Help understanding noise figure of amplifier

[mio83]

Hello,

in the past I have asked some help with calculations in this forum, and I got very good and instructive answers. So let me try again! 

I am trying to study (self teach myself) something about amplifiers.
Concretely I have an old HP 461A (1Khz to 150Mhz) linear amplifier I want to understand properly.
This is a +40db (quite flat, +/- 1 db across the spectrum) amplifier.

Regarding noise performance, the datasheet says:

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EQUIVALENT WIDEBAND INPUT NOISE LEVEL: Less than 40 uV in 40 db position.

I have a spectrum analyzer to do (more or less) precise measurements.
I did the following:

• Shorted the input of the HP 461A amplifier.
• Connected the output of the HP 461A amplifier to the input of the SA
• Verified that I get a flat noise floor across 0-150Mhz
• Using RBW of 1KHz I measured the noise floor to be -94dBm.
• Considering the -40db amplification, the amplifier is amplifying an equivalent noise of -94-40=-134dBm (again, at 1Khz RBW)
• I can deduce the density D (i.e., noise at 1Hz RBW) by subtracting by 10*log_10(1000)=30. So D = -164 dBm/Hz 
• I have verified that, from D, I can get correct predictions for the noise level read by the SW at different RBW. For example, at 1Mhz RBW I get -64dBm.

Ok, so it looks like the HP 461A amplifier has an equivalent input noise density of -164dBm/Hz.



Taking the bandwidth to be 150Mhz, I can calculate:

[/list]

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-164dBm + 10log_10(150e6) = -164 + 80.1 = -82dBm

of equivalent wideband noise input at 150Mhz bandwidth.


I can convert the power from dBm to Watts using the equation

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-82dBm = 10*log_10( P/ 0.001)
and obtain P=6.30* 10^12 Watts  (https://www.wolframalpha.com/input/?i=-82+%3D+10*log_10%28+p%2F+0.001%29%2C++v%5E2%2F50+%3D+p+)

Finally, I can convert this to RMS volts (considering a 50 Ohm input/output) using

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(v*v)/50 = P

and I obtain (https://www.wolframalpha.com/input/?i=-82+%3D+10*log_10%28+p%2F+0.001%29%2C++v%5E2%2F50+%3D+p+)

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Measured equivalent wideband (150Mhz) noise RMS voltage: v = 17uV
which is compatible with the datasheet (which says <40uV).

Are these calculations OK or am I making some mistakes?

The reason of my doubt is that, using the same style of calculations, I get a noise density expressed in V/sqrt(Hz) of only 1.4nV (https://www.wolframalpha.com/input/?i=-164+%3D+10*log_10%28+p%2F+0.001%29%2C++v%5E2%2F50+%3D+p+)


This seems too god to be true? I had a look at the datasheet of minicircuits amplifiers (made in 2020, the HP 461A was made >50years ago in 1964) and the specs for LNA amplifiers often have >14.nV/sqrt(Hz) noise density. Also the HP 461A has a measured SWR <1.1, much better than some new Minicircuits amps. Is it really possible that this old HP unit is so good?

If it is actually that good, and the calculations are OK, could I use this amplifier directly to amplify the signal received by an antenna? Ot it would not be appropriate for that task?

Thanks!

[awallin]

What SA detector mode are you using?
When testing some photodiode amplifiers I could not get the numbers from the (Siglent) SA to match at all with the default detector mode of "positive peak". I had to use the "sample" detector mode (and maybe average 10 or 100 traces) to get the numbers from the SA to match calculations.

[mio83]

I am using average (100 traces), however I am using the default "mode" (HP 8591E) which, honestly  :palm, I don't know what it is.
I'll have to check. Thanks.

[unitedatoms]

May be for more accurate reading the spectrum analyzer needs to be zeroed on its own noise first when amplifier is unpowered (cold). Then trace the added noise when amplifier is powered (hot). The added value will a catet of hypotenuse of triangle, when other catet is known noise of spectrum analyzer from datasheet.

