Semiconductor doping - doping concentration is 0?

In summary, when calculating the intrinsic carrier concentration, majority carrier concentration, and Ef-Efi for a sample of germanium with acceptor concentration Na=10^17 and donor concentration Nd=0 at a temperature of T=400, the equations used were not specified and may not be accurate. However, based on the given information and calculations provided, the intrinsic carrier concentration was found to be 9.21*10^16, the majority carrier concentration was found to be 1.55*10^17, and the value for Ef-Efi was not specified. It should also be noted that when calculating these values, some errors were found and further clarification may be needed.
  • #1
orangeincup
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0

Homework Statement


A sample of germaium has an acceptor concentration of Na=10^17 and a donor concentration of Nd=0. Calculate the intrinsic carrier concentration, majority carrier concentration, and Ef-Efi. Use T=400

Homework Equations


No*Np=ni^2
Nv=Nv*(T/300)^3/2
Nc=Nc*(T/300)^3/2
ni^2=Nv*Nc exp(Ns)

The Attempt at a Solution


1.04*10^19 * (400/300)^3/2 = 1.23*10^19 =Nv
6*10^18 * (400/300)^3/2=7.11*10^18 = Ncni=((((1.04*10^19*6*10^18)^2exp(-.66/(2*400*8.6*10^-6)))^1/2

ni=9.21*10^16

10^17/2+sqrt((10^17/2)^2+(9.21*10^16)^2)=1.55*10^17

Does this look correct? I feel like the ni value is wrong
 
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  • #2
orangeincup said:
Nv=Nv*(T/300)^3/2
Nc=Nc*(T/300)^3/2
Those equations don't make sense.
 
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  • #3
mfb said:
Those equations don't make sense.
Isn't the new Nv different at different temperatures? Are you saying it doesn't make sense because I put Nv=Nv, or because I shouldn't use it?

I wasn't sure if I had to re-calculate a new value of Nv and Nc or not, the calculations I did below that used the original values and not the ones I calculated above
 
  • #4
What is new, what is old?
Your equations could be "solved" to give T=300. If you want to indicate two different things with Nv, then use different labels.
If you use numbers not given in the problem statement, it would help to explain where they come from (I made this comment before).
Also, for long formulas it is useful to write it in terms of variables first, and then plug in numbers. A large collection of numbers is hard to decrypt, and even harder if the origin of those numbers is unclear.
 
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  • #5
Nvi=Nvf*(T/300)^3/2

Effective density of states(valence, Nv T=300 K )
Nvf is the new calculated density of states(valence, Nv T=400 K )

The Nc is the same, but for the conduction bandEffective density of states
(conduction, Nc T=300 K )

2.8x10^19

source https://www.el-cat.com/silicon-properties.htm
 
  • #6
x^2=((1.04*10^19*6*10^18)*(400/300)^3exp(-.66/(400*8.62*10^-5))
=8.50*10^14, does this look more correct? I fixed errors in my last post

My calculated ni for 300k is off by a bit though... it's suppose to be 2.4*10^13 but I calculated 2.2*10^13 using the same method above. Did I make a mistake?
 

1. What is semiconductor doping?

Semiconductor doping is the process of intentionally introducing impurities into a semiconductor material in order to alter its electrical properties. This is done to create n-type or p-type semiconductors, which have excess electrons or holes, respectively, and are essential for the functioning of electronic devices.

2. What is the purpose of doping concentration being at 0?

A doping concentration of 0 means that there are no intentional impurities added to the semiconductor material. This can occur naturally or as a result of a specific manufacturing process. In either case, it results in a pure semiconductor, which may have specific applications in certain devices.

3. How does doping concentration affect the electrical properties of a semiconductor?

Doping concentration has a significant impact on the electrical properties of a semiconductor. It determines the number of free charge carriers (electrons or holes) in the material, which affects its conductivity and other characteristics. A higher doping concentration results in a higher concentration of free charge carriers and thus a more conductive material.

4. Can a semiconductor with a doping concentration of 0 be used in electronic devices?

Yes, a semiconductor with a doping concentration of 0 can still be used in electronic devices. However, its properties may not be ideal for certain applications. For example, a pure semiconductor may have a high resistance and thus be less suitable for use in transistors. However, it may be useful for applications such as photovoltaic cells.

5. What are the potential drawbacks of a semiconductor with a doping concentration of 0?

One potential drawback of a semiconductor with a doping concentration of 0 is its higher resistance and lower conductivity compared to doped semiconductors. This may limit its use in certain electronic devices. Additionally, a pure semiconductor may be more prone to impurity contamination, which can affect its performance.

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