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 northsiderap 6th May 2005 04:36 PM

Noise Figures

I am working on an amplifier, and I am wondering how to measure and interpret noise figures for the preamp stage...

The preamplifier involves a JFET front end, and a direct-coupled class A bipolar transistor second stage.

I have run the circuit through Circuitmaker and have nice simulation results.

Now....

What the heck is pV^/Hz ? It is a noise figure, but I'm not sure what it means in the real world.

Does anyone have any information on what this is, how I could measure it on my finished amp, and what I'm looking for for noise figures?

Even a place to research this and learn it for myself would be A-OK.

Thanks

 rayfutrell 6th May 2005 05:05 PM

PV^/Hz is the noise voltage density in a one Hz bandwidth. you have to integrate it over the whole bandwidth to obtain a total noise voltage. With a preamp this is tricky because of the equalization.

Dr. Leach has a very good dissertation on noise on his website, but you now need a password for access. Possibly if enough people ask him he might make it available to the public again.

Several years ago I asked him to post the schematic of his MM phono preamp and he did it. I have a very effective grovel and beg.
Regards,
Ray

 macboy 6th May 2005 05:15 PM

Quote:
 What the heck is pV^/Hz ? It is a noise figure, but I'm not sure what it means in the real world. Does anyone have any information on what this is, how I could measure it on my finished amp, and what I'm looking for for noise figures?
The noise figure is usually expressed in units of pV/sqrt(Hz) (i.e. pico-volts per square-root of Hz), or even more commonly, nV/sqrt(Hz). It could be expressed as pV^2/Hz (this is just the square of the other one), but this has less intrinsic or intuitive meaning.

Basically, you can use this figure to determine how much noise will be present, based on amplifier bandwidth. Ex. if you have 4.5 nV/sqrt(Hz), and a bandwidth of 20 kHz, then you will have 4.5 nV * sqrt(20000) = 636 nV or 0.6 uV of noise. Note that this figure is usually referenced to the input of the amp, so now multiply it by the voltage gain of the amp, say 20 (which is 26 dB gain). Now you have 12 uV of noise at the output of the amp.

But to caculate the noise accurately, you will need to account for noise sources other than the input-referenced voltage noise. These include the input-referenced current noise (input impedance of the amp converts this into voltage noise), resistor self-noise (thermal noise), etc.

I would imagine that you could find some noise application notes or design guides at Analog Devices or TI or Linear Tech that will go into lots of practical detail about noise sources. Take a look.

 northsiderap 6th May 2005 05:35 PM

You guys (& gals) rock

Once again, this helps immensely.

The graph at the output node in Circuitmaker shows the noise level rolling up to @ 4.5 pV^2/Hz down towards 15Hz with the node designated to the signal generator and measured at the final simulated load resistor.

so I'll assume that my minimum noise level will be 5 pV^2/Hz, and my realistic voltage gain will be about 15.

that's:

25pV/Hz
25pV / 20,000 Hz (bandwidth)
=.00125

.00125*15 (gain)
=.01875 pV

That's tested with an input of 2mV and an output of over 30mV, so my theoretical noise output is @ -240db?

This seems kinda low... I'm guessing that there will me much more noise than that from other sources. Or, maybe my analysis is set up wrong...

Ok, I'll double the noise level to 10pV^2/Hz
100pV/Hz
100pV / 20,000 (bandwidth)
=.005 pV

.005 pV * 15 (gain)
=.075 pV

out of 30 mV that's still -230dB for a noise floor.

 jackinnj 6th May 2005 05:46 PM

Quote:
 Originally posted by macboy I would imagine that you could find some noise application notes or design guides at Analog Devices or TI or Linear Tech that will go into lots of practical detail about noise sources. Take a look.
Agilent should be added to that list --

 tschrama 6th May 2005 05:46 PM

I use Circuitmaker all the time, and I have used the noise simulation often to optimize designs.

