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[SOLVED] Basic question about Johnson noise for 2 parallel resistors

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fabio_807

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Hello,

Is the noise for 2 parallel resistors equal or less than their parallel equivalent. Or to put some numbers Vnoise (2Meg || 2Meg) vs Vn(1Meg).
In my estimate it is Vn(1M)/sqrt(2).

Thank you and nice to meet you guys
Fabio
 

For resistors in parallel the Johnson noise is determined by their parallel equivalent resistance.
The two 2Meg resistors in parallel have the same noise as one 1Meg resistor, not divided by the sqrt(2).
 
In my estimate it is Vn(1M)/sqrt(2)...
Click to expand...

Let us see.

The origin of this noise is thermal fluctuations. Hence the noise power will be proportional to κT where T is the absolute temp and κ is Boltzmann constant.

For a resistor R, the noise voltage (square) will be proportional to κTR. (power is v²/R).

When you put two resistors in parallel, the effective resistance will be R/2.

Hence the noise voltage (square) will be proportional to R/2

If you use noise voltage (not the square term), v(noise) (for 1MΩ) will be v(noise) for (2 MΩ)/√2

I guess it is clearer.
 
Thank you guys!
I did the math again and the actual noise is the same if I'm not mistaken.
IMG_20180803_232215.jpg
Regards,
Fabio
 

fabio_807 said:
I did the math again
Click to expand...
Simply consider by noise current not noise voltage.

In_rms**2=4*k*T*(1/R)

For R=R1//R2, In_rms**2 = 4*k*T*(1/R1)+ 4*k*T*(1/R2) = 4*k*T*(1/R)
1/R= 1/R1 + 1/R2
 
the actual noise is the same if I'm not mistaken.
Click to expand...

Well, it depends. Noise are by definition phase uncorrelated and they cannot be added just like that. If you average the noise voltage (over a long time) it will turn out to be zero (because you will get noise voltage of both signs).

The noise power will be positive definite and it will scale by σ²/N (here the N is 2).

See central limit theorem on how the averaging is done for noise (https://en.wikipedia.org/wiki/Central_limit_theorem#Classical_CLT).

There is also associated with noise a Maxwell's Demon (see, for example, https://www.auburn.edu/~smith01/notes/maxdem.htm for an easily understandable version).

In other words, you cannot reduce noise by using series-parallel combination of resistors. Often amplifier noise is specified by temp (an amplifier with a noise figure of 70K is typical).
 

c_mitra said:
Well, it depends. Noise are by definition phase uncorrelated and they cannot be added just like that. If you average the noise voltage (over a long time) it will turn out to be zero (because you will get noise voltage of both signs).

The noise power will be positive definite and it will scale by σ²/N (here the N is 2).

See central limit theorem on how the averaging is done for noise (https://en.wikipedia.org/wiki/Central_limit_theorem#Classical_CLT).

There is also associated with noise a Maxwell's Demon (see, for example, https://www.auburn.edu/~smith01/notes/maxdem.htm for an easily understandable version).

In other words, you cannot reduce noise by using series-parallel combination of resistors. Often amplifier noise is specified by temp (an amplifier with a noise figure of 70K is typical).
Click to expand...

thank you, this helps!
 

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