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How to measure 255 volts (divider/shunt?) — Parallax Forums

How to measure 255 volts (divider/shunt?)

IRobot2IRobot2 Posts: 164
edited 2009-10-30 18:54 in General Discussion
I am in the process of making an interface for an industrial battery charger using the prop. The control aspect is relatively simple with turning on a set of contactors and looking for errors and faults. The rest is merely code that controls how long the batteries are on charge, their state and at what rate they are charged.

In order to do this I need to be able to read the output of the charger (Voltage and Current) with an output around 255 VDC and 1500 A.H.. My thoughts are to use a shunt or some type of voltage divider returned to an ADC and my prop. Accuracy should be down to the volt (1-2 volts). I keep thinking this should be relatively easy to do, but I am more of a basic electronics kind of guy and never really ventured into industrial or high power before. So I am getting confused on how I can safely drop (or at least read) this higher voltage with a 0-5v ADC. I know the basic answer is to use a shunt… but even after some research I am lost on what to do with it. Any thoughts or suggestions?

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Alex Burke
"Beware of computer programmers that carry screwdrivers." -Leonard Brandwein

Comments

  • LeonLeon Posts: 7,620
    edited 2009-10-29 16:36
    I make that 382 kW, that's some battery charger!

    Leon

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    Amateur radio callsign: G1HSM

    Post Edited (Leon) : 10/29/2009 4:42:20 PM GMT
  • IRobot2IRobot2 Posts: 164
    edited 2009-10-29 16:47
    Input: 480/575 VAC, 3Ph, 60 HZ
    OutPut: 2 Circuits, 64 Cells, 1500 A.H.

    This charger is typically used in mining applications to charge large banks of batteries. It is pretty big in my opinion.

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    Alex Burke
    "Beware of computer programmers that carry screwdrivers." -Leonard Brandwein
  • LeonLeon Posts: 7,620
    edited 2009-10-29 18:26
    Are they usually charged in series?

    Leon

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    Amateur radio callsign: G1HSM
  • mctriviamctrivia Posts: 3,772
    edited 2009-10-29 18:41
    I would use a voltage divider and optoisolater. The optoisolator is optional but if you don't have and the top resister gets shorted your adc is toast.

    Make sure you do fuse the input.

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    24 bit LCD Breakout Board coming soon. $21.99 has backlight driver and touch sensitive decoder.
  • IRobot2IRobot2 Posts: 164
    edited 2009-10-29 18:44
    To make it simple they are charged in series. However, the battery units that are being charged contain around 6 -12 industrial batteries within them. Some only contain 2 large batteries. From what I gather it differs between brands and is simply looked at by the number of cells that the battery unit has and not so much as how many batteries are within it. So at times it is just one big battery. There is only one “unit” charging at time or 64 cells. However you want to look at it. But yeah, series.

    Typically there will be two types of battery units to charge. One roughly around 128 Volts and one is around 240 Volts. Any ideas on how to read the voltage that is going out of the charger?

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    Alex Burke
    "Beware of computer programmers that carry screwdrivers." -Leonard Brandwein
  • IRobot2IRobot2 Posts: 164
    edited 2009-10-29 19:42
    Thanks mctrivia, I guess you posted while I did.

    I read or heard some where about how voltage dividers become less accurate the more voltage that you drop. In this case 250 volts. Did I just misunderstand that?

    I guess I am going to have to use some pretty hefty resistors.

    Thanks for the tip on the isolator and fuse. I will be sure to do that.

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    Alex Burke
    "Beware of computer programmers that carry screwdrivers." -Leonard Brandwein
  • mctriviamctrivia Posts: 3,772
    edited 2009-10-29 20:09
    Actually if you chose large values for the resisters the current will be almost nothing. For fireworks I use less then a micro amp.

    Accuracy will go down because your adc voltage error will mean more for a larger divider but as long as the add sense current daw is significantly less then the current draw of the bottom resister it will be accurate

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    24 bit LCD Breakout Board coming soon. $21.99 has backlight driver and touch sensitive decoder.
  • kwinnkwinn Posts: 8,697
    edited 2009-10-30 04:16
    Use several 1% or better tolerance resistors in series for the top resistor and a sense resistor with a resistance value 1/100th or less than the input impedance of the ADC.

