# Non-Inverting Opamp with Voltage Source resistance

• Weaver
In summary, the conversation revolves around the use of ideal op-amps and their assumptions in circuit analysis. The first assumption of infinite input impedance leads to the conclusion that the 6.8 kΩ resistor has no effect on circuit behavior. The second assumption of equal potential at the input terminals simplifies the analysis process. The equation Vout/Vin = 1 +R2/R1 can be used for the given circuit, but it is important to be able to derive it for different circuits.
Weaver

## The Attempt at a Solution

(a) [/B]We can assume for ideal op-amps that there is:
• infinite input impedance
• low output impedance
• the potential for infinite amplification
(b) Non-inverting op amp

(c) This is the part I have trouble with.

There is three resistances. The only equation we have used before for non-inverting op amps is

Vout/Vin = 1 +R2/R1

Vin = 8V

R2 = 80K

I'm guessing the 6.8k resistor is part of R1 with 20k but I'm not sure how to add it...

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Conor_B said:
I'm guessing the 6.8k resistor is part of R1 with 20k but I'm not sure how to add it...
Given your first stated assumption about ideal op-amps, what effect do you think the 6.8 kΩ resistor is going to have on circuit behavior? (Think about how much current can flow through it and what potential drops can occur across it)

There's another assumption that can be made about the behavior of ideal op-amps that is tied to your third listed assumption, that of infinite amplification. What assumption can be made about the potential difference between the input terminals of the op-amp (when the circuit provides negative feedback, of course)? Can you analyze the circuit to determine Vout for a given Vin?

gneill said:
Given your first stated assumption about ideal op-amps, what effect do you think the 6.8 kΩ resistor is going to have on circuit behavior? (Think about how much current can flow through it and what potential drops can occur across it)

Honestly, I don't know. These assumptions were only ever mentioned in fleeting in lectures and never discussed. Being honest, I don't entirely know what effect these assumptions mean for the any given op amp circuit.

Is it that the current that could potentially flow through the resistor just be is V/R, in this case 8/6.8 approximately 1.18 A ?

gneill said:
There's another assumption that can be made about the behavior of ideal op-amps that is tied to your third listed assumption, that of infinite amplification. What assumption can be made about the potential difference between the input terminals of the op-amp (when the circuit provides negative feedback, of course)

Is that that V- = V+ ? So there would be no potential difference between the two...

gneill said:
Can you analyze the circuit to determine Vout for a given Vin?

As in nodal analysis? I' not sure how to do that with the op amp in the circuit...

Conor_B said:
Honestly, I don't know. These assumptions were only ever mentioned in fleeting in lectures and never discussed. Being honest, I don't entirely know what effect these assumptions mean for the any given op amp circuit.
Something for you to investigate then. They are important factors that greatly simplify analyzing op-amp circuits.
Is it that the current that could potentially flow through the resistor just be is V/R, in this case 8/6.8 approximately 1.18 A ?
Think about it. If the input resistance is infinite, how much current could flow through the 6.8 k resistor and into the input?
Is that that V- = V+ ? So there would be no potential difference between the two...
Yes!
As in nodal analysis? I' not sure how to do that with the op amp in the circuit...
Could use nodal analysis, yes.

Strangely enough the op-amp being there actually simplifies things:
1. Infinite input resistance, so no current drawn from the nodes they're connected to.
2. Input terminals at the same potential, so if you know one voltage you know the other, too.

Can you determine the potential at the +input terminal?

Weaver
QUOTE="gneill, post: 5632418, member: 293536"] Think about it. If the input resistance is infinite, how much current could flow through the 6.8 k resistor and into the input? [/QUOTE]

Well, V = IZ, so if Z is infinite then I has to be 0 ?

So, could I treat R1 just as 20 k and use it in the above equation?

gneill said:
Could use nodal analysis, yes.

Strangely enough the op-amp being there actually simplifies things:
1. Infinite input resistance, so no current drawn from the nodes they're connected to.
2. Input terminals at the same potential, so if you know one voltage you know the other, too.

