Solid State Physics - p-n junctions

In summary, an ohmmeter can determine the "direction" of a diode by connecting it one way and then reversing the leads. The reverse resistance to forward resistance ratio at 300K can be calculated using Ohm's Law, where V is the potential and q is the charge. The potential for reverse bias is negative, while for forward bias it is positive. A minor rounding error may occur in the calculation.
  • #1
GrantB
22
0

Homework Statement



An ohmmeter is sometimes used to determine the "direction" of a diode by connecting the ohmmeter to the diode one way and then reversing the ohmmeter leads. If the ohmmeter applies an emf of .5V to the diode in order to determine resistance, what would be the ratio of reverse resistance to forward resistance at 300K?


Homework Equations



I=I0(ee[itex]\varphi[/itex]/kBT-1) ?



The Attempt at a Solution



I know that for reverse bias, the potential [itex]\varphi[/itex] is negative, and for forward bias, it is positive. But that's basically as far as I've gotten.

I am mainly confused as to what the relationship between the potential and emf is, and how that leads you to getting the resistance.

I think I've forgotten too much of my previous physics class...

Thanks for the help.
 
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  • #2
Use Ohm's Law to write [itex]R=\frac{V}{I}=\frac{V}{I_0 e^{qV/kT}-1}[/itex]. I used V instead of [itex]\phi[/itex] and q instead of e
 
  • #3
I'm running into a problem.

When I take e-qV/kT-1 for the reverse bias it gives me a negative number.
 
Last edited:
  • #4
You also have to take V in the numerator to be negative, then you will get a positive number
 
  • #5
cbetanco said:
You also have to take V in the numerator to be negative, then you will get a positive number

Ahh, thank you.

I am getting an answer of 2.5x108 when the answer is 2.4x108. Although, I think its a rounding (error) somewhere since what I'm doing seems correct.

Thanks!
 

FAQ: Solid State Physics - p-n junctions

What is a p-n junction?

A p-n junction is a type of semiconductor device that forms a boundary between two regions of semiconductor material with different electrical properties. It is formed by joining a p-type semiconductor (which has an excess of positively charged carriers) and an n-type semiconductor (which has an excess of negatively charged carriers).

How does a p-n junction work?

A p-n junction works by creating a depletion region between the p-type and n-type semiconductors. This region is depleted of free charge carriers and creates an electric field that prevents current from flowing in the reverse direction. When a voltage is applied in the forward direction, the depletion region decreases, allowing current to flow through the junction.

What are the applications of p-n junctions?

P-n junctions are the basis for many important electronic devices, such as diodes, transistors, and solar cells. They are also used in rectifiers, which convert alternating current into direct current, and in laser diodes, which are used in a variety of applications including telecommunications and laser pointers.

What factors affect the behavior of a p-n junction?

The behavior of a p-n junction is affected by several factors, including the doping concentrations of the p-type and n-type semiconductors, the width of the depletion region, and the applied voltage. The material properties of the semiconductors, such as their band gap and mobility, also play a role in the behavior of the junction.

What is the difference between a p-n junction and a Schottky diode?

A p-n junction and a Schottky diode are both types of diodes, but they function differently. A p-n junction is formed by two different types of semiconductor materials, while a Schottky diode is made by joining a metal and a semiconductor. This difference in materials leads to differences in the behavior and characteristics of the two types of diodes.

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