In a non-inverting op-amp, the negative feedback input (V1) is calculated by (R2/Rf+R2)*Vout because it is a potential divider. However, why do we calculate the voltage across R2 and not for example Rf (i.e Rf/(R2+Rf))?
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\$\begingroup\$ Because the potential difference across Rf is not V1 but Vout - V1. \$\endgroup\$– sarthakCommented Dec 2, 2017 at 17:32
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3 Answers
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However, why do we calculate the voltage across R2 ...
Because the amplifier output becomes stable when \$ V- = V+ \$.
\$ A \$ is a very large number. Any difference in the input voltages, \$ V_+ - V_- \$ is multiplied by \$ A \$ and appears on the output.
The essence of the negative feedback (not just in op-amps) is to correct the output and reduce the difference between the setpoint (\$ V_+ \$ in this case) and the feedback (\$ V_- \$) as close to zero as the gain allows.
answered Dec 2, 2017 at 14:17
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Because that's what a voltage divider does. In this case, we calculate the voltage across R2 since that's the voltage we want to know. The voltage across R2 is the voltage being applied to the opamp negative input.
For a simple explanation and derivation of negative feedback, see my answer https://electronics.stackexchange.com/a/50472/4512.
answered Dec 2, 2017 at 14:16
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You're using a formula you memorized rather than just finding the answer with Ohm's law. The current through Rf and R2 is Vout/(Rf + R2). So the voltage at the inverting input is Vout/(Rf+R2) * R1. If you wanted to, you could also figure it as Vout - (Vout/(Rf+R2))*Rf, which simplifies to the same formula.
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\$\begingroup\$ That clears it up a little, but I'm afraid my problem is with the voltage divider part. Why does Vin equal to voltage across R2? Rf also has a line that meets R2 and then goes into the op-amp. \$\endgroup\$– FadelCommented Dec 2, 2017 at 15:09
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\$\begingroup\$ Rf is in the formula you've been using, but it's in the denominator. The alternate expression I gave in my answer: Vout-Vout*Rf*/(Rf+R2) gives the inverting input voltage as the output voltage minus the drop across Rf. There's not much point in starting off that way, though, since the alternate expression simplifies to your original formula. \$\endgroup\$– user128351Commented Dec 2, 2017 at 17:54