Why is there resistor R2 connecting output and input of op amp?

[Hyperspace2]

I have attached image. Thanks in advance.

 

[berkeman]

I have attached image. Thanks in advance.

Welcome to the PF.

That's a very basic question. What have you studied and learned about opamps and negative feedback circuits so far?

 

[Hyperspace2]

Welcome to the PF.

That's a very basic question. What have you studied and learned about opamps and negative feedback circuits so far?

I have just started to study op amp . I just know that that op amp has some triangular structure where it has inverting and non inverting input. It may have two output or one output. Some freinds in this forum have told me that the inner circuits of op amp uses differential amplifier. I know about the working principle of differential amplifier. Since I have already studied about biasing methods , I know about the negative feedback. But I couldnot relate it here. Sir , can you explain very detail manner about how negative feedback happens here due to resistor r2 connected in such a way.

 

[berkeman]

I have just started to study op amp . I just know that that op amp has some triangular structure where it has inverting and non inverting input. It may have two output or one output. Some freinds in this forum have told me that the inner circuits of op amp uses differential amplifier. I know about the working principle of differential amplifier. Since I have already studied about biasing methods , I know about the negative feedback. But I couldnot relate it here. Sir , can you explain very detail manner about how negative feedback happens here due to resistor r2 connected in such a way.

This introductory article should be a big help to you. Please read through it, and let us know if you have specific questions about parts of the article.

http://en.wikipedia.org/wiki/Opamp

.

 

[Hyperspace2]

[PLAIN]http://en.wikipedia.org/wiki/File:Op-Amp_Non-Inverting_Amplifier.svg[quote="berkeman, post: 2926745"]This introductory article should be a big help to you. Please read through it, and let us know if you have specific questions about parts of the article.

http://en.wikipedia.org/wiki/Opamp

.[/QUOTE]

I think I got it . Here vin is the function of the vout. The section of the image containing the R1,R2 and Vout and vin can be regarded as voltage divider apparatus and vin is fed at
input . I have uploaded the pic (my understanding).

 

[╔(σ_σ)╝]

I think a simple answer is that the resistor is used to tune the gain. Also when an Opamp circuit has negative feedback the negative terminal tracks the positive terminal.
Eg since in your original circuit the positive terminal grounded and there is negative feedback; you negative terminal appears to be grounded.

Without the resistor before the negative terminal, you would be shorting your input voltage and no current would flow into your feedback resistor ,thus, no amplification. :-)

 

[sophiecentaur]

Op Amps have very high voltage gain (you can often treat it as infinite).
The feedback resistor allows current to flow from output to (-) input. Whatever the input voltage is, relative to the 0V point, the output voltage 'goes the other way', about the zero voltage of the input until the feedback current cancels the current flowing through R1. This form of feedback has the action of maintaining the (-) input terminal at (or very very near)0V. It's called a virtual Earth circuit and its 'closed loop' gain is given by -R2/R1 (the negative is there because it 'inverts' the signal). It's a very neat idea because the circuit gain is pretty well independent of the gain of the actual op amp (assuming it's high enough / very high). It's very bad design practice to have circuit elements with an undefined gain so this is a useful configuration. The more feedback, the lower the gain but the more linear the amplifier is - you take your pick.

 

[Hyperspace2]

I think a simple answer is that the resistor is used to tune the gain. Also when an Opamp circuit has negative feedback the negative terminal tracks the positive terminal.
Eg since in your original circuit the positive terminal grounded and there is negative feedback; you negative terminal appears to be grounded.

