If we have ,
\(f_o=\)break frequency of the op-amp
\(f_F=\)bandwidth with negative feedback(inverting amplifier)
\(G_F=\)gain of inverting amplifier =\(\frac{AK}{1+AB}\)
\(UGB=\)unity gain bandwidth
where K=\(\frac{R_F}{R_F+R_1}{\) and B=\(\frac{R_1}{R_1+R_F}\)
Then,since the gain-bandwidth product of the single break frequency op amp is constant.
UGB=\(Af_o=G_Ff_F\).
Then \(f_F\) should be\(\frac{Af_o}{G_F}=\frac{f_o(1+AB)}{K}\)
but the book says \(f_F=f_o(1+AB)\).
Also how it is that we can use the open loop voltage gain vs frequency curve to find the bandwidth for the closed loop case , shouldn't we use a closed loop voltage gain vs frequency curve (it it exists) ?
Also if \(v_{ooT}=\)Output offset voltage of the inverting op amp
and ±\(V_{sat}=\)saturation voltages
then shouldn't \(V_{ooT}\) be equal to \(\text{(input offset voltage)}G_F\), but our book says that it is \(V_{ooT}=\frac{V_{sat}}{1+AB}\).
I understand that in the case of open loop op amp the output offset voltage is so high that it is ±\(V_{sat}\) , but in the case of closed loop inverting op amp it can be calculated using closed loop gain , since now the gain is not so high .
Please , clarify these two doubts .