How to determine numerically the diode current

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First of all, there is NO given time-dependent information. Why do you want I(D) for t<0 and t>0? I assume you want the I(D) for the diode's working point.

It would be better understood if we plug in some values. Here we go:

Circuit schematic: current source I0 in parallel with resistors R1 and R2, then series resistor R3 feeding terminal 1 and a diode with current I_D and voltage V_D, returning to terminal 1'.


Given:

[tex]R_{1}=R_{2}=R_{3}=100\Omega[/tex]

[tex]I_{0}=24\ mA[/tex]


By considering the open (oc) and short circuits (sc), you will obtain two values where:

[tex]I_{oc}=0[/tex] as [tex]V_{oc}=V_{Th}=1.2\ V[/tex]

[tex]I_{sc}=\frac{V_{Th}}{R_{Th}}=8.0\ mA[/tex]


Here is the I-V source characteristic, the red line:

I-V Characteristics graph of current I in amperes against voltage V in volts, with a red straight line falling from 0.008 A at 0 V to 0 A at 1.2 V. Axis ticks at 0.2 to 1.2 V and 0.002 to 0.008 A.



Now, if the characteristic of the diode is given by

Piecewise equation for diode current I_D(V_D): (V_D - 0.7) * 1/100 when V_D is at least 0.7, and 0 when V_D is less than 0.7.


Then plotting the characteristic on the same graph gives the green line:

Graph titled I-V Characteristics, plotting I/A against V/V: a red line falls from about 0.008 A at 0 V to 0 A at 1.2 V, and a green curve rises from near 0.7 V, crossing the red line at about 0.9 V and 0.002 A.




A. Graphically
So you know that the intersection is the working point for the diode and there you can write down the value of [tex]V_{D}=0.9\ V[/tex] and [tex]I_{D}=2.0\ mA[/tex].

B. Numerically
Find the equation of the I-V source characteristic using the famous [tex]Y=mX+C[/tex] where you will expect C = 0.008 A and and negative slope m. Then, solving the simultaneous equations proves the values obtained in A. That's simple.
 
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