I am a fresher and have been introduced to a basic electronics course at college. Recently our professor was introducing us to opamps and was referring to negative feedback as a rectification to any temperature changes in the amplifier.
Here is a schematic of a simplified opamp under -ve feedback:

enter image description here

Now our professor did Voltage gain calculation for a Dc input source:
Let the gain of the amplifier be "A".
$$ V_o= A(V_{in} -V_{o}(\frac{R_{2}}{R_2+R_1})) $$ We could write this since no current enters the amplifier due to \$\infty\$ (ideally) input resistance. On rearranging terms, we get $$ V_o= \frac{AV_{in}}{1+\frac{AR_2}{R_1 +R_2}} $$ now since A>>1 we get: $$ V_o= V_{in}\frac{R_1 +R_2}{R_2} $$ Now we can do all this for an AC source as well and algebraically the only change we get is \$V_{o}\$ is replaced by say \$V_{in}sin(wt)\$ where \$w\$ is the frequency of small signal input.
But If I consider the negative feedback logic i.e. if \$V_{in}\uparrow\$ \$\implies\$ \$V_{o}\uparrow\$ \$\implies\$ \$V^{-}\uparrow\$ \$\implies\$ \$V_{o}\downarrow\$. Which contradicts the fact that \$ V_o= V_{in}sin(wt)\frac{R_1 +R_2}{R_2} \$ can be possible by the above algebra(if it doesn't please explain) what mistake is there in my understanding? kindly explain..
Thank you

  • \$\begingroup\$ Negative feedback has little or nothing to do with temperature compensation. An ideal opamp has infinite gain, which is actually not that useful. Negative feedback allows you to get the gain you actually want. \$\endgroup\$
    – Finbarr
    Jun 3 at 12:34
  • \$\begingroup\$ @Finbarr I think it kind of does, in the sense that if due to temperature changes A increases then Vo increases hence V- increases and Vo then decreases thereby partly nullifying the effect of increase in temperature. \$\endgroup\$ Jun 3 at 12:41

1 Answer 1


If you have an AC source, the Vin term now contains the sinusoid. Vo is always a function of Vin alone, it doesn't care about time or frequency at all. If Vin changes, Vo changes in the same way, but that doesn't change the equations.

  • \$\begingroup\$ So I think I shouldn't bother about the negative feedback logic and focus on algebra right because that covers how things will be governed. \$\endgroup\$ Jun 3 at 12:42
  • \$\begingroup\$ @Damstridium How did you get that from my answer? What I'm trying to say is that you did your algebra wrong. \$\endgroup\$
    – Hearth
    Jun 3 at 12:43
  • \$\begingroup\$ here is a citation of my algebra :khanacademy.org/science/electrical-engineering/ee-amplifiers/… \$\endgroup\$ Jun 3 at 12:44
  • \$\begingroup\$ see at 4:30 he gets the same ans \$\endgroup\$ Jun 3 at 12:45
  • \$\begingroup\$ Sir I am just introduced to this concept and am willing to accept that I am wrong Please let me know how and why will Ac equations be different \$\endgroup\$ Jun 3 at 12:49

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