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Help with Isolation amplifiers

  1. Mar 30, 2012 #1
    Hello. I need help in finding the following isolation amplifiers:
    1. I'm measuring current of approximately 600 A using LEM current transducer LF 1005-S (which gives an output of 200 mA for 1000 A). LEM transducer need DC power supply, and i use bipolar 15 V dc power supply. I need isolation amplifier (200 mA - 500 mA to 10 V) that i'll connect to the DC supply and LEM transducer. Then i'll take voltage from amplifier and connect it to the acquisition card Personal DAQ 3000 USB (card has only voltage input).
    2. I have 6 Pt 100 probes connected in series with current source (1 mA). I need isolation amplifiers for Pt 100 probes (measuring temperature) which converts cca 200 mV - 500 mV to 10 V.
    Thanks
     
  2. jcsd
  3. Mar 30, 2012 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    On your first question, isn't the LEM current sensor already isolated? The datasheet says that it is galvanically isolated from the wire whose DC/AC current it is sensing...

    http://www.google.com/url?q=http://www.lem.com/docs/products/lf%25201005-s%2520sp12.pdf&sa=U&ei=8-J1T-eTAoewiQK3jr2nDg&ved=0CBAQFjAA&usg=AFQjCNEIUDLje6hxKyMaCO5jqPw84QG8RQ [Broken]

    .
     
    Last edited by a moderator: May 5, 2017
  4. Mar 30, 2012 #3
    LEM sensor has current output, and I need voltage output for personal daq 3000 card. If you connect LEM sensor with resistor there'll be too huge losses (around 1.8 W for 200 mA and 46 Ω resistor).
     
    Last edited by a moderator: May 5, 2017
  5. Mar 30, 2012 #4

    berkeman

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    Ah, I think it is just a language issue with the term "isolation amplifier". I think you are asking about how to do the current-to-voltage conversion efficiently, which has nothing to do with isolation amplifiers.

    The term Isolation Amplifier usually refers to an amplifier that gives some amount of galvanic isolation between input and output:

    http://en.wikipedia.org/wiki/Isolation_amplifier

    You would use such an amplifer if you had a sensing circuit connected to the AC Mains, and you wanted to get that sensed information into a low-voltage circuit (like a microcontroller). The isolation may be done with optoisolators or other means.

    But in your first application with the current sensor, the LEM module already gives you the galvanic isolation. I think you are asking how to convert the 200mA max output current of the sensor (which does appear to have a current mode output) into a voltage for reading by your ADC. Since it is a current mode output, you should be able to use the minimum size resistor specified in the datasheet (5 Ohms), and then just use an opamp with the appropriate gain to give you the maximum output voltage for your ADC range (which I think you list as 10V).
     
  6. Mar 30, 2012 #5

    berkeman

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    BTW, will the current you are sensing be AC, DC or a mix? If AC, then you will have to add an offset to the output of your current-to-voltage converter opamp circuit, if you want the output to be in the 0V to 10V range.
     
  7. Mar 30, 2012 #6
  8. Mar 30, 2012 #7

    berkeman

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    So you can use a load resistor R(M) of just a few Ohms to save on power.

    Let's say you use a 1 Ohm precision (0.1%) resistor. What power rating does it need to have in your application? And what does the opamp circuit look like to turn the voltage across R(M) into your +/-10V output signal?
     
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