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Regarding noise added on a signal during transmission i.e. along the long wires between the source and the amplifier, Im wondering whether in a single ended system does the input impedance of the receiver/amplifier(not the gain) has any effect on the amount of noise at the output.

For example in a single ended system; a 1V DC signal source with an output impedance Rout is coupled to an opamp which has input impedance Rin and and gain of 1 as below:

enter image description here

The output then will become roughly(neglecting the voltage divider effect) 1V plus the noise picked up due to long non twisted wires along the way. The noise could be due to EMI or ground loop.

Now where Im stuck is: Is this noise’s magnitude have any relation with the value of Rin or it is only a function of Rout?

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  • \$\begingroup\$ If the opamp has an Rin equal to that external Rsource, then a 2:1 voltage division will occur. But the current flowing thru the Rin will contribute to the total noise. \$\endgroup\$ Commented Oct 8, 2018 at 18:03
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    \$\begingroup\$ Rin is usually very high. \$\endgroup\$
    – JRE
    Commented Oct 8, 2018 at 18:05
  • \$\begingroup\$ Forget about voltage divider effect DC error, Im mostly asking about the interference on wires versus input impedance. I wrote neglect divider effect in question already \$\endgroup\$
    – cm64
    Commented Oct 8, 2018 at 18:06
  • \$\begingroup\$ Let's say that Rin is almost infinite and if Rout is big it will pick any noise. \$\endgroup\$ Commented Oct 8, 2018 at 18:09
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    \$\begingroup\$ You need to go back and re-read the answers to one of your earlier questions. The answers there apply here as well. \$\endgroup\$
    – JRE
    Commented Oct 8, 2018 at 18:49

4 Answers 4

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Yes. For high impedance situations, you have to worry about capacitive coupling into the input lines. This type of coupling is caused by voltage changes on close lines that cause a current to be injected into your amplifier input. The current flows into the amplifier input resistance, which is big, and so it causes a big interference. For low impedance situations, you have to worry about inductive coupling into the input lines. This is caused by currents that flow on close lines, that induce a voltage in series with the input signal. The voltage will then cause a large current to flow, because of the low impedance of the amplifier input.

So, the kind of interference that is a problem differs, depending on the input impedance of the amplifier. High impedance=capacitive interference, low impedance=inductive interference.

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  1. The resistor is always a source of voltage noise. If its resistance is lower than the source resistance, its noise contribution will be negligible.

  2. If the source is low impedance, then added isolating resistance larger than the source’s will lower the susceptibility for inductive pickup of external magnetic fields.

  3. Parasitics may matter more than noise, since they potentially increase susceptibility to external interference. To maintain good CMRR, the impedances of each branch of the differential input circuit should be matched as far as possible. This includes PCB layout, wiring layout, connector pin selection, etc. At a minimum, the isolating resistance should be equally split between the branches, and typically it would be added on the side of the in-amp, doubling as input protection. It’s usually much less cumbersome to have those resistors on the PCB than on the sensor assembly.

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Resistors

-Resistors Vnr = √( 4kTBR) for k=Boltzman's, T ['K], B[Hz], R[Ω]
- If source impedance = 0Ω, the input voltage noise dominates.
- If source impedance is high. e.g. xxx kΩ then noise rises* due to Johnson Noise effects
- rises* by √Rs relative to specified noise at stated R.

Active devices

-Bipolar op amps (OA) tend to have lower voltage noise than JFET ones, unless the BJTs are high speed types or the FETs are large and high capacitance types.
-Current Shot Noise = √(2I(in) q)/√Hz unless bias compensated and depends on 1/f corner
- shot noise of simple bias currents = Schottky noise
-Voltage noise is specified on the data sheet, and it isn't possible to predict it
-Current noise varies widely upon the input structure.

If the noise is uncorrelated then noise voltage adds by RMS or "root-sum-of-squares"

Summary

  • common dominant source of noise ( but not all OA's)

low R = Voltage noise of OA
medium R =Johnson Noise of R
high R = Input bias current flowing thru R

  • so it is not the bias current x OA input impedance
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  • \$\begingroup\$ You mean is that due to FET has higher input impedance than BJT? \$\endgroup\$
    – cm64
    Commented Oct 8, 2018 at 18:20
  • \$\begingroup\$ The question is about the influence of the impedance of the source and the opamp on the amount of external noise that will be picked up by the wires connecting the source with the opamp, not about the noise inherent to the particular technologies used in various models of opamp. \$\endgroup\$
    – JRE
    Commented Oct 8, 2018 at 18:42
  • \$\begingroup\$ I'll rearrange... \$\endgroup\$ Commented Oct 8, 2018 at 18:44
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Yes, apparently for current noise.

From http://www.analog.com/en/analog-dialogue/articles/what-should-i-know-about-opamp-noise.html

If the source resistance is higher, the Johnson noise of the source resistance may dominate both the op-amp voltage noise and the voltage due to the current noise; but it’s worth noting that, since the Johnson noise only increases with the square root of the resistance, while the noise voltage due to the current noise is directly proportional to the input impedance, the amplifier’s current noise will always dominate for a high enough value of input impedance. When an amplifier’s voltage and current noise are high enough, there may be no value of input resistance for which Johnson noise dominates.

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  • \$\begingroup\$ It isn't about the opamp as such. Any load would do. What the OP is trying to get at is how much influence the impedance of the source and load have on the amount of noise the circuit will pickup on the signal lines. \$\endgroup\$
    – JRE
    Commented Oct 8, 2018 at 19:40

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