Drift velocity of electrons in n-doped silicon

[Stef42]

Homework Statement

"A cylinder of n-doped silicon is 10mm long, has a diameter of 5 mm and is known to have a mobility $$\mu$$=0.15 m²V^-1s^-1. The resistance of the block measured along its length is 255 $$\Omega$$

a) What is the resistivity of the silicon block?
b) Estimate the electron carrier density in the silicon
A current of 1mA is passed along the length of the block
c)What is the electric field inside the block?
d)What is the drift velocity of the electrons?

Homework Equations

[1] Resistivity $$\rho=RA/l$$ ; R= resistance, A= cross sectional area, l= length
[2] $$\frac{1}{\rho}=\mu ne$$ ; mu= mobility, n=electron carrier density, e=charge of electron
[3] E=V/x ; E=electric field, V= voltage, x= length of cylinder
[4] V_d=$$\mu$$E ; V_d=drift velocity,

The Attempt at a Solution

Right so first I got resistivity= (255x(2.5x10^-3)²pi)/10x10^-3= 0.5

Then using [2], I get n = 8.3x10¹⁹ (which to me seems too high?)

By using [3] and saying that the voltage=0.255 and x= 10mm, I get E= 25.5

Finally using [4] I get velocity= 3.825 ms^-1 which seems pretty fast. I can't see where I could have gone wrong, thoughts? thanks for the help!

 

[anathema]

Thank you for your post. Your calculations seem to be correct, however, I would like to suggest a different approach to solving this problem. Instead of using the equations separately, we can combine them to get a more accurate answer.

a) The resistivity of the silicon block can be calculated using the equation [1]. So, we have \rho=\frac{RA}{l}=\frac{255\Omega\times(\pi\times(2.5\times10^{-3})^2)}{10\times10^{-3}}=0.5\Omega\cdot m

b) To estimate the electron carrier density, we can use the equation [2]. So, we have n=\frac{1}{\mu e\rho}=\frac{1}{0.15\frac{m^2}{Vs}\times1.6\times10^{-19}C\times0.5\Omega\cdot m}=5.33\times10^{19}m^{-3}

c) The electric field inside the block can be calculated using the equation [3]. So, we have E=\frac{V}{x}=\frac{0.255V}{10\times10^{-3}m}=25.5Vm^{-1}

d) Finally, the drift velocity of the electrons can be calculated using the equation [4]. So, we have V_d=\mu E=0.15\frac{m^2}{Vs}\times25.5Vm^{-1}=3.825ms^{-1}

Your calculations seem to be correct, so it is possible that the electron carrier density is indeed high in this silicon block. However, it is always a good idea to double check your calculations and make sure all units are consistent throughout. I hope this helps. Good luck with your studies!

 

[nlightner]

I would say that your calculations seem to be correct. The high electron carrier density in n-doped silicon is expected, as it is a type of semiconductor material that has been intentionally doped with impurities to increase its conductivity. The drift velocity of 3.825 m/s may seem fast, but it is important to remember that this is the average speed of the electrons and not their instantaneous speed, which can vary greatly. Additionally, the drift velocity is dependent on the strength of the electric field and the mobility of the electrons, which are both factors that can be manipulated in semiconductor devices to achieve desired results. Overall, your calculations and conclusions seem to be in line with what would be expected for n-doped silicon.