For the discussed inverting summing amplifier, the output voltage Vo is expressed as which of the following?

For the discussed inverting summing amplifier, the output voltage Vo is expressed as which of the following?

• A. -[4 x (V1 + V2)]
• B. -[2 x (V1 + V2)]
• C. -[4 x (V1 - V2)]
• D. -[4 x (V1 + V3)]

Answer: (A)

The key idea is that an inverting summing amplifier adds weighted versions of its inputs, with a minus sign dictated by the feedback. The output follows Vo = -Rf*(V1/R1 + V2/R2). If the two input resistors are equal (R1 = R2) and the feedback resistor is four times that value (Rf = 4R), the expression becomes Vo = -4*(V1 + V2). So the output is four times the sum of the inputs, with a negative sign because the configuration is inverting. This matches the given circuit when those resistor relationships hold. A result like -2*(V1 + V2) would require Rf/R = 2, not 4. A form with (V1 − V2) would imply one input is effectively inverted or weighted differently, which isn’t the case for two equal inputs in a straightforward inverting summer. And involving V3 would mean a third input is present, which isn’t part of this circuit.

The key idea is that an inverting summing amplifier adds weighted versions of its inputs, with a minus sign dictated by the feedback. The output follows Vo = -Rf*(V1/R1 + V2/R2). If the two input resistors are equal (R1 = R2) and the feedback resistor is four times that value (Rf = 4R), the expression becomes Vo = -4*(V1 + V2). So the output is four times the sum of the inputs, with a negative sign because the configuration is inverting.

This matches the given circuit when those resistor relationships hold. A result like -2*(V1 + V2) would require Rf/R = 2, not 4. A form with (V1 − V2) would imply one input is effectively inverted or weighted differently, which isn’t the case for two equal inputs in a straightforward inverting summer. And involving V3 would mean a third input is present, which isn’t part of this circuit.