Question about noise voltage notation

When expressing the noise voltage across, say a 50 ohm resistor, it will be called

1nV/sqrt(1Hz)

I am trying to come up with an explanation as to why this notation is used.

For example, if you want to determine the noise voltage in a 100 Hz bandwidth on the same 50 ohm resistor I think you would say that the noise voltage is 10nV. Or would you say 10nV/sqrt(100Hz)?

I am thinking that this notation is used to remind the user to take the noise voltage up by the square root of the Bandwidth , and that it does not really have any other significance than that.

Agree?

Reply to
brent
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"brent"

** It expresses the noise characteristic of the part independent of the bandwidth in use.

Makes it the most efficient way.

** No. That is stupid.
** See above.

.... Phil

Reply to
Phil Allison

No, it's a different unit, as in feet vs square feet. If you change the bandwidth, the rms noise voltage goes as the square root of the bandwidth. If you have a narrow pulse signal, the rms signal voltage goes linearly with the bandwidth.

If you want the noise voltage to come out in volts when you're done, you need the square root.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs
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Reply to
Phil Hobbs

** Try reading the OP's question again.

Your reply is completely redundant.

.... Phil

Reply to
Phil Allison

I suppose it would seem so, if you think that the only difference between a length and an area is to remind you to take the square.

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

** FFS read it again - you pompous as.

YOU MISREAD IT !!!!!!

Reply to
Phil Allison

How so? My reply quotes the relevant part of the OP's post nearly verbatim.

The physics is that for fundamental noise sources. e.g. shot noise and Johnson noise, there are no correlations between noise components at different frequencies, so the noise power goes as the bandwidth. Other wideband signals such as pulses may have quite different behaviour, because the correlations between the different frequency components are nonzero.

For instance, if you have a narrow pulse at t=0, all the frequency components have phase 0 at t=0, i.e. only the cosine components contribute.

It's a different unit, not just a mnemonic for behaviour vs. bandwidth.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs
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Reply to
Phil Hobbs

Think about what is meant by "it" here..

It's used when describing Johnson noise in a resistor, which is assumed to be "white", iow to have a constant power spectral density in the region of interest. Since the _power_ spectral density is constant, the voltage squared per Hz is constant, and thus the voltage per sqrt(Hz) is constant.

This is assumed to be true EVERYWHERE, not just from 0 to xx Hz.

IOW, if the noise voltage from 0 to 100 Hz is 10nV, it's also the noise voltage over the bandwidth 100.0kHz to 100.1kHz

Yes. 10nV (RMS)

No. That's not a voltage. Voltage is measured in volts ... or (kg*m^2)/(A*S^3).. this is a voltage spectral density.

No. Those are the proper units for this kind of noise. Dimensional analysis?

10nV/sqrt(Hz) * sqrt(100Hz) = 10 * 10 * nV * sqrt(Hz)/sqrt(Hz) = 100nV.

Best regards, Spehro Pefhany

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"it's the network..."                          "The Journey is the reward"
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Reply to
Spehro Pefhany

"Phil Hobbs = Wanker "

** GIANT Huh ????????????

Got SFA to do with you misreading it.

Imbecile.

... Phil

Reply to
Phil Allison

So you say, but you explain nothing. Don't just tell me, show me.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

(snip)

If the bandwidth (and, as Jerry mentioned, temperature) is known then, yes, quote the voltage. For a resistor or op-amp, though, the bandwidth isn't known until it is used in an actual circuit.

Part of the confusion is that voltmeters are easy to find, but spectral density meters not so common.

Now, you could use the resistor or op-amp in a variable bandwidth amplifier, and so the specifications for that would then be in spectral density terms. For a fixed bandwidth amplifier then you might as well quote voltage. (or, reasonably common, noise temperature.)

-- glen

Reply to
glen herrmannsfeldt

You can imagine connecting a noise source, like a resistor, to a bank of non-overlapping ideal bandpass filters, and dumping them into noiseless load resistors at absolute zero. Then apply conservation of energy. Then it becomes obvious that noise power is spread linearly over bandwidth.

John

Reply to
John Larkin

It isn't the same problem, though. Sending the bandpass slices to separate resistors is equivalent to forcing the phase correlations between the different frequency components to be zero--a narrow pulse and a noise waveform behave the same way in that instance.

