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I am following this reference for noise calculations of a non-inverting op-amp.

I am not able to understand why the equivalent resistor is the parallel combination of R1 and RF as shown below:

enter image description here

I couldn't find any explanation for this anywhere. All application notes just state consider the parallel combination without any explanation.

Maybe I am missing something obvious here but while trying to analyse the circuit the current source is throwing me off.

Will virtual ground still be valid here now that a current source is present between the terminals?

If not, how do we go about this then? If yes, the inverting node becomes 0V, in which case all current should have flown through R1 as it gets shorted. In neither of the cases, do we get parallel combination of R1 and Rf.

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    \$\begingroup\$ Simply apply the superposition principle for the "noise source". \$\endgroup\$
    – G36
    Commented Jun 29, 2021 at 16:32
  • \$\begingroup\$ But apart from noise source no other source is present. \$\endgroup\$
    – needbrainscratched
    Commented Jun 29, 2021 at 17:28
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    \$\begingroup\$ @needbrainscratched What's powering the op amp? \$\endgroup\$
    – Hearth
    Commented Jun 29, 2021 at 18:37
  • \$\begingroup\$ In the noise model no source is considered so that the output you get is due to noise. \$\endgroup\$
    – needbrainscratched
    Commented Jun 29, 2021 at 18:38
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    \$\begingroup\$ The op amp output is a low impedance and in your example, its VF1 is zero volts, the same voltage as your ground. So the two resistors are effectively in parallel. \$\endgroup\$
    – John Birckhead
    Commented Jun 29, 2021 at 21:09

1 Answer 1

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The reference is wrong.

Write the nodal equation at the inverting input

$$i_n = V_-(1/R_F + 1/R_1) - VF1/R_F$$

and for an ideal op amp with \$V_- = V_+\$ , if \$V_i\$ is applied to the non-inverting input, then

$$V_{out} = VF1 = V_i(1+R_F/R_1)-i_nR_F$$

You can see this without solving the nodal equations as the summing inverting amplifier simply sums the currents fed to the inverting input and multiplies them by \$R_F\$

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