# Bandwidth, Output offset voltage of Inverting Op amp.

Discussion in 'Homework Help' started by hiphop, Oct 12, 2016.

1. ### hiphop Thread Starter New Member

Oct 10, 2016
5
0

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 .

2. ### MrAl Well-Known Member

Jun 17, 2014
2,179
431
Hello there,

The key to answering this question as are many others is to analyze the circuit and try to think of what it is that you have to show, then try to invent a way to show that using the circuit analysis to prove it.

I looks like the book is 'more' correct about the formula:
fo*(1+A*B)

I have another formula but i'll have to check that later.

LATER: Redid my formula and guess what i got? Yeah: f=fo*(1+A*B)
That was using the procedure outlined below.

Because this is the homework section, to prove the above you should do the following:
1. Create an internal frequency dependent gain stage, calculate a value for C given R=1 for that stage, and calculate the transfer function of that, call it Hs, when f is equal to the cutoff frequency fo (so the output is 3db down) and the non frequency dependent gain is A. You can use a voltage controlled current source for this part. Ignore the output impedance by using a buffer with a gain of 1 to buffer the output (which simply means you dont have to include the output impedance in the gain formula with the external resistances R1 and Rf ... they are almost always much higher anyway).
2. Using that and the basic op amp formula Vout=(vp-vn)*Hs, calculate the output with K and B factors included.
3. Calculate the gain with a particular set of Rf and R1 and see that it is reasonable. Compare the result to the result using fo*(1+A*B). For a simple example, you can use Rf=R1 and that may be a good example because it is a low gain. You can also try Rf=10*R1 for example. If you care to, you can solve for the formula itself.

As a side note, note that if we start with fo*(1+A*B)/K we get a frequency that is much too low for an externally set gain of say 10 (roughly 10 times too low).

If you need more help with this we can go farther later. We can look at the output offset later too.

Also another side note, we often estimate the closed loop gain using funity/Fbw. So if we have unity gain at Fu and we want to operate at Fo, the max gain would be Fu/Fo. That is for gains that are not too low.
In your assignment however they are looking for a better answer so you might want to ignore this for now.

Last edited: Oct 13, 2016