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)
 

Thread Starter

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.
 
Last edited:

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)


 

Thread Starter

zdzislavv

Joined Jun 5, 2009
22
Thanks! I just asked college from studies and this is what he suggested me:
http://images43.fotosik.pl/145/f29e40dbdec6ada7.jpg
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!
 

Thread Starter

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!
 
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.
 
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