Circuit is:
http://img19.imageshack.us/img19/1517/scan0026a.jpg

I'm asked to find the Norton Equivalence looking into A and A'...

I can't use the v+=v- method because there is positive feedback...

I'll go with nodal analysis...

I have 3 nodes:

one at v- which I will call e1
one at vout which I will call e2
one at A which I will call e3

for e1:
(e1-V)/R + (e1-e2)/R = 0

for e2:
(e2-e1)/R + (e2-A)/R = 0

for e3:
(A-e2)/R + A/R = 0

Am I going in the right direction?

 

Strange circuit - where did you find it?

It seems to me that for any value of V1 other than zero the output would saturate to the op-amp's supply positive or negative supply rail.

 

This is the book:
http://www.amazon.com/Foundations-Electronic-Circuits-Kaufmann-Architecture/dp/1558607358

 

Perhaps the author was having a bad day.

 

Do you think the problem is just messed up?

It gives an answer in the back

i(norton) = Vin/R
R(Thevenin) = R

 

Do you think the problem is just messed up?

It gives an answer in the back

i(norton) = Vin/R
R(Thevenin) = R

Possibly messed up. I think the solution is indeterminate. I'm assuming the horizontal lines at V1 -ve terminal and A' are the circuit common or reference point.

Suppose for arguments sake that the op-amp is ideal.

Suppose also that the output Vo = 10V. What value of V1 would provide this condition?

The voltage at A would be Vo/2 = 5V. If the Op-amp can linearly drive the external circuit then the voltage at the op-amps negative terminal would also be 5V. This would mean that Vi would have to be zero volts. Pick any other output voltage (any polarity) and you get the same answer :
Vi = 0V. That's why I think the answer is indeterminate.

Maybe move on to another problem in the meantime.

Hopefully someone else has an opinion on this.

 

I think I see what you're saying...

I see there is a non-inverting setup going into the positive terminal, and an inverting setup going into the negative terminal... so the effects are simply canceling each other?

 

Hopefully someone else has an opinion on this.

The wording of this question is a little confusing, but I think this is just an example of the Howland Current Pump which has been discussed in some other threads in the past.

http://www.national.com/an/AN/AN-1515.pdf

The circuit, as shown in your original post, is marginally stable, as t_n_k correctly pointed out. However if one reads between the lines, they are asking what the Norton equivalent is between points A and A' which implies that the circuit might be used as a current source by putting a load between those points. The presence of a load resistance can stabilize this circuit.

So basically I would just analyze the circuit using your normal static equations. RonH posted an analysis previously, so I'll link to that.

http://forum.allaboutcircuits.com/attachment.php?attachmentid=7958&d=1237045783

Note that the answer for the current agrees with the book answer. Next you need to calculate the equivalent impedance looking back into A and A'

 

Well I'm a little rusty with Norton method to begin with, but when you said use the normal static equations, did you mean to carry on with the three equations from my first post?

 

I'm not even sure if I'm on the right track, but here's what I've done so far:

http://img411.imageshack.us/img411/2755/scan0027.jpg

 

I think the "for e2" equation is incorrect.

In any case to find the Norton equivalent current you have to impose a short circuit across AA' and determine the current in the short. That's the Norton current. The value can be obtained relatively easily. A solution is possible because the short circuit case makes a stable unique solution possible.

On the other hand, finding the equivalent Rnorton = Rthevenin isn't a simple matter. At least I don't think it is.

As Steveb points out, the circuit will work but only if an additional load is connected across the AA' terminals.

You can't solve it without adding a load - for the reasons of having an indeterminate solution in the original schematic - which I pointed out earlier.

So you need to do the analysis with another resistor, Rload added across AA' and in parallel with the R value already across AA'. You then have a parallel combination of R & Rload. [Steveb gave you a very useful link to a related solution which will aid in solving your problem - check it out.]

This will make a unique solution possible with a given Rload. Let's say you make Rload = R for convenience. Then do the analysis.

In this case (with Rload = R), VA = Vo/3. You'll get a solution for VA. Then you need to interpolate what will be the norton resistance keeping in mind you've solved the problem with an added Rload in place. It is possible to solve! Remember you would already have solved for Inorton at this stage.

I'd be happy to defer to others on this matter if the problem can be solved without adding Rload.

This is not a trivial exercise by the way. Good on you for attempting it!

 

What works is to calculate the impedance at A-A', allowing the opamp gain to be finite; call it Av.

I get a result of Rth = (Av * R + 2*R)/4

If Av -> ∞, then Rth -> ∞ also.

As a second case, calculate Rth with the resistor between A and A' removed, I get a result of -(Av * R + 2 * R)/(Av - 2). In this case, as Av -> ∞, Rth -> -R.

