Which statement about non-inverting summing amplifiers is false?

Which statement about non-inverting summing amplifiers is false?

• A. The differential input voltage in linear operation is kept near zero by feedback.
• B. The device can produce a weighted sum without inversion.
• C. The differential input voltage of a saturated circuit is held at 0V.
• D. They have high input impedance.

Answer: (C)

In a non-inverting summing amplifier, negative feedback drives the difference between the input terminals toward zero while the circuit operates in its linear region. This tiny differential input voltage is what allows the output to faithfully reflect a weighted sum of the inputs with positive (non-inverted) gain. The inputs are combined at the non-inverting input, and the feedback network on the inverting input sets the overall gain, so the output is indeed a weighted, non-inverted sum of the signals. The op-amp itself presents a very high input impedance, so the inputs draw negligible current, which is another hallmark of this configuration. The statement about the differential input voltage remaining at 0 V when the circuit is saturated is not correct. When the op-amp saturates, the output is pegged at a rail and the feedback can no longer force the two inputs to equal each other. The differential input voltage can then be nonzero, limited only by the device’s characteristics and finite open-loop gain.

In a non-inverting summing amplifier, negative feedback drives the difference between the input terminals toward zero while the circuit operates in its linear region. This tiny differential input voltage is what allows the output to faithfully reflect a weighted sum of the inputs with positive (non-inverted) gain. The inputs are combined at the non-inverting input, and the feedback network on the inverting input sets the overall gain, so the output is indeed a weighted, non-inverted sum of the signals. The op-amp itself presents a very high input impedance, so the inputs draw negligible current, which is another hallmark of this configuration.

The statement about the differential input voltage remaining at 0 V when the circuit is saturated is not correct. When the op-amp saturates, the output is pegged at a rail and the feedback can no longer force the two inputs to equal each other. The differential input voltage can then be nonzero, limited only by the device’s characteristics and finite open-loop gain.