[mio83]

thanks.

So if I understand your point I need to:

1) Measure SA noise floor at a given bandwidth (and normalise if it is not very flat): N_1 (in dBm)
2) Measure Amp noise floor (as I explained in my first post): N_2 (in dBm)
3) computer the difference: M = N_2 - N_1
3) Compute the powersum as:

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N = sqrt(N_1^2  + M^2) where N_1 and M in the expression above are expressed in Watts (not dBm)

Is this what you are suggesting? What does this number N represents (except from the geometric sum written above)?

[unitedatoms]

Yes. That model is possibly valid for ideal impedance match. By design the devices tend to be matching 50 ohm real with zero imaginary. If there is a mismatch then one will need a thick book go through.

I like your intent to verify the physical quantity known from from datasheet. 17uV vs 40uV. With nullifying spectrum analyzer's own noise the figure will be even lower. I can not explain why you are getting a value being so low.

Edit: I am trying to follow your calculations. The 50 Ohm resistor at room temperature is 0.9 nV/SQRT(Hz). The amplifier is 40uV per 150MHz, so it is 3.27nV/SQRT(Hz). Your sweep of -94dBm at output is according to http://wera.cen.uni-hamburg.de/DBM.shtml gives 4.4uV RMS. To convert is to SQRT(Hz) it becomes 139nV/SQRT(Hz), considering bandwidth of SA 1KHz.

Assuming 40Db amplification, the equivalent noise at input is -94-40=-134dBm @ 1kHz RBW. Or 44nV per band. Or 1.39nV/SQRT(Hz) at input.

Going back to full band of interest of 150 MHz. We get 17 uV.

I can not explain it.

edit 2: wait. I dont understand the meaning of 0.9 nV of 50 Ohm resistor. Is it for infinite band or per SQRT(Hz)? I am lost.

edit 3: ok. The wiki article about Johnson Noise helped. The figure is per Sqrt(Hz)

[mio83]

edit 2: wait. I dont understand the meaning of 0.9 nV of 50 Ohm resistor. Is it for infinite band or per SQRT(Hz)? I am lost.

edit 3: ok. The wiki article about Johnson Noise helped. The figure is per Sqrt(Hz)

Wikipedia (https://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise)  says that at room temperator (300Kelvin, which is exactly the temperature of my room these summer days) the thermal noise level is -174dbm/Hz.

This should correspond to 0.4nV per SQRT(Hz) if my calculations are correct: https://www.wolframalpha.com/input/?i=-174+%3D+10*log_10%28+p%2F+0.001%29%2C++v%5E2%2F50+%3D+p+

Where did you read the value 0.9nV for 50Ohm?


EDIT: I now see that wikipedia expresses noise at room temperature (300K) in terms of uV/sqrt(Hz) as:

• (apprximate expression) 0.13*sqrt(50) nV/1Hz = 0.91 uV/ sqrt(Hz)
• (exact expression) sqrt( 4*(1.38e-23)*300*50 )  = 0.901 uV/ sqrt(Hz)

The above values correspond to -168dBm. However it also says that it is equal to -174dBm. Weird...

Can somebody shed some light? Thanks 

[yl3akb]

Have you seen explanation of that noise definition here? https://www.hpl.hp.com/hpjournal/pdfs/IssuePDFs/1964-03.pdf

10uV noise power generated by 50 Ohm resistor used in that document, roughly corresponds to T =290K and BW=150MHz, as expected.

I like to think in noise temperatures and noise figures, instead of uV.
If you connect 50 Ohm resistor (at T = 290K) to the input of the amplifier and measure its output signal power with SA, you are measuring:
P = k*(290+ Te)*RBW*G, where
P - power in Watts,
k - Boltzmann constant,
Te - equivalent input noise temperature of amplifier [K],
G - gain of amplifier (linear),
RBW - resolution BW of SA [Hz].
 (lets assume that gain of 40 dB is sufficient for amp. to dominate over your SA noise figure).
Then rearrange to obtain Te and convert to noise figure: NF = 10*log10(Te/290 + 1). I would expect value close to '12 dB', as shown in above document example.

