Is there a way of summing two voltages without the use of an operational amplifier? http://www.electronicspoint.com/threads/add-two-voltages-without-the-use-of-a-opamp.242875/ This article said to use an operational amplifier but I want to see if its possible without one. Is there a way to make a circuit that accomplishes what the article says without an op-amp? Theoretically, is there a circuit that can sum the voltages of two voltage sources without the use of an op-amp and without hooking up the two voltage sources in series?


2 Answers 2


Well, if you can live with approximations and a big output impedance, it can be done with a simple resistive network:


simulate this circuit – Schematic created using CircuitLab

\$R_{i1}\$ and \$R_{i2}\$ are the internal resistances of the voltage sources, \$R\$ and \$R_{0}\$ are the resistors you must add. The thing work if \$R\$ is much bigger than both internal resistances, i.e. \$R >> R_{i1}\$ and \$R >> R_{i2}\$.

In fact, let's call \$V_0\$ the voltage across \$R_0\$:

$$ V_0 = V_1 \cdot \dfrac { R_0 \parallel (R_{i1} + R) } {(R_{i1} + R) + R_0 \parallel (R_{i1} + R)} + V_2 \cdot \dfrac { R_0 \parallel (R_{i2} + R) } {(R_{i2} + R) + R_0 \parallel (R_{i2} + R)} $$

But since the internal resistances in series are negligible:

$$ V_0 \approx V_1 \cdot \dfrac { R_0 \parallel R } {R + R_0 \parallel R} + V_2 \cdot \dfrac { R_0 \parallel R } {R + R_0 \parallel R} = (V_1 + V_2) \cdot \dfrac { R_0 \parallel R } {R + R_0 \parallel R} $$

So you end up with a voltage which is proportional to the sum of the sources.

The problem in this solution is that to avoid a big attenuation due to the divider ideally you would want an \$R_0\$ which is much bigger than \$R\$, but this makes the output voltage \$V_0\$ a source with higher impedance, so this may be a problem if you must connect it to another circuit which has lower impedance.


Sure there is. For a certain range you can use the same input stage as the op amp solution (voltage averager) and a voltage doubler such as the LM2765 charge pump.

\$2\cdot ({V1 \over 2} + {V2 \over 2}) = V1 + V2\$


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