TIDA-00753: Error budgeting for Inverting Amplifier - Mistake? (Appendix)

Hi Josko,

Thanks for reaching out to us again:-

My Comments :-

a) At the begining the term "gain" has been used "Because the input full-scale voltage is very low, a gain stage is required to obtain a better SNR.". Afterwards, in table 3, Reference No. 3, the Ideal Gain has been calculated as attenuation (0.01999). Is this "Ideal Gain" closed-loop gain or something else?
-> The objective is to compute the gin error due to finite Aol and mismatch in resistor
The formula is G = Aol / (1 + Aol x +ß) were ß is feedback rattio or (1 / Bclosed)
Somehow the document has error in numerical values, your observation is right
Aopen is opamp finite opamp loop gain
Bopen (should have been Bclosed) reflecting feedback rattio with mismatch
Ideal Gain is without considering resistor mismatch
* Note this is only DC gain

b) REFERENCE NO2: Does Bopen include mismatch between resistors R1 and R2? ( e.g. R2 = 49.8501k (-0,1% tolerance) and R1=1,001k (+0,1% tolerance))
-> Yes

c) REFERENCE NO 3: If R1=1k and R2=49,9k, why the IdealGain is 0,0199 instead 49,9? I mean, if Acl=Aol/(1+Aol*Beta), and the Figure 9 suggests Beta=R1/R2, than ideal closed loop gain should be R2/R1.
-> The value shown seems to incorrect, actually it should have been 50.0025, somehow the (1/Gain) is not done in computation and ß and Blcosed got interchanged - We apologize for this error

d) For the inverting amplifier, ideal closed loop gain will be -R2/R1, and that is 49,9 (if Aol is endless). Since Aol is not endless, the real closed loop gain for inv. amp will be:
Aclosed=-[Aol*R2/(R1+R2)]/[1+Aol*R1/(R1+R2)]=49,73 (tolerance of Resistors is not incluede).
(References: m.eet.com/.../24987-56190.pdf, www.ti.com/.../slyt374.pdf)
It means that Gain error due to finite Aol (15000 for LMV321) will be 0,34%. If mismatch of 0,1% resistor will be included, it will be 0,54%.
-> Acl(DC) = Aopen x (( (R2+R1)/R2 + Aopen x R1/R2 ) simplified form (R2+R1)/R2 ~ R2 hence 1 was mentioned in line item a

Any can equation in document can be used

e)REFERENCE 11: Is the input bias error calculated here as RTO value? I am asking due to noise gain (1+R1/R2) in the expression?
-> Use below equation for bias current error
Vo = - [ R3 x Ib+ + (((R3 x R2) x Ib+)/R1) - (R2 x Ib-)] can take only positive value to compute error

Regards,
Srinivasan