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Before begin asking, here the datasheet for the IC https://www.analog.com/media/en/technical-documentation/data-sheets/AD8232.pdf

At page 16 under Theory of Operation Section AD8232 architecture

What I failed to understand is how the Instrumentation Amplifier amplify the ECG signal from electrode. It has an Integrator amplifier with C1 capacitor. As far as I know, integrator will integrate the input. If the input is EKG signal then fed to integrator, doesn't it change the output? I mean we expect the output of Instrumentation Amplifier is just input amplified but still in the same form. But with the presence of integrator amplifier to my understanding it will change the output form which no longer like EKG signal we saw in the input.

Please tell me where I am wrong. I would be happy if someone could explain from the input, GM1, GM2 and integrator working principle. Thank u

Edit=== According to TimWescott's answer I try to derive the transfer function as below Derive transfer funct

The output Vo still depends on integral of the Vekg

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  • \$\begingroup\$ The op-amp is a servo amplifier driving gm2 to be the compliment of gm1. The ÷100 divider means the amplifiers output is 100x the input to gm1. \$\endgroup\$ – sstobbe Jun 7 '19 at 14:37
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GM1 and GM2 are transconductance amplifiers, per the data sheet. This means that they generate currents that are equal to \$g_m \left ( V_+ - V_- \right)\$, where \$g_m\$ is the transconductance of the amplifier, \$V_+\$ is the non-inverting input, and \$V_-\$ is the inverting input.

The output of GM2 feeds back to the op-amp. So what is integrated is the sum of:

  • The non-inverting input to the chip
  • The negative of the inverting input to the chip
  • Some constant determined by HPA
  • The negative of the output of the integrator

That last bit makes the integrator into an amplifier, with a gain fixed by the resistor network that feeds back into GM2.

The reason for all of the folderol with GM1, GM2 and the integrator is almost certainly because the chip designer found it easier to make a nicely matched pair of transconductance amplifiers than to realize the desired functionality any way else.

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  • \$\begingroup\$ Could you please elaborate why the negative of the output turns the integrator to amplifier. Could you help me by showing the transfer function of the overall instrumentation amplifier used above \$\endgroup\$ – Gidion Saputra Jun 7 '19 at 4:37
  • \$\begingroup\$ Please see the edit of the answer. I try to derive the transfer function but probably wrong at some point. \$\endgroup\$ – Gidion Saputra Jun 7 '19 at 7:31
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The problem in the second half of your question comes about because you're trying to solve a differential equation as if the left and right halves do not depend on each other.

You write: $$C_1 \frac{d\,V_o}{dt} = g_{m_1} V_{ekg}+g_{m_2} \frac{V_o - V_{ref}}{100} - g_{m_2} V_{HPA}$$

Then you integrate both sides. You cannot do that, because it is a differential equation.

So, first, you have a sign error because you forgot that the integrator is inverting; the correct equation is $$C_1 \frac{d\,V_o}{dt} = g_{m_1} V_{ekg} - g_{m_2} \frac{V_o - V_{ref}}{100} - g_{m_2} V_{HPA}$$

Second, it is a differential equation. I'll make this a bit more clear:

$$C_1 \frac{d\,V_o}{dt} = - \frac{g_{m_1}}{100} V_o + \left ( V_{ekg} + g_{m_2} \frac{V_{ref}}{100} - g_{m_2} V_{HPA} \right )$$

This is a first-order linear ordinary differential equation, which should be solvable by eye if you've taken diff-eqs, and if not you can look it up. If you have not taken diff-eqs, then you can at least solve for the condition where \$C_1 \frac{d\,V_o}{dt} = 0\$: this is the equilibrium point of the system.

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  • \$\begingroup\$ You might mean steady state, as opposed to equilibrium \$\endgroup\$ – Scott Seidman Jun 7 '19 at 19:35
  • \$\begingroup\$ Equalibrium, too, for a constant input. Or at least it is in my vocabulary. \$\endgroup\$ – TimWescott Jun 7 '19 at 19:39

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