Simple intro lvl question about Op-amps

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    Intro
In summary, the conversation is about understanding the concept of weighted summer in circuit analysis, specifically in the context of an op-amp circuit with a third resistor and a voltage source. The conversation also touches on the use of superposition principle and nodal analysis in solving circuit problems. The recommended books for introductory circuit analysis are "Hayt & kimmerly(Linear Circuit Analysis)", "Decarlo & Linn(Linear Corcuit Analysis)" and "Van Valkenburg(Network Analysis)".
  • #1
t3rom
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Homework Statement



For inverting op-amps Vout = -Vin(R2/R1)

But I've no clue what happens when you add third resistor with a voltage source to the amplifier. Is there a general gain equation for such circuit?

See the attached problem in the image.


Homework Equations




For inverting op-amps Vout = -Vin(R2/R1)

The Attempt at a Solution



I tried calculating for the R1 and R2 but the answer is wrong.

Homework Statement




I will be grateful if someone can help me understand it by solving it. Thanks!
 

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  • #2
You must have studied superposition principle in Circuit Analysis. Can you tell us what it is?
 
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  • #3
Thanks for replying!

No, I haven't. I'm only taking a basic course where they tell you an equation and you plug in the values. Theorems etc. are beyond the scope of the course.
 
  • #4
What about Nodal Analysis?. Nodal Analysis is tought in basic physics courses.
 
  • #5
Yes, I know nodal analysis.
 
  • #6
There is just one node to care for. The Op-Amps inverting input terminal

What is the Voltage at this Node?
Write its node equation?
 
  • #7
Try it by applying Kirchoff's current law. You have an ideal OP amp so no current flows into the terminals. Find expressions for the currents flowing into the node from the two voltage sources and subtract the current flowing out of the node from this. KCL states that this will be equal to zero. Find in terms of Vi or Vo.
 
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  • #8
I did this (see attached)

Please have a look at my original op-amp circuit too. Thanks!
 

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  • #9
Thats good work. There is just ine mistake. That too of just one negative sign In the Nodal analysis we take Voltage coming at the the. Why did you take In leaving the Node. Reverse the dirction of In. Your first equation will become

I1+I2+In = 0.

There is nothing wrong in taking In leaving the node. But it leads to errors as you have done in next equation itself. If you are reversed the direction of In, you have to reverse the order of voltages.
You can do either way but the better method would be reversing the direction of In.
 
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  • #10
Ah I see, it should have been -(Vn/R2-v2/R2)=0 if I stick with my current In direction, or +(V2/R2-Vn/R2) if reverse the direction of In making i1+i2+in=0. Thanks for helping me out, but mostly motivating me with hints!
 
  • #11
You are welcome.

The circuit you just solved is a 'Weighted Summer', where the output is weighted sum of input voltages. The general Vo for Weighted Summer is given by

Vo = -Rf{(V1/R1)+(V2/R2)+(V3/R3)+...+(Vn/Rn)}

Where Rf is the resistance in the feedback path and R1, R2...Rn are the resistance connected with the input voltages. In your problem n=2.
 
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  • #12
Cool! My Vo equation conforms it :)

Can you kindly recommend any good book on intro to circuit analysis which is easily accessible?

Thanks again! :)
 
  • #13
There are many good books on Circuit Analysis. It is really hard to recommend one. But the best i think is by 'Hayt & kimmerly(Linear Circuit Analysis)' It stars from basics.
But i would recommend two other books. One is by 'Decarlo & Linn(Linear Corcuit Analysis)' and the other one is by 'Van Valkenburg(Network Analysis)'.

You can refer to anyone of the three books i recommended. All three are good books.
 
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  • #14
A big thanks friend!
 

1. What is an op-amp?

An op-amp, or operational amplifier, is an electronic device that amplifies an input signal to a higher output signal. It is commonly used in electronic circuits for various applications such as amplification, filtering, and signal processing.

2. How does an op-amp work?

An op-amp consists of differential inputs, a high-gain amplifier, and output circuitry. The differential inputs compare the voltage difference between the two input signals, and the high-gain amplifier amplifies this difference. The output circuitry then converts the amplified voltage into a usable output signal.

3. What are the important characteristics of an op-amp?

The most important characteristics of an op-amp include its gain, input impedance, output impedance, bandwidth, and slew rate. The gain is the amplification factor of the op-amp, while the input impedance determines how much current is drawn from the input signal. The output impedance affects the ability of the op-amp to drive external components, and the bandwidth determines the range of frequencies the op-amp can amplify. The slew rate is the maximum rate at which the output voltage can change.

4. What are the different types of op-amps?

There are several types of op-amps, including the inverting op-amp, non-inverting op-amp, differential op-amp, and instrumentation op-amp. The inverting and non-inverting op-amps have different input configurations, while the differential op-amp has two inputs and one output. The instrumentation op-amp is a specialized type used for precise measurement and amplification of small signals.

5. What are some common applications of op-amps?

Op-amps are used in a wide range of electronic devices and systems, including audio amplifiers, signal filters, voltage regulators, and sensors. They are also commonly used in communication systems, control systems, and medical equipment. Additionally, op-amps are often used in combination with other electronic components to create more complex circuits for various purposes.

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