Why does the resistance of a semiconductor decrease with increasing temperature?

AI Thread Summary
The resistance of a semiconductor decreases with increasing temperature primarily due to the rise in the concentration of free charge carriers. As temperature increases, more electrons gain sufficient energy to transition into the conduction band, enhancing carrier density. This effect is particularly pronounced in intrinsic semiconductors, where thermal energy allows more carriers to overcome the energy gap. The discussion also highlights the relationship between current, charge density, and drift velocity, suggesting that both carrier concentration and drift velocity may change with temperature. Understanding these dynamics is essential for grasping semiconductor behavior in varying thermal conditions.
songoku
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Homework Statement


The resistance of a semiconductor decreases rapidly with increasing temperature. The main factor contributing to this effect is the rapid increase, with increasing temperature, of

a. the speed of the random motion of the free charge carriers
b. the concentration of the free charge carriers
c. the drift velocituy of the free charge carriers
d. the frequency of vibration of the atoms of the semiconductor
e. the amplitude of vibration of the atoms of the semiconductor


Homework Equations





The Attempt at a Solution


I read a little about semiconductor on wiki. I think the answer is (c). Am I right?

Thanks
 
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What's your reasoning?
 
Hi Mapes

Ups sorry, I mean the answer is (b). This is part of the article I read :

"In extrinsic (doped) semiconductors, dopant atoms increase the majority charge carrier concentration by donating electrons to the conduction band or accepting holes in the valence band. For both types of donor or acceptor atoms, increasing the dopant density leads to a reduction in the resistance."

So based on that, I think the answer is (b). But I don't know the reason why...:frown:

Thanks
 
OK, I agree with this answer, but I'd keep looking for the reason why! Why would temperature change the charge carrier concentration?
 
songoku said:
Hi Mapes

Ups sorry, I mean the answer is (b). This is part of the article I read :

"In extrinsic (doped) semiconductors, dopant atoms increase the majority charge carrier concentration by donating electrons to the conduction band or accepting holes in the valence band. For both types of donor or acceptor atoms, increasing the dopant density leads to a reduction in the resistance."

So based on that, I think the answer is (b). But I don't know the reason why...:frown:

Thanks

I think the increase is much greater with intrinsic(pure) semiconductors so I would advise you to do a little research on these.The answer comes most easily from the equation linking current with charge density, electron charge,drift velocity and conductor cross sectional area.Try a little search and come back if you can't find the equation.
 
Hi Dadface and Mapes

Sorry for taking long time to reply. Maybe this is the equation :
I=nAvq , where : n=number per volume of electrons

This is the best I can think of. The increase in temperature means that the carriers gain more energy. With this energy, some of the carriers can pass the "forbidden band" in semiconductor thus change the carriers concentration.
Another thing that pop-up in my mind is when there is flow of the carriers, the current increases. Based on the equation, there must be change in one or more variables on the right side. Maybe n will increase, but what about the drift velocity?

Thanks
 
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