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Design >> RF Design >> Noise De-Embedding
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Message started by aaron_do on Oct 26^th, 2006, 5:33am

Title: Noise De-Embedding
Post by aaron_do on Oct 26^th, 2006, 5:33am

Hi all,

can someone tell me where i've gone wrong here. I have a amplifier whose load is an attenuator as such

VDD
|
|
<
<50 ohm
< 50 ohm
------vvvvv------GND
|
|
<
<975 ohm
<
|
|
gm.Vin

In other words rather than having a 1 kohm resistive load I have a attenuator which is used for output matching for measurement purposes. So the attenuator has a voltage gain of 1/40. I can measure the S21 and the NF, and I know the source impedance is 50 ohm. Therefore I believe I can write,

NFamplifier=NFtotal - Vattn^2/V50^2*S21

which should approximately equal to NF - 2/S21 if we assume 25<<975. Vattn^2 is the attenuator output noise voltage squared which I calculated as 4kT(1/25+1/975)*25*25 based on this diagram

__4kT/975__
| |
| 975 |
-------vvvvvvv--------------
| |
< |
< 4kT/25
< |
|____|
GND

The results are completely off so i've definately made an error in theory here. Any help is greatly appreciated.

thanks,
Aaron

Title: Re: Noise De-Embedding
Post by ACWWong on Oct 27^th, 2006, 2:17am

aaron_do wrote on Oct 26^th, 2006, 5:33am:

NFamplifier=NFtotal - Vattn^2/V50^2*S21

which should approximately equal to NF - 2/S21 if we assume 25<<975. Vattn^2 is the attenuator output noise voltage squared which I calculated as 4kT(1/25+1/975)*25*25

hi Aaron,
I don't follow at all what you are trying to do in either of your equations?
Can you explain what you mean by each term in each equation ?
cheers
aw

Title: Re: Noise De-Embedding
Post by aaron_do on Oct 27^th, 2006, 2:31am

Sorry about that...let me see. The entire circuit consists of an amplifier followed by an attenuator. The attenuator provides 50 ohm matching.

Let the entire circuit's output noise power in volts squared be Pa
Let the attenuator's output noise power in volts squared be Pb
Let the source (50 ohm) noise power at the input be Pc

For the entire circuit including the attenuator,

Noise Factor, F1 = (Pa + Pc*S21^2)/(Pc*S21^2) = (Pa/S21^2 + Pc)/Pc

This is the noise figure which i can measure. I can also measure the S21. Take the attenuator output noise Pd and refer it to the input of the circuit:

Pb/S21^2

Now we should be able to say that the NF of the amplifier minus the attenuator is

F2 = (Pa/S21^2 + Pc - Pb/S21^2)/Pc = F1 - Pb/Pc*S21^2

In other words refer all noise back to the input, and take the measured noise factor and subtract the (input refered attentuator noise)/(source noise). Is this correct?

Also, the attenuator noise can be approximated as 4kT*25 while the source noise is kT*50 so the noise figure is F1-2/S21

I must have made a mistake somewhere...any help would be great

thanks,
Aaron

Title: Re: Noise De-Embedding
Post by ACWWong on Oct 27^th, 2006, 2:50am

aaron_do wrote on Oct 27^th, 2006, 2:31am:

Take the attenuator output noise Pd and refer it to the input of the circuit:

Pb/S21^2

Now we should be able to say that the NF of the amplifier minus the attenuator is

As I see it the S21 (ie gain) you are using in this statement is not the same S21 as used to the complete load of 1K... shouldn't this S21 be 0.025*S21 n??
anyway i've got to dive out to a few meeting right now, so i'll have a closer look later...

Title: Re: Noise De-Embedding
Post by aaron_do on Oct 27^th, 2006, 6:11am

hmm...i don't think so...Pb is measured at the output of the attenuator not the input. Maybe i'll put this attenuator into simulation and check the output noise to double check.

