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What goes on in a semiconductor.

  1. Jul 13, 2010 #1
    Hi Everyone!

    There are some things about semiconductors that have always puzzled me,and I think it's high time I got them cleared up!

    Here are my four questions...please help.

    1. When an impurity atom (dopant) is added to an intrinsic semiconductor,(say germanium),is the band structure of germanium atoms modified or is it the dopant atom's band structure that is modified?

    2. A hole is simply an electron vacancy...then why does a hole have a different mobility and mass than the electron?.

    3. Again,a hole is simply a vacancy of an electron,so a movement made by an electron is equivalent to a movement by a hole...then why do we add the currents due to the hole-movement and electron movement together to calculate total current?...aren't the two just different aspects of the same electron movement?

    4. In an n-type semiconductor,the minority carriers are the holes. Now,when these holes do move,they also cause electrons to move in the opposite way.....so really,it's all the same, so why do we differentiate between majority and minority carriers?
  2. jcsd
  3. Jul 13, 2010 #2
    It's true that the motion of holes is really caused by electron motion, but when we describe hole motion we are describing something different. When we say electrons move, we are referring to electrons in the conduction band. Movement of holes corresponds to the motion of electrons in the valence band. That's why you have to include both terms.

    Holes have a different mobility and effective mass than electrons. I can't give you a rigorous explanation for this, but there's no reason to expect that the motion of electrons in these two different situations should be identical. An electron in the valence band is in a very different environment than in the conduction band.

    When you add impurities to a semiconductor, you add trap levels within the band gap, but otherwise there is not generally a change in the band structure. What changes is the relative filling of the two bands, due to injection or depletion of electrons.
  4. Jul 13, 2010 #3
    Oh...I get it! So the when we talk about only electrons moving in a semiconductor,we mean conduction band electrons,which behave just like electrons in a metal.

    This is only in case of doped semiconductors though,as in intrinsic semiconductors,whenever we talk about electrons moving,they're moving in the valence band,so holes are always formed simulatneously.

    But,if this is the case,what happens in p type semiconductors? They don't have electrons in the conduction band! So which are the minority electrons in this case?

    Okay,so the semiconductor atoms' electrons are rearranged in the band structure due to the interaction with the atom of the dopant.

    Now the thing is,in a N-tpye semiconductor,the dopant atoms have the extra electrons (the fifth electron in pentavalent atoms). So since these extra electrons are what cause the improved conductivity,there seems to be no point in the semiconductor atoms' (the silicon or germanium's ) electrons being rearranged!

    On the other hand,in a p-type semiconductor,the semiconductor (silicon/germanium) atoms actually develop a positive charge as they sort of form a coordinate (dative bond) with the dopant.In this case,it is quite understandable that the electrons are rearranged.Thus in this case,it is the rearrangement of the host atoms' electrons that causes the improved conductivity,unlike in the n-type.

    Please tell me if I went wrong anywhere.
  5. Jul 13, 2010 #4
    Be careful here. Near the bottom of the conduction band, the electrons are nearly free and behave similarly to how they would in a metal. In other parts of the conduction band, the band structure has a strong impact. This is why you can get negative effective mass, for example. These electrons behave radically differently to a metal, even though they are in the conduction band.

    Even in intrinsic semiconductors, there is a population of electrons in the conduction band (and holes in the valence band) above 0 K. At room temperature, for example, there are about 10^10 conduction band electrons per cm^3 in silicon. This is 4-6 orders of magnitude less than in typical doping conditions, but still, it is these electrons in the conduction band/holes in the valence band that are reponsible for the conductivity of an intrinsic seimconductor.

    You're right that there aren't many electrons in the conduction band, but there are still some (maybe 10^5 per cm^3). These are the minority carriers. The majority carriers are the holes in the valence band, and these dominate the conductivity.

    If the temperature is high enough (room remperature for example), once the dopant has been introduced into the crystal, it will ionize and its extra electron will become delocalized in the conduction band. At this point it's no different from the electrons from the host atom. You can't tell where an electron came from, so it doesn't make sense to distinguish. Also I'm not sure what you mean by the semiconductor atoms' electrons being "rearranged".

    The semiconductor's atoms don't develop any charge. I'm not sure what you're referring to here. And again, it's not that the electrons are rearranged. You are either adding or removing electrons to the crystal, so the new ones occupy higher energy states that were previously vacant, or else previously filled states become vacant. But this doesn't correspond to a physical rearrangement of electrons in any way that we can easily visualize.
    Last edited: Jul 13, 2010
  6. Jul 14, 2010 #5
    Thanks,it's very useful to know this.

    What I learned from your previous response is that the introduction of dopants in the vicinity of the parent semiconductors atoms causes either an excess or deficit of electrons within the crystal.This situation causes more electrons to move into the conductance band(n-type) or into valence bond(p-type) ...this is what I previously called rearrangement.

    In a p-type,due to the dative bond by the parent semiconductor's,they develop a partial charge,isn't it?

    P.S Thanks for your replies johng23!
  7. Jul 14, 2010 #6
    Think of a conduction band electron as a car on a relatively empty high way. Looking down from a helicopter, the car is pretty damn mobile since there is nothing to get in its way.

