Help with CMRR calculation for Instrumentation amplifier.

Discussion in 'The Projects Forum' started by RKTim, Mar 24, 2015.

  1. RKTim

    Thread Starter New Member

    Mar 24, 2015
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    Hello out there,

    I'm working on an electronics project to design active sEMG electrodes and I have designed my first pre-amp stage (so electrodes to instrumentation amplifier). I am using the double difference method in my INA setup and inputting a 750uV differential signal to the IA's. Each INA has a gain of 2 giving overall gain of 4. I am using the AD8237 INA's that are supposed to give a high CMRR, I am also driving the reference pin with a mid-supply voltage as shown in the schematic.
    My question is in two parts;
    1, I thought I understood gain, but this project has tested me, in the data sheet for the AD8237, they specify a gain of 2 is acheived with two 49.9kO resistors. I am confused by that figure, "2". Is that a decibel measurement i.e dB or is that a linear ratio of input to output voltage?
    2, I have attached figures of my output via the Oscilloscope and have a measure for Vout, how do I then measure the CMRR of my configuration?
    Amp-stage_1.png Output_input_results.png
     
  2. #12

    Expert

    Nov 30, 2010
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    The gain number is linear, volts out for volts in.
    Testing for CMRR...I have to go back a very long way to remember that one.
    The principle is that if you apply the same signal to both inputs, a perfect amplifier would not create an output. You can't just connect the (2) input pins together because DC offset would lock the output to one of the supply voltages so you will have to use some resistance on each input pin. I swear this is in an application note, but it's in the 1960's in my memory.
     
  3. #12

    Expert

    Nov 30, 2010
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    Last edited: Mar 24, 2015
  4. RKTim

    Thread Starter New Member

    Mar 24, 2015
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    Hi #12, thanks for the timely reply. I will have a look at the links. Thanks again.

    Also, the specs of my project suggest investigating a method to eliminate the need for a reference electrode (I imagine that is what you mean by grounding the patient). Do you mean to say that with the configuration I have, I would still need a reference electrode? I plan to use a tri-polar arrangement with 3 electrodes picking up the EMG signal.
     
  5. OBW0549

    Well-Known Member

    Mar 2, 2015
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    The AD8237 is an instrumentation amplifier, not an opamp; so I don't believe he has to do anything more than tie the two inputs together and apply a varying common-mode voltage while measuring the resulting change (if any) in output.

    What you said would certainly be true for an opamp, though.
     
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  6. RKTim

    Thread Starter New Member

    Mar 24, 2015
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    @OBW0549 Thanks a lot for the reply. So my test strategy is to apply the differential input varying the magnitude of the output an then applying a common mode signal to both inputs am measuring the output and the Vo/Vi for both instances which should essentially give me my Differential gain and common mode gain then calculate CMRR from that? Is that a good enough strategy?
    Also, could you perhaps tell me about grounding the patient? Do I need a reference electrode with my design?
     
  7. OBW0549

    Well-Known Member

    Mar 2, 2015
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    That sounds right to me; as far as I know, you can check the common mode gain by shorting the two inputs together and connecting them to a variable voltage source whose other end is connected to circuit common. Your common mode gain is then delta Vout divided by delta Vin. (Just be sure to keep Vin within the allowable common mode input range for the amplifier, of course.)

    I'll pass on that; I've never worked on biomedical instrumentation.
     
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  8. RKTim

    Thread Starter New Member

    Mar 24, 2015
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    No Problem. I will wait for #12 to let me know that. Otherwise thanks for your instruction, if I run into any problems, I will let you know.
     
  9. #12

    Expert

    Nov 30, 2010
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    I forgot how to test common mode rejection. That's why I googled it. Plenty of examples on the Internet.

    As for grounding the patient, that is sometimes done with a big carbon conductive sheet applied to the thigh when doing cautery. I suppose a smaller connection would work as a ground for measuring tiny voltages. If you don't ground the patient, you will eventually deal with static charge that can be in the hundreds of volts. Your probes will be very high impedance and could take minutes to discharge the patient. Designing for a grounded patient is one way to eliminate designing for a common mode input in the hundreds of volts range. I'm trying to think of how to let the surface charge of a person drive the measuring device to the proper DC level. That won't work with a machine that plugs into the wall. The safety ground required for the power supply just won't allow it. A battery operated machine would instantly arrive at the static voltage on the patient, but that charge will be looking for a way to escape as a slow leak or a static spark discharge.

    Other people are surely more familiar with medical equipment than I am, but I have worked a few medical machines and they have always been very straight-forward in their principles of operation.
     
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