Why is my solution for real diode analysis wrong?

In summary, the conversation is about a problem involving a diode and its associated equation, I=Ise(V/VT). The person asking the question initially got the wrong solution and realized that they had ignored the VT value in the exponent. The other person then pointed out that the VT value is an approximation and can vary depending on the diode. They suggest finding the true value before continuing with the problem. Another person suggests solving for Is and VT before proceeding with the problem.
  • #1
theBEAST
364
0

Homework Statement


9XUbpS3.png


Homework Equations


I=Ise(V/VT)

The Attempt at a Solution


The solution I got is wrong. Does anyone know why?

Here is my attempt, the image was too large for this forum so here is the link:
http://i.imgur.com/4tuGxdO.jpg
 
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  • #2
You ignored VT in the exponent.

ehild
 
  • #3
ehild said:
You ignored VT in the exponent.

ehild

Isn't the VT value the same (approximately 0.025V at room temperature)?
 
  • #4
theBEAST said:
Isn't the VT value the same (approximately 0.025V at room temperature)?

The approximation is just that, an approximation. It is a good "typical" value for run-of-the-mill diodes. The actual value for a given diode depends upon the details of the diode doping. You should determine its true value for this problem.
 
  • #5
theBEAST said:
Isn't the VT value the same (approximately 0.025V at room temperature)?

Even in case it is, I1/I2 = e(V1-V2)/VT

ehild
 
  • #6
You may need to run a few simultaneous equations. Here's my suggestion: Use the generic information provided to solve for Is and VT before proceeding further. Then from there you can solve for R using the fact that both sides of the circuit must equate 10mA.
 

Related to Why is my solution for real diode analysis wrong?

1. What is a diode?

A diode is a two-terminal electronic component that allows current to flow in only one direction. It is typically made of semiconductor material and is used in various electronic circuits.

2. How does a diode work?

A diode works by having a PN junction, which is created by joining a p-type and an n-type semiconductor material. When a voltage is applied across the diode in the forward direction, it allows current to flow easily. However, when the voltage is applied in the reverse direction, the diode acts as an insulator and does not allow current to flow.

3. What is a real diode analysis?

Real diode analysis is a process of studying the behavior of a diode in real-world conditions. This includes factors such as temperature, voltage fluctuations, and manufacturing defects that can affect the performance of a diode.

4. How do you analyze a real diode?

To analyze a real diode, one must consider its characteristics such as forward and reverse voltage, forward and reverse current, and the voltage-current relationship. These can be measured and compared to the ideal diode model to determine the diode's performance in a real-world scenario.

5. Why is real diode analysis important?

Real diode analysis is important because it allows engineers and scientists to understand and predict the behavior of diodes in practical applications. This information can be used to design more efficient and reliable electronic circuits and devices.

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