CIRCUIT ANALYSIS: Non-Inverting Ideal OpAmp, 2 resistors, 1 IVS - Find Thevenin Equiv

  • Thread starter VinnyCee
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Homework Statement



Find the Thevenin equivalent at terminals a-b in the circuit below.

http://img361.imageshack.us/img361/2250/chapter5problem36gq0.jpg [Broken]


Homework Equations



KVL, KCL, v = i R, current and voltage equations for Operation Amplifier

The Attempt at a Solution



So I altered the diagram a bit.

http://img258.imageshack.us/img258/4382/chapter5problem36part2pr4.jpg [Broken]

[tex]I_1\,=\,\frac{V_S}{R_1}[/tex]

[tex]I_2\,=\,\frac{V_S\,-\,V_1}{R_2}[/tex]

KCL at [itex]V_S[/itex])

[tex]I_1\,=\,-I_2\,\,\longrightarrow\,\,\frac{V_S}{R_1}\,=\,-\frac{V_S\,-\,V_1}{R_2}[/tex]

[tex]V_1\,=\,\frac{R_2\,V_S}{R_1}\,+\,V_S[/tex]

[tex]V_{Th}\,=\,V_1\,=\,\frac{R_2\,V_S}{R_1}\,+\,V_S[/tex]

Is that right? If so, I will now put a test current of 1 Amp at the terminals to get the [itex]R_{Th}[/itex].

[tex]I_2\,=\,1[/tex]

[tex]\frac{V_S\,-\,V_1}{R_2}\,=\,1\,\,\longrightarrow\,\,V_1\,=\,V_S\,-\,R_2[/tex]

Now, using v = i R to get the Thevenin equivalent resistance.

[tex]R_{Th}\,=\,\frac{V_1}{1}\,=\,V_S\,-\,R_2[/tex]

Does that seem correct?
 
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Answers and Replies

  • #2
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Seems fine to me.
 
  • #3
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do you have to put in a test current and does it have 2 be 1??
 
  • #4


Hi, I know this is old but I have the exact same problem. I've done it the same as the guy in the first post up until finding [tex]V_th[/tex], but after that I'm confused.

When he applied a test current, shouldn't he have removed all of the independent sources? (Namely [tex]V_s[/tex]).

This would reduce the node voltage equation down to

[tex]V_p = 0 = V_n[/tex]

[tex](V_n - V_th)/R_2 = I_(test)[/tex]

[tex](0 - V_th)/R_2 = I_(test)[/tex]

[tex]-V_th/R_2 = I_(test)[/tex]

[tex]R_th = V_th/I_(test) = -R_2[/tex]

Which is invalid since it's a negative resistance, therefore you have to assume that [tex]R_th = 0[/tex]?

Can somebody explain if this thinking is valid? Since in the thread starter's solution for some reason he doesn't take out the independent voltage source when applying the test current, can someone explain why he doesn't?

Or can someone in general just explain the last step for finding [tex]R_th[/tex] since I understand the finding [tex]V_th[/tex] its just that last part that I don't understand what the original poster did exactly or why he didn't take out the independent source.

PS: For some reason my sub notation isn't working completely right, [tex]I_(test), V_th, R_th[/tex] are the test current, thevenin voltage, and thevenin resistance respectively.
 
  • #5
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I just ran across this problem I think Rth = 0 and Vth = Voc

I have a similar problem with a inverting op-amp.

The definition of an ideal op-amp is that the input has an infinite resistance and the output has an impedance of 0. The output of an ideal op-amp will always have the same voltage, it will not vary with the load.

Does anyone agree or disagree with my logic?
 

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