# Junction diode problem

1. Sep 21, 2014

### asdf12312

1. The problem statement, all variables and given/known data
(circuit attached)
If the circuit shown, D2 has 10 times the junction area of D1. What value of V results? To obtain V=50mV, what current I2 is needed?

2. Relevant equations
I = I_S * (e^(V/V_T) - 1), V_T=25mV
I(D2)/I(D1) = e^((VD2-VD1)/V_T)

3. The attempt at a solution
If D2 has 10* I_S of D1 then it has 10 times current also, but I think current through diode already given as I(D2)=2mA and I(D1)=8mA. Using node equation, V is VD1-VD2. to find this value knowing currents:

I(D1)/I(D2)=0.1 * e^((VD1-VD2)/V_T)

I included the 0.1 but wasn't sure if I should, D1 has 1/10 I_S of D2 so it seemed relevant. Solving for V=VD1-VD2:

V = 10*V_T*ln(I(D1)/I(D2))=347mV

I did this right so far?

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2. Sep 21, 2014

### rude man

Good! Use it!

Where did you get this? Makes no sense. You have the answer in the equation above. You stated the currents are 2mA and 8 mA and that is also correct.

3. Sep 21, 2014

### asdf12312

the 0.1 is because the scale current of D1 is 0.1 times scale current of D2 (the ratio I guess). when the I_S cancel out shouldn't I include this ratio in the equation?

4. Sep 21, 2014

### rude man

When you go Id2/Id1 the Is cancel out, and you do ipso facto include the ratio.

Look at those two equations, the first (right one) and the second (wrong one). They contradict each other, don't they?

5. Sep 21, 2014

### asdf12312

ok I guess that makes sense... so the information about junction area isn't needed? If I don't include the 0.1 in equation I get V=34.7mV.

6. Sep 21, 2014

### rude man

I'm sorry, I goofed you up. The 2nd equation was right, not the first.

Id1 = Is exp(V1/VT)
Id2 = 10 Is exp(V2/VT)
Id2/Id1 = 10 exp((V2 - V1)/VT) = 8/2 = 4
Solve for V2 - V1.

7. Sep 21, 2014

### asdf12312

so the answer I got was right? V=V1-V2=347mV

8. Sep 21, 2014

### rude man

Don't think so. I got V = 92.2 mV.

OK, this you had right: I(D1)/I(D2)=0.1 * e^((VD1-VD2)/V_T)
But then V = 10*V_T*ln(I(D1)/I(D2))=347mV was wrong. Just a math error. The "10" belongs inside the log argument.

Solve that correctly and you get my answer.

9. Sep 21, 2014

### asdf12312

oops... I guess I missed that. I get V=92.2mV also, and to get V=50mV the current I2 is:
I(D1)/I(D2)= 0.1 * e^(V1-V2)/V_T
I(D1)/I(D2)= 0.1*e^(50/25)

using I(D1)=10-I(D2):
I(D2)=5.75mA

Last edited: Sep 21, 2014
10. Sep 21, 2014

Correct!