How to implement 4-20mA Loop

Discussion in 'General Electronics Chat' started by RIKR09, Feb 13, 2013.

  1. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    Hello People!!!, im trying to use this XTR115 TEXAS Instruments CHIP but i dont know how to implement it. I have a voltage range between 400mV to 3V from a Non Inverter Amplfication layer. So could you help me please? Thanks a lot!!! I attached the way that im using it... but i always get an output current of 68mA :mad::confused::confused:
     
  2. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    Here are the datasheets. I might figure it out first, or not. (I'm being distracted right now.)
    I calculate 15.6 ma when the input is 3V, so the math looks about right. Miswire?
    Is it true that your high signal from the sensor is 15 microamps? That is what would be required to go through a gain of 200,000 and get 3 volts. (I've never heard of a counter that outputs 15 microamps as its "high" signal.)
     
    Last edited: Feb 13, 2013
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  3. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    Senior Thanks for your reply... and im so sorry because i attached the schematic outdated. Ia changed the Amplification Layer resistance upgrading the Gain to 100, this is because throw the sensor's layer im getting about 3.4mV to 36mV... Please look at it again, and could you explain how did you calculate the current????? i did it using a PDF from Microchip but i would like to know you way!!!! Thaks again!!!!!!!
     
  4. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    Vref is 2.5 volts. I ret is grounded at pin 3 so pin 2 must be virtual ground.
    2.5/113k = 22.1 ua.

    If the output from UA1 is a maximum of 3 volts and the input resistors sum to 22.4k,
    3/22400 = 133.9 ua

    add those together for 1561 ua, multiply by 100 and you have 15.6 milliamps.

    So now the upper half of the drawing has a gain of 10,100,000 and the counter must be supplying 297 nanoamps to get 3 volts at the output of UA1. It must be counting something very small.

    edit: I forgot the !!!!!!!!! marks.
     
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  5. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    Thanks Senior!!!!!!! :) ... It was really clear and smart your explication... Now im doing again the circuit in a protoboar as you told me about to review it ... ill being notificating you soon....!!!:eek:
     
  6. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    No :( ... It doestn work yet... Maybe i think im having problems with the way wich i supply the circuit... ... Te GND must be common right? says, the input circuit and the XTR ones? i dont understan this part of the datasheet, what does that picture means... ... Excuse my ignorance. But this get me down :(
     
  7. #12

    Expert

    Nov 30, 2010
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    6,745
    For this chip, Iret (Pin 3)is the ground pin for the sensor voltage. It is not a ground for the current loop. Your supply voltage of 7.5 to 36 volts is attached as shown in my drawing. The negative side does not connect to ground. It connects to the load which will experience 4 ma to 20 ma.

    Iret is not a ground pin for the current driver chip. You must hold the current loop ground separate from the Iret pin.
     
    Last edited: Feb 15, 2013
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  8. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    Ok Excellent explanation Senior ... So look, i took my design from
    http://www.ti.com/lit/an/sboa107b/sboa107b.pdf
    As i could see there put the IRet to ground so i did it in that way... So, i dont know what to do in my circuit with IRet :(, exactly i dont know how to hold the ground of the input circuit apart from the current driver chip. I was reading the Datasheet i watched this, but really i didnt understand it, take a look to image attached.


    Again, thanks for you Help Senior
     
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    Last edited: Feb 15, 2013
  9. #12

    Expert

    Nov 30, 2010
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    Sir, you are mistaken. The drawing you just posted does not have Iret connected to ground.

    The purpose of this chip is to drive a 4ma-20ma circuit that is NOT referenced to ground. If this does not suit your purpose, you have chosen the wrong chip. If that is the case, you are allowed to ask about a different way to achieve a 4ma-20ma circuit with a ground reference.
     
    Last edited: Feb 15, 2013
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  10. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    No im not restricted to be ground reference, i just need 4-20mA and thats all. Just i got teh XTR as an option because all the forums i have read. The question would be, what do i have to connect to IRet, according to my design?
     
