Operational Amplifier (5)

We need to configure $R_{f1}$ and $R_{f2}$ such that $\displaystyle V_{\rm out} = \sum_{i=1}^n V_i$ . First let us find how $V_{\rm in}$ is related to $V_i$ . We note that the current flowing into the non-inverting input (marked by "+") is zero, then we have:

$\begin{aligned} \frac {V_1-V_{\rm out}}R + \frac {V_2-V_{\rm out}}R + \frac {V_3-V_{\rm out}}R + \cdots + \frac {V_n-V_{\rm out}}R & = 0 \\ \sum_{i=1}^n V_i - n V_{\rm in} & = 0 \\ \implies V_{\rm in} & = \frac 1n \sum_{i=1}^n V_i \end{aligned}$

For a non-inverting OpAmp configuration, the closed loop voltage gain is given by:

$\begin{aligned} A_V & = \frac {R_{f2}+R_{f1}}{R_{f1}} \\ \frac {V_{\rm out}}{V_{\rm in}} & = \frac {R_{f2}+R_{f1}}{R_{f1}} & \small \blue{\text{Putting }V_{\rm out} = \sum_{i=1}^n V_i} \\ \frac {\sum_{i=1}^n V_i}{\frac 1n \sum_{i=1}^n V_i} & = \frac {R_{f2}+R_{f1}}{R_{f1}} & \small \blue{\text{and }V_{\rm in} = \frac 1n \sum_{i=1}^n V_i} \\ n & = \frac {R_{f2}+R_{f1}}{R_{f1}} \end{aligned}$

Therefore the answer is $\boxed{R_{f2}=(n-1)R_{f1}}$ .

This is my solution to pretty much parts 1-5 combined.

Designate the current leaving node A across $R_1,R_2$ and $R_3$ be $I_1,I_2$ and $I_3$ respectively. Also let the voltage at node A, which is equivalent to the voltage at the non-inverting input $V^+$ be $V^+$ . Using node-voltage analysis of node A:

By Kirchhoff's current law, since all the currents leave node A, and that no current flows into the OpAmp inputs, $I_1+I_2+I_3=0$ . The currents along each input branch in terms of the voltage $V^+$ :

• $I_1 = \frac{V^+ - V_1}{R_1}$
• $I_2 = \frac{V^+ - V_2}{R_2}$
• $I_3 = \frac{V^+ - V_3}{R_3}$

Then: $\frac{V^+ - V_1}{R_1} + \frac{V^+ - V_2}{R_2} + \frac{V^+ - V_3}{R_3} = 0$ $V^+(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}) = \frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3}$ $V^+ = \frac{\frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3}}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}$

Now that we know $V^+$ , we can easier calculate the output voltage, since it's just a normal non-inverting amplifier at this point. The gain for this is $G = 1 + \frac{R_{f2}}{R_{f1}}$ , so the output voltage of the non-inverting summing amplifier is: $V_{out}=G\times{V^+} = (1 + \frac{R_{f2}}{R_{f1}})(\frac{\frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3}}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}})$

If we set all the input resistances equal to each other, so $R_1 = R_2 = R_3 = R$ for some resistance $R$ :

$V_{out} = G(\frac{\frac{V_1+V_2+V_3}{R}}{\frac{3}{R}})$

$= G(\frac{V_1+V_2+V_3}{3})$

Now this is looking a whole lot like summing together the input voltages!! Notice that in order to make $V_{out} = V_1 + V_2 + V_3$ , the gain must cancel out the denominator, so the gain must equal the denominator, so $G = 3$ . This means that $1 + \frac{R_{f2}}{R_{f1}} = 3 \longrightarrow R_{f2} = 2 R_{f1}$ .

Notice that the denominator of the expression for $V_{out}$ is equal to the number of input branches! This is because for every extra input branch, you get an extra $\frac{1}{R}$ on the denominator. Now you can see that this is why you want the output resistors $R_{f2}$ and $R_{f1}$ to be equal if you have two input branches; the gain must be $$ so $R_{f2} = R_{f1}$ .

We can generalize this for any number of input branches! Notice that the node voltage at node A is always $V^+$ , and no matter how many input branches you have, the sum of all the currents leaving node A through each input branch is $0$ . The numerator of $V^+$ is the sum $\displaystyle \sum_{i = 1}^{n}\frac{V_i}{R_i}$ while the denominator is the sum of the reciprocals of the input resistances $\displaystyle \sum_{i = 1}^{n}\frac{1}{R_i}$ . This allows us to generalize to:

$V^+ = \frac{\displaystyle \sum_{i = 1}^{n}\frac{V_i}{R_i}}{\displaystyle \sum_{i = 1}^{n}\frac{1}{R_i}}$

And therefore:

$V_{out} = G\times{\frac{\displaystyle \sum_{i = 1}^{n}\frac{V_i}{R_i}}{\displaystyle \sum_{i = 1}^{n}\frac{1}{R_i}}}$

Now if we set all the input resistances equal to each other, so $R_1 = R_2 ...= R$ for some $R$ :

$V_{out} = G\times{\frac{\frac{\displaystyle \sum_{i = 1}^{n}V_i}{R}}{\frac{n}{R}}}$

$= G\times{\frac{\displaystyle \sum_{i = 1}^{n}V_i}{n}}$

As you can see, in order for $V_{out} = \displaystyle \sum_{i = 1}^{n}V_i$ (aka the sum of all input voltages) :

$G = 1 + \frac{R_{f2}}{R_{f1}} = n \longrightarrow \boxed{R_{f2} = (n-1)R_{f1}}$