# What Value of V Results in This Junction Diode Circuit?

• Engineering
• asdf12312
In summary, the diode D2 has 10 times the junction area of D1 and with a current of 5.75mA, a voltage of 50mV can be obtained.
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?

#### Attachments

• diode.png
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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:
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!

## What is a junction diode circuit?

A junction diode circuit is a type of electronic circuit that uses a semiconductor device called a diode to control the flow of electrical current. It is typically used in electronic devices such as radios, computers, and power supplies.

## What are the components of a junction diode circuit?

A junction diode circuit consists of a diode, a power supply, and a load. The diode is placed between the power supply and the load, and it controls the flow of current from the power supply to the load.

## How does a junction diode circuit work?

A junction diode circuit works by allowing current to flow in one direction, but blocking it in the opposite direction. This is due to the way the diode is constructed, with a P-type semiconductor on one side and an N-type semiconductor on the other. When a voltage is applied in the forward direction, the diode allows current to flow through it. When a voltage is applied in the reverse direction, the diode blocks the flow of current.

## What are some common problems that can occur in a junction diode circuit?

Some common problems that can occur in a junction diode circuit include overloading, overvoltage, and reverse bias. Overloading can occur when too much current is flowing through the diode, which can cause it to overheat and potentially fail. Overvoltage can damage the diode, while reverse bias can cause the diode to break down and allow excessive current to flow in the wrong direction.

## How can I troubleshoot and solve problems in a junction diode circuit?

If you encounter problems in a junction diode circuit, the first step is to check all connections and make sure they are secure. You should also check the voltage and current levels to ensure they are within the diode's specifications. If the issue persists, you may need to replace the diode or other components in the circuit.

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