here's what I think is the right way to do it

5pV^2/Hz you need to SQuarerRooT it .. for nV/sqrt(hz)

that is 70nV/sqrt(Hz) (that's quite a lot for a low noise design)

multiply with the sqrt (bandwidth) say sqrt(20e3) = 141

70 * 141 = 10uV total noise

grz,
Thijs

 macboy 6th May 2005 05:52 PM

Quote:
 so I'll assume that my minimum noise level will be 5 pV^2/Hz, and my realistic voltage gain will be about 15. that's: 25pV/Hz 25pV / 20,000 Hz (bandwidth) =.00125 .00125*15 (gain) =.01875 pV
No, 5 pV^2/Hz is not equivalent to 25 pV/Hz. If it was, they'd just express it as such, and avoid the confusion.

You need units of Volts in the result, not Volts^2, so take the square root of the noise figure you have (5 pV^2/Hz):

5 pV^2/Hz can be expressed as: 5e-12 (V^2)/Hz

sqrt( 5e-12 (V^2)/Hz ) is 2.24e-6 V/sqrt(Hz)

The voltage noise figure is 2.24 uV/sqrt(Hz). Note the significant change in the exponent; you were about 5 orders of magnitude off.

Now use that in your calculations.

 northsiderap 6th May 2005 06:10 PM

Hmmm.... Hmmm....

so 5pV^2/Hz
= SQRT .000000000005 / (SQRT Hz)
=2.236 uV / 141 (SQRT Hz)
=.01585 uV
.01585 uV * 15 (power)
=.238 uV noise

Did I do that right?

Now I'm getting @ -100dB noise figure. Yes, this is bad. I'm looking for @ -115dB. or - infinity dB of course... ;)

Well, it's my first attempt at a JFET front end, so back to the drawing board.

 macboy 6th May 2005 06:44 PM

Re: Hmmm.... Hmmm....

Quote:
 Originally posted by northsiderap so 5pV^2/Hz = SQRT .000000000005 / (SQRT Hz) =2.236 uV / 141 (SQRT Hz) =.01585 uV .01585 uV * 15 (power) =.238 uV noise Did I do that right? ...
No. The noise figure is (as you read it) "2.236 micro-volts per sqaure-root of Hz", not "2.236 microvolts divided by square-root of Hz.". Remember this is the units, not an equation! The result you are after is a noise figure in units of Volts (or microvolts, same thing), so you need to get rid of the sqaure-root of Hz part in the denominator of the units. To accomplish that, you need to multiply by something with units of sqaure-root of Hz, which is your bandwidth (in your case, 141 sqrt(Hz).). That gives you the nasty figure of 315 uV. Yuck.

Also,
The noise figure of 5 pV^2/Hz, I assume that you got this from simulations? Is this figure for noise at the output, or is it referenced to the input? This is important, because in the former case, you do not multiply it by the amplifier's gain, but in the latter case you do. Commercial opamps give noise figures referenced to the input because they have no control over how you use the amplfier (gain, etc.), but I would suspect the simulations might give a noise figure directly at the output. That makes a big difference. But probably not enough.

 rayfutrell 6th May 2005 07:34 PM

LTSpice has an example of a phono preamp using an LT1027. I modified the example by adding a phono cartridge and plotted the output noise. It was very high at 20 Hz and decreased in a concave curve to 20kHz. This is to be expected because the gain decreases about 40 dB from 20 Hz to 20kHz. What this tells you is that you need an input stage with very low noise voltage and decent noise current. The noise voltage dominates at low frequencies due to the low cartridge impedance. The cartridge impedance doubles in each octave and so does the current noise.

Suggest you use the same JFETS as in the Leach preamp at http://users.ece.gatech.edu/~mleach/. The JFETs are old but still good and readily available from many sources. In this case the low transconductance and gain of the JFETs is an advantage because you have little Miller effect capacity.
Regards,
Ray

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