# Using the superposition theorem to solve for the output voltage

I'm working on diodes again and thought I'd challenge myself to hopefully learn a bit more. I've tried solving this problem before, but the superposition theorem didn't come to mind then, but I think it'll be a good approach in this case. Either way, here's the problem:

We are given three input voltages $$\u_1,u_2,u_3\$$ whose voltages over time are described on the right. We've also assumed that $$\R_1\$$ is much smaller than $$\R_2\$$. Also, the diodes are ideal. We basically want to find $$\U_{out}\$$. So using the superposition principle, we can redraw the circuit for three different scenarios:

So, we zero all the other input voltages expect the first one. So diode $$\S_1\$$ is conducting current if and only if $$\V_{R_1} > V_1\$$. Also, we have to have that $$\ E > V_{R_1}\$$ otherwise we wont get current from anywhere flowing through the diode. But that just means that $$\ V_1 < E \$$. This only happens when $$\V_1 = 0\$$ from the diagram, but since the diode is ideal, no voltage drop over the diode exists, so $$\V_{R_1} = 0 V\$$ and therefore the output voltage $$\U_o = U_{out} = 0 V\$$ for $$\ V_1 < E \$$.

If the diode is not conducting any current, then $$\V_1 > V_{R_1} = U_{o}\$$ and $$\E =V_{R_1}\$$, so together, $$\V_1 > E\$$ which never happens according to the diagram. Therefore, the diode $$\S_1\$$ is always conducting, and $$\V_1 = U_{out}\$$ in our first case.

I'm unsure of my second argument. I also tried to use the same argument for the case when $$\u_2\$$ and $$\u_3\$$ but I couldn't really get anywhere. The final idea is to add these voltages contributed to our output voltage in a diagram. Any tips would be greatly appreciated.

Thank you once again.

• Superposition only works for linear systems. Commented Oct 16, 2022 at 12:18
• @Mattman944 Of course!!! That wasn't so clever of me. Commented Oct 16, 2022 at 18:01

It seems that you should simplify the problem by combining U2 and U3 into what I've called U4: -

Then, because of these words: -

We've also assumed that R1 is much smaller than R2

We can make another simplification: -

Can you run with it from here?

Using the superposition theorem to solve for the output voltage

To be a good EE means that one of your goals is to read circuit diagrams proficiently and, recognize the important things. This is what my answer is about; seeing simplifications and capitalizing on those simplifications. No need to use math to bludgeon your way to a solution until you have made the obvious simplifications. And, in this example, no need to use math at all; just your eyes and your brain. Final picture: -

• I understand every simplification you've made except for the last one. I can't really figure out, just by eyeballing, how you got the output voltage from the other two voltages. I've even tried it mathematically but I can't figure it out. How was your thought process when completing the last diagram? I'd really like to know since it's a clever solution by skipping all tedious math. Commented Oct 16, 2022 at 18:16
• The lowest voltage defines the output voltage hence, the output is lowest(U1, U4) @Tanamas Commented Oct 16, 2022 at 18:32
• I'm sorry again, but I can't really see why it has to be the minimum of the two voltages. Commented Oct 16, 2022 at 18:43
• Because of the direction of the diodes. Any diode input being low forces Uout to be low so, the input with the lowest voltage at any moment in time defines the output @Tanamas Commented Oct 16, 2022 at 18:49
• Now I get what happens! Thanks once again. Commented Oct 16, 2022 at 18:59