# Voltage drop across diodes

Discussion in 'Electronics Homework Help' started by nvjnj, Jan 10, 2015.

1. ### nvjnj

11
0
Sep 13, 2014
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??

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5,164
1,081
Dec 18, 2013
I can't open your picture. Can anyone else?

3. ### Supercap2F

550
150
Mar 22, 2014

5,164
1,081
Dec 18, 2013
Nope still not working Dan, but thanks for posting it.

5,164
1,081
Dec 18, 2013
Why do you think the reverse current is zero? What do you think (Is) is?

6. ### nvjnj

11
0
Sep 13, 2014
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.

5,164
1,081
Dec 18, 2013
The reverse current of one diode will be the forward current of the other. Does this make sense?

8. ### nvjnj

11
0
Sep 13, 2014
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.  