With a single load, a pulse waveform and a noise waveform with the same PSD don't have the same behaviour, because the correlations make the height of the pulse go linearly with bandwidth, whereas the noise amplitude goes as the square root.

In optical detection, that's why (for constant average power) you win SNR by going to narrow pulses when you're in the Johnson noise limit, but not when you're in the shot noise limit.

Phase correlations are also why signal averaging works.

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

To Phil, Dr. Phil (haha), Jerry, Spehro, glen,

Thanks for your replies. I've got it now. The term is a noise voltage spectral density. Of course when doing calculations you typically want a specific noise voltage based upon BW. I knew how to use the term and calculate the noise voltage correctly, but I knew something was not quite right in my understanding of the term 1nV/ sqrt(1Hz).

I am working on a noise tutorial and I want to make sure I don't screw up the explanation on this point.

thanks.

Reply to
brent

"brent"

To Phil, Dr. Phil (haha), Jerry, Spehro, glen,

Thanks for your replies. I've got it now. The term is a noise voltage spectral density. Of course when doing calculations you typically want a specific noise voltage based upon BW. I knew how to use the term and calculate the noise voltage correctly, but I knew something was not quite right in my understanding of the term 1nV/ sqrt(1Hz).

** Small but important point: the actual term is " nV/sqrt(Hz) " and should be read as:

" nanovolts per square root of ( operating ) bandwidth in Hertz "

For any real component, noise voltage is a function of temperature, current flow and the actual frequency.

IOW - it is a curve on a graph, with frequency as one axis.

..... Phil

Reply to
Phil Allison

Some of the old LTC datasheets have scope traces of the noise (i.e. time domain) with a specified test bandwidth. Of course you can use spectrum analyzers that will display the noise in root Hz. I figured the scope photos were some sort of "old school" display for people that don't trust fancy gear.

Noise does have an aura of black magic.

Reply to
miso

d

Yes, I was thinking of the 50 ohm resistor number which is 1nV/ sqrt(hz)

nt

Reply to
brent

True. But power spectral density is exactly the same concept for signal or for noise. And we don't even have to consider a "load" if we define power as x^2(t), where x(t) doesn't even have to be a voltage or current!

Pere

Reply to
o pere o

It's not difficult to go from the typical FFT display of dB (m@50R) to nV/sqrt(Hz).

Op-amps are bit of a different thing- the voltage and current noises are mostly white above the corner frequencies, but below that you have flicker 1/f noise. There are as many as 6 noise parameters for an op-amp, including the two or three corner frequencies.

Chopping + demodulation is a way to get your amplification out of the flicker noise region and into the white noise region of the circuitry.

Best regards, Spehro Pefhany

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"it's the network..."                          "The Journey is the reward"
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Reply to
Spehro Pefhany

Yes, that's true--I could have been a good deal clearer on that point. For a pulse waveform, the amplitude goes as BW (i.e. power goes as BW**2) but the pulse width goes as 1/BW, so taken over many cycles, the power in different frequencies adds (which it must).

There are a few subtleties to this that make it interesting. Say you have a signal centred at f_0, with something like a Gaussian envelope, which rolls off monotonically away from f_0, with a white noise background. From a strict linear systems viewpoint, you're better off with a very narrow bandwidth around f_0. That's because as you widen the BW, the signal is rolling off and the noise is staying the same, so the time-averaged SNR is progressively reduced.

However, if the signal is something like a tone burst (or a unipolar pulse, if f_0 is at DC), widening the bandwidth makes the signal get taller linearly, so the instantaneous signal power inside the pulse goes as BW**2, whereas the noise goes as BW. For narrow bandwidths, the pulse width goes as 1/BW, so although you lose on average SNR, you win on instantaneous SNR inside the pulse.

If you're willing to time gate the signal (ideally by multiplying it by a copy of the original pulse), you get a SNR improvement proportional to the bandwidth. Another way of putting it is that if you take the same average power and compress it into pulses or tone bursts, you get a SNR improvement of 1/(duty cycle) if you use time gating or thresholding to get rid of the noise between pulses. (Optimal time gating is a bit more complicated with tone bursts, but it isn't essentially different.)

And then there's FM radio, where the audio SNR increases linearly with increasing bandwidth, even for constant signal power. That always snookers people. ;)

(It works as long as you stay in the high-SNR limit at the discriminator input.)

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

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