With the resistor at A-A' removed, the circuit becomes a negative impedance converter: http://en.wikipedia.org/wiki/Negative_impedance_converter

These two results are consistent, since if the second case Rth of -R is paralleled with the resistor R that was removed, the resistance at that node becomes R||(-R) = ∞

Maybe the book meant for Rth to be -R, rather than R.

 

Thanks for the input guys.. I wouldn't doubt it if the question is messed up... found a lot of errors in this book already... frankly I hate the book

 

Hey guys... for what it's worth, the teacher went over the 'answer' today...

She said you're supposed to find the open-circuit voltage, short-circuit current, and thevenin resistance...

voltage into the minus terminal = Vm
voltage into the plus terminal = Vp
voltage from source = Vi
output voltage = Vo

(Vi - Vm)/R = (Vm - Vo)/R = i

v = Vp = Vm (v plus equals v minus method.. I still don't know what 'v' is)
Vo = V - iR = 2V - Vi

Open-circuit voltage = Vi
Short-circuit voltage = Vi/R
Rth = Rnorton = Open-circuit voltage/short-circuit current = R

She was just reading off the answers from her little solutions manual so ya...


Oh and by the way the answer to the successive op-amp problem was indeed 15 so I got that right, the book was wrong in saying 1.5

 

I'm with Electrician on his assertion that Rth=∞ for terminals AA'.

With respect, it appears your teacher has given an incorrect solution.

A simple means of testing the Norton equivalent circuit for AA' is to connect a load of some value to AA' and calculate the load current. Then compare this with a derivation of the current based on an analysis of the original circuit with the same load connected. The results should be the same - otherwise the Norton equivalent doesn't correctly represent the circuit behavior at the terminals AA'.

It takes a little effort, but it can be shown that within the dynamic linear range of the op-amps operation the external load current in any load connected from AtoA' will always be -V1/R, irrespective of the load. This would not be consistent with the same load being connected to the Norton equivalent derived by your teacher.

 

I'm with Electrician on his assertion that Rth=∞ for terminals AA'.

With respect, it appears your teacher has given an incorrect solution.

I agree too. This can be verified in Appendix A of the AN1515 note that I linked to earlier. If the R resistors are perfectly matched then the source impedance is infinity and you have a stiff current source with current completely independent of load resistance.

Of course, perfect resistor matching is not possible in practice, but this is an academic problem with no mention of resistor tolerance issues.

 

Thanks guys... I would highly not doubt she's wrong.. she just reads her solution manual...

This problem has been a headache for a person being introduced to circuits... I just hope its not on the test

 

I hope your instructor doesn't actually believe that analysis is correct. What is more important for you, I hope you don't believe it.

Let me show you why it can't be true.

Consider a series connection of two resistors, like this:

V1--/\/\/\/\___/\/\/\/--V2
      R1     |     R2
             Vo

Imagine that you apply a voltage at each end and calculate the output voltage, Vo. You can use superposition here. Short V2 to ground, and calculate the voltage at Vo due to V1: Vo = V1*(R2/(R1+R2))

Now short V1 to ground and calculate Vo: Vo = V2*(R1/(R1+R2))

Both fractions have the same denominator, so a general formula for the voltage at the junction of two resistors with voltages applied to the ends is:

      V1*R2 + V2*R1
Vo =---------------
         R1+R2

Now look at the circuit you started this thread with. There are two such 2-resistor strings; one with the middle node connected to Vm and the other with the middle node connected to Vp. Both strings have one end connected to Vo, but the other ends of the two strings are connected to different things. One is connected to Vin and the other is connected to ground (which is zero volts).

Since the opamps are considered to be ideal, the + and - inputs take no current; their impedance is infinite, so they have no effect on the voltage at the center node of the two resistor strings.

Now, by the formula we derived, the voltage at Vm is (Vin*R + Vo*R)/(2*R) and the voltage at Vp is (zero*R + Vo*R)/(2*R)). The only way that Vp can be equal to Vm is if Vin = zero. If Vin is anything other than zero, then Vp ≠ Vm. But since the analysis given by the book for the answer assumes that Vp = Vm, it can only be correct if Vin is zero.

So, if Vin is not zero, is is not possible for Vm to equal Vp, the opamp output will be slammed up against one of the rails and it will not be operating in linear mode; the output will be saturated.

The open circuit voltage at A will not be well defined, and it almost certainly won't be Vin.

Furthermore, if Vin = zero, ask yourself what Vo would be? Then the left end of both strings would be grounded, and whatever Vo is, it will be divided by 2 at the center node of each string, and Vm will equal Vp no matter what Vo is. Vo is indeterminate in this case.

 

Very thorough analysis... thank you...

Yes I know to never believe a word she says when she's looking at her solution manual with a confused expression on her face then asks the class if everyone understands and nobody says anything...

 

LOL!!!!!

I can just see that!!