Summing two voltages without the use of an operational amplifier?
Well, if you can live with approximations and a big output impedance, it can be done with a simple resistive network:







simulate this circuit &ndash; 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 &gt;&gt; R_{i1}\$ and \$R &gt;&gt; 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.