Title: Re: Noise De-Embedding
Post by ACWWong on Oct 30^th, 2006, 10:42am

hi aaron,

i just had a chance to look at your equations again

aaron_do wrote on Oct 27^th, 2006, 2:31am:

Yes I agree... sorry for the aberation in my last post !
Noise Factor = 1+ (Circuit noise power in/ noise power in)
Circuit noise power in = Pa/S21^2 - Pb/S21^2
so yes F2=F1-Pb/(Pc*S21^2)

aaron_do wrote on Oct 27^th, 2006, 2:31am:

Also, the attenuator noise can be approximated as 4kT*25 while the source noise is kT*50 so the noise figure is F1-2/S21

The noise power only becomes an input noise voltage of kT*50 if the input is conjugately matched. Is your input conjugately matched ?
Any luck with working back from simulation results ?

cheers
aw

Title: Re: Noise De-Embedding
Post by aaron_do on Oct 30^th, 2006, 5:28pm

Hi aw,

thanks for your help. The matching is very good. S11 is better than -30 dB at the operating frequency. Also I have had no luck in working back with simulation results. Well I guess I'll keep trying. Thanks again for the help

Aaron

Title: Re: Noise De-Embedding
Post by aaron_do on Oct 31^st, 2006, 5:07am

Ok I kept trying and kept failing...maybe someone can help. Here's some simulation results based on the following figure:

VDD
|
|
<
< 71 ohm
< 50 ohm
------vvvvv------GND
|
|
<
< 1284 ohm
<
|
|
LNA : gm.Vin

Simulation at 2.5 GHz
NFtotal, Overall Noise Figure: 11.3 dB
NFlna, Noise Figure at LNA output node: 4.4 dB
S21: -15.95 dB
S22: -16.2 dB
S11: -14.02 dB (resonates at 2. 7 GHz but measured LNA resonates at 2.5 GHz)
Rin: 44.88 ohm
Rn : 176.5 ohm (by the way does anybody have the definition for this?)

The resistor network using 4kTR should contribute an output noise of 693 pV/sqrtHz, but when simulating the resistor network alone, it showed an output noise of 577 pV/sqrtHz. I'm not sure why. I simulated the same network using ideal resistors and the output noise was around 683 pV/sqrtHz which is close enough.

the equation I think that should work is,

Flna = Ftotal - AttenuatorOutputNoise/kT50*S21

From the above simulated results, the equation is not even close. In fact the answer is negative i think....any help at all is greatly appreciated.

thanks,
Aaron

Title: Re: Noise De-Embedding
Post by ACWWong on Nov 1^st, 2006, 4:01am

what happens if you do this is in simulation:

VDD
|
<
< 1313 ohm port 2
<
|
LNA=gm.vin

with the LNAmatched to 50 ohm port 1, what S21 (LnaAv should equal this S21+10lg(1313/50) and NF do you get ? Confirming these numbers should help towards understanding your initial result. Also i guess you are using ideal VDD (ie GND and VDD are the same for RF purposes with no impedance?)

cheers
aw

Title: Re: Noise De-Embedding
Post by aaron_do on Nov 1^st, 2006, 5:10am

Hi aw,

yeah i checked that...I used the transfer function analysis for the LNA output node. XF gain is 17.01 dB and S21 is -15.95 dB so the difference is 44 V/V times which is correct. Also I am using ideal sources.

This calculation has really got me stumped...

Its really late here but i thought i'd throw this out before i leave...just wondering if the fact that noise cannot be attenuated below the noise floor has any relevance here. i.e. since the gain is negative, the signal can be attenuated but the noise cannot...this might affect the calculation...just brain storming

thanks for the help,
Aaron

Title: Re: Noise De-Embedding
Post by ACWWong on Nov 1^st, 2006, 6:58am

The noise figure of an x dB attenuator is x dB...
so using Friis nois factor power cascaded lineup

NFtotal=NFLNA+[(NFattn-1)/S21LNA]

power gain S21LNA = 17dB =50
NFLNA=4.4dB=2.75
NFtotal=11.3dB=13.5

This then means the NFattn must be 27dB, so it power loss is 27dB.
This attnuator voltage gain is 29/(29+1284)= -33dB. The 6dB dicrpenacy is to mismatch between 1284 and 29.

what do you think to this train of thought ?
cheers
aw

Title: Re: Noise De-Embedding
Post by aaron_do on Nov 1^st, 2006, 5:26pm

Hi aw,

the numbers i'm giving you are simulation results :-) so there's no mismatch. good brain storming though and thanks for so much help

Aaron

Title: Re: Noise De-Embedding
Post by aaron_do on Nov 1^st, 2006, 7:29pm

hmmm...i think i got it...i think the LNA noise gets lost in the noise floor after the attenuator. Therefore it cannot be de-embedded...someone please tell me i'm wrong...haha

Aaron