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Source: Schaums Outline of Electronic Devices and Circuits, Second Edition by Jimmie J. Cathey

In the given solved example from a textbook on electronic devices, the author has use the voltage divider rule after assuming \$V_{s2}=0\$. Clearly for equation 1 to be true, the current in the non inverting terminal \$i\$ must be 0.

Why is this true? In few other examples current is taken as zero in the inverting input terminal. I wish to know the reason behind this assumption and how to know out of the two which input terminal has zero current? Can the current be ever taken zero in both of them?

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    \$\begingroup\$ FYI: EE uses \$ instead of just $ to start and end inline math. \$\endgroup\$
    – The Photon
    Commented Aug 12, 2020 at 4:24
  • \$\begingroup\$ In an ideal OP-AMP input resistance is infinitive and output resistance is 0, also gain is infinitive as well. Thus this infinitive input resistance let to 0 current in input terminals, so input current of both inverting and non-inverting terminals is equal to 0. It worth mentioning its not the case in practical circuits, because there is no infinitive input resistance in real OP-AMPs. \$\endgroup\$
    – arsalan ghasemian
    Commented Aug 12, 2020 at 4:32
  • \$\begingroup\$ So it's always the input terminal (the one that is not grounded), that has zero current? \$\endgroup\$
    – Exulansis
    Commented Aug 12, 2020 at 4:33
  • \$\begingroup\$ @Exulansis Theses are IDEAL mathematical models, not what you will find in real-world conditions. Ideal models are used to keep the math simple. Including bias and offset currents would greatly complicate the simple math lesson. \$\endgroup\$
    – user105652
    Commented Aug 12, 2020 at 4:36
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    \$\begingroup\$ @Exulansis, there are two input terminals on an op-amp, and in an ideal op-amp both have zero current. \$\endgroup\$
    – The Photon
    Commented Aug 12, 2020 at 4:39

2 Answers 2

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It's one of the ideal op-amp assumptions that the current into either input pin is zero.

In even a crummy real op-amp it's generally less than 1 uA. In FET-input op-amps it can be as low as femtoamps.

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  • \$\begingroup\$ How do I know which one has zero current? \$\endgroup\$
    – Exulansis
    Commented Aug 12, 2020 at 4:32
  • \$\begingroup\$ @Exulansis, both inputs have approximately zero current. \$\endgroup\$
    – The Photon
    Commented Aug 12, 2020 at 4:37
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Can the current be ever taken zero in both of them?

Pretty much most of the time. Your circuit shows a passive mixer of two voltage sources (VS1 and VS2) feeding an op-amp and, the most common application for this is in audio. That means you can reliably take the current into the inputs as being zero. Even though it isn't exactly zero, it will have little consequence for audio mixing if you assumed that it is.

Why is this true?

Strictly speaking it isn't true but it makes little difference to assume that it is.

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