If both inputs of the op amp shown in Fig. 1 are used simultaneously, then from equation, the output voltage is given by:

V_o = A_o(V₂ − V₁)

When V₂ > V₁ then V_o is positive, its maximum value being the positive supply voltage + V_s, which it has when (V₂ > V₁) ≥ V_s/A_o. The op amp is then saturated. For example, if V_s = +9 V and A_o = 10⁵, then saturation occurs when

i.e. when V₂ exceeds V₁ by 90 μV and V_o ≈ 9 V. When V₁ > V₂, then V_o is negative and saturation occurs if V₁ exceeds V₂ by V_s/A_o i.e. around 90 μV in the above example; in this case, V_o ≈ −V_s = −9 V.

A small change in (V₂ − V₁) therefore causes V_o to switch between near +V_s and near to −V_s and enables the op amp to indicate when V₂ is greater or less than V₁, i.e. to act as a differential amplifier and compare two voltages. It does this in an electronic digital voltmeter.