From: butts#@shocker.ee.twsu.edu (Ronald W. Butts -SUPER ACCT)
Subject: Re: HELP! op-amp problem
Organization: Electrical Engr. Dept, Wichita State Univ.
Distribution: usa
Date: Fri, 9 Oct 1992 07:34:44 GMT
X-Posted-From: gegan1.ee.twsu.edu
NNTP-Posting-Host: sol.ctr.columbia.edu

In article <1992Oct8.160444.15278@eng.ufl.edu> zzang@stat.ufl.edu (zzang) writes:
>From: zzang@stat.ufl.edu (zzang)
>Subject: HELP! op-amp problem
>Date: Thu, 8 Oct 92 16:04:44 GMT
>Hello,
>I have a hard time in analyzing op-amp circuit.
>I recently come across an exercise in a book.
>the following is from the spice program I use to
>solve my problems.(but, sure it doesn't :-{)
>
>Simple Circuit Analysis
>*
>*                     
>*               +-----www----+     
>*  V            |    -       |         +------||-------+ 
>*  o----www-----+----|\ 6    |         |    -          |
>*  1            2    |  >----+---www---+----|\ 7       |
>*                   0|/ 0    3         4    |  >-------+-----o y(t)
>*                                          0|/ 0       |5
>*                                                     www
>*                                                      |
>*                                                      x
>* all resistances are equal to R, the capacitance is C, and R*C=1
>* 0 means ground
>the question is when you connect node x to the node 2,
>what is the voltage output y(t) ?
>the answer in the book is y'(t)= -y(t) + V
>but my solution is y'(t)= y(t) + V

Your solution is what I get.

>
>here is how my solution comes out:
>we consider the voltage at the node 3 V(3)
>and assume the op-amp is ideal, so the inverting voltages
>V(2) and V(4) are 0.
>1. look to the right
>V(3)/R = -Cy'(t)
>2. look to the left
>V/R + y(t)/R = -V(3)/R
>combine 1. and 2., we have y'(t) = y(t) + V
>
>my 2nd question is what the y(t) will be if we connect x to 4 ?
>following the above argument, I got y'(t) = -y(t) + V
>but if the op-amp are ideal, node 2 and 4 are considered grounded,
>why these two connections come out different solutions?

With output connected to node 4, you can now look at first OP-amp unit as 
inverting amp with gain of -1. So node 3 = - V.

Writing KCL at (-) input of second op-amp (knowing its potential is ideally 
zero, since + input is grounded)
	[0 - (-V)]/R + (0 - y)/R + (0 - y)'C = 0
	Which simplifies to diff equation for y(t):
	RC*y'(t) + y(t) = V

In the case where the output is fed back to node 2, you will find the diff 
eq to be:
	RC*y'(t) -y(t) = V. 
which when RC=1 reduces to your answer you obtained.

>
>could someone give me some helps on this?
>
>Thanks in advance!
>.WIDTH OUT=80
>.TEMP 35
>R1 1 2 5K
>R2 2 3 5K
>R3 3 4 5K
>R4 5 4 5K
>C1 4 5 0.0002F
              ^^^ Be careful here! SPICE will only ignore
		  added letters if they are not a metric suffix.
		  "F" will be taken as "FEMTO" (x*1E-15).
		   Likewise, appending "A" for "Ampers" after the
	           value for a current source will, with most SPICE
 		   species, be treated as "atto" (x*1E-18)--not good.

>X1 0 2 0 6 3 LM324
>X2 0 4 0 7 15 LM324
>V1 1 0 DC .006
>.PRINT DC V(3)
>.LIB EVAL.LIB
>.END
>
>-- 
>Zhuo Zang[~{j0WA~}]
>Department of Statistics
>University of Florida            Email: zzang@stat.ufl.edu