I used xf simulation to simulate my amp in open loop and closed loop(unity gain feedback) with the same DC bias condition
I used a huge series inductor and a huge capacitor to ground to block the AC path(I hope it is correct)
Compared to the open loop configuration, in a closed loop configuration, Vout/Vindiff, Vout/Vincm, Vout/Vss, Vout/Vdd, all these gains decrease around 50dB
In the end, when I calculate CMRR=Vout/Vindiff/(Vout/Vincm)
PSRR+=Vout/Vindiff/(Vout/Vdd)...
All these final values of CMRR, PSRR remain the same as the open loop results
I am a litte amazed by that
Can it be because Acl=Aol/(1+AolF), so in a closed loop configuration, all of the parameters reduced by the same factor (1+AolF), but the relative ratio of CMRR; PSRR remain the same
Re: Got the same CMRR, PSRR no matter open loop or closed lo
BillQ said:
... Compared to the open loop configuration, in a closed loop configuration, Vout/Vindiff, Vout/Vincm, Vout/Vss, Vout/Vdd, all these gains decrease around 50dB
In the end, when I calculate CMRR=Vout/Vindiff/(Vout/Vincm)
PSRR+=Vout/Vindiff/(Vout/Vdd)...
All these final values of CMRR, PSRR remain the same as the open loop results
I am a litte amazed by that
I'm not: If you decrease all these ratios by the same ratio, the ratios of these ratios must stay constant - math tells you the truth ;-) .
The mistake is your wrong definition of CMRR & PSRR !
CMRR ≠ Vincm/Vindiff but CMRR = Vincm/Voutcm = Vincm/Vout
PSRR+ ≠ Vdd/Vindiff but PSRR+ = Vdd/Vout (>1 , to get positive dB values)
All used values are ac values, of course.
The way how you calculate the CMRR resp. PSRR values must result in those constant values.
CMRR=Adm/Acm=(dVout/dVin)/(dVout/dVcm) ≠ dVcm/dVin
Why? Because dVout received from a dVin change is not the same as the dVout resulting from a dVcm change, so you cannot simply cancel them against each other! Similar pb. with the PSRR calculation.
So if you want to use the 'complicated definition' (s. above) of these rejection values, you must evaluate the respective differentials and their ratios separately. Then your calculation should give you the correct results - if your original equations are correct.
I didn't check them, because I prefer the more straightforward definitions which I gave you above.