# operational amplifiers

#### mik3

Joined Feb 4, 2008
4,846
The gain of the first op amp is:

G=R2/(R1//C)

it is easier to transform parallel resistances to conductances and add them

G=R2/[(1/R1)+(jωC)]=(R1*R2)/(1+jω/ωh)

To obtain the Bode plot you have to transform to this:

G=(ωh*R1*R2)/(ωh+jω)

#### zdzislavv

Joined Jun 5, 2009
22
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#### zdzislavv

Joined Jun 5, 2009
22
http://images50.fotosik.pl/144/2082963a176f7f0c.jpg
I've got still some difficulties with understanding it. I remember from the classes that for a) inverting op-amp Uin we've got two resistors and Uin connected to minus, b) noninverting op-amp resistors are connected to minus and Uin to plus. I check it on Wikipedia and that's the same, ok. From the classes I also remember formulas for a) inverting k=-R2/R1, b) noninverting k=1+R2/R1. I check on Wikipedia and I've got for a) G(closed loop)=Uout/Uin=-Rf/Rin, b) K=R1/(R1+R2), Uout/Uin=G/(1+G*K)=1/[(1/G)+K], G(large)=>Uout/Uin->1/K=1+R2/R1. So it looks like for a) inverting op-amp the formula which I know, indicated by k is the same as formula for G. So what is the difference between k and G? And for b) formula which I know for k is the same as formula for 1/k. So what is it finally k or 1/k? And as I understand in this exercise the first op-amp is non-inverting, isn't it?
Greetings!

#### Jony130

Joined Feb 17, 2009
4,930
In case b.
Uout/Uin=G/(1+G*K)
G is a open-loop gain of a opamp ( forward gain ).
K - feedback factor ( feedback gain ), sometime we use letter "β" for feedback gain.

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#### zdzislavv

Joined Jun 5, 2009
22
http://images45.fotosik.pl/145/80e562c34fc07717.jpg
I still don't understand this whole calculation with G and whether 1+R2/R1 is K or 1/K for non-inverting amplifier. Is there any way to calculate it only with the use of K (as in that second picture in my first post)?
Greetings!

#### Jony130

Joined Feb 17, 2009
4,930
In your example the G (open-loop gain) is like for all ideal opamp
equal G=infinity
Uout/Uin=G/(1+G*K)
For G=infinity
Uout/Uin=1/K
K for non-inverting amp
K=R1/(R1+R2)
thus,
Uout/Uin=1/K=(R1+R2)/R1=1+(R2/R1)

#### zdzislavv

Joined Jun 5, 2009
22
Thanks! I just asked college from studies and this is what he suggested me:
I understand everything up to the point where he calculates logarithms. 1) Why is it 20log(R1-R2), not 20log[(R2+R1)/R1]? 2) What is this 20log[jωC(R2R1/(R2+R1))]? It looks like jω*(1/ω1) so what is it for? 3) Why is there 20logR1?
Greetings!

#### zdzislavv

Joined Jun 5, 2009
22
OK, now I understand that we just add two logarithms 20log[(R2+R1)/R1] and 20log[jwC(R2R1/(R2+R1))]. For the second plot I have to take into account that for increase the value is equal to 45 degrees (3.14/4). How to write for the second graph equation with arctan (to write it in the symbolical form, not only as a graph)?
Greetings!

#### The Electrician

Joined Oct 9, 2007
2,724
One way to do it is to plot the arctan of the imaginary part of your transfer function divided by the real part of the transfer function.

In other words:

Phase = arctan(IM(tfr)/RE(tfr))

Be sure that you use an arctan function that returns its result in degrees, not radians. If you only have one that returns radians, then multiply the result by 180/Pi.