Solve Junction Diode Problem: V & I2 for D2=10xD1 Area

In summary, if the circuit shown, D2=10 times the junction area of D1. What value of V results? To obtain V=50mV, what current I2 is needed?
  • #1
asdf12312
199
1

Homework Statement


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If the circuit shown, D2= 10 times the junction area of D1. What value of V results? To obtain V=50mV, what current I2 is needed?

Homework Equations


i(D)=Is(eVD/VT-1)

The Attempt at a Solution


I know junction area is related to diode current, so D2 has 10 times current as D1. But I still don't really know what to do. I tried KCL equation so 10/11mA -> 0.91mA going through D1. Then current thru D2 = 10*I(D1)=9.09mA this satisfies the KCL equation since 10mA is going on. But I didn't use the equation for diode current I listed so I have a feeling I did this all wrong. Then my other question is how does I2 determine V? It would help if I knew how to find voltage through the diodes but I don't want to assume that it is something like VD=0.7V. the problem didn't say anything about it neither.
 
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  • #2
Choosing I1 and I2 effectively dictates the currents through both diodes. So you can set the currents. Now, the area ratio presumably shows up as differences in the Is for each diode. Is that so? If diode D1 has a saturation current Is, what would be the saturation current for D2?

Can you take your Relevant Equation and write expressions for VD for each diode?
 
  • #3
i(D1)=Is(eV/VT-1)
i(D2)=10*Is(eV/VT-1)

because Is increases by same value as area junction A. and then there is the assumptions V=0.7 and VT=25mV (room temperature). But I don't see what else I can do, because part of the current I1 goes through both diodes. I think I have too many unknowns (diode current, Is, V)
 
  • #4
asdf12312 said:
i(D1)=Is(eV/VT-1)
i(D2)=10*Is(eV/VT-1)

because Is increases by same value as area junction A. and then there is the assumptions V=0.7 and VT=25mV (room temperature). But I don't see what else I can do, because part of the current I1 goes through both diodes. I think I have too many unknowns (diode current, Is, V)

Have faith! Things will cancel.

You've written expressions for the diode currents given their potential drops (V, should be VD. You already have a variable V which stands for the output voltage. Don't confuse things by duplicating meanings for variables!). That's fine, but you'll need to rearrange them so you have expressions for VD in terms of the diode current for each. Note that you're looking for a particular "output" voltage V. So you'll be dealing with the voltages across components. That's why you want to rearrange the expressions so you have the diode voltages in terms of their currents...

You know EXACTLY how much current goes through each diode as soon as you specify I2. KCL applies! What are the two currents iD1 and iD2 in terms of I1 and I2? Can you write then write expressions for the diode voltages in those terms?
 
  • #5
Are the values i had in my OP right? I(D1)=I1/11 and I(D2)=10(I1)/11. I could just be overthinking this...if I2=2mA does that mean I(D2) is 2mA and I(D1) is 8mA? Doesn't seem right because I(D2) is supposed to be 10 times I(D1).
 
  • #6
asdf12312 said:
Are the values i had in my OP right? I(D1)=I1/11 and I(D2)=10(I1)/11. I could just be overthinking this...if I2=2mA does that mean I(D2) is 2mA and I(D1) is 8mA? Doesn't seem right because I(D2) is supposed to be 10 times I(D1).

There's no choice. If I2 is 2mA then 2mA MUST be flowing through D2. That leaves 8mA for D1 by KCL. So those would be the currents IF I2 is chosen to be 2mA. You're looking to find some value of I2 to satisfy a particular V.

So the value of I2 is not known at the outset, you have to find it. In order to do that you need to find the V that results from a choice of I2.

The area relationship for the junctions only tells you what the relative saturation currents are for the diodes. It does not tell you what the actual currents are; those are imposed by the rest of the circuit. The actual currents will produce particular VD's across the diodes according to the diode equation. Those VD's will set the output voltage V. Your mission is to find an I2 that will set the VD's in such a way as to produce that V.

So. Write the expressions for the VD's in terms of the diode currents.
 

FAQ: Solve Junction Diode Problem: V & I2 for D2=10xD1 Area

What is a junction diode?

A junction diode is a semiconductor device that allows current to flow in one direction but blocks it in the opposite direction. It is made up of a P-N junction, where one side is doped with a material that has an excess of positive charge carriers (holes) and the other side is doped with a material that has an excess of negative charge carriers (electrons).

How do I solve for V and I2 in a junction diode problem?

To solve for V (voltage) and I2 (current) in a junction diode problem, you will need to use the diode equation, which is V = Vt * ln(I/Io + 1), where Vt is the thermal voltage (typically around 26 mV), I is the current through the diode, and Io is the reverse saturation current. You will also need to use Ohm's Law (V = IR) to solve for the current in the circuit.

What is the relationship between D2 and D1 in a junction diode problem?

D2 is the area of the second junction diode and D1 is the area of the first junction diode. The ratio of D2 to D1 is important in determining the current and voltage in the circuit. A larger D2 will result in a larger current and voltage, while a smaller D2 will result in a smaller current and voltage.

What is the purpose of solving for V and I2 in a junction diode problem?

Solving for V and I2 allows you to understand the behavior of the junction diode in a circuit and determine the current and voltage levels at different points. This information is crucial in designing and analyzing electronic circuits.

What are some common applications of junction diodes?

Junction diodes have a variety of applications in electronic circuits, such as rectification, voltage regulation, and signal processing. They are also commonly used in power supplies, solar cells, and LED lights. They can also be used in combination with other components to create more complex electronic devices, such as transistors and integrated circuits.

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