently bought a clamp meter that does DC-Amps, pretty clever and seems to be very accurate when compared with an in-line Amp meter.

So here's the issue. If I clamp around one wire from the PS when both the heated bed and nozzle are drawing current I get 11A on one wire and 9A on the other. If I clamp a round both wires I get about 14A so I am confused.

Can I clamp around both wires for a total current reading? If I clamp one, is it showing a true reading?

ASCII Art

0-vdc ----+-+ +-+----- 12vdc [ PS | | | | [ measure each 11A and 9A | | | | [ measure both 14A +---+ +---+ | | | | [ Junction block +---+ +---+ | | | | [ measure each 11A and 9A | | | | [ measure both 14A Hotbed --+ | | +-- Nozzle ----+ +----

Thoughts?

Thanks

Reply to
Dave, I can't do that
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On Wednesday, March 7, 2018 at 4:44:03 PM UTC-5, Dave, I can't do that wrot e:

on +/- rails with three screw connectors for positive and three for negativ e. The printer has two sets (+/-) of wires coming from the PS to a junction block. Then there are two sets of wires (+/-) going to two screw terminals on the controller card.

s to be very accurate when compared with an in-line Amp meter.

e are drawing current I get 11A on one wire and 9A on the other. If I clamp around both wires I get about 14A so I am confused.

So what is the make and model of the meter? Might just as well include the make and model of the power supply.

Dan

Reply to
dcaster

I recently bought a clamp meter that does DC-Amps, pretty clever and seems to be very accurate when compared with an in-line Amp meter.

So here's the issue. If I clamp around one wire from the PS when both the heated bed and nozzle are drawing current I get 11A on one wire and 9A on the other. If I clamp around both wires I get about 14A so I am confused.

Can I clamp around both wires for a total current reading? If I clamp one, is it showing a true reading?

ASCII Art

0-vdc ----+-+ +-+----- 12vdc [ PS | | | | [ measure each 11A and 9A | | | | [ measure both 14A +---+ +---+ | | | | [ Junction block +---+ +---+ | | | | [ measure each 11A and 9A | | | | [ measure both 14A Hotbed --+ | | +-- Nozzle ----+ +----

Thoughts?

Thanks

===================

If you pass all the power supply wires through the clamp the reading should be zero, whether the printer is on or off, on AC or DC.

Clamp meters are intended to read the magnetic field around a single conductor. If more than one wire passes through the clamp they should show the difference between the currents, which is zero if the same current passes through in opposite directions.

The Hall Effect sensor in a DC clamp-on is very susceptible to zero drift, from temperature changes (your hand) and on sensitive ranges from orientation in the Earth's magnetic field. My UT210E which resolves to 1mA DC varies by 25-30mA depending on whether the clamp is perpendicular or parallel to Earth's field lines. I need to hold it in the measuring position, zero it, then clamp around the wire and read within a few seconds before it drifts.

-jsw

Reply to
Jim Wilkins

I should have written the algebraic sum of the currents, with the sign depending on the direction of current flow.

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-jsw

Reply to
Jim Wilkins

What happens when you do the same experiment with the inline DC amp meter?

What happens when you reverse the direction of one of the wires so that the currents subtract?

Is it really DC current? If the current is not really constant, the meter has an algorithm that converts what it sees to a DC current value for display. Average? RMS? other? The waveform of the AC components of the currents will affect the converted number that the meter calls DC depending on what algorithm the meter implements and the shapes of the two waveforms.

You also have to worry about saturation in the meter core. Your loads are likely pulse-width modulated and have a peak current well in excess of the average...maybe... If the peak current exceeds the range of the meter, the top end will get compressed.

Put the biggest caps you can find on the load end of the printer connections. Measure the current in the wires between the caps and the PS. If the caps are BIG, the current will look more like DC and the readings should converge toward what you expect.

Doesn't look like it here, but watch out for paths that circumvent your measurement. Ground here is not the same as ground over there. It's possible to have current go in directions that don't include the wire you're enclosing.

Reply to
mike

Thanks guys, I have just decided to stay with the single wire. I was being lazy as it is a bit of a struggle to get enough room for the clamp between them.

This morning it (correctly) measured the 11A and 9A but the combined was 19 A, so obviously it is confusing the meter with a variety of probable causes as was mentioned in your replies.

I will just stop the laziness and use the single wire from the junction blo ck to the control board. That is always accurate when checked with the pass

-through meter. It just confused me a little as I was pretty sure 11+9 did not equal 14 and then this morning with the meter quite cold, it showed 19. That could have easily been confused as 11+9 is close enough to be 19 and thus assumed that measuring both was near-enough. That could have been nast y. :)

When combined it probably has something to do with the temps (19C at the mo ment) as was also mentioned. The single wire clamp and pass-through reading s seem accurate despite the temp. It was about 25C yesterday when I got the 14A double-wire reading.

Lesson learned and thanks anyway.

Reply to
Dave, I can't do that

Perhaps a "rounding error"? If the 11 was really 10.6 and the 9 was realy 8.6, the REAL current would be 19.2, and the combined reading would round to 19

Reply to
Clare Snyder

If the measurement circuit could resolve to tenths the maker would add another digit to the display, and to the price.

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The standard assumption is that the displayed reading can be off by at least 1 count. "11" could be 10 or 12, "9" could be 8 or 10, so the sum could be from 18 to 22. Notice that the sum of two 10% errors is still a 10% error.

-jsw

Reply to
Jim Wilkins

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