50A DC Current Sensing

I have both a 50A DC hall effect IC with with the same specifications as Allegro's ACS750LCA-050 (

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) and an old analog simpson guage with a 0-50A plate which I presume to be a

50mV meter. The output on the IC is linear enough for my purposes outputting 2.5V @ 0A and 4.5V @ 50A with a Vcc of 5V. Is it even possible to mate both of these? If so, what would be the easiest way/design? If any more information is needed, feel free to ask.

Thanks, Brian

Reply to
Brian
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Brian, this is an easy task, and I imagine you'll get plenty of people describing various ways to do it. For example, the Simpson meter could be placed between the 750LCA-050 IC's output and a resistive divider connected between 5V and ground, plus a trimpot to remove the 5% zero error. We need to know the meter's internal resistance, which you can measure with a multimeter. And just in case it was really meant to work with 100mV shunts, etc., it would be helpful to know its full-scale current (or voltage), which you can measure with a multimeter, 1k resistor and a bench supply.

--
 Thanks,
    - Win
Reply to
Winfield Hill

yes it is.. for the - side of your meter, use a Voltage divider with a small range pot on it in the middle of the divider to calibrate the zero point.. just a pot with Resistors from each outer leg for the voltage divider. what you need to do is set this divider to 2.5 volts from the

5 volt supply. next. you have 4.5 volts for full 50Amps, since you already have 2.5 volts gone, your net voltage would be 4.5 - 2.5 = 2 volts. now you know that you have only a 0..2 volt range in reality for the + side of your meter. you need to know the current of your meter at full scale. or simply use an ohm meter to test the meter coil and do the math. I := E/R = 0.050/R = Im ( Im= meter current). now, since we have 2 volts to work with you now need to subtract the meter voltage, Es (voltage sum) = 2.00 - 0.050 = 1.950 volts. now the series Resistor needed for this from the output of the hall detector to the + input of the meter would be 1.950/Im

P.S., when selecting resistors for the voltage divider on the

- side of the meter, use low value ones so that you don't get a noticeable effect in the math. something like maybe a 470 ohm on each outer leg of a 100 ohm pot. the center tap of the pot is what goes to the - meter side one wing of the R goes to common, and the other to your 5 volt supply. happy camping and i hope it is close enough for your needs.

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
Reply to
Jamie

Obviously you need to buck that Vcc/2 zero current output with the same. Then you need to two trims: one for the 1-5& sensor + components error, and another for possibly 15% gain error deviation from 40mV/A.

Reply to
Fred Bloggs

I tried measuring the meter's internal resistance and the resistance started out at a higher value and dropped down to 50 Ohms when the meter hit full scale. I just connected my multimeter leads to the leads on the meter. If I did this correctly and my math is correct, then the meter draws 1mA at full scale which would mean I would need some form of resistance equaling 1950 Ohms in series between the IC output and the meter and then the voltage divider on the other end of it according to Jamie's post. Is this all correct?

Thanks, Brian

Reply to
Brian

No!- You were told to use a limiting resistor and bench power supply to determine the current for full scale. Not only are you clueless about the basics, but you can't even follow the simplest of directions.

Reply to
Fred Bloggs

First off I was trying to follow Jamie's instructions of "or simply use an ohm meter to test the meter coil and do the math." Second, I am cluless, thats why I asked for help. If your point in posting was just to flame its pointless and unproductive. Not all of us are electrical engineers and some (like me) have very limited knowledge.

Reply to
Brian

OK, now you know the meter is a 50mV type meant for use with shunts, as you thought, with 1mA full scale and a 50-ohm resistance. You've calculated that you need 1950 ohms of additional series resistance, which is correct, except you'll ant the +/-15% as Fred suggested.

Start with two 1k resistors from Vcc to gnd, to establish the Vcc/2 midpoint with a 500-ohm Thevenin output resistance. Add a 1k pot from Vcc to gnd, and take its wiper through 47k to the Vcc/2 node, to provide a +/-26mV zero-trim range from the pot's +/-2.5V swing. Your meter goes from the Vcc/2 node to the 750's output through the 1950 ohms you calculated, less 495 ohms for the 1k dividers and zero adj ckt, and less about 300 ohms to provide for a 1k Fred's gain pot. You can use a 1k series resistor with the 1k gain pot, that's close enough. Have fun calibrating your 50A Hall-effect electronic meter.

