Am I doing this ideal op-amp problem correctly?

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In summary, the conversation discusses solving a circuit problem using voltage division and the node-voltage method. The resulting equation for V0 is 5/2V3-2V1-2V2, but there is some confusion about the value of V3 and how it affects the output. The conversation suggests taking one input at a time and using superposition to find the corresponding output, with the other voltage sources being set to zero volts.
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Homework Statement


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Homework Equations


[tex]Vp=Vn[/tex]
[tex]Ip=In=0[/tex]

The Attempt at a Solution


By voltage division:
[tex]Vp = 8/(8+8) V3[/tex]
[tex]Vp = 1/2V3[/tex]
[tex]Vn = 1/2V3[/tex]

By node-voltage method at the inverting input terminal:
[tex](Vn-V1)/4 + (Vn-V2)/4 + (Vn- V0)/8 = 0[/tex]
Multiply equation by 8
[tex]2(Vn-V1) + 2(Vn-V2) + Vn-V0 = 0[/tex]
[tex]V0 = Vn+2Vn-2V1+2Vn-2V2[/tex]
[tex]V0 = 5Vn-2V1-2V2[/tex]
Substitute in 1/2V3 for Vn

[tex]V0 = 5/2V3-2V1-2V2[/tex]

Is this the correct answer? I'm pretty sure about the V1 and V2 part, because it matches the summing amplifier circuit equation of -(Rf/R1*V1+Rf/R2*V2). However, for V3, I don't see how its resistor composition would yield an answer of 5/2. Could someone clarify?

Thanks
 
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  • #2
One easy way to work problems with multiple inputs is to take one input at a time, get the corresponding output, then add all three outputs. So for your worrisome V3 input the output is

Vo3 = V+(1 + Rf/Ri) = 0.5V3(1 + Rf/Ri) = 0.5V3(1 + 8/2) = 2.5V3

Also,
Vo1 = -8/4 v1
Vo2 = -8/4 v2

Then by superposition,
Vo = Vo1 + Vo2 + Vo3.

Note that when you do that the other voltage sources are set to zero volts (grounded), not left 'floating'.
 

1. How do I know if I have chosen the right op-amp for my circuit?

To determine if you have chosen the right op-amp for your circuit, you should first check the specifications of the op-amp and make sure it meets the requirements for your circuit. This includes the input and output voltage ranges, current capabilities, and bandwidth. Additionally, you may want to simulate your circuit using the chosen op-amp to see if it performs as expected.

2. Is there a specific process for solving ideal op-amp problems?

Yes, there is a general process that can be followed for solving ideal op-amp problems. This includes identifying the inputs and outputs of the op-amp, applying the appropriate ideal op-amp rules (such as virtual short and virtual ground), and using Kirchhoff's laws to solve for the unknown variables. It is important to also double check your work and make sure your solution makes sense in the context of the problem.

3. How do I handle negative feedback in an ideal op-amp problem?

In an ideal op-amp problem, negative feedback can be handled by assuming that the op-amp is in a closed loop configuration. This means that the output of the op-amp will be adjusted to keep the input voltages at the same level. This assumption allows us to use the ideal op-amp rules and simplifies the circuit analysis.

4. Can I use real-world op-amp models for solving ideal op-amp problems?

No, real-world op-amp models should not be used for solving ideal op-amp problems. The whole point of an ideal op-amp problem is to simplify the circuit analysis by assuming that the op-amp has infinite input impedance, zero output impedance, and infinite gain. Using a real-world op-amp model would introduce additional complexities and defeat the purpose of solving an ideal op-amp problem.

5. What if my calculated values do not match the expected values in an ideal op-amp problem?

If your calculated values do not match the expected values in an ideal op-amp problem, it is important to double check your calculations and make sure you have correctly applied the ideal op-amp rules. It is also possible that the given circuit has non-ideal elements or the assumptions made for the ideal op-amp problem are not applicable. In this case, you may need to adjust your approach or use a different method to solve the problem.

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