# Opamp as voltage follower

1. Sep 14, 2008

### vvkannan

i have problem in understanding the basics of op-amp
let's consider an opamp as voltage follower (i.e) the output is connected to inverting terminal directly and if V1(say 3v) is applied as input voltage ,to the non-inverting terminal, the output gets saturated immediately +v-cc(power supply,say +12) which is fedback to inverting terminal(V2).now the inverting terminal voltage(12 against the 3 in non-inverting) is more so the output should saturate to -Vcc(say -12) which is fedback to inverting terminal.now V1 -V2 is positive and hence again we get +Vcc (due to very high gain).in this way it again shud act in a non linear way.
but why and how it maintains the voltage nearer to input, given at non-inverting terminal at output)?

2. Sep 14, 2008

### Redbelly98

Staff Emeritus
As the output voltage is changing from +Vcc to -Vcc, at some point it will reach 3V and stay there.

3. Sep 14, 2008

### Averagesupernova

What the OP is describing is an oscillation. vvkannan, are you questioning why op-amps don't simply go into oscillation?

4. Sep 15, 2008

### vvkannan

iam not saying an op-amp shud go into an oscillation supernova.i have trouble in understanding how it maintains the exact (slightly less than exact input) input when connected as a voltage follower.

5. Sep 15, 2008

### Averagesupernova

Well that is the nature of negative feedback. Consider an op-amp with no feedback has a gain of voltage gain of 100,000 or more. The more negative feedback combined with very high open loop gain, the more linearly the output will follow the input. I guess I am having a hard time explaining it (as usual). Think of negative feedback as a correction. A positive voltage on the non-inverting input tells the output to swing in a positive direction. But there is feedback coming to the inverting input. So this 'throttles' the op-amp so-to-speak.
-
The only analogy I can think of that is VERY simplified is a person who has lost a significant part of their hearing. The brain (non-inverting input) tells the mouth to speak. Their ears hear their own voice and send the signal back to the brain. The brain then judges how loud they should be speaking based on what you would call negative feedback from the ears. When a person loses significant hearing ability, there is less feedback to the brain from the ears and they usually end up speaking louder.

6. Sep 16, 2008

### Chandra214

negative feedback helps linearize a system, fine.
similarly here Vo stabilizes at making the difference between the inputs ( inverting and non inverting) to 0.
to have 0 volts between these lines ( virtual Ground) the feed back should be same potential as it is at the other input but with reverse polarity. so as to say, we need two potentials pulling each other to attain equilibrium.

7. Oct 12, 2008

### lukmannet

Emiter follower has gain 1. V(in) = V(out). What for? Why?
Well, an Op Amp functioning as an emiter follower would have VERY HIGH input impedance and VERY LOW output impedance.
Let me give you an illustration:
Imagine you have the project of measuring the terminal voltage of a 100 Mohms resistor in series with other components. If you are using a conventional multimeter with low impedance in parallel with the resistor, absolutely you will never get the exact terminal voltage of the resistor. Why?
Because as soon as you stick your multimeter pins to the resistor terminals, the elctric current then flow into your multimeter causing the voltage to drop.

But if you are using emiter follower, it will not relatively happen as you are measuring the output of the emiter follower which has very low impedance.

Well, I hope you can get my thought.

Regards,

Luky

8. Oct 12, 2008

### Corneo

I think you meant to say large impedance inside the voltmeter. Remember that for an ideal voltmeter the parallel impedance is infinite. Where as the impedance for an ideal current meter is 0.

9. Oct 13, 2008

### lukmannet

Nope. I wrote correctly, low impedance for the instance. Think of resistors in parallel.
As far as I know so far, there are no ideal either voltmeter or amperemeter. They are all theoritic.

10. Nov 18, 2008

virtual short concept

[u have good answer for the voltage follower
but what is virtual short?
is it that the voltage at noninverting terminal is equal to one of the potential of one of the points on the inverting input
because of which the difference(v2-v1) becomes 0
but its doubt ful?????

11. Nov 18, 2008

### Proton Soup

nothing happens instantaneously. the amp has limited bandwidth. or, it may be easier to think of simply the slew rate (V/s). as soon as the output gets a little above 3V, Vi- is greater than Vi+ and the gain reverses direction. this compensation happens long before the output hits "the rails" (power supply voltage).

12. Nov 20, 2008

### vvkannan

Yes now i understand it better .