diode question

Look at http://everycircuit.com/circuit/6526407893843968 - same as yours but an additional scope trace. You can see that the green trace has a value halfway between zero and negative amplitude. Both diodes together act like a voltage divider from very large resistors - but as soon you put even a very small load on that voltage, it's gone (toggle switch). Diodes are not perfect isolators, and in reality you may not be able to measure this voltage with a Voltmeter because the Voltmeter is already to much of a load.

@Bushmills, Good reply.

as always

Thank you! So essentially. I am just seeing how a diode could operate in a less than ideal situation when there is no Load. Is that accurate?

I think you could say that, yes - but consider that no real component is ideal. A simulation may present components as if they were ideal, but where they do, the simulation is flawed. In case of diodes does EveryCircuit not model them as ideal components - which is why the tiny currents through them cause that voltage between diodes. Try this: take a very large resistor, I think 500 GigaOhm is the largest value you can get with EveryCircuit, a diode and a DC source of, say, 10 Volts. Change time base of scope to maximum time per division (the diode also has a capacitance, and you want it charged quickly). If the diode capacitance is charged, the whole 10 Volts should drop over the diode if it was a perfect isolator, and therefore, 0 Volt over resistor. But you'll find that there's still a small voltage drop over the resistor. From that voltage drop, and the known resistance can you calculate current through resistor, and therefore through diode. With that current, and the voltage over diode you calculate the resistance used in the simulation for the diode. Twice that resistance in series gives a voltage divider.