 ### Network # non inverting amplifier

Feb 6, 2014
137
Looks o.k.

OpAmp theory is well described here.

You need to perform a small signal analysis (.AC in SPICE) of the circuit to see the amplitude and phase vs. frequency.
k.thanks for your reply #### LvW

Apr 12, 2014
604
here have modified my circuit..please check this
and i know that GBWP=Av * f
this is theoretical explanation
practically how the gain affect the bandwidth of an opamp?

This answer concerns the relation between gain and bandwidth for a non-inverting opamp-based amplifier:

In most cases, we can use a first order lowpass function for the real frequency dependence of the open-loop gain:
Aol=Ao/[1+s/wo]
Thus, based on the expression for the closed-loop gain (Hfb=Feedback factor)
Acl=Ao/(1+Hfb*Ao) we can write
Acl=1/[(1/Ao)+(s/(wo*Ao*Hfb))].

With 1/Ao<< Hfb and 1/Hfb=(1+R2/R1) we arrive (after suitable re-arranging) at
Acl=(1+R2/R1)[1/(1+s/(wo*Ao*Hfb))]
The expression in brackets again is a first order lowpass function having the corner frequency
w1=wo*Ao*Hfb.

Hence, due to negative feedback the bandwidth wo (open-loop gain) is enlarged by the factor Ao*Hfb.
More than that, we can write
wo*Ao=(w1/Hfb)=w1*(1+R2/R1)

This is the classical constant "Gain-Bandwidth" product (GBW=wo*Ao) which can be written also as
w1/wo=Ao/Acl(ideal)
with Acl(ideal)=1+R2/R1.

Last edited:

#### LvW

Apr 12, 2014
604
With 1/Ao<< Hfb and 1/Hfb=(1+R2/R1) we arrive (after suitable re-arranging) at
Acl=(1+R2/R1)[1/(1+s/(wo*Ao*Hfb))]
The expression in brackets again is a first order lowpass function having the corner frequency
w1=wo*Ao*Hfb.

I like to be a bit more detailed (only the middle part of my former answer) :

With 1/Ao<< Hfb and 1/Hfb=(1+R2/R1) we arrive (after suitable re-arranging) at
Acl=(1+R2/R1)[1/(1+s/(wo*Ao*Hfb))]
The first part is the IDEAL closed-loop gain Acl(ideal)=(1+R2/R1)
and the second part (expression in brackets) again is a first order lowpass function having the corner frequency
w1=wo*Ao*Hfb.

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