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  1. #1
    Join Date
    Jun 2008
    Location
    NC
    Posts
    84

    Using resistor to convert to 4-20ma to 1-5v

    When placing a resistor in the 4-20ma to convert it to voltage is it placed in series or parallel?

    When placing it in parallel i get no voltage at the inputs. I run the 4-20ma signal into the analog input, with the common of the 24vdc to the common of the controller, then place the resistor between the common and analog input of the controller. This results in nothing

    When placing the 4-20ma signal into one end of the resistor and out the other end i got 7 volts which makes no sense since its a 250ohm resistor. I should be reading around 1.5-2vlts (ma output is 6.21) Either the resistor is bad or ??

    Also, how do you figure out the wattage of the resistor to use??

    Also had to just flip dipswitches to convert prior to this

    thanks!

  2. #2
    Join Date
    Sep 2008
    Location
    New England
    Posts
    137
    Ive never used a resistor for a AI usually only on AO. It seems like you had it installed correctly.
    "The bitterness of poor quality remains long after the sweetness of low price is forgotten". --Benjamin Franklin

  3. #3
    Join Date
    Mar 2005
    Location
    Georgia
    Posts
    387
    Quote Originally Posted by Shockwave View Post
    When placing a resistor in the 4-20ma to convert it to voltage is it placed in series or parallel?

    When placing it in parallel i get no voltage at the inputs. I run the 4-20ma signal into the analog input, with the common of the 24vdc to the common of the controller, then place the resistor between the common and analog input of the controller. This results in nothing

    When placing the 4-20ma signal into one end of the resistor and out the other end i got 7 volts which makes no sense since its a 250ohm resistor. I should be reading around 1.5-2vlts (ma output is 6.21) Either the resistor is bad or ??

    Also, how do you figure out the wattage of the resistor to use??

    Also had to just flip dipswitches to convert prior to this

    thanks!
    Is this 4-20 with a two wire connection?

  4. #4
    Join Date
    Jun 2008
    Location
    NC
    Posts
    84
    Quote Originally Posted by chadtech View Post
    Is this 4-20 with a two wire connection?
    Yes a two wire

    wattage i figured out, max volts squared divided by resistor value = watts

  5. #5
    Join Date
    Jul 2009
    Location
    Kansas
    Posts
    375

  6. #6
    Join Date
    Feb 2013
    Location
    Phoenix, AZ
    Posts
    198
    Here is the normal hookup for a 2-wire 4-20ma instrument converted to 1-5V at the controller (resistor is 250 ohm):

    Name:  4-20.jpg
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  7. #7
    Join Date
    Mar 2012
    Location
    WV
    Posts
    21
    I use a 500ohm and just go across the two wires at the device that is controlling the device.One side on positive one side on negative. If that makes sense. Works everytime. Also this will not work the other way around

  8. #8
    Join Date
    Mar 2012
    Location
    WV
    Posts
    21
    Oh sorry buddy.. that for 4-20 to 0-10VDC I read your post wrong

  9. #9
    Join Date
    Dec 2008
    Location
    Ontario Canada
    Posts
    546
    500 ohm resistor will work, however most manufactures I believe use 570. This will give 2-10VDC not 0-10. Wattage shouldn't matter, as amp draw is almost nonexistent. It's a resistive circuit, so ohms law applies.

  10. #10
    Join Date
    Feb 2013
    Location
    Phoenix, AZ
    Posts
    198
    Quote Originally Posted by LKJoel View Post
    500 ohm resistor will work, however most manufactures I believe use 570. This will give 2-10VDC not 0-10. Wattage shouldn't matter, as amp draw is almost nonexistent. It's a resistive circuit, so ohms law applies.
    If Ohm's law applies, (and it does), then 570 ohms will not give the expected results:

    At 4ma:

    E = I R
    E = 0.004 X 570
    E = 2.28V

    At 20ma:

    E = 0.02 X 570
    E = 11.4V

    The load resistance needs to be 500 ohms for 2-10V and 250 ohms for 1-5V.

    wattage is:

    P = I^2 X R

    P = 0.02^2 X 250
    P = 0.1 W

    P = 0.02^2 X 500
    P = 0.2 W

    In both cases wattage is less than 1/4W, but the 500 ohm resistor is close, and will get warm (at 20ma).

  11. #11
    Join Date
    Apr 2007
    Location
    Amarillo by mornin'
    Posts
    853
    Name:  4_20_05.PNG
Views: 918
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    "It's not that I'm smart, it's that I stay with the problem longer”
    Albert Einstein

  12. #12
    Join Date
    Dec 2008
    Location
    Ontario Canada
    Posts
    546
    Quote Originally Posted by DDC_Dan View Post
    If Ohm's law applies, (and it does), then 570 ohms will not give the expected results:

    At 4ma:

    E = I R
    E = 0.004 X 570
    E = 2.28V

    At 20ma:

    E = 0.02 X 570
    E = 11.4V

    The load resistance needs to be 500 ohms for 2-10V and 250 ohms for 1-5V.
    (at 20ma).
    I am we'll aware of how ohms law works. And of course the "correct" answer is 500 ohms. I do however know that I have seen in several applications where manufacturers have used a 570 ohm resistor to preform the same function. Don't know why, if anybody does know I would appreciate an answer

  13. #13
    Join Date
    Feb 2013
    Location
    Phoenix, AZ
    Posts
    198
    Quote Originally Posted by LKJoel View Post
    I do however know that I have seen in several applications where manufacturers have used a 570 ohm resistor to preform the same function.
    Could you give an example of one? I have not run across this, would be interesting to examine.

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