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

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    \$\begingroup\$ stevenvh's answer to your previous question answer this well. \$\endgroup\$ – MikeJ-UK May 15 '12 at 13:31
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    \$\begingroup\$ Did you try to research about it? \$\endgroup\$ – clabacchio May 15 '12 at 16:02
  • \$\begingroup\$ @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... \$\endgroup\$ – abdullah kahraman May 15 '12 at 16:03
  • \$\begingroup\$ 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 \$\endgroup\$ – user1083734 May 15 '12 at 16:30
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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:

enter image description here

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_-) \$

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

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