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The circuit above has 4 IDEAL diodes and there are 16 cases to check. Is there any shortcut to fiqure out which case to check first (to guess which diodes are on or off)?

  • \$\begingroup\$ Do you need to consider loads attached to Vo? \$\endgroup\$
    – The Photon
    Commented Nov 24, 2017 at 10:13
  • 2
    \$\begingroup\$ Just start at the left. 1st diode is on, with ground sourcing current \$I\$ and resistor sinking same. Add next stage, sourcing current \$\frac{I}{2}\$, which lowers sourced current from ground but doesn't violate prior assumptions. Add 3rd stage, sinking \$\frac{I}{4}\$; etc. In short, the ground node at left winds up sourcing the difference between the pull up resistors and the pull down resistors, or 5 parts of 10. The lower two resistors sinking 10 parts and the upper two resistors sourcing 5 parts. Make sense? \$\endgroup\$
    – jonk
    Commented Nov 24, 2017 at 11:18
  • \$\begingroup\$ @jonk unfortunately, "D1 is on", though true in this case, is not guaranteed for all resistor values. SO as a method.. it's not that simple. \$\endgroup\$
    – Trevor_G
    Commented Nov 24, 2017 at 15:00
  • \$\begingroup\$ Which version of "ideal" diode is this? \$\endgroup\$
    – Trevor_G
    Commented Nov 24, 2017 at 15:23
  • 1
    \$\begingroup\$ @Trevor Was merely responding to this example. Had I wanted to respond generally, I'd have written an answer! :) \$\endgroup\$
    – jonk
    Commented Nov 24, 2017 at 19:26

3 Answers 3


The first thing I usually do when facing a problem like this is re-arrange it!


simulate this circuit – Schematic created using CircuitLab

This re-arrangement should make it abundantly obvious that the nodes enclosed by the red rectangular box are pulled towards voltages that are more positive than the nodes enclosed by the blue rectangular box.

This immediately suggests the idea that all the ideal diodes may be ON. Whether or not they actually are will depend on if you can make the currents sum up correctly. In this particular case, it's best to just start with the assumption here and see if the details work out.

Given that these are ideal diodes (which I take in this case to mean that they have zero voltage drop across them when ON, regardless of current) and taking the assumption for now that all the diodes are ON, the nodes enclosed by the red rectangular box are at the same potential as the nodes enclosed by the blue rectangular box. See note to the right side pointing this out. If all our assumptions are correct about the diodes being ON, then this voltage is determined by the grounded node at the left side.

Note that in the end it does turn out that the currents all balance out. I had earlier pointed out in a comment (made when I was nearly asleep last night and barely conscious) that the ground node would supply 5 parts out of 10. You can easily see how this divides out because I've also added the currents I was imagining last night -- nicely spelled out for you.

The above worked out nice because all the currents were in concert with each other. Let's keep the topology and just randomly change the resistor values so that the concert becomes cacophonous, rather than so cutely worked out.


simulate this circuit

The nodal equations look like the following:

$$\begin{align*} \frac{V_1}{R_1} &= \frac{-8\:\textrm{V}}{R_1} + I_{D_1} + I_{D_2}\\\\ \frac{V_2}{R_2} + I_{D_2} + I_{D_3} &= \frac{8\:\textrm{V}}{R_2}\\\\ \frac{V_3}{R_3} &= \frac{-8\:\textrm{V}}{R_3} + I_{D_3} + I_{D_4}\\\\ \frac{V_4}{R_4} + I_{D_4} &= \frac{8\:\textrm{V}}{R_4} \end{align*}$$

Where we know the following:

if D1 is ON then V1 = V0 else ID1 = 0
if D2 is ON then V1 = V2 else ID2 = 0
if D3 is ON then V2 = V3 else ID3 = 0
if D4 is ON then V3 = V4 else ID4 = 0
... and in all cases ID1 >= 0, ID2 >= 0, ID3 >= 0, and ID4 >= 0

So just start walking through from the left. Assume \$D_1\$ is ON, but all the rest are off. Then from the first equation (the others don't yet apply) and our rules, we have:

$$\begin{align*} V_1 = V_0 &= 0\:\textrm{V}\\\\ I_{D_2} &= 0\:\textrm{A}\\\\ \frac{V_1}{R_1} &= \frac{-8\:\textrm{V}}{R_1} + I_{D_1} + I_{D_2}\\\\ \therefore\quad I_{D_1}&=\frac{V_1+8\:\textrm{V}}{R_1}= 0.\overline{36}\:\textrm{mA} \end{align*}$$

Since \$I_{D_1}\ge 0\$, our assumptions are good so far.

