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OP-Amplifier question - uni. problem

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Perky123

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Hello guys,

this is the second thread I am posting in this great forum, hoping that you will help me just like the last time few weeks ago (special thanks to Boncuk and eblc1388). This time - Op Amplifier diagrams.

I have the following arrangement and values given: (problem.jpg)

the questions are:

a) Draw the graph of U2
b) Draw the graph of U3
c) Draw the graph of U4

The solution is given in solution.jpg.
So my issues are:

a) more then clear that U2=U1. So - ok. (U1 graph is given!)
b) Using the formula for an "Integrator" (integrator.jpg) which according to my sometimes pure mathematics knowledge yields U3=(-1/RC)*U1*t, I calculate that for the time interval 1<t<3, U3 changes from 0 to -6V (t=2ms, U1=3V).
So far so good. Doing it the same way but setting (t=3s and U=-1V) for the time interval 3<t<6 I get that U3 is supposed to change to 3V... unfortunately the solution given says -3V!?? How come!?
And how come for 6<t<8 U3 contiues to be -3V, while U1 is 0V!?? What am I doing wrong!?

c) absolutely no idea how to combine U2 and U3 to draw this :confused:

Thanks a lot in advance!!

Veni
 

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Last edited:
am... I didn't really get that (about the subtraction) - neither graphically nor logically. Doesn't make much sence to me :(

non-English deleted - moderator
 
Last edited by a moderator:
Perhaps the best way to understand what is going on is to work out the math. U2 is on the non-inverting input. There is negative feedback in place, which means that the voltage on the non-inverting input is impressed on the inverting input. You also know that the current through R4 is equal to the current through R3. Set that equality up, then solve for U4.

U4 will be expressed in terms of a gain on the level of U3 subtracted from a gain on the level of U2. Make a plot of the two terms then add/subtract them graphically to get U4.
 
ok guys... obviously I am making a mistake somewhere, beacause I get the following for U4:

U4 = U2(1+R4/R3) - U3(R4/R3)

which actually is almost the same as what Willbe says, only that he has + and me - minus!! Can't really understand why he takes -40/30 !??

That's how I do it:

U2 = U+ = U-
I3 = (U3 - U-)/R3 = (U3 - U2)/R3
I4 = (U4 - U-)/R4 = (U4 - U2)/R4
I3 + I4 = I- = 0 ==> I3 = -I4
U3/R3 - U2/R3 = U2/R4 - U4/R4
U4 = U2*R4/R4 + U2*R4/R3 - U3*R4/R3 = U2*(1+R4/R3) - U3*(R4/R3) = U2*(1+4/3) - U3(4/3).

can anyone correct me please!?
 
The standard opamp derivation is a special case that assumes that the input and output are ground referenced.

The inputs are both changing so the standard Vo=-ViRf/Ri does not apply.


Sorry I am at work and can not give thorough replies
 
Last edited:
No need to correct. You have the right answer for U4 in terms of U2 and U3.

Ops! Actually yes :D :D You are right... I guess I was too tired when I was doing the math :D :D
ok - so U4 is clear now!

Now the only mystery left is U3:

Using the formula for an "Integrator" (integrator.jpg) which according to my sometimes pure mathematics knowledge yields U3=(-1/RC)*U1*t, I calculate that for the time interval 1<t<3, U3 changes from 0 to -6V (t=2ms, U1=3V).
So far so good. Doing it the same way but setting (t=3s and U=-1V) for the time interval 3<t<6 I get that U3 is supposed to change to 3V... unfortunately the solution given says -3V!?? How come!?

And how come for 6<t<8 U3 contiues to be -3V, while U1 is 0V!??
 
Last edited:
it is working properly... it is not supposed to change *to* 3V it is supposed to change a *total* of 3V.

The integral of zero is zero.

Again, look at currents. the opamp is trying to keep 0V on the - input. that puts a fixed voltage of 0V (U1) across R5, so if you have a good cap you do not get a change in U3 from T6-8.
 
it is working properly... it is not supposed to change *to* 3V it is supposed to change a *total* of 3V.

The integral of zero is zero.


this actually makes sence... never thought of it that way.

Thanks a lot!! I'll solve it all once again but I think it is all clear now!
 
Hi Veni,

I guess you already got the solution to this problem so no need for comment
on that, but it's nice to see a woman who is involved with electronic
circuits...dont see that much.
 
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