# Designing voltage summer without op-amp

The above circuit is from Operational Amplifier Adder.

Now remove feedback resistor and op-amp, and get the output. From the answer of the question, I know how the end result for Vout would be, assuming load resistance exists.

But now suppose that there are N inputs, and there is some load resistance RL. Can anyone show me a simple way or equation of quantifying how load resistance makes the sum result diverge from the ideal value? (The question comes from the fact that adding parallel resistances involve ||, which when expressed into final output takes a lot of time to expand and I still do not know whether there exists a general formula for any N inputs.)

• As far as I know there IS no easy/simple way to solve this. With a load resistor in place of the opamp the circuit will ONLY work if the Rload as significantly smaller in value than R1, R2, etc. Only then the voltage across Rload will remain small so R1,R2 etc convert V1, V2 etc into a current which is summed up and flows through Rload. If the resistors can be anything you will have to solve all network equations resulting in something complex. Sep 9, 2015 at 14:05

The output voltage is

$$\begin{split} V_{\text{OUT}} &= R_1||R_2||..||R_n||R_L \cdot \sum{(V_i/R_i)} \\ &= \biggl(R_1||R_2||..||R_n\biggr)||R_L \cdot \sum{(V_i/R_i)} \\ &= \frac {R_K R_L}{R_K+R_L}\cdot \sum{(V_i/R_i)} \end{split}$$

where $R_K = R_1||R_2||..||R_n$

If you consider the ideal output voltage is when $R_L = \infty$, then

$$\frac {V_{\text{OUT}}}{V_{\text{IDEAL}}} = \frac {R_L}{R_K + R_L}$$

To look at it intuitively, $R_K$ is simply the Thévenin equivalent source resistance of the divider without the load resistor.

As Spehro shows a passive summation circuit can be a simple way to combine signals but one disadvantage is that the summed signal will backfeed into each of the sources. This could be an issue if you wanted to use the source unmixed as well as combined - you would get crosstalk.

The amount of crosstalk will depend upon the impedance of the source and a low enough output impedance this wouldn't be noticed.