Without the resistor before the negative terminal, you would be shorting your input voltage and no current would flow into your feedback resistor ,thus, no amplification. :-)

Op Amps have very high voltage gain (you can often treat it as infinite).
The feedback resistor allows current to flow from output to (-) input. Whatever the input voltage is, relative to the 0V point, the output voltage 'goes the other way', about the zero voltage of the input until the feedback current cancels the current flowing through R1. This form of feedback has the action of maintaining the (-) input terminal at (or very very near)0V. It's called a virtual Earth circuit and its 'closed loop' gain is given by -R2/R1 (the negative is there because it 'inverts' the signal). It's a very neat idea because the circuit gain is pretty well independent of the gain of the actual op amp (assuming it's high enough / very high). It's very bad design practice to have circuit elements with an undefined gain so this is a useful configuration. The more feedback, the lower the gain but the more linear the amplifier is - you take your pick.

Thanks for the reply people. I have made following conclusion from your quotes
1) The R1 and R2 network which is voltage divider basically fed into the negative terminal will
have negative feedback- Vgain= A( V+ - V-) , (because V- is subtracted from V+ , overall it has effect of decreasing the gain)
2) It tries to maintain the neagtive terminal at 0V .
3) Thus making the gain independent of A and making it dependent of R1 and R2.

 

[staticd]

I must say that I am thoroughly confused after reading this thread. Although this discussion is very conceptual. I think that is you did the math, you would find the reason to have the R2 resistor.

The open-loop gain of an op amp (OA) is infinite (~100,000 V/V).

ONE of the configurations of an op amp is the inverting op-amp, the one that is first pictured. Th R2 resistor provides a negative feedback to the op amp and brings the gain down to a much more manageable -R2/R1.

Note : We use 0 to represent the voltage at the negative input terminal because as the gain approaches 0, the voltage between the terminals approaches zero (virtual ground).

--> i1 = Vi-0/R1 = Vi/R1
--> Vo = 0 - i1R2
--> Vo = -(Vi*R2)/R1
-->Vo/Vi = -R2/R1

Think of it as stomping on the pedal of a car with no R2 and then knowing when to accelerate or decelerate with R2 there (negative feedback).

 

[Hyperspace2]

Note : We use 0 to represent the voltage at the negative input terminal because as the gain approaches 0, the voltage between the terminals approaches zero (virtual ground).

--> i1 = Vi-0/R1 = Vi/R1
--> Vo = 0 - i1R2
--> Vo = -(Vi*R2)/R1
-->Vo/Vi = -R2/R1

Thanks for the math. It brings a clear picture.

Think of it as stomping on the pedal of a car with no R2 and then knowing when to accelerate or decelerate with R2 there (negative feedback).

Good analogy. Thanks!

 

[sophiecentaur]

Thanks for the reply people. I have made following conclusion from your quotes
1) The R1 and R2 network which is voltage divider basically fed into the negative terminal will
have negative feedback- Vgain= A( V+ - V-) , (because V- is subtracted from V+ , overall it has effect of decreasing the gain)
2) It tries to maintain the neagtive terminal at 0V .
3) Thus making the gain independent of A and making it dependent of R1 and R2.

I wouldn't disagree too much with that except with the word "tries". OK in casual conversation but, of course, no circuit actually "tries" to do anything (far too dumb). Substitute a phrase such as "has the effect of".

Feedback is a 'loop' and cause-and-effect are inter related. If you think in terms of "what happens if the output voltage were to change a bit?" and then say what will happen to the current flowing into the (-) input terminal and then how that would affect the voltage at the output terminal i.e. pulling it back to its original value, that could be a way of getting to grips. Personally, I am a great believer in following the maths from an initial mathematical model of the circuit.

 

[staticd]

Agreed. I think that if you really want to learn how op amps work, you have to do the math. Otherwise, you should just know that you can modify your input and output impedances, your voltage or current gain based on the configuration. I am sure there are plenty of tables out there that do just that...

 

[Hyperspace2]

OK in casual conversation but, of course, no circuit actually "tries" to do anything (far too dumb).

Yeah, circuits are not that intelliegent.

Personally, I am a great believer in following the maths from an initial mathematical model of the circuit.

Agreed. I think that if you really want to learn how op amps work, you have to do the math.

Maths