. ________________________________________________________ Vcc . | Allegro | | . __|__sensor ,------, 1.0k | . | | _ V | | 1k . | |---|_|--- 1k --- 1k pot ---+-----+--- 47k ---> ZERO . |_____| meter GAIN | pot . | 1mA 1.0k | . ____|____________________________________|______________| gnd

--
 Thanks,
    - Win
Reply to
Winfield Hill
[crosspost: sci.electronics.design,sci.electronics.basics, followups-to sci.electronics.basics] On Mon, 13 Feb 2006 04:30:27 -0800, Brian googled:

Well, don't use an ohmmeter to measure the internal resistance of a meter movement - you stand a very real risk of damaging the meter.

Two, learn to quote context - don't click "Reply" at the bottom of the post you're replying to, click "Show Options" at the top, and when the options pane opens up, click _that_ "Reply" link. Then trim appropriately, and bottom post.

Three, what is the model of the Simpson, and tell us everything that's knowable about the instrument - like, _everything_ on its name plate, its model number, etc.

The way to determine the internal resistance of an analog meter is to take a 1.2 or 1.5V cell, put a pot in series - some meters have 50 MICROamp movements, so make it at least a megohm or more, and adjust it until the meter reads full-scale. Then, take another pot, and put it directly in parallel with the meter movement, adjust it until the meter needle is at exactly half-scale, disconnect everything, and measure the resistance of the pot that was in parallel with the meter. That will be equal to the internal resistance of the meter. From there, you can determine how to use it in your circuit.

I've crossposted to s.e.basics, and set followups there, because that's really where a question like this belongs. :-)

Good Luck! Rich

Reply to
Rich Grise

Those old Ammeter moving coil movements were quite often temperature compensated for operation from a low-resistance source, and 1950 ohms would look like a current source.

The job might be easier if you just go and look for a 1 milliohm ammeter shunt. There is at least one still available in the UK Radiospares catalogue (at £25 though), but there are now cheaper pc-mount alternatives. eg, The IRC OAR5 series is 1%,5W, about 80 UK pence each, and is available with a minimum value of 5 milliohms. 5 in parallel would do the job.

--
Tony Williams.
Reply to
Tony Williams

Temperature compensation for a low resistance source would mean the compensation was against change of the movement coil resistance. So unless there is additional compensation required against deflection versus current, no compensation for a current source should be needed.

Reply to
Fred Bloggs

Sounds right. Tony?

So it's a large 1-milliohm shunt with a 50mV 2.5W drop, versus a small 0.12-milliohm Hall current sensor with a 6mV 0.3W drop, plus associated wiring losses of course. Standard shunts come with convenient flat surfaces with threaded bolt holes for the 50A cables, whereas the Hall sensor is meant for PCB mounting. Tony's set of IRC OAR-3 or OAR-5 5-milliohm 5% resistors would allow us to make a relatively-small PCB-mounting shunt, just to complicate the decision tree. If Allegro had a bolt-hole version of their Hall sensor, we'd have a full range of choices. One appealing aspect of the shunt approach is the lack of offset and gain adjustments, which would greatly simplify installation.

. ________________________________________________________ Vcc . | Allegro sensor | | . __|__ACS750LCA-050 ,------, 1.0k | . | ___ | _ V | | 1k . ||___||---|_|--- 1k --- 1k pot ---+-----+--- 47k ---> ZERO . |_____| meter GAIN | pot . | 1mA 1.0k | . ____|____________________________________|______________| gnd

------------- vs. ----------------

. 0.001 1% shunt or CSL . ----+--/\\/--+---- 50A cable . | _ | . '--|_|--' . 50mV meter . . --- or --- . . 0.005 5% OAR, PLO or LOB . ,--/\\/--, . +--/\\/--+ . ----+--/\\/--+---- 50A cable . +--/\\/--+ . +--/\\/--+ . | _ | . '--|_|--' . 50mV meter

IRC also offers the CSL series of Kelvin-terminal PCB-mounting sense resistors, available from 0.25 to 2.5 milliohms, 5W, 1%.

--
 Thanks,
    - Win
Reply to
Winfield Hill
[snip]

That's the point about 'cheap' sensors, (of any sort), they rarely have an absolute calibration, and getting it calibrated is an unwarranted expense for a 1-off.

If a Hall effect sensor is wanted (for say isolation), then use one with a big aperture, which allows many turns to be temporarily wound-on for calibration at a much lower current.

Your dwg modified slightly.

Make the assembly so that the Kelvin connection holes bolt directly onto the meter studs. It's two short lengths of brass.

--
Tony Williams.
Reply to
Tony Williams

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Ummm.... that URL works on one computer but not the other?

Anyway, you are right Fred and my memory was faulty.

The Weston Instruments company improved the D'arsonval movement so that there was complete compensation.

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Tony Williams.
Reply to
Tony Williams

In message , Tony Williams writes

Try

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Tony Williams
Reply to
Tony Williams

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