Can you help me solve this Op-Amp problem?

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In summary, the problem involves finding the output voltage (vo) of a non-inverting op amp circuit with two input resistances (R1 and R2). The equation for vo is vs * ((R1+R2)/(R2)), where vs is the input voltage. The question is whether vo will be independent of another resistance, Rs. To solve the problem, the voltage at the positive input (vp) is first found and then used to create a "new" circuit for finding vo. The final output voltage is determined to be 9000 mV.
  • #1
Inept
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Homework Statement



npi9za.png



Homework Equations



vo = G*vs

where G = ((R1+R2)/(R2)

therefore,
vo = vs * ((R1+R2)/(R2)

would vo be independent of Rs ?

The Attempt at a Solution



I assumed it was a Non Inverting Op Amp...but I haven't come across one with (two?) input resistances? I'm not sure where to go with it. I began to use the equation I have above, but I don't think it's right. I think (or I keep telling myself) that I have to use Rs or Rx in part a somehow.

For part b) I believe I have to figure out the linear range of Rx, which should be doable. So I'm not worried about that, assuming what I'm thinking is correct...

If anyone can help thanks! If not, thanks anyways!
 
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  • #2
One question that may help - What is the voltage at the + input of the Op Amp. Is it 750 mV, or something else?
 
  • #3
Inept said:

Homework Statement



npi9za.png

Homework Equations



vo = G*vs

where G = ((R1+R2)/(R2)

therefore,
vo = vs * ((R1+R2)/(R2)

would vo be independent of Rs ?

The Attempt at a Solution



I assumed it was a Non Inverting Op Amp...but I haven't come across one with (two?) input resistances? I'm not sure where to go with it. I began to use the equation I have above, but I don't think it's right. I think (or I keep telling myself) that I have to use Rs or Rx in part a somehow.

For part b) I believe I have to figure out the linear range of Rx, which should be doable. So I'm not worried about that, assuming what I'm thinking is correct...

If anyone can help thanks! If not, thanks anyways!

For starters, what is the voltage at the + input (in terms of Vs)?

EDIT: phyzguy had the same thought.
 
  • #4
I solved it, I had to solve for a non-inverting (positive) input voltage (vp) first and then created a "new" circuit. Using that circuit I found the output voltage using that equation I listed above while replacing vs with vp . I got vp = [(80k/(80k+20k)]*750 mV = 600 mV

then vo = [(3.3k+46.2k)/3.3k]*600 mV = 9000 mVIt feels good to figure my own answer out...lolEDIT: I didnt refresh, so I didn't see your guys replies. Thanks all! :)
 
  • #5


Sure, I would be happy to help you solve this Op-Amp problem. It seems like you have already made some progress by identifying the circuit as a non-inverting Op Amp and writing out the relevant equation for the output voltage. To answer your question, yes, the output voltage (vo) will be independent of Rs as long as the ideal assumptions of the Op Amp are satisfied (infinite input impedance and zero output impedance). This means that the input resistor Rs will not affect the output voltage.

To solve part a), you can simply plug in the given values for R1, R2, and vs into the equation you have written and solve for vo. To solve part b), you can use the linear range equation, which is vo(max) = Vcc - Vee, where Vcc is the positive supply voltage and Vee is the negative supply voltage. This will give you the maximum output voltage that the Op Amp can provide.

I hope this helps you solve the problem. If you need further assistance, please let me know. Remember to always check your circuit assumptions and make sure they are satisfied before solving any Op Amp problem. Good luck!
 

1. What is an Op-Amp?

An operational amplifier, or Op-Amp for short, is a type of electronic circuit component commonly used in analog signal processing. It is a high-gain voltage amplifier with two inputs and one output, and is typically used to amplify and manipulate signals in a variety of applications.

2. How does an Op-Amp work?

An Op-Amp works by taking the difference between its two input voltages, amplifying that difference, and outputting the result. The high gain of the Op-Amp allows for small changes in input voltages to result in large changes in output voltages. It also has a high input impedance, meaning it draws very little current from the input sources.

3. What are the common applications of Op-Amps?

Op-Amps have a wide range of applications, including as voltage amplifiers, filters, oscillators, comparators, and more. They are commonly used in audio equipment, signal processing circuits, and instrumentation amplifiers.

4. What is the difference between ideal and non-ideal Op-Amps?

Ideal Op-Amps have infinite gain, infinite input impedance, zero output impedance, and zero offset voltage. They also have a flat frequency response and infinite bandwidth. Non-ideal Op-Amps, on the other hand, have finite values for these parameters and may exhibit other characteristics such as noise and distortion.

5. What are some important considerations when using Op-Amps?

When using Op-Amps, it is important to consider parameters such as gain, bandwidth, and input and output impedance. It is also important to ensure proper power supply and grounding, and to be aware of any potential sources of noise or interference. Additionally, understanding the limitations and non-idealities of Op-Amps can help in designing more accurate and reliable circuits.

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