Instrument amplifier help

Discussion in 'The Projects Forum' started by persian, Mar 26, 2010.

  1. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    hi there,

    I am having a problem with my circuit.I am using the same amplifier as the other guy was using. I am using a single supply voltage about 12 v and using the voltage regulator to give me 5v on to a quarter bridge. I have been using a 10 ohm/25turn potentiometer as my strain gauge plus 340 fixed resistor. I have been checking that for each turn on the strain gauge the voltage should change in a certain amount.but my results are half off.

    calculation =((4.98*0.4*#of turns)/(2*(2*365+0.4*# of turns)))

    results after each turn starting from 0

    trial 1 trial 2 actual amount calculated
    0 0 0
    0.437 0.492 1.36006
    1.2 1.226 2.718632
    2.08 2.091 4.086433
    2.57 2.62 5.445599
    3.35 3.41 6.803279
    4.05 4.13 8.159476
    4.73 4.79 9.519476
    5.32 5.35 10.86743

    Moderator's note: the request for an interested party's email address has been removed. These are public forums, and we insist all business be conducted openly in them.
     
    Last edited by a moderator: Mar 26, 2010
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Persian,
    Even though you are using the same instrumentation amplifier, your situation is different.

    You should have started a new thread/topic about your situation; providing a link to the original thread would have been helpful.

    It is very likely that a Moderator will break off your post into it's own thread.

    We do not help people via E-mail or PM's.

    Please attach an image of your schematic. .PNG format is preferable, as it is compact and not "lossy" like .jpg format files are.
     
    Last edited: Mar 26, 2010
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
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  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    To the OP: please use this thread for your questions.
     
  5. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    I have created a new thread please check it out
    thank you
     
  6. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
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    Sorry - this one has been open for your use. Please confine your question to this one.

    We might remind you that the schematic has yet to get posted.
     
  7. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    Hi there,

    I am having a problem with my circuit.I am using the INA125 amplifier. I am using a single supply voltage about 12 v and using the voltage regulator to give me 5v on to a quarter bridge. I have been using a 10 ohm/25turn potentiometer as my strain gauge plus 340 fixed resistor. I have been checking that for each turn on the strain gauge the voltage should change in a certain amount.but my results are half off.

    calculation =((4.98*0.4*#of turns)/(2*(2*365+0.4*# of turns)))

    results after each turn starting from 0

    trial 1 trial 2 actual amount calculated
    0 0 0
    0.437 0.492 1.36006
    1.2 1.226 2.718632
    2.08 2.091 4.086433
    2.57 2.62 5.445599
    3.35 3.41 6.803279
    4.05 4.13 8.159476
    4.73 4.79 9.519476
    5.32 5.35 10.86743

    I set each potentiometer to 25 each so that r1 to r3 is 365 while the r4 is 370. I balance the circuit and then I turn the 10 ohm potentiometer set as 5.I am using 1000 gain. I am not using a load between 11 and 5 only because the voltmeter it self has resistance
     
  8. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Lots of variables in that setup. What's the linearity of that 10 turn pot, and how does the IA respond to changes in it plus a fixed resistor (a voltage divider circuit).

    If you hold one IA input at ground and the other at 1 mv, how close to 1 volt is the output?
     
  9. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    You have too much of a load on your reference output.

    See page 12, figure 4 in the datasheet for a simple emitter follower to boost reference current.
     
  10. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
  11. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
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    By calculation, yes. What is it, actually? You are trying to account for some discrepancy in real vs. calculated response. That means it's a good time to check on circuit elements.
     
  12. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    how is there too much load for the output because i am just using the voltmeter to measure it
     
  13. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    i tried my circuit without any potentiometer in my circuit and the output is correct

    the only mistake that is involve is the potentiometer but i am not sure what
     
  14. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Which potentiometer did you remove? There are several.

    If you removed a potentiometer from the bridge, then you removed the load from the reference output.

    If the bridge is complete, you will have roughly a 15mA load on the reference output. In my opinion, that is a bit too much. If you incorporate the transistor voltage follower shown in Figure 4 on page 12 of the datasheet, you will eliminate most of the load from the amplifier itself. It is a very simple and inexpensive modification to try.

    You do not necessarily have to use a TIP29 transistor; any general-purpose NPN transistor such as a 2N2222, BC547, etc - as long as it has an hFE of 30 to perhaps 300, and an Ic rating of 100mA or more.

    If that does not help with the problem, then we can explore other options.
     
  15. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    I removed all the potentiometer.

    With the potentiometer my results are 50 percent off.

    I read in this book called "nondestructive testing book vol 2"

    it says that it is possible to temperature compensate a potentiometric circuit, with a loss of 50 percent of the maximum sensitivity, if a dummy gage is used
     
  16. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Did you try this suggestion, which I've made twice before?

     
  17. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    hi there,

    i did the calculations of the circuit from the amplifier of page 13 figure 6. Only the part below the voltage regulator. I have attached my calculations if there is anything wrong please let me know because i get the gain as 30k not 60k. If this is correct then the gain facto is half off. Then my results are correct
     
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  18. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    hi there,

    never mind I found my problem on the equation going to buy the transistor
     
  19. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    hi there,

    I would like to let you know that the transistor did work it proving that there was not enough current through the amplifier. The only problem is now is that the turn ability of the strain gauge potentiometer is still 50 percent off. But im am going to make sure with different calculations by setting the whole circuit to to balance out to 5 V and set the strain gauge potentiometer to 0.Then all i will be doing is turning the potentiometer strain gauge and see the voltage change
     
  20. persian

    Thread Starter New Member

    Mar 26, 2010
    12
    0
    Hi, there

    I checked the gain and it works probably by setting the Wheatstone bride to 0.1Volts and setting the gain to be 50.4. The output voltage was 5.04.

    Do you have any suggestions to check the circuit is working properly
     
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