lna mixer noise figure

our chip integrates LNA, MIXER and LO blocks all on chip.
When I do mixer noise figure simulation using the standard setup(see attached figure) I find that noise figure is different if I remove the 75ohm resistor which is in parallel with the port9. but why?

My thinking is that there is no matching problem between LNA and mixer since the interface is on chip. So I should remove that 75ohm in the noise figure setup and just give port9's resistor a value equals to the output resistor of LNA.

any comments?

Thanks.

Designing for Optimum Noise Figure means not only to find the Optimum Source Impedance and the Source Reflection coefficient (Γs), but also is to determine the Optimum load-reflection coefficient (ΓL) needed to properly terminate the LNA?s output.

thanks vfone,

since the mixer and LNA are all on chip. Do we really need on chip stage-stage matching. There should be no matching problem at all. And I think only voltage gain rather than power gain counts for on-chip blocks.

Let me do some hypotesis:
-your Mixer is 'high impedance' (much larger than 50 or 75 Ohm).
-you get rougly 6 dB more gain if you remove the 75 Ohm res.

If this is the case, so the answer could be:
the spectre port is designed to transfer power into a load and not a voltage. So if you put a certain amplitude value to make, for example, an AC sim and do not terminate the port with proper impedance, you'll measure a different voltage at port's nets.
Noise figure calculations in spectre are done in this sequence: sim output noise and gain, transfer output noise at input (simply dividing by gain), calculate noise figure REFERRED TO PORT IMPEDANCE.
So the proper procedure in your case, could be (supposing you need 50 Ohm referred noise figure).
Terminate the port with 50 Ohm NOISELESS resistor (if the resistor is not noiseless the simulator will add its noise contribution to Mixer noise figure) and connect a vcvs with the gain=2 to Vp and Vn. Connect the mixer to output of vcvs.
Simulate and get the result.
Verify that gain obtained by noise figure simulation is the same of the one obtained with PSS and Pac.
Be careful that you need to have clear the difference between SSB and DSB noise figure in mixers.
I hope it can help.
Mazz

Thanks Mazz,

You are right the RF and LO input is AC coupled to the mixer(DC generated inside). I donot check the gain via RF port source. Rather I use voltage source to check the gain so I am not sure whether there is 6dB lose if I remove the 75ohm resistor.

I did find a 6dB difference in noise figure if I replace the vcvs egain from 2 to 1,i.e.,if egain=1, NF is 6dB higher. Could you give me an idea why you choose egain=2 rather than 1?

I am also try to verify the cadence noise figure calculation by hand,using the formula NF=10*log(1+Nout/Nsource). Nout(unit v^2) is the integration of the pnoise reported output refered noise((in unit v^2/Hz) over the bw(bandwidth) and Nsource=KT*75ohm*bw*(Av^2). The resulted hand calculated NF=18dB

It turns out if egain=1, simulated NF=24dB,
if egain=2, simulated NF=18dB
so egain=2 did make sense in NF calculation. why is this way?

Thanks!

The reason is that noise voltage generated by port (its input resistor indeed) is divided by two between resistor itself and the R4 (in your picture). So you need to double it to apply a correct (for simulator calculations) noise voltage to your mixer.
Be aware that you are fixing the neg pin of port to gnd and so your input signal could not be perfectly differential.
I suggest you to use 2 different vcvs (each with gain=1) and drive the first one (A)at positive input and the second one (B)at neg input. Connect together pos output of (B) and neg output of (A) the other two output to mixer input.
If you want more detail, you can go to http://www.designers-guide.org/, in the analysis pages you'll find an interesting guide to generate differential signals (with common mode control).
Did you put noise to zero in R4?
Another suggestion: never trust to simulators (they do good jobs, but maybe we do not use them well!).
Check every time that you are applying right signals and above all, do CALCULATION BEFORE SIMULATION.

I hope it can help.

Thanks Mazz,

I understand your point. But I still donot have a consistant conclusion. Say we assign a noise source with Vn^2=4kTRs*bw to the voltage source(Vs) inside the port, then notationally Vs=Vn. After the 1/2 voltage divider and egain=2, we have Vn as the input to the mixer. This is right. Then I can calculate NF=10log(1+Nout/Nsource) with Nsource=4KTRs*bw*(Av^2) since Vn^2=4kTRs*bw.
The resulted NF=24dB rather than 18dB.

Is there anything wrong with my reasoning?

Your calculations are correct.
But, forgive me, I do not understand why you don't have a conclusion.

If I have well understood:
-calculation from pnoise results to NF gives you 18 dB
-simulations using my way gives you same result (18 dB)

Let me suggest you to measure the input noise in a different test bench putting 1 or 2 in the vcvs and verify the 4kTRsBw formula, so you will be sure about the noise signal you are applying at mixer input. Be aware of the '4' coefficient in matched systems.

I hope it can help.
Mazz