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Explain to me this circuit with biopotential signal

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smiles

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Here is one part of the circuit that I found in a document and also a part of explanation followed: "A biopotential signal detected by a bioelectrode is coupled to the noninverting inputs of the first-stage amplifier and the shield driver amplifier. The input impedance is given mostly by the input impedance of the front-stage op-amps, yielding 100M? paralleled with 100pF. R1 limits the current that can flow through the input lead, while diodes D1 and D2 shunt to ground any signal that exceeds their zener voltage ..."
Could you explain me the red statement ?
Thank you !!!
 

Diode question

I think D1 and D2 is to limit the input range to from -Vz to +Vz.
1N914 is a FAST SWITCHING DIODE.
The Vz is the forward biaed voltage of the diode. It is about 0.7V, so the input range is about -0.7v to 0.7v.
 

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    smiles

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Diode question

uhm ... the symbol tell that it is normal diode, not Zener ???
"zener voltage of diode" seems strange to me, search with google just found 12 results :(
 

Diode question

smiles, I don't like this design for a sensitive bioelectrode, 1n914 is for overvoltage protection so it dose not damage your opamp but its leaky and can introduce considerable voltage error over 20k resistor. I would use an ultra low leakage diode for such design.
 

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    smiles

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Diode question

Thanks !!! How about the capacitor ? Why it is there ?
 

Diode question

in conjunction with 20k resistor they are a low pass filter with cutoff~36kHz
 

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