Review my calculations.
The diode voltage is now 2V. The diode current is 4μA.
The voltage across the 750kΩ resistor is 3V. The voltage adds up to 5V.

Thanks, MrChips this was most helpful it seems that eventually the Rd value settles at a current which satisfies the limiting resistor too ?

 

Thanks, MrChips this was most helpful it seems that eventually the Rd value settles at a current which satisfies the limiting resistor too ?

Yes, it has to do that. Basically, you are solving two simultaneous equations, otherwise it would either oscillate (negative feedback with time delay) or saturate (positive feedback).

 

Thanks, MrChips this was most helpful it seems that eventually the Rd value settles at a current which satisfies the limiting resistor too ?

There really isn't any "eventually" about it: within a few microseconds, the voltage at the junction of the 750 kΩ resistor and the Schottky diode becomes stable; the process by which it does so is of no concern and has no explanatory value.

Trying to understand what's going on in terms of a time sequence of events (voltage going up, voltage going down, etc.) just adds unnecessary confusion. It'd be far better for you to look at this circuit as a simple voltage divider: on the top leg you have your 750 kΩ resistor which conducts 1.33 μA for every volt across it, and on the bottom you have a nonlinear element with a more complex V-to-I relationship in which current increases rapidly within the first few hundred mV, then gradually levels out to a more or less constant value. The best way to view it is via a graphical "load line" analysis, as @Bordodynov did in post #19: the voltage at which the current through the resistor and the current through the diode become equal, is the equilibrium voltage.

 

There really isn't any "eventually" about it: within a few microseconds, the voltage at the junction of the 750 kΩ resistor and the Schottky diode becomes stable; the process by which it does so is of no concern and has no explanatory value.

Trying to understand what's going on in terms of a time sequence of events (voltage going up, voltage going down, etc.) just adds unnecessary confusion. It'd be far better for you to look at this circuit as a simple voltage divider: on the top leg you have your 750 kΩ resistor which conducts 1.33 μA for every volt across it, and on the bottom you have a nonlinear element with a more complex V-to-I relationship in which current increases rapidly within the first few hundred mV, then gradually levels out to a more or less constant value. The best way to view it is via a graphical "load line" analysis, as @Bordodynov did in post #19: the voltage at which the current through the resistor and the current through the diode become equal, is the equilibrium voltage.

That is quite a nice summation to this problem, however I like to try and understand everything about a process and the failure to do so doesn't spoil the enjoyment of trying. Whereas, accepting that things are what they are eventually leads to big gaps in understanding.

 

there might be some "cryptic math" coverage about in the papers involving the improvement and the desig of the Schottky barrier diodes
https://ecee.colorado.edu/~bart/book/book/chapter3/ch3_2.htm

 

I have worked twice with ultrasensitive pin-diode detectors, once it was pin diodes for detection of alpha and beta particles and second was for 2GHz analog-optical signal reading (Hamamatsu certain product) Last need a minus 250 V reverse bias and first minus 25 Volts. Then the reverse current is between picoamperes and femtoamperes. If to diminish that voltage, the current is INCREASING (become larger!!).

 

There is a junction capacitance that has to be charged when any voltage is applied for the first time. For an example in the 1N5817 the capacitance is around 0.25nf, so there would be a slight delay in the time before the circuit reaches equilibrium.

Aside from that, the most significant diode part of that diode has theoretical equation:
i=(e^(V/(Vt*N))-1)*Is
and we have approximately:
Is=2e-5
N=1.24
and for reverse currents Rs is negligible.
We can set the thermal voltage to 0.026 volts.
So using the above equation we have:
i=(e^(V/(0.026*1.24))-1)*2e-5
and for simplicity this can be written approximately as:
i=2e-5*(e^(31*V)-1)

So we have in the test circuit a voltage Vs, large resistance Rs, and this diode.
The circuit equation for the diode voltage is thus:
V=Vs-Rs*i
and written out:
V=Vs-Rs*2e-5*(e^(31*V)-1)
and if we make Rs=100k we have:
V=Vs-100000*2e-5*(e^(31*V)-1)
or:
V=Vs-2*(e^(31*V)-1)

And say the first test voltage is -10 volts, then we have:
V=-10-2*(e^(31*V)-1)

and note we have the diode voltage V on both sides and it may not do much good to simplify this any more so we solve this numerically and we get:
V=-8 volts.

So the diode voltage is -8v and the drop across the 100k resistor is -2v because the test voltage source is -10 volts, which means the Rs and diode currents are both -20ua.

The time constant to charge up to this equilibrium level is about 25 microseconds so after about 125 to 150 microseconds we get very close to that level of current and voltage.

 

The time constant to charge up to this equilibrium level is about 0.025 nano seconds so after about 0.125 to 0.150 nano seconds we get very close to that level of current and voltage.

For a 100 kΩ resistor and 0.25 nF junction capacitance I get a time constant of 0.025 millisecond, not 0.025 nanosecond.

Minor difference...

 

For a 100 kΩ resistor and 0.25 nF junction capacitance I get a time constant of 0.025 millisecond, not 0.025 nanosecond.

Minor difference...

Well the problem with the way you did it is as follows...

You did it right
You've got to stop doing that! (ha ha).
I'll correct my previous post right after i post this one.

Oh i just saw that "minor difference" also funny