What Value of V Results in This Junction Diode Circuit?

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Discussion Overview

The discussion revolves around a junction diode circuit where participants are trying to determine the voltage (V) across the diodes given their current values and junction areas. The conversation includes attempts to solve for V and the current needed to achieve a specific voltage of 50mV, involving the application of diode equations and relationships between the currents and voltages across the diodes.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that if D2 has 10 times the junction area of D1, it should also have 10 times the saturation current (I_S), leading to a proposed relationship between the currents and voltages across the diodes.
  • Another participant challenges the use of a specific equation, suggesting that the ratio of the currents should not be included in the way it was presented, indicating a misunderstanding of the equations involved.
  • Some participants discuss the implications of the junction area information, questioning its relevance to the calculations.
  • There is a correction regarding the placement of the "10" in the logarithmic equation, which affects the calculation of V.
  • Multiple participants arrive at different values for V, with one suggesting 347mV and another correcting it to 92.2mV based on a mathematical error in the previous calculations.
  • A later reply provides a method to find the current I2 needed to achieve V=50mV, using the established relationships and current values.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct application of equations and the resulting values for V. There is no consensus on the final value of V, with different calculations leading to different results.

Contextual Notes

Participants reference the saturation current (I_S) and its relationship to the junction area, but the implications of these relationships remain unresolved. The discussion includes corrections and clarifications that highlight the complexity of the calculations involved.

asdf12312
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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?
 

Attachments

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asdf12312 said:

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.
 
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?
 
asdf12312 said:
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?
 
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.
 
asdf12312 said:
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.
 
so the answer I got was right? V=V1-V2=347mV
 
asdf12312 said:
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.
 
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:
  • #10
asdf12312 said:
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!
 

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