Testing instrumentation amplifiers

Discussion in 'Homework Help' started by nothing909, Dec 4, 2017.

  1. nothing909

    Thread Starter Member

    Nov 5, 2016
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    I just have some very simple questions to ask. I'm testing a basic 3 OP instrumentation amplifier

    INA.png


    I have to short V2 to ground and test V1 from -2V to 2V, in 0.25 increments. I then have to short V1 to ground and test V2 from -2 to 2V in 0.25 increments. the results i have are for example:

    results.PNG

    i then have this equation to find the gain: ∂vout/ ∂vin1 when i'm doing this equation, i'm not undestanding the ∂. do i add up all the results and then divide it by all the vin1's added up like: 12.2+10.3+9.2.../2+1.75+1.25... or will i just take one result like 10.3/1.75. i know this is stupid question, i just wanna know if i'm doing it right. either way i do it, i get like ~6. when i do both the tests, each gain is ~6. do both the gain of each of the buffer amplifier add up 6+6=12 and 12 is the overall gain for the INA?
     
  2. WBahn

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    Mar 31, 2012
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    The ∂ means that you are not dividing vout by vin, but rather the change in vout caused by the change in vin.

    Mathematically, ∂vout/∂vin1 is known as the "partial derivative of vout with respect to vin1", which is how much does vout change as a result of a tiny change in vin1, all else being held constant.

    Seems odd they would give you that equation if you've never seen partial derivatives before.
     
  3. nothing909

    Thread Starter Member

    Nov 5, 2016
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    can you mathematically show me how i'm working the gain for V1 or just explain it more with numbers, i'm a little confused. i just to make sure i'm doing it correctly.
     
  4. WBahn

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    It's nothing more than the CHANGE in the output voltage divided by the CHANGE in the input voltage.

    If you get 1.5 V out when you have 0.25 V in and you get 2.3 V out when you have 0.75 V in, then the change in the output was (2.3 V - 1.5 V) = 0.8 V while the change in the input is (0.75 V - 0.25 V) = 0.5 V. So the gain in this region is 0.8 V / 0.5 V or 1.6.

    In looking at your table, why don't you have any measurements between -1 V and +1 V (other than 0 V)?
     
  5. nothing909

    Thread Starter Member

    Nov 5, 2016
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    ok, thanks for clearing that up. just to clarify the answer for V1 would be Vin = -2-2 = -4 Vout = -12.2-12.1 = 24.2 so the gain ~ 6?

    i have one more question. when you look at both the tables, the results are basically just the same but the signs are opposite, yes. so when you calculate the gain for both of them and they are both ~6, the overall gain of the circuit is 12?

    oh, and the table is incomplete, i just was giving an example of some of the results, i was too lazy to write in the rest xD
     
  6. WBahn

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    Mar 31, 2012
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    You don't just take the extreme values (otherwise there was no point in taking all that data).

    You want to plot the gain at various inputs levels. If the amplifier is nice and linear, then you will get the same value for the gain at all points. But if it isn't, then this is something you want to know about.

    If the direction moves in the opposite direction for one input relative to the other, then they don't have the same gain. One is positive and one is negative. There is a HUGE difference between having a gain of +6 and a gain of -6.

    To get the differential gain of the amplifier, you take the output voltage and divide it by the difference in the input voltage. Does that give you a gain of 12?

    As before, you can calculate the differential gain at a number of operating points to see how consistent the amplifier is.
     
  7. nothing909

    Thread Starter Member

    Nov 5, 2016
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    okay, so i'll just take the results for the 2 and 1.75 inputs, the difference between the outputs and difference between the inputs for they points will give me the gain.

    i'm saying that when you set both the buffer amplifiers, at say 2V, the gain for the V1 is -6 and the gain for V2 is 6 so the overall gain is the difference between them which is 12.

    i'm sure everything i just said there is correct, yes?
     
  8. crutschow

    Expert

    Mar 14, 2008
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    No.
    You need to think more about what "gain" means.
     
  9. nothing909

    Thread Starter Member

    Nov 5, 2016
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    can you expand a bit, what am i not doing correctly?
     
  10. WBahn

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    Mar 31, 2012
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    Assuming that we accept that the outputs will superimpose (we hope they they will, but then we hope the gain will be independent of input voltage, too), if you have V2 = 2 V and V1 = -2 V and if the output is 24 V (assuming it could be if it didn't saturate), then what would the calculated gain be? What is the output voltage? What is the differential input voltage? Is the ratio of the two yield a value of 12?
     
  11. nothing909

    Thread Starter Member

    Nov 5, 2016
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    Well, if V2 was 2V and V1 was -2V, the difference in input is 4V.

    gain = output/differential input voltage = 24/4 = 6?

    i'm sorry to keep going on asking the same question but i'm still confused about this gain. what is 6 the gain of? is it the gain of the overall instrumentation amplifier, is it the gain of one of the buffer amplifiers, or both?.... or am i just thinking about everything the wrong way?
     
  12. WBahn

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    Mar 31, 2012
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    Think of that that entire circuit as being in a black box with four terminals. It has two input terminals (where V1 and V2 are applied) and two output terminals (Vout and GND). Another name for this kind of circuit is a diff-amp (differential amplifier). A multimeter in voltmeter mode is an example of this, except the output is a display of the voltage between Vout and GND -- the reading only depends on the difference between the voltages at the two terminals, so V1 = 2 V and V2 = -2 V should produce the exact same reading as V1 = 5 V and V2 = 1 V.

    Mathematically, what you have is

    Vout = GAIN * (V2 - V1)

    This is the ideal case. But in the real world the output may depend on the average of the two voltages as well as the difference. So we actually have two gains -- the one above is the "differential gain" and the one that depends on the average is the "common mode gain".

    This might help:

    https://forum.allaboutcircuits.com/xfa-blog-category/circuit-analysis.37/view-entries
     
  13. nothing909

    Thread Starter Member

    Nov 5, 2016
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    so i think i've figured out a little more about gain and i understand the difference between common mode and differential.

    above when i calculated 12.2+12.2/2+2 = ~6 <--- this value is the DIFFERENTIAL gain of the circuit? i'm sure thats correct
     
  14. WBahn

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    Mar 31, 2012
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    Note that 12.2+12.2/2+2 is NOT ~6, it is 20.3

    You need to start paying attention to order of operations. What you meant to say was

    (12.2+12.2)/(2+2) = ~6

    And, yes, this is the differential gain of the amplifier. I think you are getting that.

    Paying attention to the notation is vital. If you don't, then you leave open how to interpret and evaluate your work, either by others or by yourself in a later step. I can't begin to count the number of times I've seen people write something like

    6+9/3 instead of (6+9)/3 and then, usually after a few steps on some other part (or because these were symbols and they plugged numbers in later) they came up with 9 instead of 5.

    More to the point, you are almost certainly going to be typing equations into programming languages, simulators, spreadsheets, and other environments were your expression need to be in text form. If you persist in being sloppy here, I guarantee you will be sloppy there and the very fact that you are used to interpreting sloppy notation based on what was meant and not what was written will bite you because the program you are using cares only about what you actually wrote.
     
  15. nothing909

    Thread Starter Member

    Nov 5, 2016
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    1
    when testing an instrumentation amplifier practically vs simulating, i get a 6 gain, but when testing practically, the gain is like 5.53. why is the gain not 6, what factors cause it to drop slightly?
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    All of your components, particularly your resistors, have tolerances.

    Look at the equation that gives you your gain as a function of the component values (for the case when each resistor might be different). Now consider how much the gain can vary if some components are at the upper end of their tolerance limit and others are at the lower end.
     
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