Me and my friends have been working on this for days, and none of us are even close to the answer. We are supposed to calculate the voltage drop (Vd) across the diode, while the saturation current is equal to 3x10^-16 A. R1 = 1kΩ and Is = 1 mA The circuit can be found at: Any help would be appreciated.
Simply convert this circuit to a voltage source + resistor + diode Vs = Is*R = 1V And know you can easily find Vd and Id using Iteration and Shockley diode equation. http://en.wikipedia.org/wiki/Diode#Shockley_diode_equation
But shouldn't Vs = (Ir + Id) * R where Ir is resistor current and Id is diode current. Let me clarify, in my image, Io is the current supply, while Is is meant as saturation current. Does your method apply under these conditions?
And I using google and this code Code ( (Unknown Language)): (ln( (1 - 0.6)/300E-15) -1 )) * 0.026 find this solution Vd = 0.693005758 http://en.wikipedia.org/wiki/Diode_modelling#Iterative_solution
This is a diode specific problem, so simple circuit equations aren't enough. You need to take into account that That way you have that equation and You can solve that system with some recursions starting from an initial value. The solution will be approximate.
I understand now , I just did not get the Vs part, but that now I get it. I answered my own question when I said Vs = (Ir + Id) * R, which is exactly what you did, but with the equivalent circuit. I guess I have to read up on my Thévenin's theorem. Again thank you very very very much!!!
i cant seem to get this exact number i keep getting 0.699 even when i use goole calculator instead of my own. and when i increase the source current, Vd becomes greater than 0.7 can anyone think of what im doing wrong?
Well very strange When I put this on google Code ( (Unknown Language)): (ln( (1 - 0.6)/300E-15) -1 )) * 0.026 I get 0.699886283 (1) 0.692416403 (2) 0.693055624 (3) 0.693001535 (4) Later on I check this with hand + calculator.
Interesting. I wrote a little Basic program to do the iterative calculation, and got VD=0.71689909. It checked out on my HP15C calculator. Code ( (Unknown Language)): Vt=.026 Vs=1 Is=3e-16 R=1e3 Vdnew=.6 Do Vd=Vdnew Vdnew=Vt*log((Vs-Vd)/(R*Is)-1) Print "vd=";Vd Print "Vdnew=";Vdnew Loop until Vdnew=Vd end Note that "log" is the natural logarithm, at least in Just BASIC v1.01. What did I do wrong? Or what did you guys do wrong?
Well I think I know where I made the error. I forget one bracket in the google code Code ( (Unknown Language)): (ln(( (1 - 0.6)/300E-15) -1 )) * 0.026 VD = 0.716899093V And I also use Mathematica software and I get this solution Vd = 0.7168990929699331
with the corrected bracket I got 0.725886 in excel In google ... the result is (ln(( (1 - 0.6)/300E-15) -1 )) * 0.026 = 0.725886283 The constant e in excel is 2.71828182845904 I'll be checking it again later ... with my TI-83.
Take care not to mix the decical logarithm (log) with the natural logarithm (ln). I think the formula calls for the natural logarithm.
"log" in basic is the natural log. Log10 in basic is the decimal log. Log10 is also used in excel for decimal log. Much to my surprise, the TI-83 agreed with google and excel at approximately 0.7258862829 The TI-83 uses 2.718281828459 for e. I am at a loss for the error at this moment.
You didn't run the iterations. The answer you got, 0.7258862829, is the first step in the iteration, and you only get that if you use VD=0.6 as your first guess. You need to plug that number into VD in the right side of the equation, and re-evaluate. Plug the new answer into VD and evaluate again, ad nauseum, until the evaluated value doesn't change. See my Basic program in post #14, Code ( (Unknown Language)): Vt=.026 Vs=1 Is=3e-16 R=1e3 Vdnew=.6 Do Vd=Vdnew Vdnew=Vt*log((Vs-Vd)/(R*Is)-1) Print "vd=";Vd Print "Vdnew=";Vdnew Loop until Vdnew=Vd end and the link posted by Jony130 in post #7.