Voltage drop across diodes

[nvjnj]

I have the following circuit for which the voltage drop across each diode is required...**Assume both diodes are identical.

Following date is also given:
Is=5μA
q=1.60217646*10^-19 C
k=1.6806503*10^-23 m^2kgs^-2K^-1
T=300 K
n=2

If = Is (e^(qv/nkt)-1) equation is to be used.

So far...
Since D2 is reverse biased, no current is supposed to flow (If=0), hence this equation cannot be used as far as I think...Am I mistaken??

 

[Arouse1973]

I can't open your picture. Can anyone else?
Adam

 

[Supercap2F]

I can't open your picture. Can anyone else?
Adam

Click on it's name, or..

 

[Arouse1973]

Nope still not working Dan, but thanks for posting it.
Adam

 

[Arouse1973]

I have the following circuit for which the voltage drop across each diode is required...**Assume both diodes are identical.

Following date is also given:
Is=5μA
q=1.60217646*10^-19 C
k=1.6806503*10^-23 m^2kgs^-2K^-1
T=300 K
n=2

If = Is (e^(qv/nkt)-1) equation is to be used.

So far...
Since D2 is reverse biased, no current is supposed to flow (If=0), hence this equation cannot be used as far as I think...Am I mistaken??

Why do you think the reverse current is zero? What do you think (Is) is?
Adam

 

[nvjnj]

Why do you think the reverse current is zero? What do you think (Is) is?
Adam

Not the reverse current, I said that (I think) the forward current (If) is zero as diode D2 is reverse biased, hence only reverse saturation current flows across D2.

However, that knowledge is insufficient for me to derive voltage drop across each diode, or I can't seem to bridge the knowledge.

 

[Arouse1973]

The reverse current of one diode will be the forward current of the other. Does this make sense?
Adam

 

[nvjnj]

The reverse current of one diode will be the forward current of the other. Does this make sense?
Adam

Yes, that part is understood.....
So substituting the shockley equation as If=Is and evaluating for voltage(v) gives the voltage drop across D1 as 43.7mV, while voltage across D2 is (12V-43.7mV)=11.96V is the answer??
I just thought that since the forward biased voltage drop of D1 is so low(mV range), I may have made a mistake.

 

[Arouse1973]

No your right. The voltage drop across a diode is dependant of the current through the diode. There is a misconception that a diode will always have a 0.6-0.7 volt drop across it, the volt drop is dependant on current.
Adam