NF from pnoise wrong?
I have build the behavioral model of a simple amplifier to have a NF of 3dB. I can verify this with noise analysis and the result is NF=3dB.
Now I add an ideal passive switch, driven by an ideal square wave and enable pss+pnoise to obtain NFdsb (mixer etc does NOT contribute noise).
If I set the "number of maximum sidebands" to 1 I get again the expected result: 3dB. However, when I increase this number, it gets larger and larger and converges to something like 3.965 dB (e.g. for 50 sidebands).
I do understand noise folding. However, I do not see why the noise figure should increase if the mixer does not contribute any noise. Because the noise figure is defined as "(total noise at the output)/(noise caused by the input source)". So if there would be noise folding it would also affect the input noise and hence cancel out.
Simple example: Source resistance Rs, input resistance of amplifier matched to Rs, amplifier gain A, output referred noise of the amplifier vo^2. Noise folding factor of the passive mixer (due to harmonics): beta. NF of amplifier (linear): nfamp. Total noise figure: nftot.
As I said, I set vo^2 in a way such that the NF of the amplifier has a certain value (e.g. 3dB):
nfamp = 1 + vo^2/(kT*Rs*A^2) --> vo^2 = (nfamp-1)*kT*Rs*A^2
Noise at output due to input: kT*Rs*A^2*beta
Noise at output due to amplifier noise: vo^2*beta
nftot = (kT*Rs*A^2*beta + vo^2*beta)/(kT*Rs*A^2*beta) = 1 + vo^2*beta/(kT*Rs*A^2*beta)
= 1 + (nfamp-1)*kT*Rs*A^2/(kT*Rs*A^2) = 1 + nfamp - 1 = nfamp
So independent of the noise folding the total noise figure should match the noise figure of the amplifier ... even for pss+pnoise.
Where am I wrong?
I remember having similar struggles with SPECTRE's NF calculations on N-path passive mixers similar to this work. The conversion loss (and thus the minimum DSB NF) of a simple two-path ideal mixer is in fact 3.92 dB (when the impedances on the RF and IF ports are broadband), so if anything your results are too optimistic. "Noise folding" is one way of explaining it, but I like the way Andrews explains it in terms of harmonic reradiation. They represent the effect of the mixer by introducing a virtual shunt impedance on the IF port which leads to a very simple LTI model of the mixer with the correct gain and NF. Ultimately I was able to get simulation results matching their theory very well (using DSB NF, not IEEE NF).
Hi, oh, I am aware of this paper (tried it out some time ago).
But I am not sure if this is my point. My point is: No matter what happens after the amplifier (whether it's a mixer or amplifier - as long as it's noiseless) it shouldn't do anything to the NF because it would apply to the source noise AND the amplifier noise and hence cancel out.
I went one step ahead and made the entire schematic noiseless. The noise summary shows me 100% for rn in the source port (source noise). Still the NFdsb that I get from pnoise is something like 0.85dB ...
Only if the amplifier is before the mixer (and it's gain is very large). I assumed you were talking about the mixer being before the LNA?
When is the mixer before the LNA? (except maybe for mixer-first but then I wouldn't call it LNA but more baseband amplfier).
In any case without the mixer (where I get NF=3dB as expected) I have
PORT -> Balun -> Amplifier -> PORT
With the (ideal) mixer and pss/pnoise (where I get NF=3dB only for sidebands=1 and up to 3.9 otherwise) I have
PORT -> Balun -> Amplifier -> Mixer -> PORT
Okay then it's possible that your ~4 dB result is accurate according to the friis formula. Can't say for sure since you didn't give the gain of the amplifier. Have you tried simulating the gain and NF of the mixer alone? If you do you should find that it does have conversion loss and has NF>0, even when it is ideal.
Correct, all the noise is coming from Rn, but in the case of a mixer that doesn't mean F=1, because of harmonic reradiation/folding/whatever you like to call it. That is, according to the standard definition of F. The IEEE definition should give F=1 as you expect. But nobody uses the IEEE definition except for academic purposes, it's the "normal" F which matters.
