# Inverting/Noninverting Amplifier

1. Sep 12, 2009

### uzair_ha91

I know this is a school type question, but it's an example problem whose solution is already given and it's the solution I am unable to understand....
Find the current gain of the following circuit.
http://img17.imageshack.us/img17/2389/opamp.png [Broken]
Solution: As the input signal Vi is connected to non-inverting input, so the op-amp acts as a non-inverting amplifier. We have:
R1=$$\infty$$ (infinite) and R2=0
Gain=1+R2/R1=1

My first question is that as you can see in the figure, both inverting input and the noninverting input are connected to the output circuit, how does this affect things?
Second, please explain this: R1=$$\infty$$ (infinite) and R2=0

Last edited by a moderator: May 4, 2017
2. Sep 12, 2009

### Redbelly98

Staff Emeritus
R1 and R2 do not appear in the figure, which seems to be the confusing part.

Does the book have a figure of a general non-inverting op-amp circuit? Perhaps R1 and R2 refer to resistors in that circuit, and when they are set to ∞ and 0, respectively, you get the circuit you show in post #1.

EDIT: I am thinking of a circuit like this:

3. Sep 12, 2009

### waht

By connecting the output to the input, you are setting up a negative feedback mechanism. This is a very important concept when dealing with op-amps and transistors.

The op-amp has a very large gain (could be 20,000 in practice). This means that a tiny voltage between the non-inverting and inverting inputs will be amplified by gazillion times at the output. In practice the voltage at the output will swing to the supply voltage because it is the maximum available. To keep the op-amp from always swinging that high we have to restrain it so to speak. When the output is hooked up to the input, an equilibrium can be reached and prevent that output from going out of control. And hence a very precise and controlled gain can be achieved with op amps, among many other things.

This is the definition of an op-amp.

4. Sep 12, 2009

### Bob S

Redbelly98's illustration in post #2 is an excellent example of a voltage follower with gain.
The equations are:

1) Vout = A(V+ - V-) [ where opamp gain A is arbitrarily large]
2) Vin = V+
3) V- = R1*Vout/(R1 + R2)
4) Vout/A = Vin - R1*Vout/(R1+R2)
5) Vin = Vout(1/A + R1/(R1+R2))
Let A ==> infinity
6) Vout = [(R1+R2)/R1]*Vin

Bob S

Last edited: Sep 12, 2009
5. Sep 12, 2009

### dlgoff

Actually the output circuit is not connected to the non-inverting input. Maybe the arrow showing the voltage of the output Vo is confusing you. It's not a connection.

6. Sep 14, 2009

### uzair_ha91

Sorry for answering late......Ok I checked a figure of a general non-inverting op-amp circuit...R1 is beside the "triangle" part at its left, and R2 above it

I googled the term "voltage follower" and found this matching figure::

But I am still confused.... what makes the gain in this case unity?

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7. Sep 14, 2009

### Staff: Mentor

In the unity gain follower configuration, the opamp will be doing all it can to keep the two inputs at the same voltage. If the V+ input is slightly above V-, then the output will swing positive, which will tend to raise the V- input to match the V+ input. The high gain of the opamp is what makes the input difference voltage so small.

(BTW, never anthropomorphize opamps. They hate it when you do that.)

8. Sep 14, 2009

### waht

Opamps amplify the voltage difference V2 - V3 by thousands of times

(V3 - V2) * Gain = Vout

Gain = 20,000 (for 741 chip)

So, V3 - V2 must very small if Vout is to equal V3.

In a voltage follower, as the output voltage increases

V3 - V2 starts to become small, and thus lowering the output voltage, the tug of war will continue until V3 - V2 becomes so small that when amplified thousands of times will equal the input voltage.

9. Sep 14, 2009

### Mr.Green

Hi every body
I think the non-inverting opamp is more understandable in figure below:

The gain is given by:
G = 1 + R2/R1
The proof should be available in any book.

For a voltage-follower:

R1 = Infinity [open circuit between inverting input and GND]
R2 = Zero [short circuit between input and output]

So, G = 1 + zero = 1

OR... look at it this way:

The inverting input and the output is short circiuted, so the output voltage is the same voltage of the inverting input. However, the voltage of the inverting voltage is approximately equal to that of the non-inverting voltage (due to the negative feedback). Eventually, the output voltage is equal to the non-inverting voltage (which is the input voltage).

Vin = Vout

10. Sep 14, 2009

### uzair_ha91

I think I'm beginning to understand...thanks.