Find Thevenin Equiv. at a-b in Circuit Below

In summary, the original poster is trying to find the Thevenin equivalent resistance of an op-amp, but is confused about why Rth = 0. They assume that Rth = 0 because the output voltage will always be the same. Another poster offers a valid solution that assumes Rth = 0.
  • #1
VinnyCee
489
0

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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  • #2
Seems fine to me.
 
  • #3


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


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?
 

1. What is Thevenin's theorem?

Thevenin's theorem is a method used in circuit analysis to simplify complex circuits into an equivalent circuit that contains only a voltage source and a resistor, known as the Thevenin equivalent. This makes it easier to analyze and solve circuit problems.

2. How do you find the Thevenin equivalent circuit?

To find the Thevenin equivalent circuit, you need to follow these steps:

1. Disconnect all the loads connected to the two terminals (a-b) in the original circuit.

2. Calculate the open-circuit voltage (Voc) across the two terminals (a-b).

3. Calculate the equivalent resistance (Req) between the two terminals (a-b) when all the voltage and current sources are turned off.

4. Draw the Thevenin equivalent circuit with the calculated Voc and Req values.

3. Why is Thevenin's theorem useful?

Thevenin's theorem is useful because it simplifies complex circuits into a single voltage source and resistor, making it easier to analyze and solve circuit problems. It also allows us to determine the behavior of a circuit without having to know all the details of the internal components.

4. What is the difference between Thevenin and Norton equivalent circuits?

The main difference between Thevenin and Norton equivalent circuits is that Thevenin uses a voltage source and a resistor while Norton uses a current source and a resistor. These equivalent circuits are mathematically equivalent, meaning they will give the same results in terms of voltage and current values.

5. Can Thevenin's theorem be used for any circuit?

No, Thevenin's theorem can only be applied to linear circuits, which are circuits that have linear relationships between voltage and current. This means that the components in the circuit must follow Ohm's law and have constant resistance values. Additionally, Thevenin's theorem can only be applied to circuits with two terminals.

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