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https://designers-guide.org/forum/YaBB.pl Design >> RF Design >> Noise De-Embedding https://designers-guide.org/forum/YaBB.pl?num=1161866031 Message started by aaron_do on Oct 26th, 2006, 5:33am |
Title: Noise De-Embedding Post by aaron_do on Oct 26th, 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 27th, 2006, 2:17am aaron_do wrote on Oct 26th, 2006, 5:33am:
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 27th, 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 27th, 2006, 2:50am aaron_do wrote on Oct 27th, 2006, 2:31am:
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 27th, 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 30th, 2006, 10:42am hi aaron, i just had a chance to look at your equations again aaron_do wrote on Oct 27th, 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 27th, 2006, 2:31am:
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 30th, 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 31st, 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 1st, 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 1st, 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 1st, 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 1st, 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 1st, 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 |
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