How current ADC's work

Discussion in 'General Electronics Chat' started by sharanbr, Jan 12, 2018 at 3:50 AM.

  1. sharanbr

    Thread Starter Well-Known Member

    Apr 13, 2009
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    I am pretty new to Analog. I have exposure in Digital.

    I assume, regular ADC's take voltage as input and digitize that at the output. I would like to know how current ADC's work.
    I am seeing a datasheet from TI and I am seeing reference to integrator as the first stage before ADC stage.
    http://www.ti.com/lit/ds/sbas370a/sbas370a.pdf

    Kindly share your comments.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Did you see "General Description" or "Basic Integration Cycle"?

    Did you read them?

    If not, please do and then ask for clarification on what you don't understand.
     
  3. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    That A/D is specialized. You might take a look here: http://www.chem.purdue.edu/courses/chm621/lecture/electronics/8.7 Gated Integration.pdf for general info on gated integration and what it's used for. A pico-couloumb is a very small quantity of charge to measure.

    With one set-up I did, i had to measure very small currents near the resolution of the instrument (an Electrometer), so i would typically put the instrument in charge mode, then zero it and then take a reading 30 sec later and divide by the time. 1-2 pA became much easier to measure.
    1 pA = 1E-12 Amps.
     
  4. sharanbr

    Thread Starter Well-Known Member

    Apr 13, 2009
    79
    1
    Dear WBahn, KeepitSimpleStupid,

    let me apologize if my question was confusing. My intention is not to understand how the specific chip from TI works.
    I have seen regular ADC where voltage is measured or converted but I have generally not seen current ADC.
    That's when I searched and found the TI chip.

    Definitely, I don't want to get into intricacies of integrator unless it is necessary to even appreciate current ADC.
    I hope I have clarified my position clearly now.

    Having said that, I continued my search and what I have inferred so far is that in order to implement current ADC, the following scheme is followed. This is very simplistic view though:

    1) the incoming current (say Ia) is passed through a resistor to convert it to voltage (Say Va). Let us assume that the resistor is Ra.
    So, effectively, Va = Ia*Ra
    2) this voltage is converted into digital output (say Vd). Now, I can get current value from this by using Id = Vd/Ra

    I guess this is the method. Please provide your comments.
     
  5. smooth_jamie

    Member

    Jan 4, 2017
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    This is the traditional way, or you can use the hall effect type sensors instead which are non-invasive. From Figure 5 in the datasheet it doesn't indicate a sense resistor, (signal goes straight into integration op-amp?) but I'd imagine there is something before it to convert the current into a voltage signal (since theoretically no current can flow into or out of the input of an op-amp).
     
    Last edited: Jan 12, 2018 at 8:16 AM
  6. MrChips

    Moderator

    Oct 2, 2009
    15,124
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    As you have already described, most current measurements are made by measuring the voltage across a load resistor.

    There are various ways of converting a current to voltage. One such method is the integration amplifier which integrates the current over time. Other methods use charge-coupled amplifiers which converts very small currents (picoamps) into voltage.

    Why are you interested in measuring current?
    Describe what you are trying to do so that we can better advise you.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    That approach can work and is used. The problem with it as described is that the resistance, Ra, changes the current being measured unless it is much, much smaller than the impedance of the current source being measured (just as the input impedance of a voltmeter affects the measurement unless it is much, much larger than the impedance of the voltage source being measured). One way to deal with this is to buffer the input current so that it always sees a very low resistance and then the measurement resistance is placed in the path of the buffered current.

    To measure very small currents, you would need very large resistors to get enough of a voltage signal to measure. But if you put the current into a capacitor for a fixed amount of time, then the charge on the capacitor will grow for as long as you want (within limits). This is called integration because the integral of current is charge and you are converting a current into a charge on a capacitor, whose voltage then serves as an indirect measure of the current (since you have control over the amount of time that was used). That's how that TI part works. To keep the part linear you again buffer the current so that the current entering the capacitor continues to match the input current even as the capacitor voltage changes.
     
  8. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    2,227
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    This is the opposite of what you asked for, but..

    In the industrial world voltage and current out are used a lot. See http://www.analog.com/en/products/i...terface-devices/industrial-4-20ma-driver.html

    The problem with voltage out is "ground loops". With current out, you can place a resistor at the voltage input of a device and eliminate the loop.

    So, 0 to 20 mA and 4 to 20 mA are common. 4 mA allows a device to draw power as well and it becomes a 2 wire device.
     
  9. sharanbr

    Thread Starter Well-Known Member

    Apr 13, 2009
    79
    1
    smooth_jami, mrchips, wbahn, keepit...,

    thanks a lot for the comments. I think now I understand the principle of current ADC.
    Please note that I am new to analog world and my exposure is what I studied in college during my engineering course.
    I am trying to brush up the basics as some of my projects involve mixed signal designs where analog blocks are integrated.
     
  10. Titanpick

    New Member

    Sep 9, 2017
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