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Help in semiconductors: minority and majority carriers!

  1. Jun 13, 2012 #1
    I am currently taking up a course in electronic devices and circuits. Our first topic was about semiconductors and I'm having trouble understanding the doping process. We didn't have any prerequisite solid state course or any other course related to it that's why my professor explains the concepts qualitatively. We didn't tackle concepts i see in the internet like mobility, fermi energy.............and i have no idea what those are. I want to try and study those concepts by myself, but i suppose they require higher levels of physics, than what i currently know. It would consume too much time and it might affect my performance in my other subjects.
    My problem is about minority and majority carriers:
    If silicon is doped with a pentavalent atom, the four valence electrons of silicon will form covalent bonds with four valance electrons of the pentavalent atom and there is an unpaired electron from the pentavalent atom. It acts as a free carrier electron. What i don't understand is why should there be minority carriers. With all those covalent bonds there are initially only unpaired free electrons and i think there should be no holes . A hole appears when the free electron moves but then the number of free electrons must be the same as the number of holes. My question is how can we say that the electrons are the majority carriers and the holes are the minority??

    I have the same problem with silicon doped with a trivalent atom

    Please help, Thanks in advance =)
     
  2. jcsd
  3. Jun 13, 2012 #2

    Simon Bridge

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    An "undoped" semiconductor has an equal number of electrons and holes. A semiconductor that has been doped to be "n-type" has had extra electrons added to it. Since there are now more electrons than holes, the electrons are the majority charge carriers and the holes are the minority charge carriers.

    A semiconductor that has been doped as "p-type" has essentially had extra holes added to it, so the holes are now the majority charge carrier and the electrons are the minority charge carriers.

    "Minority" and "majority" just refer to whether a semiconductor has more holes than electrons (p-type), or more electrons than holes (n-type).
     
  4. Jun 15, 2012 #3
    You have raised an interesting point. If all four electrons are covalently bonded and one extra electron is given by n-doping, why should there be minority carriers?

    The answer is there will be minority carriers due to thermal energy. The semiconductor is in room temperature and average energy of carriers(electrons) will be 26 meV. [This energy comes from the energy associated with the degree of freedom of electrons]. Now, you know some energy is required to break a bond. The thing is 26 meV is actually too little to break covalent bonds. But as this amount of energy is actually an average - some electrons will have way higher energy than this .. and some have too little. Those electrons which have very high energy will break covalent bonds and generate electron hole pairs. Thus you'll always have minority carriers in doped-semiconductor ...

    except.. in 0K temperature the average thermal energy is also 0 eV. Thus no bond breakage and no minority carriers. In this temperature an intrinsic semiconductor will not even have any carriers. This phenomenon has a name - carrier freeze out.
     
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