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0 to 5V to 4 to 20mA

Discussion in 'General Electronics Discussion' started by kurtclark, Feb 2, 2012.

  1. kurtclark

    kurtclark

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    Feb 2, 2012
    I am generating a 0-5V signal from a PWM and need to convert it to 4 to 20mA, I tried a few circuits that I found online but it does seem to work as I thought it would. These circuits generally give 0 to 20ma output.

    What I was looking for is 4mA for 0V input and 20mA for 5V input.
     
  2. Resqueline

    Resqueline

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    Jul 31, 2009
    See if this thread gives you some clues to the workings & requirements of such circuits.
     
  3. kurtclark

    kurtclark

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    Feb 2, 2012
    It seems that the op-amps in that circuit are powered by 20V + , I would like the op-amps to be powered by 5V and only the output transistor or mosfet to be powered by 24Vdc.
     
  4. Resqueline

    Resqueline

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    Jul 31, 2009
    That should make it even simpler, provided the load is allowed to hang from +24V instead of being connected to ground.
     
  5. kurtclark

    kurtclark

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    Feb 2, 2012
    Could it be possible with the load being connected to ground?
     
  6. Resqueline

    Resqueline

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    Jul 31, 2009
    Anything is possible but the 5V op-amp requirement makes that more of a challenge. What's the reason behind that requirement/restriction?
     
  7. kurtclark

    kurtclark

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    Feb 2, 2012
    I want to power the device from the current loop, so I have to ensure that circuit current is within the 3.5mA range.
     
  8. Resqueline

    Resqueline

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    Jul 31, 2009
    Ok, so it's to be a loop-powered sensor/transmitter.
    That implies that the circuit initially receives almost 24V, but has to consume less than 4mA as you say. What's the load resistance?
    It also seems to to me that it'll be floating. Are you sure about the ground requirement? Have a look at this circuit example from Maxim.
     
  9. kurtclark

    kurtclark

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    Feb 2, 2012
    Generally speaking the load would be 250ohm so you would get 1V for 4mA and 5V for 20mA, but I would like to make the circuit to work suitably upto 1000ohm load.
    The supply voltage is typically 28VDC.
    I'm sorry I didn't understand your comment about the ground requirement, yes the ground would be have to be self created.
    Have a look at this circuit from Analog Devices it is loop powered, but it works on an input of 0 to 1.6V.
    http://www.analog.com/en/circuits-from-the-lab/CN0145/vc.html

    I do not need galvanic isolation per se even a non isolated loop transmitter would be fine.
    Thanks for the link it gives me a good insight of loop powered systems.
     
  10. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    If you have a working circuit that gives you 0 mA... 20 mA output, scale it down so it gives 0 mA...16 mA output.
    Then add a fixed current source (4 mA) in parallel. The total current will be 4 mA + (0 mA ... 16 mA) = 4 mA ... 20 mA.

    You don't even need a second current source. Just the 4mA jus generate the power for the measurement circuit from it.

    Harald
     
    Last edited: Feb 3, 2012
  11. kurtclark

    kurtclark

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    Feb 2, 2012
    Wonderful suggestion.
    I will try this out.
     
  12. kurtclark

    kurtclark

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    Feb 2, 2012
    I'm sorry I didn't understand that why I didn't need a second current source?
     
    Last edited by a moderator: Feb 3, 2012
  13. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    I'm sorry, I didn't express that clearly.
    I thought along the following lines:
    You have a circuit (let's call it A) that measures some property and controls at it's output a current source 0...16 mA. Now, somehow this circuit needs to be powered. The beauty of a 4...20 mA intzerface is, that there is always at least 4 mA available. You can use this idle current to generate the supply voltage for your circuit.
    Look at the schematic:
    [​IMG]

    And of course, yes, you need a second current source for that, too. I hadn't thought that through thoroughly.


    Regards,
    Harald
     

    Attached Files:

  14. kurtclark

    kurtclark

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    Feb 2, 2012
    Excellent explanation Harald, and thanks for the drawing that indeed was quite nice!
    But I have a question, Lets says my circuit takes only 1.2mA for its operation what to do?
    how do I manage the remaining 2.8mA?
     
  15. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
  16. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    In answer to your question:
    Don't be afraid, no current will be lost. Since the sum of all current in node is 0 - Kirchhoffs law - not yet overruled by any parliament of the world :) - all current that is not required to operate your circuit will flow through the lone resistor and develop some voltage drop across it (2.8 V at 1 kOhm). By adjusting the resistor you can adjust this voltage. Adding a capacitor and a zener diode parallel to the resistor will help you to stabilize and smooth out the voltage.

    Harald
     
  17. kurtclark

    kurtclark

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    Feb 2, 2012
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