Voltage drop across opposite diodes in series

Question

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:

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?

Answer 1

(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!

Answer 2

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}

Answer 3

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.

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.

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

Answer 4

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.