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I want to design a voltage divider circuit to bias an op amp, this circuit will be a part in another circuit to measure a biomedical signal. Is there a noise issue arises from using voltage divider to provide biasing? if I choose R1 = 10K and R2 = 10K, then will it make any difference if I choose R1 = 10M and R2 = 10M?
In a paper I read the following:

Using a simple resistive bias network is impractical from a reliability and noise standpoint.Although a biasing resistor can be bootstrapped to the required resistance, minimizing its current noise contribution requires an impossibly high value

I don't know if they mean a voltage divider by "simple resistive bias network", but I assumed that.

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  • \$\begingroup\$ Yes it will make a difference, having effectively the Johnson noise of the two resistors in parallel. Alternatively you can decouple the resistive divider to keep it quiet, and connect the "thing needing biassing" to it via a lower value (quieter) resistor - or directly, if that isn't shorting your input out! \$\endgroup\$ – Brian Drummond Jan 23 '16 at 15:01
  • \$\begingroup\$ Your input resistor divider/bias/feedback network should usually have values that are significantly lower than the OP AMP input resistance if they are voltage mode inputs (typicaly). Check out the aplication circuits for the part in question from a couple of manufacturers and other similar parts. They will indicate typical starting values. \$\endgroup\$ – KalleMP Jan 23 '16 at 16:27
  • \$\begingroup\$ @BrianDrummond, I've just finished reading about Johnson noise. I conclude from formula that if we have an infinite resistor then there will be an infinite voltage noise, and zero current noise! it's obvious that I have contradiction here since the open circuit(which is infinite resistor) isn't produce an infinite noise. Also, The equation tells that if you have a short circuit with infinite bandwidth then the noise will approach to infinity. Could you help me here :) \$\endgroup\$ – hbak Jan 23 '16 at 17:53
  • \$\begingroup\$ The paper also says "in non-contact capacitive coupling, the input resistance must be in excess of 1TΩ". Where are R1 and R2 in your circuit, and how are you biasing the input? \$\endgroup\$ – Bruce Abbott Jan 23 '16 at 18:19
  • \$\begingroup\$ @BruceAbbott Exactly. R1 and R2 are the two resistors that used in a voltage divider. Actually, I'm asking about the paper itself and I'm not designing anything, it's just to keep the question more generalized. The authors of the paper who want to bias the input because: LT6078 requires an input bias- ing network to provide a DC current path to counteract leakage currents and fix the DC input voltage to a mid-rail level for maximum output signal range \$\endgroup\$ – hbak Jan 23 '16 at 18:26
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It is unlikely that Johnson (thermal) noise is what is being referred to. However high impedance resistor dividers are more susceptible to picking up noise from external sources, and are less stable (because they are more susceptible to leakages).

Use 10k's and add a decoupling capacitor (say 1-100 uF) to reduce noise to insignificant levels.

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  • \$\begingroup\$ But what about "minimizing its current noise contribution requires an impossibly high value". They said that a high impedance resistor, reduce the noise. If you think that I missed something to add in my quotation, tell me to add more to the question. \$\endgroup\$ – hbak Jan 23 '16 at 17:31
  • \$\begingroup\$ Perhaps the context from your quote is different ? (Johnson) noise from a DC resistor divider can be effectively filtered with a capacitor. \$\endgroup\$ – jp314 Jan 23 '16 at 18:50

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