Op-amp summing node currents — does current flow “into” a virtual ground? Consider the inverting-input summing node of an op-amp with negative feedback. What correctly describes the current at that node?

Introduction / Context:
Students sometimes think a “virtual ground” is a magical sink for current. In ideal op-amp analysis, input bias currents are approximately zero, so KCL at the summing node tells us that currents arriving via input resistors leave via the feedback network, not into the op-amp input pin. This clarifies how inverting gain equations are derived and why the op-amp input node must not be used as a power return.

Given Data / Assumptions:

• Ideal op-amp behavior: infinite input impedance, zero input current.
• Negative feedback keeps the inverting node at approximately the noninverting input voltage.
• Linear operation (no saturation).

Concept / Approach:
At the summing junction, KCL applies: the algebraic sum of currents is zero. With i_in− ≈ 0 A, the current through each input resistor is balanced by the current through the feedback path. The node voltage being near 0 V does not imply the node can accept arbitrary current; it merely simplifies the gain relationship v_out = −(Rf/Rin)*v_in for the inverting stage.

Step-by-Step Solution:

Verification / Alternative check:
Measure input bias in datasheets (nA to pA). SPICE shows node current into the input terminal is orders of magnitude smaller than through Rin and Rf. Therefore, practical designs account for tiny bias only in precision applications (add bias-cancellation resistors).

Why Other Options Are Wrong:

Common Pitfalls:
Using the summing node as a return for other circuits; neglecting bias current errors in precision designs; ignoring finite open-loop gain which makes the node only approximately at 0 V.

Final Answer:
Branch currents enter and leave the node through resistors, but essentially none flows into the op-amp input terminal