Below is the simulation of a 1N4148 that shows the voltage reaches 2V at near 1.9A (blue trace).
But at that current, the power dissipated is about 3.8W (yellow trace), which will very rapidly zap the diode.

View attachment 193645

Thanks for your explanantion, I got it

 

Below is the simulation of a 1N4148 that shows the voltage reaches 2V at near 1.9A (blue trace).
But at that current, the power dissipated is about 3.8W (yellow trace), which will very rapidly zap the diode.

View attachment 193645

very nicely explained. Thank you. Could you tell me how did you add that arrow in the plot? and how did you plot I vs V?

 

hi T,
Right click on the Plotted area, see image, Draw [ choose line style and colour]
E

 

Adding this feature will complicate the program algorithm and increase the time of scheme simulation. And this is only to help a very limited number of users of the program. A competent engineer will study the currents and powers of the elements himself. Some elements withstand tens and hundreds of times the nominal currents for a short time. And it complicates the decision of a question - will let a smoke or will explode a radio element or not.

At least one of the spices I use (Tina-TI) has maximum power ratings for all components, although the student version doesn't use them. I presume the professional version has such a feature - a warning, at least, if not an exploding component animation. It also has a variety of Monte Carlo options for varying component values within tolerances, also not available in the student version., and also very time-consuming to perform And simulation over a range of temperature, which is something else that makes real-world semiconductors unlike the simple spice versions.

 

hi T,
Right click on the Plotted area, see image, Draw [ choose line style and colour]
E
View attachment 194788

Thank you

 

Regarding the two diodes reversed biased in series in circuit 1 as explained they will have a small reverse bias current that first hand device physics shows is proportional to the square root of the applied voltage up to a certain limit of course (Semiconductor Device Fundamentals by Pierret page 272). Because the same current reverse bias current runs through both diodes the crossing point where both there i-v curves intersect is the solution.