# Multiple diode circuit analysis

1. Feb 19, 2014

### etf

It is well known that in order to solve diode circuits we must assume state of diodes, replace diodes with appropriate model (0.7V voltage drop) and solve circuit. Then we check result and if it agrees with initial assumption, we successfully solved our circuit.
If we mark number of diodes in circuit with n, there will be 2^n possible combinations of diode states. What to do in situation with 5 diodes, for example? You will agree that it will be quite impossible to check all combinations.
I was trying to find easiest way to solve multiple diodes circuits so I came up with this idea: I assume that all diodes in circuit are conducting so I replace all of them with 0.7V voltage drop and solve that circuit. Then I form another circuit which consist of 0.7V voltage drops instead of diodes for which I proved that conduct in previous step and open circuit for diodes I proved that doesn't conduct in previous step. Then I solve that circuit. Does my idea make sense? :)

Last edited: Feb 19, 2014
2. Feb 19, 2014

### .Scott

Sure.

And if your second circuit doesn't work out, create a third with corrections based on what you discovered in the second. Continue until you have a solution - but watch what you are doing.

--- Edit to replace the problem case --

as long as you're dealing with passive circuits, I can't find a problem case that's any worse than trying all combinations.

Clearly, if you end up going back and forth between two combinations, you need to replace your simple 0.7V or nothing with a different model.

Last edited: Feb 19, 2014
3. Feb 19, 2014

### etf

I thought that second step would be last step and there will be no more solutions since I found right diode states in first step. I will analyse your example. Thanks for reply!

4. Feb 19, 2014

### etf

I will upload one example which I solved to ilustrate my method.

5. Feb 19, 2014

### etf

You can see here that everything is completed in second step.

6. Feb 19, 2014

### .Scott

You need to check the results after the second step.
There are cases where they won't check. For example, in the circuit below, you would conclude that the diodes are both off. But on checking you would turn one of them back on again. In fact, they're probably both partially on.

----- Edit to provide a better example.
Code (Text):

┌──Diode──┬──Diode──┐
+1V ──1KΩ──┤         │         ├─ Gnd
└───1MΩ───┴──100KΩ──┘

Last edited: Feb 19, 2014