Kral said:
The inverting configuration has lower bandwidth because the feedback factor is smaller.
Let
. Rf = Feedback Resistor
. Ri = Input resitor for Inv configuration, Resistor to ground in Non-Inv configuration.
. a = feedback factor (fraction of the output that is fed back to the input).
.
For the Inverting configuration,
. Rf = 2 Ri
. a = Ri/(Rf + Ri) = Ri/(2Ri+Ri) = Ri/3Ri = 1/3
For the Non-Inverting configuration
. Rf = Ri
. a = Ri/(Rf + Ri) = Ri/(2Ri) = 1/2
.
At any frequency (above DC), the gain error due to limited open loop bandwidth will be higher for the inverting configuration than for the non-inverting configuration.
Regards,
Kral
For a Feedback system the Gain of the system is given as:
G = Aol /( 1 + β Aol) .... (1)
where Aol = Open loop Gain
β = Feedback Factor ( I don't know if Kral uses this to define his feedback factor, because if he does then his value of feedback factors seem wrong to me)
also Aol β is together called the loop gain, which can be calculated by breaking the loop at a point and injecting a test signal and calculating the gain of the signal at the other side of cut.
So assuming that the opamp has a dominant pole and its gain function can be describes as:
A(s) = Ao / ( 1 + s/wo) ....(2)
where Ao is the DC gain of the opamp and wo is the 3dB Bandwidth of the opamp.
FOr Non Inverting Configuration:
Aol = A(s)
β = Ri/(Ri + Rf) = Ri/2Ri = 1/2
For Inverting Configuration:
Aol = -A(s) Rf/(Ri + Rf) = -A(s) 2/3
β = -Ri/Rf = -1/2
So in general case Aol = K A(s) and β is same for both(only sign difference), K = 1 for Non Inverting and K = -2/3 for Inverting.
Put this and equation (2) in eqn (1) we have:
G = KA(s) /(1 + β K A(s))
=
K Ao /(1+s/wo)
------------------
1 + β K Ao/(1+s/wo)
on simplification we have:
G =
KAo/(1+βKAo)
------------------
1 + s/[wo(1+βKAo)]
So BW is: wo(1+βKAo)
therefore for Non Inverting case: BW = wo ( 1 + Ao/2)
for Inverting case : BW = wo(1 + Ao/3)
Hope this makes things clear