G4((G1*X) + (G2*Y) + (G3*Z))
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'.Hi there, I need help on this question! If you can help, that will be great!!
Thanks, Waiting 4 ur Reply
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.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'.
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.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!!!!!!!!!!!!
If I do your work for you can I have your diploma?........................................
So What is the Answer!
I need to find out a.s.a.p! Or give an equation!
Well, lets look a bit closer. Lets 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.Nonsense! It is not bistable.
How does an amplifier with positive feedback act as a follower?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.
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.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.
I agree with you there.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......
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