  11. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    Iret is connected to the ground of the sensor board, but not connected to the current loop. The voltage supply for the current loop does not connect to ground. It is separate. It must be an isolated loop. It must not share a ground with the sensor board. You must have a second power supply for the current loop and it must not connect to the ground of the power supply for the sensor board.
     
  12. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    Finally and according about you told me and the reading i make the following circuit. so far I have not had good results; but when i have it ill post inmediatly the solutions... Thanks senior and principally for your patience.
    :)
     
  13. #12

    Expert

    Nov 30, 2010
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    6,745
    You might have burnt the chip because the datasheet says anything over 45 ma will burn it. Try a new chip.
     
  14. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    Yes, one more time you were right!!! I had to change my XTR and using the same configuration as you can see in the schematic attached, im getting variations of voltage while the potetiometer does the same. I did my calculations as follows: for 400mV the current through R9 must be 4mA and in the moment of test it, the resulst are satisfactory; likewise for 4V the current must 20mA but im having 25.5mA with the corresponding voltage of 6.26V, so i dont why is this. What can i do to solve this inconvenient?

    Regards Senior, and thank you because your help!
     
  15. #12

    Expert

    Nov 30, 2010
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    6,745
    I calculate 24.53 ma, like this:

    5V/(10k + 12.4k) = 223.2 ua
    2.5V/113k = 22.1 ua
    Sum = 245.3 ua
    The chip multiplies by 100 to get 24.53 ma.

    The chip does not care what size R9 is, as long as it is not too much resistance for the voltage to drive 20 ma through it. When you want to adjust your output, you look to the input current. If your chip is off by less than 1 ma, build a circuit to adjust your input current. You are not going to convince the inside of the chip to adjust.

    There is still the question: Is the chip off by less than 1 ma or is my meter off by less than 1 ma? Well..did you use 1% resistors? Is your meter good quality? These are mere polish. I'm sure you can do this from here.
     
  16. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    well... im getting the supply voltage from de VReg pin ... Something that i noticed is that when i measure this pin referenced to IRet the voltage is 4.4V or 4V sometimes. Besides of this, i measure the current 5V/(10k + 12.4k) = 100.4 ua aprox ... And 2.5V/113k = 22.1 ua (Perfect), everything this according to the schematic attached before. So my problem is at the input as again you told me... What can i do in this case? The resistor wich im using now are the Through hall ones beacuse im mounting it in a protoboard....
     
  17. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    Vreg to Iret is not an important voltage measurement. It changes according to the current in the output loop. Voltage does not matter to this chip. Current matters.

    5V/22.4k is not 100.4 ua. It is 223.2 ua.
    This tells me your ability to measure is off a bit.

    One way to go about this is to accept the idea that the chip is not perfect and include adjustments to compensate. The 1% resistor values that were designed in are very good, but not perfect. The actual values that I calculate are 112.5k and 22,500 ohms. I put them in my first drawing. Then I made a drawing that has adjustments and added that.

    Still these adjustments can not compensate for your ability to measure 5V/22.4k as 100.4 ua.
    I think your measuring is much more problem than the chip or the 1% resistors.
     
    Last edited: Feb 16, 2013
  18. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    Thank you #12.

    I know how to measure, the thing is that i dont know why the REAL current is aprox 100uA, i know that according to Ohm's Law this is imposible but im getting this value, maybe is the tester wich im using or really i dont know what it could be, so i ask you... Ill probe just right now what you told me!

    Thanks again!
    Regards, Ricardo
     
    Last edited: Feb 18, 2013
  19. RIKR09

    Thread Starter New Member

    Feb 13, 2013
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    Effectuvely #12, Today i got a new Tester and measure the current and the values got it are corresponding to calculated ones. I made some testing and the only thinq that i should do now is to adjust the currents values wich corresponding to the Zero and Span one's of my Sensor, with the Potentiometer you told me.

    Now i can say Thank You!!!All my doubts and misconceptions about this topic are resolve!!

    Regards, Ricardo
     
    Last edited: Feb 20, 2013
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