# Types of amplifier - what is the difference?

What is the difference between an open loop amplifier and a voltage follower (unity gain buffer)?

• stevenvh's answer to your previous question answer this well. – MikeJ-UK May 15 '12 at 13:31
• Did you try to research about it? – clabacchio May 15 '12 at 16:02
• @clabacchio nope, why research when you can ask people questions cheaply and have the answer with no effort, right? Isn't it logical, I think it is the correct thing to do... – abdullah kahraman May 15 '12 at 16:03
• I posted this after recieving a reply on a previous question: electronics.stackexchange.com/questions/31888/…. I did google both amplifiers but wanted a direct comparison, which I could not find. stevenvh had not answered any of my questions at the time of this question's posting. Please do not assume I am trying to cheat answers out of fellow users. I am new to electronics.stackexchange.com and am a novice at all things electronic. I apologise for the way my question appears to lack consideration. I would delete this question if it was not already answered – user1083734 May 15 '12 at 16:30

I just posted this as an answer to another question, but I can reuse it here :-) (without the part about the non-inverting amplifier).

edit
Oops, I hadn't noticed the question is from the same person as the other one.

Let's look at the most simple feedback situation: The opamp will amplify the difference between $V_+$ and $V_-$:

$V_{OUT} = 100 000 \times (V_+ - V_-)$

Now $V_+ = V_{IN}$ and $V_- = V_{OUT}$, then

$V_{OUT} = 100 000 \times (V_{IN} - V_{OUT})$

or, rearranging:

$V_{OUT} = \dfrac{100 000}{100 000 + 1} \times V_{IN}$

That's as good as

$V_{OUT} = V_{IN}$

This is a voltage follower, a $\times$1 amplifier, which is mostly used to get a high input impedance and a low output impedance.

The feedback reduces the very high amplification to $\times$1. Note that the high amplification is needed to get $V_{OUT}$ as close as possible to $V_{IN}$.

The open loop amplifier will have the high amplification you can see in the transfer function

$V_{OUT} = 100 000 \times (V_+ - V_-)$

• People learn much here right? It's like you are having a lesson with your grasshopper :) – abdullah kahraman May 15 '12 at 16:02
• @abdullah - I think he asked this without even waiting for answers to the other question... – stevenvh May 15 '12 at 16:03
• Ah, this is so lame.. Anyways, you get a +1 from me for all your efforts. – abdullah kahraman May 15 '12 at 16:04
• @abdullah - Thanks! OK, this one was not so hard, mostly copy and paste, but the other answer I typed out completely all by myself! ;-) – stevenvh May 15 '12 at 16:09
• @stevenvh I really appreciate you helping this beginner out. Thanks – user1083734 May 15 '12 at 16:31