EEG RnD – Choosing the Right Op-Amp

The key element here is that they are measuring charge displacement. In a similar way than a high pass filter where when a signal is moving on one plate of the capacitance of your C-R filter and can be picked up on the resistor side.

You can also think of as that electrode acting like an antenna picking the signal emitted from the brain.

In our case, several considerations here:

• You want to have the cutoff frequency of your 'C - R filter' electrode system to be as low as possible since the typical brain activity frequency - which I am supposing is ~few Hz. Which means large C and large R.
• The large resistance is usually obtain using op-amp, instrumentation operational amplifier in particular.
• The C is given by the capacitance between your brain and the electrode. Your body tissues (bone, skin, hair) act as dielectric here just as in a regular capacitor.
• Typical value of coupling between the electrode and the brain can be very low especially if you have a lot of hair, keep in mind that the value of the capacitor is to first order 1/(distance to brain), so you want to make that distance as small as possible.
• A bigger plate of metal will also give you a higher capacitance.
• However a bigger plate will also average brain activity over a bigger region.
• The signal from the brain activity is quite small, therefore low noise, high gain electronics is required amplify the signal from the electrode.
• Because the signal from the brain is so small your electronics has to be sensitive to very small charge displacement is therefore very sensitive to pick-up from any other electromagnetic field in the vicinity (probably a good idea to have a notch filter in your readout electronics). The instrumentation amplifier helps here too (for common mode rejection in particular).

For question 1: \$V_S^i\$ will appear at the negative terminal of OA2. At high frequency, the impedance of \$C_b\$ is low, so the voltage on both sides of the diodes approach the same value, thus this is no current, an effectively high input impedance. For low frequencies, input that differs by more than a diode drop from \$V_{bias}\$ will cause current through at least one of the diodes.

For question 2: For true common mode signals, \$ V_s^i = V_{cm}\$. Thus, \$V_{cm}\$ also appears at the negative input of OA1. With the same voltage at both ends of the series combo of Rf and Cf, there is no current.

For question 3: I think this may be the same answer as for question 1, with the point just made a slightly different way.

Question 4: As long as there are no saturations -- yes, and that is central to the analysis of the circuit.