0
\$\begingroup\$

For an ideal opamp the unity gain buffer has a voltage gain of exactly 1; so Vin = Vout:

enter image description here

But if we neglect all non-idealities except the DC open loop gain Aol, how can we mathematically/analytically demonstrate the effect of the Aol to the error between the input and the output voltage? I have read once that if Aol=infinity then Vin = Vout but if Aol=10000 then Vin is not exactly equal to Vout. But I cannot derive myself why neither I can find a source in internet for derivation of this.

\$\endgroup\$
2
  • \$\begingroup\$ I am really surprised - You did not "find a source in internet for derivation of this"? If I had time and motivation enough - I could give you at least 100 sources....how many minutes did you spend for searching? \$\endgroup\$
    – LvW
    Mar 18, 2020 at 11:10
  • \$\begingroup\$ I guess I was searching with wrong keywords. I should have googled "negative feedback" probably. \$\endgroup\$
    – pnatk
    Mar 18, 2020 at 11:39

1 Answer 1

1
\$\begingroup\$

But I cannot derive myself why neither I can find a source in internet for derivation of this.

Consider this: -

enter image description here

At the point marked with the red X we can say: -

$$(V_{IN}-V_{OUT})\cdot A_{OL} = V_{OUT}$$

$$A_{OL}\cdot V_{IN} = V_{OUT}\cdot(1 + A_{OL})$$

$$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{A_{OL}}{1+A_{OL}}$$

Can you take it from here and plug some numbers in?

\$\endgroup\$
1
  • \$\begingroup\$ Yes this was exactly I was looking for. \$\endgroup\$
    – pnatk
    Mar 18, 2020 at 10:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.