Now, change the assumption about \$D_2\$ and turn it ON. Then from the first two equations (the others don't yet apply) and our rules, we have:

$$\begin{align*} V_2 = V_1 = V_0 &= 0\:\textrm{V}\\\\ I_{D_3} &= 0\:\textrm{A}\\\\ \frac{V_1}{R_1} &= \frac{-8\:\textrm{V}}{R_1} + I_{D_1} + I_{D_2}\\\\ \frac{V_2}{R_2} + I_{D_2} + I_{D_3} &= \frac{8\:\textrm{V}}{R_2}\\\\ \therefore\quad I_{D_1} &= \frac{8\:\textrm{V}}{R_1}-\frac{8\:\textrm{V}}{R_2} = -2.\overline{30}\:\textrm{mA}\\\\ I_{D_2} &= \frac{8\:\textrm{V}}{R_2} = 2.\overline{6}\:\textrm{mA} \end{align*}$$

It's clear now that \$I_{D_1}\$ violates our requirement of being ON, so it must be off. Solving for that case yields:

$$\begin{align*} V_2 &= V_1\\\\ I_{D_1} = I_{D_3} &= 0\:\textrm{A}\\\\ \frac{V_1}{R_1} &= \frac{-8\:\textrm{V}}{R_1} + I_{D_1} + I_{D_2}\\\\ \frac{V_2}{R_2} + I_{D_2} + I_{D_3} &= \frac{8\:\textrm{V}}{R_2}\\\\ \therefore\quad V_1 = V_2 &= 8\:\textrm{V}\cdot\frac{R_1-R_2}{R_1+R_2} = 6.08\:\textrm{V}\\\\ I_{D_2} &= \frac{16\:\textrm{V}}{R_1+R_2} = 640\:\mu\textrm{A} \end{align*}$$

Since \$I_{D_2}\$ is positive, we can now say that if \$D_1\$ is OFF and \$D_2\$ is ON then we have satisfied things up to this point and can move on to the idea of now turning on \$D_3\$. As a double-check, you can see that \$V_1=V_2\$ is quite a bit more positive than the ground reference on the other side of \$D_1\$, so this should also confirm for you that \$D_1\$ really is OFF -- up to this point, anyway.

I'll stop at this point, since by now you have a recipe for continuing on your own. At least you can see that you don't have to make an exhaustive table of all 16 permutations. But.. there still is work to be done, too.

By incrementally solving the problem you can work out the right answer without having to exhaustively test all permutations. It merely requires a logical application of rules, one step at a time, allowing the pieces to fall where they may, as you move forward. You may need to walk backwards on your determined states if, upon adding the next stage some prior condition reverses itself. (As happened in this new case I "randomized" up, where the prior assumed state of \$D_1\$ had to be changed when the 2nd state was added.) When it does happen, change it and recheck the results, and if that satisfies things then advance again by adding in the following stage.


Looking at your schematic, you have two three rails. +8V, -8V, and ground.

It is fairly self evident that current is flowing downwards in all the resistors. The question becomes, is there current flowing out of ground through D1.


simulate this circuit – Schematic created using CircuitLab

So let us remove D1 and see what happens to the voltage at the top of R1.

You now have this...


simulate this circuit

It should be obvious to you that D2 is on since the cathode will always be a lower voltage than the anode. Similarly D4 is on since the anode will always be a higher voltage than the cathode.

That leaves us with D3.


simulate this circuit

Looking at that, even without doing the math, you should be able to see, that when off, the voltage on the left of D3 would be a equal to the right. So D3 is at the verge of turning on.

However, we now know that the top of R1 will in fact be significantly below ground if D1 is off.

As such, we now know D1 MUST be on.

If D1 is on, there is more current flowing through R1 which will raise the voltage at the top of R1. This then, finally, pushes up the voltage on the left of D3, so it will be on too.

Curiously, with "ideal" diodes, depending on your definition of that, the output is ground on this circuit if you take the diode forward voltage as zero.

On a more general note. I do not think there is any one rule fits all methodology for figuring out diode circuits, other than starting at the outside(s) and working your way in. It get easier with practice, but even the experienced EE will have to look at it for a while to find the best approach / starting point.


Taking the case of D4 off then Vo is going to be rail Voltage. it is just a logic equation then which determines Vo to be between -8V and +8V D1 on D2,3 and 4 off then -8V It is just a logic equation so write a truth table.

  • \$\begingroup\$ You could improve your answer by showing how to write a truth table for this analogue circuit. Many of your readers would be interested. \$\endgroup\$
    – Transistor
    Commented Nov 24, 2017 at 17:00
  • \$\begingroup\$ if we take d1 on and D2,3,and 4 off the voltage around D2 is going to be 16 witch is in contradiction with D2 being off. :| \$\endgroup\$
    – mahdi
    Commented Nov 24, 2017 at 20:19
  • \$\begingroup\$ You are correct mahdi, I must engage brain first. \$\endgroup\$
    – LateDev
    Commented Nov 24, 2017 at 21:15

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