Working principle of AD8232 instrumentation amplifier

Question

Here is the datasheet for the AD8232 IC.

On page 16 under Theory of Operation Section:

What I failed to understand is how the instrumentation amplifier amplifies the ECG signal from an electrode. It has an integrator amplifier with C1 capacitor.

As far as I know, an integrator will integrate the input. If the input is an ECG signal fed to an integrator, doesn't it change the output? I mean, we expect the output of an instrumentation amplifier to be just the input amplified, but still in the same form. But with the presence of an integrator amplifier, in my understanding, it will change the output form which no longer like the ECG signal we saw in the input.

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

According to TimWescott's answer I try to derive the transfer function as below:

The output Vo still depends on integral of the Vecg.

Answer 1

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.

Answer 2

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.

Answer 3

The output is the low pass filtered input. This can be seen from your equation. As long as the input spectrum is within the filter’s bandwidth, the output will be the input multiplied by the filter’s gain.