I know that a silicon diode with forward bias drops 0.7V, and I also know the two resistors should have the same voltage drop because they have the same resistance and same current (because the branch with the opposite diodes in series should have a current of almost 0). Knowing that I can calculate that the resistors drop 4.3V and the 2 forward bias diodes drop 0.7V, and I expected the other diodes to drop 0V. But simulating this circuit on CircuitLab gives the following voltage drops:

Circuit Lab simulation

I don't understand why the opposite diodes in series drop 2.5V each. If anything I would think the forward biased diode should drop 0.7V and the reverse biased diode should drop the remaining 4.3V, but that doesn't seem to be the case. Why is this?

These are "ideal diodes" on CircuitLab but I assume they will behave the same. Is this a correct assumption to make? If not, what would change if they were silicon diodes in real life?

  • 5
    \$\begingroup\$ It would depend on the leakage current through the reverse-biased diode, and how the "forward-biased" one (which won't be very forward biased, with only a few nanoamps of forward current) responds to such a low current. \$\endgroup\$
    – Hearth
    Commented Mar 12, 2021 at 14:10
  • 4
    \$\begingroup\$ a forward biased silicon diode dropping 2.5v would have so much current flowing through it, that it would burn up. \$\endgroup\$ Commented Mar 12, 2021 at 14:10
  • \$\begingroup\$ CircuitLab is silly here. You really may wish to try this circuit in real life, and do some measurements. That's the problem with simulations: they are only the first step, never meant to replace testing physical circuits. If you don't have a multimeter with a decently high input impedance, a FET-input op-amp will do well as a buffer. You could also try a better simulator - LTspice gets it right on the first try. \$\endgroup\$ Commented Mar 13, 2021 at 23:27
  • \$\begingroup\$ The voltage drop at a conducting diode is 0.7 V. This is valid if a current is flowing. When the diode is reversed, no current is flowing and the voltage drop may be higher than 0.7 V \$\endgroup\$
    – Uwe
    Commented Mar 14, 2021 at 15:38

4 Answers 4


(Update: we've fixed the convergence issue described below and the simulation now converges just fine!)

(CircuitLab simulator developer here.)

You appear to have discovered one of the drawbacks of Ideal Diodes. Their V-I curves are piecewise-linear and continuous, but not smooth. Not smooth means the derivatives are discontinuous.

This discontinuity can cause convergence issues, especially since you have two of these back-to-back. You'll see this noted with a yellow circle at the bottom right of the editing window, or with a message WARNING: nonlinear convergence failed. Be suspicious of results. at the bottom-right of the window if you explicitly click "Run DC Solver".

I get this convergence failure warning when I try to simulate your original circuit:


simulate this circuit – Schematic created using CircuitLab

There are two possible solutions.

First, you can use P-N Junction Diodes, which are both continuous and smooth and will not experience this kind of convergence problem:


simulate this circuit

Alternatively, you can continue to use Ideal Diodes, but add some resistors that help the solver find a solution:


simulate this circuit

Either approach will eliminate this convergence issue, but using the smooth P-N Junction Diodes seems easier to me!

  • 3
    \$\begingroup\$ Wouldn't it be better to use a tanh function in the model to simulate the step but with arbitrary (and adjustable) sharpness? \$\endgroup\$ Commented Mar 12, 2021 at 18:13
  • \$\begingroup\$ @SpehroPefhany tanh() may be expensive in terms of computation, but low poly approximations can be made easily. (3x-x^3)/2 comes to mind for [-1,1]. This has a second derivative. For more, integrate (1-x^2)^n and adjust accordingly. \$\endgroup\$ Commented Mar 13, 2021 at 19:31
  • \$\begingroup\$ Without even looking at the code, I bet you that the piecewise function is no faster than a suitable continuous approximation. Furthermore, if the JIT doesn't produce a separate machine code implementation for each instance of the diode, the branch predictor will be constantly mispredicting inside the diode model. Depending on how slow the surrounding code is, it may get buried, but it would be a rather typical problem in C/C++ simulators, unless the compiler turned the code branchless. \$\endgroup\$ Commented Mar 13, 2021 at 23:34
  • \$\begingroup\$ Update: we've fixed the convergence issue described below and the simulation now converges just fine! (Moderators: should I entirely rewrite the post to reflect that, or leave it as-is?) \$\endgroup\$
    – compumike
    Commented Mar 25, 2021 at 17:45

The voltage is not distributed equally across the two anode to anode diodes.


simulate this circuit – Schematic created using CircuitLab

This could be the mistake in simulation


simulate this circuit


It is easy to make an error on the most trivial observations. I thought I was looking the two series diodes in reverse bias when you reported CircuitLab has the 4V drop split in half. I would NOT trust that simulator for anythig nor any one for reverse leakage, unless you know the simulator assumptions. (Including Falstad's ( I believe the models for 1N4148 and 1N4004 were reversed so I swapped the labels )

In any case, when you look at datasheets the graphs are ALWAY nominal and the TABLES are TYP and WORST case for some conditions like 25'C.

Simulators rarely pay much attention to thermal temp rise and leakage changes, or batch differences in tolerance of leakage so neither should you. Yet you must consider this if it's important.

enter image description here enter image description here

The silicon diode with the highest voltage rating when used at low voltage is likely to have a lower leakage current.

We are always taught a diode is 0.7V yet if you take a dozen different diodes , that won't be true. But if you take the same different Silicon diodes and bias them with 1mA you will get almost 600 mV because the current isn't high enough for the bulk resistance differences to have much effect.

Schottky diodes are different and have more leakage.

Just simulating here different pairs of 2 series reverse/forward diodes and reverse/reverse. Go ahead and (left click Swap Terminals and see the changes and connect a fake pot of 100 GOhm for leakage dust.

enter image description here

When doing real work, remember Murphy's Law. If anything can be reversed, it will be.

  • \$\begingroup\$ In your fallstad link, you accedentially swapped the labels of the first 4 diodes, the left 2 diodes are a 1N4148, the middle 2 diodes are a 1N4004 \$\endgroup\$
    – Ferrybig
    Commented Mar 13, 2021 at 12:20
  • \$\begingroup\$ No I intentionally swapped them because I know from the Vf Vr vs I characteristics they are wrong. Trust me I verified this. But don’t , verify yourself in a Test. I also indicated this in my answer. \$\endgroup\$ Commented Mar 13, 2021 at 14:24

If a diode is reverse biased the total voltage drop on the diode is equal to the voltage drop parallel to it. Since no current flows through the reverse biased diode and you have a 100 ohm resistor which has a voltage drop of 4.3V and a diode forward biased which has a voltage drop of 0.7V parallel to the reverse biased diode, their sum of the voltage drops must be equal to the voltage drop of the reverse biased diode (ideally).

One mistake you may have done is to put two voltmeters in series, parallel to the reverse biased diode and the forward biased diode which is in series with it, and since voltmeters have ideally infinite resistance you would find that the voltage drop on each voltmeter is half the voltage drop of the loop.

  • \$\begingroup\$ So would you expect it to be distributed as 4.3V and 0.7V between the forward and reverse biased diodes or is it more complicated than that? \$\endgroup\$
    – Param
    Commented Mar 14, 2021 at 3:59
  • \$\begingroup\$ No.Since no current flows from the branch of the reverse biased diode the voltage drop of the forward biased diode will be 0 V and the reverse biased diode will have a voltage drop of 5V. \$\endgroup\$
    – Miss Mulan
    Commented Mar 14, 2021 at 4:04

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