See http://www.designers-guide.org/Forum...num=1162399761
http://www.designers-guide.org/Forum...num=1239925339
Indeed, NFieee gives me ~0 dB NF for a noiseless circuit.
However, I am still confused. Can you help me in deriving this result (NFdsb=0.887dB for completely noiseless circuit, NFdsb=3.9 dB for the case with the 3dB NF amplifier).
Since the NF is defined as "total output noise"/"noise at output DUE to input noise" I am confused how this can be more than 1 when there is only the input source because then "total output noise"="noise at output DUE to input noise". (maybe it's because a mixer is a nonlinear system and linearity breaks which is somehow an assumption?)
Anyway, let's start with the most simple case: Ideal source with Rs and a hypothetical mixer with matched input resistance and conversion gain 1.
The noise power at the input of the mixer is then k*T*Rs. What is it after the mixer? Say, due to downconversion it is gamma*k*T*Rs. But again, "total output noise"=gamma*k*T*Rs and "noise at output DUE to input noise"=gamma*k*T*Rs, so NF=0dB!
So even if not, the difference I am seeing (0.887dB) corresponds to a factor of 1.2267. Where does it come from? It is not 2/pi, pi/2, 2/pi^2, pi^2/2 etc. .... but it should be something that's related to the summation of the Fourier coefficients of the square wave (i.e., 2/pi*(1+1/3+1/5+...) or similar).
What I wanted to add: The analysis of the Andrews paper is not entirely appropriate for this case because it assumes sampling mixing (i.e. a huge capacitance).
"total output noise"/"noise at output DUE to input noise" is basically the IEEE definition of NF, not the standard definition (which gives the same result as the y factor method). The standard definition can be more specifically described as "total output noise"/"noise at output DUE to input noise in the bandwidth of interest." The presence of a mixer means that noise from outside the bandwidth of interest is converted to your IF port, causing the effective source noise to actually increases above 4KTRs.
It has nothing to do with nonlinearity, since a passive mixer is ideally linear.
All mixers are "sampling" mixers. The capacitance at the IF port is just there to determine the IF bandwidth and suppress oob blockers. The presence of the capacitance doesn't affect the baseband NF.
Did you try simulating the mixer alone?
Ok, thanks I get the idea now.
Yes, as written I simulated the mixer only and got 0.887dB instead of 0dB.
So what I am looking now is an analytical way to derive this result. It must be something like:
nf = 4*kT*Rs*alpha^2/(4*kT*Rs)
or for amplifier + mixer:
nf = (4*kT*Rs*alpha^2*A^2 + alpha^2*voamp^2)/(4*kT*Rs*A^2)
where alpha is an aliasing factor describing the noise folding due to mixing. I.e.all the noise appearing before the mixer would have the aliasing factor included whereas the denominator would not have it included. A square wave has Fourier coefficients 4/pi (1+1/3+1/5+...) and I think it should be related to that - but it is not.
I forget which paper goes through the derivations of NF for passive mixers, I know it can be found in the references of the Andrews papers... will search tomorrow.
Analog devices has a good derivation of the voltage gain of an ideal N=2 passive mixer. They gloss over the derivation of NF, but it can be found it's equal to the conversion loss, as expected. As for your 0.887dB I can't really guess without seeing the specific design.
I went ahead and found the derivation, out of boredom. Given an input noise spectral density Nin, the total noise on the output and the output noise contributed from the source at the RF frequency of interest will be:
This assuming that the LO and RF frequencies are different (not a direct conversion receiver), which is why the coefficients are 2/(pi*n) instead of 2/(pi*n). The sum uses positive and negative n to account for image frequencies. So as you can see, the total output noise is identical to the input noise (as expected for a lossless mixer), but only a fraction of that output noise is contributed by the bandwidth of interest. The ratio of the two is 2.467, so NFssb=3.92dB.
This is the case when Nin is constant vs frequency. If you have filters on the RF port, then the results change. But NFssb is defined as being with a broadband source.