Op Amp Experiment

Discussion in 'General Electronics Chat' started by MikeD_72, Dec 6, 2008.

  1. MikeD_72

    Thread Starter Active Member

    Nov 11, 2008
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    I've got what is probably a simple question from an experiment I did. I hooked up a battery to a potentiometer, fed the signal through a voltage follower, and then through an inverting op amp with a gain of about 5.5. Schematic below (R4 & R5 as simulating the 1k pot I used):

    [​IMG]


    I tested a range of input voltages from -1.5 to 1.5V. I had to flip the battery's terminals around to achieve the negative input voltages. I plotted a graph of my measurements (Vout vs Vin):

    [​IMG]


    Everything was pretty much as I expected except for the voltages at which the actual gain curve levels off. I knew they would be appreciably lower than the rail voltages, but what surprised me was that they weren't equal in magnitude. This prompted me to simulate the circuit in Spice. Why is there a difference in the levelling off voltages between the Spice model and the actual measurements? Or maybe the better question is, why are the magnitudes of the levelling-off voltages different?

    Thanks!
     
  2. nanovate

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    May 7, 2007
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  3. SgtWookie

    Expert

    Jul 17, 2007
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    I'll bet that you manually plotted your + side first, and then the - side.

    When you started out, your battery was relatively fresh. As your testing and recording session continued, your battery became more depleted, thus your results were offset by the lesser voltage from the battery.

    Try your manual measurements using a regulated bench power supply instead of a battery. You can make a bench supply very cheaply using a salvaged ATX form factor computer power supply and a few parts. Google "ATX bench supply" for lots of ideas.
     
  4. MikeD_72

    Thread Starter Active Member

    Nov 11, 2008
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    Nanovate, does this quote summarize what I'm observing?
    "...the output swing of a bipolar op amp depends on its load current." (Page 5)
    I couldn't make a whole lot of sense out of the article because I know nothing about transistors.

    Sgt Wookie,
    You're correct that I tested the positive inputs first, but I don't see how this could affect my data for two reasons:

    1) I am powering the op amps with regulated +/-6Vdc from a wall source
    2) I was able to reach the levelling-off voltage for both positive and negative inputs
     
  5. eblc1388

    Senior Member

    Nov 28, 2008
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    The SPICE simulation looks acceptable to me.

    Perhaps the question you should be asking yourself is why the actual measured gain is not linear in the active range of the LF347. You know 100% it should be linear, as indicated by calculation or via SPICE simulation.
     
  6. MikeD_72

    Thread Starter Active Member

    Nov 11, 2008
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    The results do indicate a linear behaviour. Some error is generated because I am using an analog multimeter with weak resolution.
     
  7. eblc1388

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    How can you be so sure that the voltage difference measured is not caused by the error of your analog meter?
     
  8. MikeD_72

    Thread Starter Active Member

    Nov 11, 2008
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    I have 4 data points for each level region of the graph, which gives me strong confidence in the values I am measuring. Unless you can explain to me how my multimeter might start showing a discrepancy of 0.8V between readings that should be theoretically identical, I don't think that the original question I posed can be chalked up to experimental error.
     
  9. eblc1388

    Senior Member

    Nov 28, 2008
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    1. Is your meter accurate? How do you know?

    2. Of course all these questions is just to distract you from the real question of: Why would you have decided that SPICE simulation would be a 100% representation of what will happen in the real world? How about the Opamp model you've used? Does it taken good care of the saturated region? How would you know?
     
    Last edited: Dec 6, 2008
  10. SgtWookie

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    Jul 17, 2007
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    Decent analog meters have an impedance of 10,000 Ohms per volt. Cheap analog meters are much worse.

    Even cheap DMM's have practically infinite input impedance. The amount of error they introduce is so small as to be negligible in all but the most delicately balanced or HF circuits.

