0
\$\begingroup\$

I don't understand how Sedra/Smith got \$R_{id} = 2R_1\$. Why doesn't any of the current go through the grounded \$R_2\$?

Pg. 75 of Sedra/Smith Intro to Microelectronics 6E

\$\endgroup\$
2
\$\begingroup\$

The current does flow through the lower R2, but the opamp returns it (via the power supplies) to the upper R2, which is exactly equivalent to shorting the two inputs of the opamp together.

\$\endgroup\$
  • \$\begingroup\$ Thanks you are right!! I drew the diagram, it makes sense now. \$\endgroup\$ – hesson Apr 27 '14 at 18:08
0
\$\begingroup\$

Input current doesn't go to ground via R2 because there is no return path to allow it. If you want to model the input as a differential voltage (Vd) riding on top of a common mode voltage (Vcm), then you do have a return path, and the common mode impedance is R1 + R2. But the differential input impedance remains 2R1.

\$\endgroup\$
  • \$\begingroup\$ Thanks, but how could \$i_1\$ flow through the virtual short? Technically there is no return path for \$i_1\$ either since the opamp terminals should be open-circuited. \$\endgroup\$ – hesson Apr 27 '14 at 17:29

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.