CIRCUIT ANALYSIS: Non-Ideal OpAmp with 2 resistors - Find Thevenin Equivalent

In summary, the conversation discusses finding the general Thevenin equivalent circuit seen at the terminals of a non-ideal, non-inverting op-amp. The solution involves using KCL and voltage and current markers to derive equations for V_0 and V_2, which can then be solved for to determine the Thevenin equivalent circuit. There is some confusion in the conversation about the correct equation to use and whether V_d should be equal to V_in or negative V_in.
  • #1
VinnyCee
489
0

Homework Statement



There is a non-inverting op-amp below.

http://img291.imageshack.us/img291/8242/chapter5lastproblemea1.jpg [Broken]

The op-amp is NOT ideal. We assume that [itex]R_i\,=\,\infty[/itex], [itex]R_0\,>\,0[/itex] and A is finite.

Find the general Thevenin equivalent circuit seen at the terminals.


Homework Equations



KCL, v = i R


The Attempt at a Solution



I changed the diagram to use a model given for a non-ideal op-amp.

http://img144.imageshack.us/img144/6144/chapter5lastproblempartyx5.jpg [Broken]

Now I add some voltage and current markers.

http://img166.imageshack.us/img166/5948/chapter5lastproblempartsk9.jpg [Broken]

[tex]V_d\,=\,-V_{IN}[/tex] <-----Right?

[tex]V_1\,=\,-A\,V_{IN}[/tex]

Now, the current equations)

[tex]I_1\,=\,\frac{V_1\,-\,V_0}{R_0}\,=\,\frac{-A\,V_{IN}\,-\,V_0}{R_0}[/tex]

[tex]I_2\,=\,\frac{V_0\,-\,V_2}{R_2}[/tex]

[tex]I_3\,=\,\frac{V_2}{R_1}[/tex]

KCL at [itex]V_0[/itex])

[tex]I_1\,=\,I_2\,\,\longrightarrow\,\,\frac{-A\,V_{IN}\,-\,V_0}{R_0}\,=\,\frac{V_0\,-\,V_2}{R_2}[/tex]

Solving for [itex]V_0[/itex])

[tex]V_0\,=\,\frac{-R_2\,A\,V_{IN}\,+\,R_0\,V_2}{R_0\,+\,R_2}[/tex]

KCL at [itex]V_2[/itex])

[tex]I_2\,=\,I_3\,\,\longrightarrow\,\,\frac{V_0\,-\,V_2}{R_2}\,=\,\frac{V_2}{R_1}[/tex]

Solving that equation for [itex]V_0[/itex])

[tex]V_0\,=\,\frac{R_2\,V_2\,+\,R_1\,V_2}{R_1}[/tex]

But which do I use? Are they both right? ONe wrong? Or all wrong?
 
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  • #2
I think you only did a partial KCL at node 2. Summing currents into the node you should get:

[tex]\frac{V_d-V_2}{R_i} + \frac{V_0-V_2}{R_2} + \frac{0-V_2}{R_1} = 0 [/tex]

and for the V0 node:

[tex]\frac{AV_d - V_0}{R_0} + \frac{V_2 - V_0}{R_2} = 0[/tex]

Now you can solve for V_0 and use the V_in=V_d constraint, though you seems to think is otherwise. You might have a reason that I don't see, but I think V_d = V_in, and not the negative.
 
  • #3


First of all, great job on setting up the problem and showing your work. It's clear that you understand the concepts involved in analyzing an op-amp circuit.

To find the Thevenin equivalent circuit, we need to find the equivalent resistance and voltage seen at the terminals of the op-amp. The equivalent resistance can be found by shorting out the voltage source and finding the total resistance seen by the input. In this case, since Ri = ∞, we can ignore it and just look at the resistors R1 and R2 in parallel. So the equivalent resistance is R1 || R2.

To find the equivalent voltage, we can use the voltage divider formula, since the op-amp is in a non-inverting configuration. So the equivalent voltage is Veq = V0 = (R2/(R1+R2)) * V2.

Putting these together, the Thevenin equivalent circuit at the terminals is a voltage source with voltage Veq and a resistor with resistance R1 || R2.
 

1. How do I find the Thevenin equivalent of a circuit with a non-ideal OpAmp and two resistors?

To find the Thevenin equivalent of a circuit with a non-ideal OpAmp and two resistors, you will need to follow a few steps. First, determine the open-circuit voltage of the circuit by setting the input of the OpAmp to zero and solving for the output voltage. Next, calculate the equivalent resistance by removing the non-ideal OpAmp from the circuit and finding the equivalent resistance between the two resistors. Finally, combine the open-circuit voltage and equivalent resistance to get the Thevenin equivalent voltage and resistance.

2. What is the purpose of finding the Thevenin equivalent in this type of circuit analysis?

The Thevenin equivalent allows you to simplify a complex circuit with a non-ideal OpAmp and two resistors into a more manageable circuit with one voltage source and one resistor. This makes it easier to analyze and understand the behavior of the circuit.

3. Is the Thevenin equivalent always an accurate representation of the original circuit?

The Thevenin equivalent is an approximation of the original circuit and is only accurate under certain conditions. It assumes that the load connected to the circuit is much larger than the equivalent resistance, the circuit is linear, and there are no dependent sources in the circuit.

4. Can the Thevenin equivalent be used to solve for any variable in the circuit?

No, the Thevenin equivalent can only be used to solve for the voltage or current at the load connected to the circuit. It cannot be used to solve for any other variables in the circuit, such as the voltage or current at a specific node.

5. Are there any limitations to using the Thevenin equivalent in circuit analysis?

There are a few limitations to using the Thevenin equivalent. It only works for linear circuits and cannot be used for circuits with dependent sources. Additionally, it may not be accurate for circuits with high-frequency components or circuits that have a non-linear behavior.

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