[yl3akb]

You shoudl not short  (or leave open) the input of the amplifier. The specification is only valid in 50 Ohm environment.

[mio83]

Have you seen explanation of that noise definition here? https://www.hpl.hp.com/hpjournal/pdfs/IssuePDFs/1964-03.pdf

 (lets assume that gain of 40 dB is sufficient for amp. to dominate over your SA noise figure).
Then rearrange to obtain Te and convert to noise figure: NF = 10*log10(Te/290 + 1). I would expect value close to '12 dB', as shown in above document example.

Yes I had read that document.
But I am not sure exactly where I should modify my calculations:

Probably what I need is an explicit way to move from Noise Power/Hz (in dBm) to nV/sqrt(Hz) ?
I have been using:

• convert N dBm/Hz to M Watts/Hz using the equation: 10*log_10(M/0.001) = N 
• convert M Watts/Hz to V nV/sqrt(hz) using the equation: (V^2)/50 = M

Is this wrong?

Regarding the equation you suggest:

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NF = 10*log10(Te/290 + 1)

it takes value NF=12 (the one you expect) for Te = -271 (see https://www.wolframalpha.com/input/?i=10*log10%28x%2F290+%2B+1%29+%3D+-12: ) which is less than 2Kelvin above absolute zero (-273 Kelvin)... 

[mio83]

You shoudl not short  (or leave open) the input of the amplifier. The specification is only valid in 50 Ohm environment.

Thanks, I will redo my tests with a 50 Ohm terminator at the input (this seems weird to me, it adds the noise of the 50ohm terminator to the internal noise of the amplifier, but ok).

Anyway, just to make sure my calculation steps are correct, given the datasheet's 40uV wideband input equivalent noise (150Mhz bandwidth) what should I theoretically expect for equivalent values of:

• noise density expressed in  dBm/Hz
• noise density expressed in nV/sqrt(Hz) 

thanks!

[yl3akb]

Can you say the power value you are measuring with SA when input of amp is loaded with 50 Ohm (not shorted)? (and RBW setting). I will try the calculation.

NF = 12 dB (or lineary F = 10^(12/10) = 15., corresponds to Te = (15.8-1)*290= 4300 Kelvins. I dont understand where You get -271.

[yl3akb]

You shoudl not short  (or leave open) the input of the amplifier. The specification is only valid in 50 Ohm environment.

Thanks, I will redo my tests with a 50 Ohm terminator at the input (this seems weird to me, it adds the noise of the 50ohm terminator to the internal noise of the amplifier, but ok).

Anyway, just to make sure my calculation steps are correct, given the datasheet's 40uV wideband input equivalent noise (150Mhz bandwidth) what should I theoretically expect for equivalent values of:

• noise density expressed in  dBm/Hz
• noise density expressed in nV/sqrt(Hz) 

thanks!

With Te You will be able to calculate input noise power (P = k*Te*BW), and from that noise voltage in case of 50 Ohms if You want that.
It general is expected that impedance of source feeding the amplifier will be 50 Ohms. In this case, the source is 50 Ohm resistor. And it is known variable in this case, which helps to calculate noise of amplifier itself.

P.S.
1. As for "noise density expressed in  dBm/Hz'', I would expect P = 10*log10(1.38064852e-23*(290+4300)*1)+30 = -162 dBm/Hz
2. As for "noise density expressed in nV/sqrt(Hz)", I got U = sqrt(1.38064852e-23*(290+4300)*1*50) ~ 1.8 nV/sqrt(Hz)  <--WRONG. Correct formula is P = k*T = U^2/(4*R) => U = 2*sqrt(k*T*R) = 2*sqrt(1.38064852e-23*(290+4300)*1*50) = 3.6 nV/sqrt(Hz) which gives value close to theoretical voltage of U = 3.6*sqrt(150e6) = 40 uV @ 150 MHz (well I got 44 uV, maybe the assumed noise BW is not 150 MHz exactly, but little smaller?? )

P.P.S. This example if for total input noise (including resistor noise). You probably meant noise of amp. itself - In this case simply dont add 290 K in above examples (but result wll be similar duet to high the noise figure).

