Diode voltage divider problem

Question

For the voltage divider below, I am asked to find the Q-Point for both ideal diodes in forward-bias mode (i.e., they are both short circuits with one-way current):

In my work, I found \$i_1\$ and \$i_2\$; however, I am worried that my assumption that \$i_1 = i_{D_1}\$ does not hold and \$i_1 = i_{D_1} - i_{D_2}\$ instead.

My question is: does the superposition of mesh currents to obtain branch currents (and by extension, KCL) still apply when it comes to diodes, given, using a mesh analysis procedure in this example, \$i_2\$ shouldn't be permitted to pass through the cathode side of \$D1\$?

I hope this question about my assumption of current makes sense. My work is below:

@Hearth You are correct, thanks for noticing this! I will edit the question to include: please don't consider that an issue, just tell me conceptually whether the branch current through \$D1\$ is superposition of the mesh currents or the mesh currents themselves — Anish G.
– Anish G., Commented Sep 3 at 15:39
@Circuitfantasist You're right. I just realized (with Hearth 's initial comment) that the original circuit I attached was a circuit from the second part of the question (which is unrelated to this post). I have since fixed this post with the correct circuit schematic that matches my work — Anish G.
– Anish G., Commented Sep 3 at 16:16

brhans · Accepted Answer · 2025-09-03 21:16:42Z

My question is: does the superposition of mesh currents to obtain branch currents (and by extension, KCL) still apply when it comes to diodes, given, using a mesh analysis procedure in this example, i2 shouldn't be permitted to pass through the cathode side of D1?

You wrote a pretty intense decoy for that actual question.

Yes, of course that holds. Where should the currents go, if they don't form lightning beams?

I think your whole problem was that you were assuming something paradoxical about your diodes being in forward bias, due to confusing two variants of a circuit.

(note that it makes very little sense to talk about Q-points of ideal diodes; that's a bit boring, considering how their I/V "curves" look like. Maybe that contributed to the confusion; we did very little of this graphical solving for such simple circuits, for good reason, I think. I find it neither overly intuitive, nor is it applicable for anything complicated, whereas you can write down the circuit as system of equations with exponentials if you leave the domain of "ideal" diodes, and then numerically solve that, which actually works (and is how every circuit simulator works. There's not a little program inside that draws curves on virtual paper and marks a crossing point). The core takeaway, that an operating point happens when the constraints from multiple equations are met, can be easily communicated, imho, by showing such a graph once; no use training students in the not-really-working-for-anything-but-toy-examples-or-incredibly-complicated graphical technique from the early 20th century.)

Thanks for your thorough answer. My follow-up question is: Are the mesh currents in mesh analysis an "abstraction" of sorts? Because if it were the case that there was an actual mesh current \$i_2\$ flowing through the right-most (or bottom-most) branch, would it not pass through the opposite direction of \$D_1\$? That's where my confusion occurs, because I understand that the branch current should be flowing in the direction enforced by the diode, but I don't know if the enforced current direction from the diode negates the mesh current entirely (kinda like an ideal current source) or not — Anish G.
– Anish G., Commented Sep 3 at 16:57
no, not an abstraction. This is all measurable. I'm not sure what your point here is; you can superimpose currents, full stop. You're imposing constraints here that aren't based in anything. — Marcus Müller
– Marcus Müller, Commented Sep 3 at 17:28
of course, if you calculate things right, then the currents will be physically sensible. But just because the sum of say 12 currents is positive doesn't mean that every individual current needs to be. — Marcus Müller
– Marcus Müller, Commented Sep 3 at 17:29

Circuit fantasist · Accepted Answer · 2025-09-03 18:06:12Z

Let's find the answer using the simulator. It might not help you successfully submit your homework, but it will help you intuitively understand what is happening in the circuit. Here is a possible scenario.

Scenario

1. We draw the circuit diagram in a way that is easy to understand intuitively: the "positive" elements are above the zero voltage line (ground), and the "negative" ones are below it.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

2. We connect a voltmeter to the output and see that it shows zero volts. This makes us assume that diode D1 is 'on' (forward-biased); obviously, D2 is as well.

schematic

^{simulate this circuit}

3. To be sure, we disconnect D1. We see that the output voltage becomes 3.9 V. This means that D1 was indeed forward-biased.

schematic

^{simulate this circuit}

4. We replace the diodes with a piece of wire (short circuit). To measure the current, we connect ammeters. We can simplify the circuit diagram if we replace the resistors with real ammeters with the same resistance. In place of diode D1, we connect an ideal ammeter.

schematic

^{simulate this circuit}

5. We read the currents from the ammeters.

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