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Consider an electronic circuit consisting of linear components plus a number of ideal diodes. By "ideal" I mean they can either be forward-biased (i.e. \$v_D=0\$ and \$i_D\geq 0\$ ) or reverse-biased (i.e. \$v_D\leq 0\$ and \$i_D=0\$).

These circuits can be calculated by arbitrarily declaring each diode either forward-biased or reverse-biased, and setting \$v_D=0\$ for every forward-biased diode and \$i_D=0\$ for every reverse-biased diode. After the resulting linear circuit has been calculated, we have to check whether at every forward-biased diode \$i_D\geq 0\$ and at every reverse-biased diode \$v_D\leq 0\$ is satisfied. If yes, that's our solution. If not, we have to try another set of choices for the diodes. So, for \$N\$ diodes, we can calculate the circuit by calculating at most \$2^N\$ linear circuits (usually much less).

Why does this work? In other words, why is there always one choice that leads to a valid solution and (more interestingly) why are there never two choices that both lead to valid solutions?

It should be possible to prove that on the same level of rigor with which e.g. Thevenin's theorem is proven in textbooks.

A link to a proof in the literature would also be an acceptable answer.

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    \$\begingroup\$ Because a physical circuit can be in a single state at a time only. It's not quantum mechanics... \$\endgroup\$ – Eugene Sh. Jul 25 '16 at 13:39
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    \$\begingroup\$ @EugeneSh.: That's true, but that isn't what the OP is asking. Some circuits can be in any one of a number of different states given identical external conditions. The question is to prove that there is only one such state for the class of circuits that the OP is describing. \$\endgroup\$ – Dave Tweed Jul 25 '16 at 13:49
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    \$\begingroup\$ @Eugene Sh.: e.g. a flip flop (or any bi-stable circuit) is a counterexample of a circuit that has more than one solution. If there are no "same inital condition" given, you have to assume any conditions and look which stable solutions are available and then you find that some circuits have only one no matter what initial conditions (e.g. linear circuits) and others have more than one. \$\endgroup\$ – Curd Jul 25 '16 at 14:29
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    \$\begingroup\$ @EugeneSh. The point here is to prove that the diode circuit's steady state behavior does not depend on initial conditions, there is only one stable solution. Unlike a flip-flop, which has multiple stable solutions and can be used as a memory element (the "initial conditions" is a memory write). \$\endgroup\$ – Evan Jul 25 '16 at 14:40
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    \$\begingroup\$ @EugeneSh. The point is not that a non-linear circuit can be in a well defined state given the initial conditions, but just the opposite. The theorem to which the OP refers guarantees that there is just one solution regardless of initial conditions, which is rather peculiar for a nonlinear circuit. \$\endgroup\$ – Lorenzo Donati supports Monica Jul 25 '16 at 15:18
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I assume this is for a contrived problem where there is a circuit with known passives and some I's and V's given and spots marked for diodes of unknown direction. My answer is:

Hopefully the creators of the problems have constrained themselves to instances where their assumptions lead to their conclusions.

It could be theoretically unsolvable by having a diode be extraneous; consider grounding both sides of a diode. There could be non-trivial cases using virtual grounds or other equal voltages that could be hard to spot.

There surely could exist valid circuits that only differ by the direction of a diode for any value of "valid circuit" that includes diodes. Consider modeling switches using those ideal diode rules, how can you decide if a switch was meant to be on or off? Hopefully the given currents and voltages give enough hints. And hopefully they haven't given you conflicting hints.

This shifts the question to "How can you tell if an instance has enough information to be unique?" I remember the answer being something like you need one independent given for each independent unknown, but I'm sure I couldn't prove that or come up with a general test for the independence of either.

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For ideal diodes, there can be multiple solutions.

Trivial counterexample: Take any circuit containing ideal diodes which you have solved. Now replace one of the ideal diodes with, if forward-conducting, a pair of diodes connected in parallel, or if reverse-biased, a pair in series, maintaining the orientation in either case. How do you solve for the distribution of current or voltage between the two? You can't, the ideal diode model leads to a convex hull of equally-valid solutions.

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    \$\begingroup\$ You're really stretching the definition of "circuit" here. Two reverse-biased ideal diodes in series create an isoated node between them, and two forward-biased ideal diodes in parallel create an isolated loop. This is not useful in the context of the question. \$\endgroup\$ – Dave Tweed Jul 25 '16 at 14:15
  • \$\begingroup\$ @DaveTweed: How is the circuit post-modification any less a circuit than it was before the change was made? \$\endgroup\$ – Ben Voigt Jul 25 '16 at 14:17
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    \$\begingroup\$ It isn't, but your modification doesn't create a useful distinction. If two ideal diodes join a pair of circuit nodes, the only thing that matters is the total voltage or total current between those nodes; the distribution of the voltage or current between the diodes individually is of no consequence whatsoever. And throwing in an irrelevant term like "convex hull" is just pure technobabble. \$\endgroup\$ – Dave Tweed Jul 25 '16 at 14:24
  • \$\begingroup\$ This is very useful, as it shows there is no hope of a proof of uniqueness without further assumptions. Of course the next question is whether it is sufficient to exclude two diodes in a row and two diodes in parallel, or if there are counterexamples of greater complexity. \$\endgroup\$ – Stefan Jul 25 '16 at 17:34
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I don't have a rigorous proof, but the general idea is that as long as the components of a circuit have V-I curves that are single-valued functions (this includes diodes as well as linear components), there can be only one solution to the circuit overall.

