• posted

I have been pondering the notation for noise voltage. The noise voltage across a 50 ohm resistor is expressed as 1nV/sqrt(1Hz)

As I ponder this it seems to me that one is concerned about the noise voltage period. For instance, if you have a 1Hz BW signal you would say the noise voltage on a 50 ohm resistor is 1nV.

If you have a 100 Hz signal you would say that the noise voltage is 10 nV (on a 50 ohm resistor with a 100 Hz BW)

Would you say 10 nV/sqrt(100Hz) ? (I don't think so)

It seems to me that the notation of 1nV/sqrt(1Hz) is really just a reminder that when calculating the noise voltage that the noise voltage goes up by the square root of the Bandwidth, and is not really relevant other than for the reminder.

Agree?

• posted

Please ignore this post and answer the other post on this. Google took over an hour to post the first on and so I posted again....

• posted

I have been pondering the notation for noise voltage. The noise voltage across a 50 ohm resistor is expressed as 1nV/sqrt(1Hz)

As I ponder this it seems to me that one is concerned about the noise voltage period. For instance, if you have a 1Hz BW signal you would say the noise voltage on a 50 ohm resistor is 1nV.

If you have a 100 Hz signal you would say that the noise voltage is 10 nV (on a 50 ohm resistor with a 100 Hz BW)

Would you say 10 nV/sqrt(100Hz) ? (I don't think so)

It seems to me that the notation of 1nV/sqrt(1Hz) is really just a reminder that when calculating the noise voltage that the noise voltage goes up by the square root of the Bandwidth, and is not really relevant other than for the reminder.

Agree?

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There are two formulas.

the "one sided power spectral density" is given by

vsd^2 = 4*kb*T*R

But this equation is given per hertz of bandwidth so to be precise we write it as

vsd^2 = 4*kb*T*R nV/Sqrt(Hz)

In a sense this equation does not take into account the bandwidth. For 10 hz we have sqrt(10) times more noise.

Your formula, does not give you the noise voltage but the nosie voltage per hertz. If you want the noise voltage it is

v = sqrt(4*kb*T*R*BW) = sqrt(4*kb*T*R) * sqrt(BW) = vsd*sqrt(BW)

Note how we are multiplying your formula(which is per sqrt(Hz)) by sqrt(BW) and this gives us only volts left.

The idea is rather simple. Suppose we have a BW of 1 Hz for a 50Ohm resistor at room temp. The nV/Sqrt(Hz) = 1nV/Sqrt(Hz).

Now if we are applying that to a bandwidth of 100 herts we have 10nV of noise. This is because each "band" or each "hz" contributes adding some noise.

If you low pass the resistor then you can reduce it's bandwidth. Theoretically a resistor has infinite noise. In reality every resistor has finite bandwidth and limited noise.

Take a 50Ohm resistor with 1nV/sqrt(Hz). If the bandwidth is from 0 to 1Mhz then the noise is 1uV. If the bandwidth is from 100khz to 1Mhz then it is

0.86uV. if it is from 0 to 900khz it is 0.86uV.

The best way I can explain it is that 1nV/sqrt(Hz) is the density of noise and when we multiply(or rather integrate) this over our bandwidth we end up with the total noise over that bandwidth(which is generally what we want).

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