AD536A help - to measure the current of a solenoid through a shunt resistor.

Discussion in 'General Electronics Chat' started by adwina2, Aug 6, 2015.

  1. adwina2

    Thread Starter New Member

    Jul 14, 2015
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    My end goal is to measure the current of a solenoid through a shunt resistor, gaining the output and using an AD536A to give RMS values. The control signal to the solenoid will be a variable PWM with dither. I am using the AD536 with a single ended 12V supply and I plan on gaining the input to the AD536 to 5V full scale. When I put in a sine wave straight to the 536, I get the correct rms value. I put an input cap in series with the function generator input and it does not matter if I put an offset or not on the sine wave, I still get the rms value corresponding to a sine wave without the offset(ampl * .707). This alone seems wrong because the DC offset should increase the rms value. When I put in a square pulse, with offset to make it positive, I get a lower value than expected, which is something I can correct with gain. But, when I increase the duty cycle, the rms output goes down, and rms voltage out goes up when I decrease the duty cycle. This is opposite of what it should do. Does anybody have any experience with this chip and using it single ended?? I used an AD536A with a +/-12V supply and it seems to work well. Any suggestions are much appreciated.


    Thanks in advance,

    adwina2
     
  2. ronv

    AAC Fanatic!

    Nov 12, 2008
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    You need to bias it for single supply operation and AC couple the input like fig. 17.
    http://www.mouser.com/ds/2/609/AD536A-245585.pdf
     
  3. adwina2

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    Jul 14, 2015
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    Yeah, this is the way I already set it up.....
     
  4. ronv

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    If the frequency of the pulse is very low input cap must be very large. What is the frequency of the pulses?
     
  5. ronv

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    A
    A hh, I see your problem. Once you AC couple it you loose the DC information. :(
     
  6. crutschow

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    Mar 14, 2008
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    If you want the DC plus AC RMS value then you need to DC couple the input, as show in Figures 13, 14, or 15 (depending upon the package type).
     
  7. ronv

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    But then he needs a +/- supply voltage I think.
     
  8. crutschow

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    Well, sometimes you can't have everything. :(
     
  9. PeterCoxSmith

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    Feb 23, 2015
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    well you could use a single supply, create a mid-rail VCOM voltage applied to COM pin, which means the output will be wrt VCOM. It may also mean you have to level shift the input so it is referenced the the same VCOM voltage.
     
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  10. ronv

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    Yes, sad but true. :(
     
  11. crutschow

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    That should work.
    You could use a rail-rail type instrumentation amp to shift the voltage from the ground reference to the mid-rail reference voltage (instrumentation amp REF output connects to the mid-rail voltage).
     
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  12. adwina2

    Thread Starter New Member

    Jul 14, 2015
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    I believe that is what fig. 17 in the data sheet is doing, correct? Pin 10 is the mid-rail, and I am using the same ground throughout. Do I need to use the pin 10 mid-rail as my output reference ground? When I supply a 5Vpp pulse at 50% duty cycle I get the following across RL:
    When I use pin 10 as my reference I get approximately -1.5V
    When I use the shared common of the supply I get around 2.5V.

    Also, I am using around 1K Hz signal with a .1uF cap for Cav. I wanted to have a cap to fit a large range, and this works well for the values I am using.

    When I use the input cap, the RMS values go down whether it is a square or sinewave input. But, the all the rms values are low, regardless of the cap and signal combination.

    I suppose leaving the signal low from the shunt(maybe just amplify it to 2.5 Vmax or so), and treating the +12V supply as a +/- 6 volt supply by level shifting the input would work. Is that what you mean Peter?
     
  13. adwina2

    Thread Starter New Member

    Jul 14, 2015
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    Thanks for your replies!
    This made more sense when I went back to the circuit. I referenced the input to pin 10 COM, leaving the mid-point divider, and did not use the input cap and the value is still a little low(3.2Vrms instead of 3.53Vrms), but I can make this work by adjusting the scales. BUT....it works! As the duty cycle increases the Vrms increases! Thanks for the help!
     
    Last edited: Aug 7, 2015
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  14. PeterCoxSmith

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    Feb 23, 2015
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    I suppose leaving the signal low from the shunt(maybe just amplify it to 2.5 Vmax or so), and treating the +12V supply as a +/- 6 volt supply by level shifting the input would work. Is that what you mean Peter?

    Yep!
     
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