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  • \$\begingroup\$ Kind of induction on a superposition. Base case would be a single-diode circuit, which is easy to show having single solution. Then the induction step to show the combination of the base circuits are having single solution. \$\endgroup\$ – Eugene Sh. Jul 25 '16 at 13:56
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    \$\begingroup\$ However, the ideal diode discussed in the equation doesn't have a single-valued I-V curve. \$\endgroup\$ – Ben Voigt Jul 25 '16 at 14:05
  • \$\begingroup\$ @BenVoigt: When dealing with ideal components and the associated zeros and infinities, you have to be careful. The concept of limits is crucial: the forward resistance is infinitesimal but not zero, and the reverse conductance is also infinitesimal but not zero. When considered in this fashion, the equation is indeed single-valued. \$\endgroup\$ – Dave Tweed Jul 25 '16 at 14:08
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I think ist quite simple:

you can treat the forward biased ideal diodes as shorts and the reversed biased ideal diodes as open circuits. So in any case you get circuits with only linear components (because all diodes either resolve to open circuits or shorts) and those linear circuits are known to have exactly one solution.

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    \$\begingroup\$ But each of those circuits will have a solution -- how do you prove that only one is self-consistent? \$\endgroup\$ – Ben Voigt Jul 25 '16 at 14:14
  • \$\begingroup\$ @Ben Voigt: ok, I understand. That is not yet proven (and probably is the main work) \$\endgroup\$ – Curd Jul 25 '16 at 14:20
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From Wikipedia load lines entryFrom Wikipedia load lines entry

There is only one unique solution because of the nature of the problem. This is best illustrated graphically, in the form of load lines. The diode has an equation that describes the relationship between current through it (y axis), and voltage across it (x-axis). Here, the x-axis is the voltage across the diode.

Look what happens to the current across the resistor as the voltage across the diode changes. If the voltage is Vdd across the diode, then there would be no voltage drop across the resistor, as the voltage across the resistor and the diode must sum to Vdd), and there would thus be zero current across the resistor (Ohm's Law). Similarly, if there were a zero voltage drop across the diode, there would be Vdd across the resistor, and the current through the resistor would be Vdd/R.

Now, we know those to be unrealistic situations, as the current in the diode and resistor must be equal. Given the equation for the resistor (linear) and the equation for the diode (non-linear, but monotonic increasing), we can see on the graph that this can only happen at one unique point, the intersection of the two curves.

Thus, the simultaneous solution of three equations (the resistor, the diode, and the fact that the two currents must be equal) give use one unique solution.

This method will work for all circuit elements.

It's a bit different for reverse-current diodes, as the resistor current goes the other way, and a quadrant needs to be added to the graph.

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  • \$\begingroup\$ The diode I-V-curve you show is not the I-V-curve of an ideal diode. \$\endgroup\$ – Curd Jul 25 '16 at 14:16
  • \$\begingroup\$ @Curd: Given the lack of scale factors, it's close enough. See my comment to Ben Voigt. \$\endgroup\$ – Dave Tweed Jul 25 '16 at 14:17
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    \$\begingroup\$ This is a good explanation for the case with one diode, but my actual problem is the case with several diodes. \$\endgroup\$ – Stefan Jul 25 '16 at 17:22
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The 'proof' of this would only work for certain circuits. If you have some gain and the only nonlinear elements are the diodes themselves you can have multiple possible states. For example (may not be the simplest possible example).

This circuit will work with an ideal perfectly linear op-amp and the output never goes off to infinity or saturates, yet with 0V in it can be about +6 or about -6 at the output, with one pair or the other of diodes conducting. It will also work with 'almost ideal' diodes that have a forward drop when on and no other nonidealities.

schematic

(and of course tunnel diodes are a special case with their non-monotonic I-V curve).

The proof would probably have to require only passive elements such as resistors (no dependent current or voltage sources). Or perhaps only with ideal diodes with 0V Vf.

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  • \$\begingroup\$ Isn't it clear that the class of circuits we're talking about here excludes anything with gain, such as any 3-terminal devices or negative-resistance devices? \$\endgroup\$ – Dave Tweed Jul 25 '16 at 17:14
  • \$\begingroup\$ @DaveTweed No it isn't. The original question says 'linear components' which is not restrictive enough, at least for diodes with forward drop. Typical textbook questions have only independent voltage and current sources and resistors and ideal or somewhat ideal diodes. Real and useful circuits usually involve op-amps, IME. \$\endgroup\$ – Spehro Pefhany Jul 25 '16 at 17:18
  • \$\begingroup\$ I did mean what you describe as typical textbook questions. \$\endgroup\$ – Stefan Jul 25 '16 at 17:41
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    \$\begingroup\$ You're right, the question should say "passive" if it means to exclude active-but-linear elements. \$\endgroup\$ – Ben Voigt Jul 25 '16 at 17:42
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This is not a complete proof, but perhaps it will get you on track:

If there are multiple solutions, there is at least one diode that can be either forward or reverse biased. Consider one such diode. In a given solution, it is either forward or reverse biased. Let's define the voltages at its terminals, Va and Vb, such that if it is forward biased, Va >= Vb, and if it is reverse biased, Vb >= Va. In either the forward- or reverse-biased case, the Rest of the Circuit (RotC) produces these voltages at the terminals of the diode.

Since you stated that the circuit consists of linear elements and diodes, either the RotC is a purely linear network, or it includes more diodes.

If the RotC is a purely linear network, it has only one solution, and the only solution to the constraints Va >= Vb and Vb >=Va is that Va = Vb.

If the RotC includes more diodes with multiple possible solutions, consider the next such diode. Again, it is either connected to a linear network, or a network with more diodes with multiple possible solutions.

If we assume there are a finite number of diodes in the circuit...

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