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So if I have a summing amplifier circuit like the one below: Summing Amplifier

I know the Vin1, Vin2, and Vout value and the circuit is designed such that Vout = -(Vin1+2Vin2) but how do I calculate the gain? I know on a normal op-amp circuit with one input source the gain is just Vout/Vin. But if I have two input sources then will my gain be Vout/(Vin1+Vin2)?

On PSPICE I set the DC and ACMAG values for the input voltages to 2V, and Vout is -6. So it works out right, but still what is the gain?

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    \$\begingroup\$ \$V_{out} = -(V_{in1}\frac{R_3}{R_1} + V_{in2}\frac{R_3}{R_2})\$. You can have different gains for different input sources, so to define "gain" you have to specify which source you're talking about. \$\endgroup\$ – Bitrex Feb 17 '13 at 1:49
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Gain is set for each \$ V_{in} \$.

\$ A_{Vin1} = -\dfrac{R_3}{R_1}\$.

\$ A_{Vin2} = -\dfrac{R_3}{R_2}\$.

and so on, \$ A_{Vin N} = -\dfrac{R_3}{R_N}\$.

If you were to analyze this circuit using superposition, you would find that for each input signal, it is just an inverting amplifier. The signals are amplified or attenuated individually. Therefore, gain is only a meaningful quantity with respect to each individual signal. Since gains are all independent of each other, summing them does not yield a meaningful result.

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  • \$\begingroup\$ Is there such a thing as total gain? If so, is it \$ A_{total}= -\frac{R_3}{R_1} - \frac{R_3}{R_2} \$? \$\endgroup\$ – Richard Feb 17 '13 at 2:07
  • \$\begingroup\$ The signals are being amplified or attenuated individually and summed. As bitrex said his comment, the gain in the summing amplifier can only be quantified with respect to a single signal. Trying to do the same for the entire system yields a meaningless result. \$\endgroup\$ – Matt Young Feb 17 '13 at 2:17
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The purpose of a summing amplifier is to generate an output that is a weighted sum of its input signals. Because of the high open loop gain of the operational amplifier, pin 2 is a virtual ground, i.e. the voltage between pins 2 and 3 is close to zero. Because of this, the 2 input signals are effectively isolated from each other. The gain to the output of each input is determined by the ratio of the feedback resistor, R3, to the particular signal's resistor. Thus the gain for input 1 is -R3/R2 and the gain for input 2 is -R3/R2. The isolation allows each of the input signals to have different gains. (More than 2 signals can be summed by simply adding another resistor with another input signal) The overall output is the sum of these two signals. There is no quantity as "total gain" since there are 2 separate input signals.

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If you put 2V on each input and get -6V out, and R1=R2, then the ratio of R3 to R1 is 1.5 to 1. The gain of each input is -1.5, negative due to the inverting nature. Instead of DC inputs, if you put a 1V AC signal on both inputs (both in-phase and same freq), then you'll get 3V AC out.

Since you show +/-7V supplies, if you put much more than 2V DC in, the output will saturate (limit at ~ -6.5V).

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