Question 16.3: Calculate the voltage gain of the following circuit....
Calculate the voltage gain of the following circuit.
The analysis is similar to that used in earlier examples. First we note that, as in earlier circuits, the negative feedback forces V_{-} to equal V_{+} , and therefore
V_{-} = V_{+}Since no current flows into the inputs of the op-amp, V_{-} and V_{+} are determined by the potential dividers formed by the resistors.
V_{-} is easy to calculate and is given by
V_{-} = V_{0} \frac{10 KΩ}{10 KΩ + 20 KΩ} = \frac{V_{0}}{3}V_{+} is slightly more complicated to compute, since it is determined by the two input voltages. However, applying the principle of superposition, we know that the voltage on V_{+} will be equal to the sum of the voltages that would be generated if each input voltage were applied separately.
If V_{1} is applied while V_{2} is set to zero, then the resistor connected to V_{2} effectively goes to ground and is in parallel with the existing 10 kΩ resistor that goes from V_{+} to ground. Therefore,
V_{+} = V_{1} \frac{10 KΩ//10 KΩ}{10 KΩ//10 KΩ + 10 KΩ} = V_{1}\frac{5 KΩ}{5 KΩ + 10 KΩ} =\frac{V_{1}}{3}If now V_{2} is applied while V_{1} is set to zero, we have a directly equivalent situation and clearly, because of the symmetry of the circuit,
V_{+} = V_{2} \frac{10 KΩ//10 KΩ}{10 KΩ//10 KΩ + 10 KΩ} = V_{2}\frac{5 KΩ}{5 KΩ + 10 KΩ} =\frac{V_{2}}{3}Therefore if both inputs are applied simultaneously we have
V_{+} = \frac{V_{1}}{3} + \frac{V_{2}}{3}Now since
V_{-} = V_{+}we have
\frac{V_{0}}{3} = \frac{V_{1}}{3} + \frac{V_{2}}{3}and
V_{o} = V_{1} + V_{2}Thus the circuit is a non-inverting adder. This circuit can be extended to have any number of inputs (see Appendix C).