    You can pick up an adequate DMM from Harbor Freight Tools for under $4 when they're on sale. They're remarkably accurate, particularly for their low cost. I tested one against a Fluke 5100 Calibrator that itself had just been calibrated, and was genuinely suprised to discover that the readings I obtained were within 1% of the Calibrator's outputs.
     
  11. MikeD_72

    Thread Starter Active Member

    Nov 11, 2008
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    I see what you are saying. I wasn't aware of this information to begin with, but I inspected my multimeter and found it was 20kOhm/Volt. Given that I was measuring the output across at 10kOhm resistor and that the multimeter resistance was 2.5*20k=50kOhm, I can see that the equivalent resistance is going to drop significantly. But how does this affect my reading? The op amp has a low impedance output, thus the voltage drop from the output to ground virtually has to match the theoretical value. :confused:

    Edit: I should add that I will try to get this test done at work with a bench-top PSU and digital multimeter to improve accuracy.
     
  12. eblc1388

    Senior Member

    Nov 28, 2008
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    The output voltage across opamp's output 10KΩ reach 5V or more so you should be using a meter range of 5V or more. This gives the meter internal resistance to be 5*20KΩ = 100KΩ or more. But as you've said, it does not matter because the Opamp will take care of the additional load.

    If you have connected the analog meter at the output permanently, then your source of error is the meter pointer does not move for a slight change in input voltage. For example, if Opamp output voltage changed from 4.0V to 4.2V, the meter pointer might possibly not moving to indicate the new voltage. This is due to the static friction of meter movement.

    That's why engineers used to tap any analog meter slightly before taking any reading.
     
  13. nanovate

    Distinguished Member

    May 7, 2007
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    Yes, also the output stage of the LF347 is a NPN/PNP type. NPN typically have lower saturation voltages -- so output gets closer to the rail.

    One thing to remember about SPICE, it is only as good or complete as the model. At the extremes it may do funny things like if you overdrive the inputs the SPICE may not agree with an actual device (which has physical tolerances and variances).
     
  14. Audioguru

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    Dec 20, 2007
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    The first opamp is driving only 180 ohms. Its allowed minimum load is 1k ohms or 2k ohms.
     
  15. MikeD_72

    Thread Starter Active Member

    Nov 11, 2008
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    eblc1388, you're right - I made a small error in that calculation. The resistance of my meter was actually 100k. I was only using one multimeter for the experiment, so I didn't have any problems such as static friction in the needle. I will be interested in seeing the difference when I run the experiment with digital multimeters.


    Nanovate, I like what you are suggesting. As I understand it, different saturation voltages would lead to different voltage drops across the transistors - thus different behaviour at the positive and negative rails.

    Through my experiences in trying to model mechanical systems, I know that a complete model is never attainable, and that at it's best it is only a good approximation of what to expect.


    Audioguru, I would like to know where you found out that the LF347's minimum load is >1kOhm, so that I can find something like this for myself in the future! I don't see any mention of such a condition on my datasheet, but a few measurements (Gain is one of them) are measured with R_L=2k. Also, correct me if I'm wrong, but doesn't the second op amp act a load on the first?


    Thanks everyone. Hopefully I will be able to post my experiment results tomorrow evening. :)
     
  16. Audioguru

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    Most opamps are spec'd with a minimum load of 2k ohms because their minimum output current is about 20mA and at such a high current (!) the voltage loss is high.
    The 180 ohm input resistor of the second opamp is the load of the first opamp.
     
  17. eblc1388

    Senior Member

    Nov 28, 2008
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    But I have already changed it into 1800Ω along with the correction on the first voltage follower stage several posts back.
     
  18. Audioguru

    New Member

    Dec 20, 2007
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    Sorry.
    I saw the red correction but not the blue one.
     
  19. MikeD_72

    Thread Starter Active Member

    Nov 11, 2008
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    I re-ran the test, this time wiring the op amp into the comparator configuration. The output voltages were more symmetrical this time: 4.91V when high, -4.63V when low. This despite the fact that the rail voltages weren't very good to begin with: 5.69V and -6.120V.
     
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