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 SeanPool 1st August 2005 04:46 AM

Resistor Noises: How to reduce them

I just figured if we parallel n exact same resistors together,
we will get 1/n times the noise power of each resistor.

For example, we we want 1 ohm resistance, we can
parallel 2 pieces of the 2-ohm resistors and get the noise
power half of what we'd get had we used a 1 ohm resistor.
(Assuming 1-ohm and 2-ohm resistors generate equal noise
power).

Is this already well known besides the fact that n parallel
resistors would reduce the tolerance error by n times?

Any opinions?

-Sean

 bocka 1st August 2005 07:25 AM

Quote:
 For example, we we want 1 ohm resistance, we can parallel 2 pieces of the 2-ohm resistors and get the noise power half of what we'd get had we used a 1 ohm resistor.
Paralleling two ideal 2 ohm resistors gives exactly the same noise figure as one 1 ohm resistor.

 jackinnj 1st August 2005 11:02 AM

when designing regulators balance the amount of current drawn in the error sense circuit with the theoretical noise -- i.e. if you have enough current to spare, 1 milliamp drawn by the divider will be less noisy than 100 microvolts -- seems simple enough.

 ilimzn 1st August 2005 11:41 AM

Resistor voltage noise En=2*SQRT(k * T * B * R),

k = Bolzman constant
T = temperature
B = bandwidth of interest
R = resistance in ohms

Depending on material you also have other noise sources, usually proportional to current.

Paralleling n resistors does not reduce noise or tolerance by a factor of n, but by a factor of SQRT(n) because the noise in n resistors is not correlated.
If you use that fact and include it in the equation above, you will see that n parallel resistors of X ohms generate exactly the same noise voltage as a resistor of X/n ohms.

I would say you have assumed a bit too much about your facts ;)

 Circlotron 1st August 2005 01:37 PM

Re: Resistor Noises: How to reduce them

Quote:
 Originally posted by SeanPool (Assuming 1-ohm and 2-ohm resistors generate equal noise power)

Quote:
 Originally posted by ilimzn Resistor voltage noise En=2*SQRT(k * T * B * R) R = resistance in ohms
High value resistors are noisier than low value ones, so what is gained by summing/cancellation of non-correlated noise from the paralleled resistors is is lost by having to use larger (noisier) resistors to begin with.

Blasted numbers! :rolleyes:

 johnthetweeker 1st August 2005 02:03 PM

Yes, lower resistence means lower noise, whether you use a single resistor or paralleled bunch. But there is some dependency on the type of resistor you use. Wirewound resistors are quieter than the other types.

Take a look here:
http://www.aikenamps.com/ResistorNoise.htm

To reduce thermal noise, you could try cooling down the circuit with a peltier block or something. But practically, you may have to cool down to liquid nitrogen temperatures to get significant reduction in noise.

John.

 SeanPool 1st August 2005 06:32 PM

Your opinions are much appreciated. I was wrong :o I see now that my assumption that the noise power was independent of the resistance is way too much. I agree with ilimzn:

Quote:
 Resistor voltage noise En=2*SQRT(k * T * B * R),
Yes, the "average" noise power generated by paralleling n resistors will be the same (not larger, not smaller) as using only one resister. (The average noise power is found using the autocorrelation of the voltage across the resistor, assuming the voltages across the resistors are uncorrelated and have zero mean).

-Sean :o

 SeanPool 1st August 2005 06:40 PM

Thanks John.. The web is very informative.

-Sean

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