ECG Circuit Analysis: Calculating Contribution of Eint to Vout

[Ariastity]

What is the contribution of E_int
to the output voltage V_out?

Homework Equations

The Attempt at a Solution

To calculate the contribution of E_int, I removed the other two voltage sources.

Then, I divided them into three potential dividers, giving me:

V_body= (Z_C2/Z_C2+Z_C1) x E_int;
V₁=(Z_in1/Z_B1+Z_in1) x V_body;
and V₂=(Z_in2/Z_B2+Z_in2 )x V_body

Lastly, Vout= Vbody- V1-V2.

Is this method valid?

 

[gneill]

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.

 

[Ariastity]

Thanks for your reply!

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

I tried using nodal analysis only to get V_body without applying voltage divider, and it gives me:
i_couple1=(E_int-V_body)/Z_couple1
and i_couple2=V_body/Z_couple2,
with i_couple1=i_couple2+i_body.

I don't really know how to go on from there and quantify V_body in terms of E_int, as further nodes only give me equations that depend on the value of V_body.

 

[gneill]

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:

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.,