Multiple diode circuit has several solutions: which one to choose?

Question

Here is very interesting discussion on this topic: Proof that every circuit with diodes has exactly one solution

And this is my example:

Let's start. We have 3 generic diodes with cut-off voltage = 0.7 V. Voltage Vb=-0.7 V, Va = 0 V.

I assume D1 is off, others are on. Then the current across R1 is the same as across D2. Equation for the node B:

I(D2) + I(D3) = I(R3);
(5-0)/8k + I(D3) = (-0.7+5)/2k;
I(D3) = (-0.7+5)/2k - (5-0)/8k = 2.15 - 0.625 = 1.525 mA.

The current through D3 has positive value, also, the assumption is correct.

Now I assume that D2 is off, others are on.

(5-Va)/8k = (Va+10)/4k;
Va = -4.53 V.

I(D3) = (-0.7+5)/2k = 2.15 mA.

The current is positive. Assumption is correct. LTSpice confirms it.

My question is: if any given multiple diode circuit has several solutions, which one must be chosen as a correct one?

P.S. It was suggested in the answers that in case if several solutions were found, then the consistency of the whole circuit must be checked for each solution. It's obviously doable for the circuit with 2-3 diodes, but how should one proceed if the circuit contains dozens of diodes and meshes? Is there some industrial/academic method which allows to discard automatically all the inconsistent solutions?

Answer 1

Start with \$D_3\$. It "looks" forward-biased. So let's assume it is active (on.) This means that \$V_\text{B}=-700\:\text{mV}\$. Now, if \$D_2\$ is also active and forward biased, this implies that \$V_\text{A}=0\:\text{V}\$. But if this is the case, then it follows that \$I_{R_1}=\frac{5\:\text{V}-0\:\text{V}}{8\:\text{k}\Omega}=625\:\mu\text{A}\$. But we know that \$I_{R_3}=\frac{-700\:\text{mA}-\left(-5\:\text{V}\right)}{2\:\text{k}\Omega}=2.15\:\text{mA}\$ -- half of which must be present in \$D_2\$ and half of which must be in \$D_3\$. But if \$I_{D_2}=1.075\:\text{mA}\$, then the voltage drop across \$R_1\$ must be greater than \$8\:\text{k}\Omega\cdot 1.075\:\text{mA}= 8.6\:\text{V}\$ or \$\gt 8.6\:\text{V}\$. But that would mean that the voltage at \$V_\text{A}\$ is well below \$0\:\text{V}\$. This contradicts the assumption that \$D_2\$ is active and on. Therefore, we must conclude \$D_2\$ is off and therefore all of \$I_{R_3}=2.15\:\text{mA}\$ must flow through \$D_3\$ (which is on and confirms our earlier assumption about \$D_3\$ being on and active) and none of it through \$D_2\$. Since \$D_2\$ is off, it isolates the circuit into two parts, both of which are independent from each other and easily analyzed, now.

Answer 2

have I made some errors in my first assumption?

Of course you did. Just because you can make one true statement about the circuit, doesn't mean that the assumption was valid. You need to be able to solve the entire circuit without contradicting that assumption.

In other words, you must find ALL the voltages and currents throughout the whole thing, and verify that, say, the voltage across D1 doesn't contradict your original assumption that it is not conducting.

In fact, it should be obvious that since V2 is the most negative potential in the entire circuit, if the anode of D1 is connected to any other point, it must be conducting!