P.P.P.S. But in datasheet formula they refer to "Total output noise power (referred to input)" so you should add that resistor noise as shown above. So specified noise includes noise of 50 Ohm resistor + equivalent input noise of amplifier.

[mio83]

Can you say the power value you are measuring with SA when input of amp is loaded with 50 Ohm (not shorted)? (and RBW setting). I will try the calculation.

NF = 12 dB (or lineary F = 10^(12/10) = 15., corresponds to Te = (15.8-1)*290= 4300 Kelvins. I dont understand where You get -271.

Thanks for helping!
I have made two pictures:

• SA (RBW=1Mhz, 0db of internal attenuation, 100avg) without anything at its input. Pretty flat noise floor at -87dBm 
• SA (same settings) connected to the output of the amplifier (the amplifier has a 50 ohm resistor at its input): It's pretty bumpy, but at 75Mhz it reads -64dbM, which is exactly the same reading I had measured (see first post of this thread) with the input of the amplifier shorted.

pic1: https://www.dropbox.com/s/x56g7kpeg1fk43f/amp0.png?dl=0

pic2: https://www.dropbox.com/s/07850rj8b3s1wc2/amp1.png?dl=0


NOTE: The amplifier has a very flat +40dBm amplification, when it actually amplifies something (not thermal noise), with +/-1 db across the 150Mhz (datasheet and verified).


EDIT: In the pictures it is visible that I am using "SP detection" mode, that is (from manual): In sample mode, the instantaneous signal value at the present display point is placed in memory. Sample detection is activated automatically for noise level markers, during video averaging, and for FFT measurements.

[yl3akb]

Thanks for the graphs. So you say, that with shorted input you get flatter response?
Well, You have to understand that amplifier will amplify noise and signal exactly the same way. In this case the signal is the noise generated by resistor. Maybe the resistor is bad?

But anyway, lets assume you got -64dBm with RBW=1MHz and G = 40dB.
We can calculate total input noise spectral density: P = -64 - 10*log10(1e6) - 40 =  -164 dBm/Hz => or at BW=150MHz: P = -164+10*log10(150e6) = -82 dBm.
Converting to linear power in Watts and then voltage: P_lin = 10^(P_log-30)/10 = 6 pW, and you know what, here I got the same value as You in beginning: U = sqrt(P_lin*50) = 17uV. Then I realized that the formula for noise voltage I used is wrong. The correct would be (in matched 50 Ohm load case): P_lin = U^2/(4*R) => U = sqrt(P_lin*4*R) = 34 uV, which is much closer to theoretical 40 uV. Sorry for the confusion, learning every day myself.

[yl3akb]

Corrected my previous noise voltage calculation as well

[xmo]

Have you tried using the Marker Noise function of your spectrum analyzer?

Also, HP-A-K have an application note you may find helpful.  You can probably find more than one version on the internet.  Sometimes earlier versions are more helpful when newer documents cover newer instruments.

Agilent Application Note 1303  "Spectrum and Signal Analyzer Measurements and Noise"

[mio83]

But anyway, lets assume you got -64dBm with RBW=1MHz and G = 40dB.
We can calculate total input noise spectral density: P = -64 - 10*log10(1e6) - 40 =  -164 dBm/Hz => or at BW=150MHz: P = -164+10*log10(150e6) = -82 dBm.
Converting to linear power in Watts and then voltage: P_lin = 10^(P_log-30)/10 = 6 pW, and you know what, here I got the same value as You in beginning: U = sqrt(P_lin*50) = 17uV. Then I realized that the formula for noise voltage I used is wrong. The correct would be (in matched 50 Ohm load case): P_lin = U^2/(4*R) => U = sqrt(P_lin*4*R) = 34 uV, which is much closer to theoretical 40 uV. Sorry for the confusion, learning every day myself.

