The Designer's Guide Community Forum
https://designers-guide.org/forum/YaBB.pl Simulators >> Circuit Simulators >> PSRR and CMRR for fully differential amp https://designers-guide.org/forum/YaBB.pl?num=1073321544 Message started by Winson on Jan 5th, 2004, 8:52am |
Title: PSRR and CMRR for fully differential amp Post by Winson on Jan 5th, 2004, 8:52am hi, Does anyone know how to setup up for simulations measurements for fully differential op amp? to measure: 1.PSRR 2.CMRR Thanks. |
Title: Re: PSRR and CMRR for fully differential amp Post by Mighty Mouse on Jan 5th, 2004, 9:08am Winson, You would use the test bench described in A test bench for differential circuits, which can be found at http://www.designers-guide.com/Analysis/. Then you would measure PSRR and CMRR in the same way you would for a single-ended amplifier. In particular, if you were using Spectre, you would run an XF analysis with the differential output signal being identified to the XF analysis as the output of the circuit. Then you would measure the gain from the differential input to the output (Adm), from the common-mode input to the output (Acm), and from the power supply to the output (Aps). They all can be measured in one XF analysis. Then, if I remember correctly, PSRR = Aps/Adm CMRR = Acm/Adm |
Title: Re: PSRR and CMRR for fully differential amp Post by wstyeah on Jan 5th, 2004, 7:21pm hi mighty mouse.. thank you for your replied. ;) but what i am thinking is.. the PSRR of fully differential is PSRR= (Adm)/(Asp - Asn) ? same for CMRR, CMRR = (Adm)/(Acmp -Acmn) ? Asp is supply gian for positive output terminal and so on. |
Title: Re: PSRR and CMRR for fully differential amp Post by Mighty Mouse on Jan 6th, 2004, 12:18am The beauty of this test bench is that it completely separates the differential and common mode signals at both the input and the output, so there is no need reconstruct the desired signal by addition or subtraction. This approach eliminates a lot of confusion. Having said that, I believe your definition is the reciprocal of mine, which I believe is wrong. |
Title: Re: PSRR and CMRR for fully differential amp Post by wstyeah on Jan 7th, 2004, 7:34am hi mighty mouse.. thanks again for replying :) erm.. do you mind to draw out the configuration for me? Thanks first. Warmest Regards, Winson |
Title: Re: PSRR and CMRR for fully differential amp Post by Mighty Mouse on Jan 7th, 2004, 10:31am Winson, You would use the test bench as shown in Figure 3 (or 5) in A Test Bench for Differential Circuits (though you probably do not need the resistors). This requires that you get your hands on an ideal balun. You can find one in analog lib in the more recent versions of Artist. If you cannot find it, you can create one as described in the paper (use the Spectre version rather than the Spice version if you can). Finally, if you are using Spectre, perform an xf analysis with the differential output signal (d from the output balun) passed into the xf analysis to define the output for the analysis. The xf analysis will directly compute Adm, Acm, and Aps as the transfer function from Vcm, Vcm, and Vdd to the declared output. |
Title: Re: PSRR and CMRR for fully differential amp Post by jeffyan on Feb 13th, 2004, 10:51pm For fully differential op amp, to get the right results of PSRR,CMRR, i think, the mismatchings should be added to differential pairs,right?? according to the opinion of Mr.Mouse, PSRR and CMRR of fully differential amp are identical to single-ended amp's respectively,right? ??? |
Title: Re: PSRR and CMRR for fully differential amp Post by Mighty Mouse on Feb 14th, 2004, 3:49pm Yes, generally you need to model mismatch in differential circuits to get any meaningful measurements for PSRR and CMRR. If you use the balun to separate out the differential and common-mode signals, then the measurement prodedure for measuring differential and common-mode PSRR and CMRR are the same (but of course the results will be different). |
Title: Re: PSRR and CMRR for fully differential amp Post by erikwanta on Feb 14th, 2004, 11:25pm Run spectre XF analysis and plot the following: CMRR db20(1/DATA("/Vic/PLUS" "xf-xf")) PSRR- db20(1/DATA("/Vn/PLUS" "xf-xf")) PSRR+ db20(1/DATA("/Vp/MINUS" "xf-xf")) Where Vic is the input common mode voltage, Vn is the negative supply, and Vp is the positive supply. |
Title: Re: PSRR and CMRR for fully differential amp Post by jeffyan on Feb 14th, 2004, 11:31pm i don't know whether it is right: for a differential amp op,the value of CMRR or PSRR, is infinite if the circuit is symmetrical perfectly(if no mismatching added)! because we are concerning the differential signal. so i think the definition of PSRR&CMRR are PSRR= (Adm)/(Asp - Asn) and CMRR = (Adm)/(Acmp - Acmn), as same as Mr wstyeah's. but they are different to the results we get from the new test bench. need mighty mouse's help! best regards jeff. |
Title: Re: PSRR and CMRR for fully differential amp Post by Mighty Mouse on Feb 15th, 2004, 11:59am Jeff, You are correct, the rejection becomes infinite as the undesired gain (either the common mode gain or the gain from the supplies to the output) goes to zero. Therefore PSRR+ = Adm / Add PSRR- = Adm / Ass CMRR = Adm / Acm So what this says is that there are two PSRRs, one for the positive supply and the other for the negative supply. You would not combine them unless you had reason to believe that the interferrence from the supplies would be balanced relative to ground, which seems unlikely. Referring to Figure 5 in http://www.designers-guide.com/Analysis/diff.pdf, if an XP analysis were performed with the output being Vod, then these metrics are calculated using PSRR+ = Vid / Vdd PSRR- = Vid / Vss CMRR = Vid / Vic (Here I am assuming that Vdd and Vss represent the positive and negative power supplies and that Vxx represents the gain from source xx to the output). Erik was on the right track with his formulas, but he gave the gain from the supplies to the output, not the rejection ratio. |
Title: Re: PSRR and CMRR for fully differential amp Post by richloo on Feb 6th, 2005, 11:11pm erikwanta wrote on Feb 14th, 2004, 11:25pm:
I've tried the analysis sweep until GHz, but i only get a straight line with -6.4K dB. I wonder this is a setup problem. can u teach me how to setup this analysis? Thanks in advance! |
Title: Re: PSRR and CMRR for fully differential amp Post by lurx on Sep 4th, 2008, 9:57pm it should be PSRR+ = Vdd/Vid PSRR- = Vss/Vid CMRR = Vic/Vid |
The Designer's Guide Community Forum » Powered by YaBB 2.2.2! YaBB © 2000-2008. All Rights Reserved. |