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Differential amplifier

Discussion in 'General Electronics Discussion' started by johnchew, Oct 23, 2015.

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  1. johnchew

    johnchew

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    Oct 23, 2015
    I am currently doing a mini project for one of the module in school, where i encountered problem with the accuracy of the differential amplifier output. The signal of interest is from a amplified and filtered IR sensor, the sensor outputs a nominal voltage of about 3V and varies in the mV range when object moves toward or away from it. Therefore I have tried to used a differential amplifier to remove the dc offset, however, by feeding the sensing signal into the positive input and 3Vdc into the negative input, I still got a output voltage of about 1V with a unity gain configuration (just to test out the circuit).

    The differential amplifier that i used is connected as the following. And the output voltage should be govern by
    Vo = R3/R1 (V1-V2) but that is not what I'm getting. I'm hopping to get a varying range of few voltmeter that I could use to control the duty cycle of a buzzer.

    also, Is there a way to control the frequency of my buzzer using the output voltage that I will be getting?

    Thank you.
    upload_2015-10-23_22-0-55.png
     
  2. Ratch

    Ratch

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    Mar 10, 2013
    Doesn't R2 and R4 have anything to say about what the voltage will be? I think you should reconsider what the voltage formula is.

    Ratch
     
  3. johnchew

    johnchew

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    Oct 23, 2015
    I understand that R2 and R4 works as a voltage divider. Currently I'm connecting 4 similar resistor as my R1, R2, R3, R4. By calculation, it shld give me Vo = - V1+V2 or am I missing something over here?
     
  4. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    What are the resistor values you are using?
    Adam
     
  5. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Are you using a split supply and what opamp are you using?
    Adam
     
  6. johnchew

    johnchew

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    Oct 23, 2015
    I'm using 5.6k resistors also prior to the resistor I already have my signals pass through a voltage follower thus it shld also remove any impedance matching issues
     
  7. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    And the second question please?
     
  8. johnchew

    johnchew

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    Oct 23, 2015
    I'm using LM385 cos that's what I been given. And for this project I can only use component given by my school
     

    Attached Files:

  9. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    You mean LM358?
     
  10. Ratch

    Ratch

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    Mar 10, 2013
    I agree that if all the resistors are equal, then Vo = - V1+V2.
    johnchew.JPG

    Ratch
     
  11. johnchew

    johnchew

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    Oct 23, 2015
    Yup it is lm358
     
  12. BobK

    BobK

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    Jan 5, 2010
    You have not told us what wrong output you are getting with what inputs. And you have not answered the question about whether you are using a split supply. How do you expect us to help?

    Bob
     
  13. johnchew

    johnchew

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    Oct 23, 2015
    Sorry I'm using a single 5v supply to power my op amp
     
  14. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    And please answer Bobs first question.
     
  15. johnchew

    johnchew

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    Oct 23, 2015
    Sorry all, here are the details regarding the circuit inputs/outputs.

    -The op-amps I am using are LM358. Currently I am only using a single supply. +9V to pin8 and Ground to pin4 as I do not need negative outputs.
    -The differential amp I have built with the LM358 is the same as the picture I have posted in the first post.
    -I have use the following configuration for the differential amp, R1=R2=R3=R4, all resistor are 5.6k ohms.
    -V1 is the output from a voltage buffer (follower) with a voltage of 3.200V and V1 is fixed.
    -V2 is the output from another voltage buffer with a voltage range of 3.2xxV to 5.5xxV.
    -V2 is an amplified signal from an IR detector.
    -For testing, I would stand infront of the IR detector and move either closer or further away from the IR detector. V2 changes accordingly.
    -The signal V2 is non-linear with distance. Meaning to say, within 15cm, V2 changes from 3.5xxV to 5.5xxV. However beyond 50cm, only the last two digits, 3.2xxV changes.
    -As I am doing a distance measurement device for the blind, I am interested in the distance of 0.5m to 2m away from the IR detector itself.
    -The readings for V2 I get from the IR detector is not always exactly the same, but it is quite similar during each test.
    -I would stand at 2m infront of the IR detector, and V2 would be around 3.211V, as I move closer to the IR, say 1.5m, V2 would be 3.230V, then I move ever closer to the IR, say 1m, V2 would be 3.270V.
    -And finally within the 15cm range, V2 increases exponential to 5.5xxV as my hand moves closer and closer to cover the IR.
    -Hence, I am interested in only the last two digit of V2.
    -I set up a differential amp to get V2-V1, in which V1 is 3.200V.
    -Example, at 1.5m, V2-V1, i would get 3.230-3.200=0.030V.
    -I would then amplify this 0.030V by 100times to get a useful output to drive my buzzer.
    -However the differential amplifier is giving me a output of, V2-V1, 3.230-3.200 = 1.xxxV
     
  16. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Ok so I expect the opamp you are using cant resolve the low voltage you need. You may need to look at a rail to rail opamp. The LM358 is quite old now.
    Adam
     
    davenn likes this.
  17. BobK

    BobK

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    Jan 5, 2010
    The LM358 cannot reach 0V when driven by a single supply. It has a miniumum output of something like 40mV. Since the voltages you are trying to get are in that range it is not suitable.

    However, you still should not have gotten a voltage of over 1V for the case you mention. Did you actually measure both inputs and the outputs with stable voltages on them? This seems wrong to me.

    Also, I question how accurate this IR is for distance sensing in the range you specify. It's output voltage would have to be accurate to about 0.1% in order to get any resolution between 3.21 and 3.27.

    Actually, if all you are trying to do is discriminate a single distance, a comparator is a better circuit to use.

    Bob
     
  18. Laplace

    Laplace

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    Apr 4, 2010
    In case the school did not provide a comparator in the bag of components to use, the LM358 makes a fair comparator in a circuit such as this. You might consider generating the reference voltage directly from the sensor itself by using a very long time constant RC filter, if sensor drift is a problem.
     
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