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ECG Circuit Analysis

  1. Dec 9, 2016 #1
    What is the contribution of Eint
    to the output voltage Vout?

    Picture1.jpg
    2. Relevant equations


    3. The attempt at a solution
    To calculate the contribution of Eint, I removed the other two voltage sources.

    Picture2.png
    Then, I divided them into three potential dividers, giving me:
    Picture3.png

    Vbody= (ZC2/ZC2+ZC1) x Eint;
    V1=(Zin1/ZB1+Zin1) x Vbody;
    and V2=(Zin2/ZB2+Zin2 )x Vbody

    Lastly, Vout= Vbody- V1-V2.

    Is this method valid?
     
  2. jcsd
  3. Dec 9, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    You can isolate the contribution of Eint by suppressing the other sources as you've done. That's fine.

    The Vbody voltage divider is not isolated. It's loaded by the two other voltage dividers. So you can't rely on the voltage divider equation to produce the true value for Vbody. That mucks up the calculation of V1 and V2 since they both depend on Vbody.

    Use nodal analysis for the three nodes (V1, V2, Vbody) to handle the dependencies.

    You haven't yet specified what the gain of the amplifier is, so it's not possible to say what Vout will be given V1 and V2.
     
  4. Dec 10, 2016 #3
    Thanks for your reply!

    I tried using nodal analysis only to get Vbody without applying voltage divider, and it gives me:
    icouple1=(Eint-Vbody)/Zcouple1
    and icouple2=Vbody/Zcouple2,
    with icouple1=icouple2+ibody.

    I don't really know how to go on from there and quantify Vbody in terms of Eint, as further nodes only give me equations that depend on the value of Vbody.
     
  5. Dec 10, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    If you have three nodes you need to write three node equations. They will be be interdependent (simultaneous equations) that need to be solved together.

    In this case, since it is assumed that the amplifier input impedance is being represented by the external impedances Zin, you could ignore V1 and V2 for now and treat the ZB and Zin pairs as branches of the Vbody node. You'd have to go back and solve for V1 and V2 later. But the important thing is that you can't ignore the loading effects of these branches on the Vbody node when you solve for Vbody. The single-node version of the circuit would look like this:

    upload_2016-12-10_10-6-30.png

    Note how the two ZB branches parallel the ZC2 impedance. You can write a single node equation to solve for Vbody in this case, taking into account all the branches.,
     
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