kinda messy maths/electronics question HELP?

M

Mr X X

New member
E=supply voltage=2v
R=resistance=50ohms
Ib= 'reverse bias current' = 1*10^-9 Amps (1nA)
k=Boltzman's constant = 1.38*10^-23
T=temp of diode=300K
Q=charge=1.6*10^-19
Vr = voltage across resistor (unknown)
Vd = voltage across diode (unknown, doesnt really matter, not trying to find it yet)
By ohms law, E=Vr+Vd
= I*R+Vd
and I = Ib*(e^(Q*Vd/T*k) -1)
wish i could write it neater, but i need to find current I so that i can find Vr.
Since Vd is unknown, im guessing i sub in "E-I*R" in the place of Vd in that final equation? how do i then solve for I?

or is there an easier way to do it that i'm missing lol, just using v=ir doesnt work, it ends up Vr=2 which is wrong as E=2=Vr+Vd
 
Assuming the diode and resistor are in series, he current across the diode is the same as the resistor.
The Voltage drop across the resistor is calcualted by V =IR
and the Volage (reverse) across the diode is E -Vr
 
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