Thanks. Could you explain why the 4R factor in your equation?

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U^2/4R = P_lin?

I don't see where it comes from.

It seems to me like you are confirming the correctness of my calculations for:
1) Power density in dBm/hz  (P)
2) Power density (linear) in Watt/hz (P_lin)

but you are saying the formula for calculating the equivalent nV/sqrt(Hz) from P_lin is:

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U^2/4R = P_lin?

Note: this link from Keysight https://www.keysight.com/main/editorial.jspx?ckey=1000001959:epsg:faq&id=1000001959:epsg:faq&nid=-11143.0.00&lc=fre&cc=FR

suggests instead to using the equation:

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U^2/R = P_lin?

we both used at first.

[yl3akb]

At first I was thinking (And You as well, I guess) of voltage drop on load resistor on which we are evaluating dissipated power, and nothing wrong there. But it turns out that noise voltage is defined as voltage of ideal voltage source (U = sqrt(4*k*T*R)=2*sqrt(k*T*R), see https://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise ), which is in series with noiseless source resistor. When we load this simple circuit with matched load, this voltage is divided equally and on load resistor we get voltage of U_load = U/2  = sqrt(k*T*R). Since we now dissipated power density from measurement using SA (P = k*T) on this resistor and want to obtain the U, calculation would be: U/2 = sqrt(P*R) => U = 2*sqrt(P*R). The key here was this missing factor of '2'. Of course, we can rearrange it to get P = U^2/(4*R).

Its hard to comment on the Keysight formula, there is too little details.

[yl3akb]

Here is a good illustration of the above:
https://www.analog.com/media/en/training-seminars/tutorials/MT-052.pdf

[mio83]

Thank you very much for this detailed answer.

I am not entirely certain there is a wide agreement about this.
For example, here: https://www.harald-rosenfeldt.de/2018/08/19/measuring-rf-noise-with-an-rf-spectrum-analyzer-or-how-to-convert-dbm-into-v-sqrthz/

it talks about estimating the noise floor of a SA  (with input terminated by a 50ohm load).
The data it shows is valid for RBW=10Khz.
The relation used to move from power density W/1Hz to nV/sqrt(Hz) is:

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P_lin with BW=1Hz = V^2/50 with sqrt(1Hz)=1Hz
as suggested by the Keysight page linked earlier.

Anyway, thanks a lot again for the time spent helping me with this.
Perhaps others can add their knowledge too. 

[mio83]

Have you tried using the Marker Noise function of your spectrum analyzer?
(Attachment Link)
Also, HP-A-K have an application note you may find helpful.  You can probably find more than one version on the internet.  Sometimes earlier versions are more helpful when newer documents cover newer instruments.

Agilent Application Note 1303  "Spectrum and Signal Analyzer Measurements and Noise"

Thanks a lot! I didn't know about this functionality of my HP 8591E!
I will read the application note carefully.

[mio83]

Following xmo's suggestion I used the built-in Marker Noise functionality of my HP 8591E to do the measurements.

Here are the results (+40gain already factored in the SA settings) with the input of ampifier terminated with 50ohm (however I noticed no differences at all when leaving the input open: in particular the noise floor is still bumpy, so it's not generated by the imperfections of the 50ohm terminator).

1) Noise density in dBm/Hz :

2) Noise density in nV/1Hz: 

I think the developers of the HP used the right math to do the conversions, so I expect this to be correct.

Conclusion: it looks like indeed this >55years old unit has an average noise density (it varies across the 150Mhz spectrum) of -160dBm/Hz or, equivalently, about 2nV/sqrt(Hz). Also, it have a very low SWR and very good flat response of +40db (+/- 1 across all the spectrum).

