# I need Help! Urgent!

Discussion in 'Homework Help' started by gunes, Nov 21, 2006.

1. ### gunes Thread Starter New Member

Nov 21, 2006
8
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Hi there, I need help on this question! If you can help, that will be great!!

2. ### Papabravo Expert

Feb 24, 2006
10,137
1,786
U1 is a noninverting stage with a gain G1 greater than 1
U3 is an inverting stage with gain G2
U4 is an inverting stage with gain G3
U2 is a summing configuration that computes the following sum with gain G4:
Code ( (Unknown Language)):
1.
2. G4((G1*X) + (G2*Y) + (G3*Z))
3.
Your task is to figure out the value of the gains G1 thru G4.
Hint: They will involve ratios of the resistors in the input and feedback circuits of each operational amplifier. Second hint is that G4 is equal to one because R7=R8=R9=R10= 1 K Ohms. Let us know what you get.

3. ### pebe AAC Fanatic!

Oct 11, 2004
628
3
It is diffucult to arrive at an expression for W in terns of X because U1 is a bistable, so the answer will depend on which way U1 output has 'flipped'.

4. ### gunes Thread Starter New Member

Nov 21, 2006
8
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U1 is a non-inverting output, but i still dont understand this man!!!

What has this got to do with computers! This has got nothing to do with wat i'm studying but its one of the units in the course!

Need More help!!!!!!!!!!!!

5. ### Papabravo Expert

Feb 24, 2006
10,137
1,786
Nonsense! It is not bistable. It is a noninverting follower. The output acts in such a way as to make the input at the plus terminal equal to the input at the minus terminal.

6. ### Papabravo Expert

Feb 24, 2006
10,137
1,786
Stop screaming and calm down. This is a basic analog computer circuit. If you don't understand the behavior of an operational amplifier just say so. I can't help you with the disconnect between what you are studying and the content of the course. Take that up with the instructor. It's your tuition nickel.

An operational amplifier is a very high gain amplifier. The open loop gain could be on the order of 10^5. That open loop gain factor is applied to the voltage difference between the two terminals. When there is feedback from the output back to one of the inputs then the output behaves in such a way as to make the voltage difference between the two terminals equal to zero.

Are you with me so far?

7. ### gunes Thread Starter New Member

Nov 21, 2006
8
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........................................
........................................

I need to find out a.s.a.p! Or give an equation!

8. ### Papabravo Expert

Feb 24, 2006
10,137
1,786
If I do your work for you can I have your diploma?

9. ### gunes Thread Starter New Member

Nov 21, 2006
8
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just wanna check the answer for this

is it

w = -(10x + 10y + 72z)

10. ### Papabravo Expert

Feb 24, 2006
10,137
1,786
No
Hint
G1 = 1 + (72K/7K) = +10.28
G3 is not equal to 10
G4 = -(72K/1K) = -72

11. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,157
Papabravo,

You might want to revisit the G1 answer.

12. ### pebe AAC Fanatic!

Oct 11, 2004
628
3
Well, lets look a bit closer. Lets add a parameter that is not shown, but is necessary in any practical circuit  supply voltages. Let us assume that the ground shown is at 0V and the supply rails are +10v and 10v.

Now assume X is at 0v and the non-inverting input pin1 is more negative than X. The output will then go low taking pin1 even lower and the action will culminate in the output going to its maximum negative excursion, somewhere near the ve supply rail. Pin1 will then be at -0.89v.

Now lets see what effect X has on pin3. Within the limits 0v to 0.89v it will not change. When X is reduced to below that on pin1, U1 output will go high  almost to the +ve supply level. Feedback will take pin1 to +0.89V.

Again, nothing will happen until X is increased to exceed the voltage on pin1, ie. +0.89v. Then the output will go low again.

The action is such that X has no effect on U1 output while it is in the range 0.89v to +0.89v. Once outside that range, it will make U1 output toggle between the supply rails. Clearly, U1 has two stable states  its output is either high or low, with no definable state in between.

That is the definition of a bistable.

Nonsense? I dont think so!
How does an amplifier with positive feedback act as a follower?

13. ### Papabravo Expert

Feb 24, 2006
10,137
1,786
I apologize gnetlemen. It seems that I was focused on the topology of the circuit and missed the input symbols. The circuit, as drawn, makes very little sense either as a practical circuit or a pedagogical tool.

14. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,157
The original poster, a computer major, posed the question yet doesn't see the purpose of the circuit ... other than it's another problem the instructor wants them to solve.

I can see papabravo's solution as being valid for low level signal applications and yet, I can see pebe's solution as being valid as an event trigger in the computer industry.

The original poster hasn't described the application for which this circuit is used, maybe, just maybe, validating an application in the computer field would help the original poster more.

15. ### pebe AAC Fanatic!

Oct 11, 2004
628
3
U1 does not behave linearly as an attenuator or an amplifier would, nor does its action follow any law. So it is impossible to determine W in terms of X.

Gunes, I suggest you go back to the questioner and ask him to explain the reasoning behind his question.

16. ### Papabravo Expert

Feb 24, 2006
10,137
1,786
...nor does its action follow any law...

When I simulate U1 it looks very much like a schmitt trigger. The thresholds are not symmetric about ground and the saturation voltages depend on the opamp chosen and the supply voltages. It is nonlinear for sure but a long ways from indeterminate.

17. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
Hi,

Taking closer look at the pin numbers cited in U1, my guess it that it is a poorly written question. No op amp has that pinout. I might guess that it's supposed to be a follower with gain, rather than the mess shown. That is really not the way to make an inverter with gain. Convertion has the non-inverting input above the inverting. U1 makes sense (ignoring the pin numbers) that way.

Sometmes, instructiors don't proofread their stuff.

18. ### pebe AAC Fanatic!

Oct 11, 2004
628
3
I agree with you there.

19. ### richbrune Senior Member

Oct 28, 2005
106
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Ttrick question:
U1's input has positive feedback. Any positive signal applied to the negative input will cause the output to increase, which will in turn feed back into the positive input, causing DEVICE DESTRUCTION. I've tried it.
Rich

20. ### Papabravo Expert

Feb 24, 2006
10,137
1,786
I don't think the device destroys itself. The output stage saturates and stops there. Most opamps will not even go all the way to the rail and they can stay in saturation indefinitely as long as they are not trying to drive a low impedance load.