Please make me sure (or show another way to solve this exercise) that my set of equations to calculate the diode voltage and current Id are right.
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I don't think there's an analytical solution. You have to solve by iteration and convergence.So, the only right way is to solve the attached system of equations? Isn't there any approximation to make it easier? Is that diode in reverse or forward state?
Yes, there is one. We treat the diode as an open circuit when the diode is in reverse biased.Isn't there any approximation to make it easier?
My professor gave me this task.Yes, there is one. We treat the diode as an open circuit when the diode is in reverse biased.
Who gave you this problem?
The diode in this case is reverse biased. Let's take things in two chunks.So, the only right way is to solve the attached system of equations? Isn't there any approximation to make it easier? Is that diode in reverse or forward state?
If I understand the plot give, it's saying that the diode voltage is about 1005 V. Does that make sense?Do you know the n and thermal voltage Vt ?
And see the numerical solution for Vt = 25mV and n = 2 because you cannot solve it by using just algebra.
https://www.wolframalpha.com/input/?i=(5++x+)/1000000+=+1*10^3*(E^(x/(2*0.02586492))+1)
Oh crap, I didn't see that! I wouldn't have done well on this test.The diode in this case is reverse biased.
Because of the fact that the diode is reverse biased.Why does your exponential equation have a negative sign in the exponent?
Look at the diagram. Vd is defined as the forward voltage across the diode. So if it is reversebiased, Vd will be negative. If you then go and artificially change the sign in the equation, you are saying that magnitude of the current goes UP as the voltage across the diode becomes more negative.Because of the fact that the diode is reverse biased.
So are you really claiming that if you put a 5 V source across a 1 MΩ resistor in series with this diode such that it is reverse biased that Vd will be +1005 V across the diode?Hi,
Yeah it is reverse biased with a reasonably high reverse voltage that's how the exponential becomes near zero and leaves us with 1 inside the brackets.
I went back and looked at your Wolfram link and see that it shows two solutions  I was only seeing a single solution (the 1005 V) when I looked before.Because of the fact that the diode is reverse biased.
Actually, in this case you might have. The diode is being operated so far down the curve that it is pretty linear.Oh crap, I didn't see that! I wouldn't have done well on this test.
Hi,So are you really claiming that if you put a 5 V source across a 1 MΩ resistor in series with this diode such that it is reverse biased that Vd will be +1005 V across the diode?
Huh? Why on earth would it have to have a rating of 5.5 MA?Hi,
Seeing as this diode must have a rating of something like 5.5 Meg amps i took the resistor R to be 0.001 ohms not 1 megohm.
Why not? The current will be about 5 μA and the voltage across the diode will be a fraction of a millivolt.I dont think it works with a 1 megohm resistor in series.
No, it's not. Not in diodes that we come across in normal every day life. But since Is is proportional to area, it is certainly conceivable to build one, if not as a single device, then by putting a sufficient number of them in parallel.Or else there is something else wrong. I dont think it is typical at all to see Is=0.001 amps.
Doesn't a reverse current of 5 μA seem a LOT more reasonable than claiming that, somehow, 1000+ V is going to magically appear across a diode connected to a 5 V source?In all fairness for a diode rated that high i might be inclined to take it as a short circuit when used with a resistor of high value like 1Megohm. That would make the reverse current Id equal to 5/1000000 amps which is 5ua.
Now we see what happens when textbooks do not give any numerical results to compare.