Basic Op-Amp Question

[bittware]

Hello experts,
Can you explain which circuit has the more bandwidth? And why?

 

[ahmad_abdulghany]

I think both have same BW .. as the BW depends basically on the Feed back impedance .. if both feedbacks are the same the BW will be the same ..
you can use simple program to simulate this as a check .. if you need further help i am ready to help you ..

 

[electronics_kumar]

SINCE WE KNOW FOR A AMPLIFIER "GAIN X BANDWIDTH SHOULD BE A CONSTANT"

SO, WHAT I AM GOING TO SAY IS SINCE BOTH CONFIGURATION IS DIFFERING BY A VALUE OF ONE
GAIN= R1/R2
&
GAIN= (1+R1/R2)

CERTAINLY BANDWITH WILL ALSO DIFFER FROM EACH OTHER BY FACTOR OF ONE

 

[Kral]

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

 

[nxing]

These two config should are different types of feedback, aren't they?

 

[bittware]

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

Could you explain the conclusion in more detail?

 

[electronics_kumar]

yes KRAL could you explain it for one more time so that i can understand it clearly

 

[aryajur]

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

 

[jiangwp]

If the two structures have the same gain.

For inverting feedback:

A1(s)=Rf/Ri
β1=Ri/(Ri+Rf)

For noninverting feedback:
A2(s)=(Rf+Ri)/(Ri)
β2=Ri/(Ri+Rf)

so, the unity gain for feedback system is

UGB=A(s)•(β•ω)

corresponding the formula, the noninverting feedback has more larger bandwidth.