What Value of V Results in This Junction Diode Circuit?

[asdf12312]

Homework Statement

(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?

Homework Equations

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

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?

 

[rude man]

Homework Statement

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

Good! Use it!

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

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.

 

[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?

 

[rude man]

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?

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?

 

[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.

 

[rude man]

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.

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.

 

[asdf12312]

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

 

[rude man]

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

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.

 

[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

 

[rude man]

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

Correct!