Can you determine the potential at the +input terminal?

Ok, but isn't it from nodal analysis that the equation Vout/Vin = 1 +R2/R1 comes from?

So, could I use that or do I have to derive a specific one for this circuit?

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Conor_B said:
Well, V = IZ, so if Z is infinite then I has to be 0 ?
Yes. So that input resistor has no effect.
So, could I treat R1 just as 20 k and use it in the above equation?
I suppose so. You haven't specified exactly where R1 and R2 are or shown why you've chosen the formula you did, but as it turns out, your choices are fine.
Ok, but isn't it from nodal analysis that the equation Vout/Vin = 1 +R1/R2 comes from?

So, could I use that or do I have to derive a specific one for this circuit?
Well, the equation certainly can be derived using nodal analysis, but there are other ways too. You can use the formula if you're sure it's applicable (it is in this case). But you should always be prepared to derive the formula for a given circuit. Sometimes you'll be given a circuit that isn't exactly like one of the "standard" layouts and you'll have to do the work. And, you could always suffer a memory lapse and forget which formula applies in a given situation. You should be prepared to derive it quickly.

gneill said:
I suppose so. You haven't specified exactly where R1 and R2 are or shown why you've chosen the formula you did, but as it turns out, your choices are fine.

The choice for R1 and R2 was based of my lecture notes as was the equation choice.

(I realize now, that the notes suggest no current flows into the terminals of the amplifier, so by extension of that the resistance in the Vs is unimportant, like you pointed out)

gneill said:
Well, the equation certainly can be derived using nodal analysis, but there are other ways too. You can use the formula if you're sure it's applicable (it is in this case). But you should always be prepared to derive the formula for a given circuit. Sometimes you'll be given a circuit that isn't exactly like one of the "standard" layouts and you'll have to do the work. And, you could always suffer a memory lapse and forget which formula applies in a given situation. You should be prepared to derive it quickly.

Ok. I'll certainly keep that in mind for the future.

So in this case Vout = (1 + 80k/20k)(8) = 40 ?

Conor_B said:
The choice for R1 and R2 was based of my lecture notes as was the equation choice.

View attachment 109715

(I realize now, that the notes suggest no current flows into the terminals of the amplifier, so by extension of that the resistance in the Vs is unimportant, like you pointed out)

Ok. I'll certainly keep that in mind for the future.

So in this case Vout = (1 + 80k/20k)(8) = 40 ?
Yes indeed!

gneill said:
Yes indeed!

Ok great. Thank you very much !

## 1. What is a Non-Inverting Opamp with Voltage Source resistance?

A Non-Inverting Opamp with Voltage Source resistance is a type of operational amplifier (opamp) circuit that uses a voltage source as a feedback resistor. This circuit is commonly used in electronic devices to amplify and filter signals.

## 2. How does a Non-Inverting Opamp with Voltage Source resistance work?

In this circuit, the input signal is applied to the non-inverting input of the opamp, and the output is fed back to the inverting input through a voltage source as a feedback resistor. This creates a virtual ground at the inverting input, which results in a high input impedance and low output impedance. This allows for a high gain and accurate amplification of the input signal.

## 3. What are the advantages of using a Non-Inverting Opamp with Voltage Source resistance?

Some of the advantages of this circuit include high input impedance, low output impedance, and high gain. It also has a linear response to the input signal, making it suitable for precision applications. Additionally, this circuit is easy to design and has a simple configuration.

## 4. How do I calculate the gain of a Non-Inverting Opamp with Voltage Source resistance?

The gain of this circuit is determined by the ratio of the feedback resistor (voltage source) to the input resistor. The gain can be calculated using the formula: Gain = 1 + (Rf/Rin), where Rf is the feedback resistor and Rin is the input resistor.

## 5. What are some common applications of Non-Inverting Opamp with Voltage Source resistance?

This type of circuit is commonly used in audio amplifiers, instrumentation amplifiers, active filters, and voltage regulators. It is also used in signal conditioning circuits, such as in sensors and transducers, to amplify and filter weak signals before further processing.

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