    If the input impedance of the ADC is 100K, then use a 1K or lower resistor as the last resistor (sense resistor) in the chain.
    Assuming a 1K sense resistor the other resistors would need to be 50K total. For safety I would use 10 5K resistors in series to prevent a single resistor failure damaging the circuit. With these resistor values 1/4 W resistors would be adequate.

    The alternative would be to use the string of resistors to provide a current to the inverting input of an op amp that outputs a voltage proportional to the current.
  • IRobot2IRobot2 Posts: 164
    edited 2009-10-30 12:30
    Thanks kwinn, that really helps me further along. I was actually wondering if I could use 1/4 W resistors, and the comment of using several resistors in series as a sort of fail safe is an awesome idea.

    Thanks for helping me out every one.

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    Alex Burke
    "Beware of computer programmers that carry screwdrivers." -Leonard Brandwein
  • mctriviamctrivia Posts: 3,772
    edited 2009-10-30 12:53
    1% will give you a couple volts potential error. Calibration can correct in software or you can use .1%

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    24 bit LCD Breakout Board coming soon. $21.99 has backlight driver and touch sensitive decoder.
  • IRobot2IRobot2 Posts: 164
    edited 2009-10-30 14:40
    Is there any possible way that I can determine what the current is on the output of the battery charger using this circuit? Since I now know the voltage, and I have a known resistance should I not be able to easily figure out what the current is? Or would this only tell me the current of what my circuit is drawing and not the output of the charger?

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    Alex Burke
    "Beware of computer programmers that carry screwdrivers." -Leonard Brandwein
  • mctriviamctrivia Posts: 3,772
    edited 2009-10-30 14:46
    No you can't

    To calculate charge curent place a small resister between battry negative and charger negative

    Measuring voltage acroSs that will give you curent

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    24 bit LCD Breakout Board coming soon. $21.99 has backlight driver and touch sensitive decoder.
  • kwinnkwinn Posts: 8,697
    edited 2009-10-30 15:21
    mctrivia is correct regarding current. A .01 ohm resistor will give an output voltage of 10mV/amp so assuming the charge rate is 10% of the amp hour battery rating you would have a 1.5V output at 150A. Power dissipation in the resistor would be 225W so an alternative measurement method may be desirable.
  • IRobot2IRobot2 Posts: 164
    edited 2009-10-30 15:31
    Yeah, the 150A and 225W is a little more than I can probably deal with. I am probably going to have use a current transformer and figure out some matching circuit. A simple solution would be to use a current transducer. But those get pretty pricey. However the simple 0-5v output would make that part a breeze for me.

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    Alex Burke
    "Beware of computer programmers that carry screwdrivers." -Leonard Brandwein
  • kwinnkwinn Posts: 8,697
    edited 2009-10-30 17:59
    I wonder if there is a hall effect sensor that could be used for DC current measurement.

    Take a look at the attached data sheet. Available from Digikey, Newark, and Arrow. Cost under $10.00. Probably similar devices available from other manufacturers.

    Post Edited (kwinn) : 10/30/2009 6:32:24 PM GMT
  • IRobot2IRobot2 Posts: 164
    edited 2009-10-30 18:11
    Well I have been looking at this DC Hall Effect Current Transducer. But I believe it is in the 3 - 4 hundred dollar range [noparse]:([/noparse] This is what actually sent me on my quest to find something cheaper.

    http://www.crmagnetics.com/newprod/ProductView.asp?ProdName=CR5210

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    Alex Burke
    "Beware of computer programmers that carry screwdrivers." -Leonard Brandwein
  • mctriviamctrivia Posts: 3,772
    edited 2009-10-30 18:26
    TGHGCR0050FE $27

    Max current you can use is 141A will drop 0.7005V at that current.

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    24 bit LCD Breakout Board coming soon. $21.99 has backlight driver and touch sensitive decoder.
  • kwinnkwinn Posts: 8,697
    edited 2009-10-30 18:31
    Check my earlier post.
  • IRobot2IRobot2 Posts: 164
    edited 2009-10-30 18:54
    @kwinn, That ACS758ECB seems to be exactly what I am looking for. I need to go through the data sheet again but I think that will do the trick. It also didn't look to complicated to implement either. I think the hardest thing will be trying to get it in line with the large output cable.

    Thanks so much kwinn and mctrivia for all your help. It is really appreciated!

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    Alex Burke
    "Beware of computer programmers that carry screwdrivers." -Leonard Brandwein
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