AD620 circuit problem //Output

Discussion in 'Analog & Mixed-Signal Design' started by sohamkul, Jul 2, 2016.

  1. sohamkul

    Thread Starter Member

    Jun 24, 2016
    43
    2
    Hello everyone,

    I am building a data acquisition system,i am using AD620 (instrumentation amplifier), have read data sheet thoroughly and i have used the following circuit which i obtained from internet (image attached ). I have also use two zero pots to offset zero on the inverting terminal of AD620.(images attached)

    Problems:
    1) even when the differential signal is not applied the output of AD620 jumps to 3.5 to 3.87volts
    What is that i am doing wrong ??. i have tried various methods even then i am unable to amplify the differential signal
    should i add resistors on both input terminals of AD620? please help .

    p.s. I have grounded the reference pin
    Thank you in advance.


    (Internet image)
    . Capture2.PNG IMG-20160703-WA0006.jpg Capture.PNG

    Thank you in advance.
     
  2. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,145
    204
    I can make a few comments only:
    1) Your zero source is yucky and is high Z, fed from power rails
    2) Type of power supply? (linear/switching?
    3) Bypass caps (power rails, zero source)?
    4) Does your signal have a PATH to ground.
     
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  3. sohamkul

    Thread Starter Member

    Jun 24, 2016
    43
    2
    1) i am using an SMPS
    2) i have also used decoupling capacitors on each IC (0.1micro f)
    3)sorry i do not have much knowledge about analog electronics can you please help me out to refine the circuit .

    Thank you
     
  4. dannyf

    Well-Known Member

    Sep 13, 2015
    1,809
    361
    Could be nothing, all depending on what you meant by "not applied", and where the offsets are.

    And the way you adjust the offsets is funny. Re-read the datasheet.
     
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  5. crutschow

    Expert

    Mar 14, 2008
    13,009
    3,233
    You cannot leave an input open (floating) to the amp. It needs a path for its input bias current, otherwise the input floats to some arbitrary voltage.
    Connect a resistor across the two inputs if it's not connected to the signal source.
     
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  6. sohamkul

    Thread Starter Member

    Jun 24, 2016
    43
    2
    That was exactly the problem !! I connected both input terminals to ground and the circuit works like a charm !! Thankyou !
     
  7. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,145
    204
    Note: In my post #2:

    4) Does your [input] signal have a PATH to ground?
     
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  8. sohamkul

    Thread Starter Member

    Jun 24, 2016
    43
    2
    Exactly i didn't get it the first time you said it(dumb of me) but that was the problem ! now i have connected both inputs to ground and reference to ground as well
    circuit works well !!
    1)but can you please tell me as of to how should i connect inputs should i connect them directly to input pins or should i use resistors ?
    2)Can you please suggest a zero pot assembly ?

    Thankyou Man !!:rolleyes::rolleyes:
     
  9. crutschow

    Expert

    Mar 14, 2008
    13,009
    3,233
    What is the value of the signal source voltage?
    Is it a single-ended or differential source?
    How much zero adjustment range do you need?
     
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  10. sohamkul

    Thread Starter Member

    Jun 24, 2016
    43
    2
    signal is differential and output is basically 0-2 mV and i want the output to be 0-2Volts.
    what i want to do is using offset pot i want to set base of output at 1V so that output varies between
    1v-3v i am going to connect this assembly to ADS1115 which has 3.3 volt input range only .


    thank you !
     
  11. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,145
    204
    Then use REF. Supply the REF terminal with a LOW Z source from an OP-amp.

    Another comment I have to make is:
    a) Is your A/D going to ratiometric?
    1 and 3 is different than 1 and Vss of 3.3 V

    When you make the reference of your A/D equal to the supply voltage, your pretty much assuming ratiometric operation. When you use a precision reference to the A/D, then you get stuff like 0-5V.

    So as the power supply varies with load and noise for that matter 50% is always 50%*Vs. Sometimes Vs is say 5 and other times it's 5.1 and 50% is either 2.5 or 2.55 This is the way most automotive sensors work, but it always reads as 50% no matter what the supply voltage is.

    So, if you set your A/D reference to the supply voltage, you want REF of the opamp to be a % of the supply.
    e.g. A resistive divider amplified x1.

    Then based on OP-amp selection, you can decide what to do with the smaller offsets like the IA and the REF amplifier.
     
  12. crutschow

    Expert

    Mar 14, 2008
    13,009
    3,233
    As KISS noted, you want to use the REF input to control the output offset.
    Use a resistive divider and an op amp follower to generate +1V for the REF input.
     
  13. sohamkul

    Thread Starter Member

    Jun 24, 2016
    43
    2
    I am Extremely sorry for late reply guys (KISS and crutschow) i was busy building other components of the circuit .
    Guess what the circuit worked like a charm !!!!:):):):cool::cool:
    Thank you so much for your help ~~ :)
     
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