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If you connect a delicate sensor directly to a heavy load, the sensor voltage drops instantly because the load draws too much current. How do we boost signal voltage while drawing practically zero current from the source?
An operational amplifier (op-amp) is an integrated circuit that amplifies the voltage difference between two inputs: the non-inverting terminal (V+V^+) and the inverting terminal (Vβˆ’V^-).
An ideal op-amp has infinite input resistance, meaning zero input current (I+=Iβˆ’=0Β AI^+ = I^- = 0\text{ A}), and infinite open-loop gain.
In open-loop operation, the op-amp responds purely to its differential voltage dependence: Vout=A0(V+βˆ’Vβˆ’)V_{out} = A_0(V^+ - V^-), where A0A_0 is the open-loop gain (typically 10510^5 to 10610^6).
In open-loop operation, the op-amp exhibits extreme differential voltage dependence described by Vout=A0(V+βˆ’Vβˆ’)V_{out} = A_0(V^+ - V^-), where A0A_0 is the open-loop gain (typically 10510^5 to 10610^6).
Without feedback, even a microvolt difference forces the output to swing fully to the positive supply rail (+Vs+V_s) when V+>Vβˆ’V^+ > V^-, or down to the negative supply rail (βˆ’Vs-V_s) when V+<Vβˆ’V^+ < V^-.
Because A0A_0 is enormous, if V+>Vβˆ’V^+ > V^-, the output swings instantly to +Vs+V_s; if V+<Vβˆ’V^+ < V^-, it swings to βˆ’Vs-V_s.
What happens when we route a portion of the output voltage back to the inverting input?

The Non-Inverting Topology

In a non-inverting amplifier, the input signal VinV_{in} connects straight to the non-inverting terminal (V+V^+), ensuring the signal source experiences an ultra-high input impedance.
A voltage divider formed by feedback resistor RfR_f and ground resistor R1R_1 feeds a fraction of the output voltage back into the inverting terminal (Vβˆ’V^-).
In 1927, Bell Labs engineer Harold Black discovered that feeding an amplified signal back into an amplifier's opposite input eliminates distortion and stabilizes gain.