Question on semiconductor device,

[run_margaret]

Homework Statement

I have a question on semiconductor device,,

A drift current density of Jdrf=150 A/cm2 is required in a semiconductor device using n-type silicon with an applied electric field of E=25V/cm. Determine the imjpurit doping concentration that will achieve this specification.

i don't really understand this topic, can someone help me please..

the answer is, If Nd = 3.13 * 10power 16 per cmcube,, then impurity doping concentration will be 1200..

i don't know how can i find Nd.. sind n has about the same value as Nd..

Homework Equations

Jdif = e(μn)nE

The Attempt at a Solution

(μn)n = 3.75 * 10 power -19


how can i find n or Nd ??

i know that (μn) = 3.75 * 10 power -19 / n

 

[KataKoniK]

Hello,

To find the impurity doping concentration (Nd) that will achieve a drift current density of 150 A/cm2, we can use the formula for drift current density:

Jdrf = e(μn)nE

Where:

- Jdrf is the drift current density (given as 150 A/cm2 in this case)
- e is the electronic charge (1.6 * 10^-19 C)
- μn is the electron mobility (given as 3.75 * 10^-19 cm2/Vs)
- n is the electron concentration (unknown)
- E is the applied electric field (given as 25 V/cm)

To solve for n, we need to rearrange the equation:

n = Jdrf / (eμnE)

Now, we can substitute in the given values and solve for n:

n = (150 A/cm2) / (1.6 * 10^-19 C * 3.75 * 10^-19 cm2/Vs * 25 V/cm)
n = 2.5 * 10^16 electrons/cm3

Since n is the electron concentration, we can assume that it is equal to the impurity doping concentration (Nd) for an n-type semiconductor. Therefore, the impurity doping concentration that will achieve a drift current density of 150 A/cm2 is 2.5 * 10^16 cm^-3.

I hope this helps clarify the topic for you. Let me know if you have any other questions or need further assistance. Good luck with your studies!

 

[Mene]

but i don't know how to find n or Nd


Hi there,

I would be happy to help you with your question on semiconductor devices. First, let's define some terms that may be helpful in understanding this problem:

- Semiconductor device: A device made from a semiconductor material, such as silicon, that can be used to control the flow of electricity.
- Drift current density: The flow of charge carriers (electrons or holes) through a semiconductor due to an applied electric field.
- N-type silicon: A type of semiconductor material where the majority of charge carriers are electrons.
- Impurity doping concentration: The amount of impurities, such as boron or phosphorus, intentionally added to a semiconductor material to alter its electrical properties.
- Nd: The doping concentration of donor impurities, such as phosphorus, in n-type silicon.

Now, let's break down the problem step by step:

1. Determine the drift current density (Jdrf) required in the semiconductor device. The problem states that Jdrf = 150 A/cm2.

2. Determine the applied electric field (E). The problem states that E = 25 V/cm.

3. Use the equation Jdrf = e(μn)nE to solve for the product of the mobility of electrons (μn) and the doping concentration of n-type silicon (n). This equation relates the drift current density to the mobility of the charge carriers, the doping concentration, and the applied electric field.

4. Plug in the given values for Jdrf and E. The equation now becomes 150 = e(μn)n(25).

5. Use the given value of the mobility of electrons (μn) to solve for n. The problem states that (μn)n = 3.75 * 10-19 cm2/Vs. Therefore, we can divide both sides of the equation by 3.75 * 10-19 to get n(25) = 40.

6. Solve for n by dividing both sides by 25. This gives us n = 40/25 = 1.6 x 1016 cm-3.

7. Finally, since n is the doping concentration of n-type silicon, we can say that the impurity doping concentration needed to achieve a drift current density of 150 A/cm2 is Nd = 1.6 x 1016 cm-3.

I hope this helps to clarify the problem and