0
\$\begingroup\$

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))?

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Because the potential difference across Rf is not V1 but Vout - V1. \$\endgroup\$ – sarthak Dec 2 '17 at 17:32
1
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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\$ – Fadel Dec 2 '17 at 15:09
  • \$\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\$ – stretch Dec 2 '17 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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