 

[TimFox]

The noise figure (log of noise factor) depends on the source impedance.  In general, a practical amplifier has an optimum source impedance (which may be complex) that gives the lowest noise figure.  If one has a transformer, or tunable transformer-like network, between the actual source (with its impedance) and the amplifier input, it may be possible to achieve this optimal value.
For RF amplifiers, the "noise match" that gives the best noise figure (and, therefore, best signal-to-noise ratio) is not, in general, equal to the impedance that gives the maximum signal output.
A good noise-figure meter usually has a switched diode noise source that has a good 50 ohm output impedance, and can be used to tune an amplifier for best noise figure from 50 ohms.  \
In my MRI days, we used an -hp- noise figure meter with a wide-range pi-network box (with high-Q components) to find the noise match:  after getting the best noise figure, we terminated the pi-network with 50 ohms (where the noise diode used to be) and measured the impedance that presented at the output (where the DUT used to be).

[mio83]

The noise figure (log of noise factor) depends on the source impedance.  In general, a practical amplifier has an optimum source impedance (which may be complex) that gives the lowest noise figure.  If one has a transformer, or tunable transformer-like network, between the actual source (with its impedance) and the amplifier input, it may be possible to achieve this optimal value.
For RF amplifiers, the "noise match" that gives the best noise figure (and, therefore, best signal-to-noise ratio) is not, in general, equal to the impedance that gives the maximum signal output.
A good noise-figure meter usually has a switched diode noise source that has a good 50 ohm output impedance, and can be used to tune an amplifier for best noise figure from 50 ohms.  \
In my MRI days, we used an -hp- noise figure meter with a wide-range pi-network box (with high-Q components) to find the noise match:  after getting the best noise figure, we terminated the pi-network with 50 ohms (where the noise diode used to be) and measured the impedance that presented at the output (where the DUT used to be).

Thanks for the informative answer. However, if I understand it correctly, the measurement setup I made can only produce higher noise density power results (because the match is not ideal) than what could be obtained with matching networks as you suggested.

I am still a bit surprised by my measurements. Again (see pictures posted above) with a 50ohm terminator at the input of the amplifier, my HP 8591E measures about -160dbm/Hz of noise density power (calculations done entirely by the SA with the Noise Marker functionality, so I trust them).

Expressed in Watts/Hz, this is exactly 1e-19 Watts (https://www.wolframalpha.com/input/?i=-160+%3D+10*log_10%28p%2F0.001%29)

From this I can (using the formula from https://en.wikipedia.org/wiki/Noise_temperature) calculate the Noise Temperature T=24.1  (https://www.wolframalpha.com/input/?i=1e-19+%3D+t*300*k%2C+k%3D1.381*10%5E%28%E2%88%9223%29)

From this (again using the equation in https://www.wolframalpha.com/input/?i=-160+%3D+10*log_10%28p%2F0.001%29) I obtain a Noise Factor of 1.08 (https://www.wolframalpha.com/input/?i=%28300%2B24.1%29%2F300).

Taking 10*log_10 of the Noise Factor, I obtain a Noise Figure of 0.33.


All these numbers seem quite impressive.

Again, is this possible? Granted this HP 461A amplifier did cost >300 USD in 1964 (more than 2k of todays USD, accordingly with https://www.usinflationcalculator.com/)... but is it possible that it is really so good, even compared to today's amplifiers?



Anyway, perhaps it's more useful to forget about my concrete measurements and the HP 461A, and simply ask:

GENERAL QUESTION: If I read in a datasheet that a LNA has noise figure of 0.33, how much should I expect (from purely mathematical calculations) when I plug a 50ohm resistor at the amplifier's input, and measure the noise power density (in dBm/Hz) of its output with a modern spectrum analyzer?

Thanks again!

EDIT: I hope I don't look/sound pedantic with all these numbers. I bought this amplifier precisely to use it to learn these things, and I am simply having a lot of fun with all of this