    Now consider a traffic jam on that highway, with all the cars bumper-to-bumper except for one little space that a car can fit into (this being our hole). When a car enters that spot, another car can enter the spot occupied by the previous car. And this continues so forth as people slowly make their way home.

    Looking down from a helicopter, it appears that the hole in the traffic jam is moving backwards even though, in reality, it's just the movement of cars (electrons) from place to place that closes one space and opens up another. Would you expect this hole in the traffic jam to be as mobile as the car on the empty highway?
  8. Jul 14, 2010 #7
    I'm not quite sure if we can treat electrons as cars in a highway. As far as I know, mobility is related to the effective mass (the lower the mass, the higher the mobility), and hence the shape of the E-k diagram.

    In the valence band, there are usually two degenerate bands at k=0, the light hole and heavy hole bands. Which have different effective mass and mobility.

    I do not know if there is any case in which hole mobility is larger than electron mobility in crystalline semiconductors, but I do not see any fundamental reason why this would not be possible. If there is any, please let me know.
  9. Jul 14, 2010 #8
    Does it have anything to do with the fact that the valence electrons are actually involved in bonding, while the conduction electrons are delocalized? Or how about the fact that impurity scattering increases with proximity of the electron to the scatterer. Can we say the valence electrons are actually closer to the atom cores?

    Look at the chart half way down the page. http://en.wikipedia.org/wiki/Effective_mass_(solid-state_physics [Broken])
    It says that in Ge, the hole effective mass is smaller than the electron effective mass.
    Last edited by a moderator: May 4, 2017
  10. Jul 14, 2010 #9
    It could be related to the fact that the valence band electrons have bonding character. However, both conduction and valence band electrons are delocalized.

    Also, it is usually the case that the effective mass for electrons is smaller that for holes, which I think it is not related to impurity scattering. So even if the scattering was the same, the mobilities would not.

    This table is the effective mass for density of states, which for a single spherical band would be the same as the effective mass for mobility. But in general they are not.

    The density of states effective mass is just the mass that allows you to use the equations of a single spherical band to calculate the density of states.

    The effective mass used for the mobility is actually a tensor, so that mobility may be different in different directions (although this is not the case in Si and Ge due to the symmetry)
    Last edited by a moderator: May 4, 2017
  11. Jul 15, 2010 #10
    From as much as I've understood,this analogy is very effective! I think I understand now,that electrons come in two types--in one type,they are associated with holes and are resstricted to the valence bond and in the other type,they move around in conduction bands,which has nothing to do with holes.This point was the source of most of the confusion.

    Please could some one confirm my statement that I put in my last post,regarding if the host semiconductor atoms in a p-type crystal does develop a partial charge due to the formation of a dative bond with the dopant?
  12. Jul 15, 2010 #11
    Just my two cents, but I think you should stay away from this analogy. As I posted above, the holes in Ge are actually more mobile than the electrons, so it's not so simple. Better to keep the question unanswered than to adopt a potentially misleading explanation.

    I'm not sure what you mean by associated. The holes exist in the VB, so if that's what you mean, then ok. But you could also say the CB electrons are more associated with the holes, since the two are constantly being generated in pairs and recombining. Furthermore, when a electron is excited into the CB, it can remain attracted to the hole that it left due to the Coulomb force. The CB electron and VB hole can remain in a pair (this is called an exciton).

    You certainly can't say that CB electrons have "nothing to do" with holes. The number of CB electrons and VB holes are directly related by the expression np = ni^2, where ni is the intrinsic carrier concentration. This is true in all semiconductors regardless of doping.
  13. Jul 15, 2010 #12
    I agree with you that this analogy is not convenient, but actually mobility of electrons in germanium is larger than that of holes (take a look at http://www.ioffe.ru/SVA/NSM/Semicond/Ge/electric.html" [Broken]).

    The density of states effective mass is different that the effective mass for mobility.
    Last edited by a moderator: May 4, 2017
  14. Jul 15, 2010 #13

    As I understand it, by dative bond you mean that the dopant atom shares two electrons with a host atom, both of which come from the host. This may not be the most useful way to think of it.


    Let's just say the lattice is Si. On the left, the hole is with the dopant atom it came from. There are three bonds with adjacent Si atoms, and one broken bond. None of these will be considered dative bonds. Now the hole moves down and over to the right. Ok, the bottom bond of the dopant could be called dative because we know that both electrons involved came from the Si. But this bond is indistinguishable from the other three bonds between dopant and Si. Why call it something different? After a while, you have no idea which electrons came from the dopant. There are about a million Si atoms for each dopant atom. Clearly, NONE of the dopant electrons will be anywhere near the dopant. Are you going to say that it has four dative bonds? That's misleading since it contributed 3/4 of its own bonding electrons.

    But yes, you're right that the atomic core (nucleus plus valence electrons) is negatively charged relative to the atomic core of the host.
    Last edited: Jul 15, 2010
  15. Jul 15, 2010 #14
    Ok, thanks for that. I wasn't aware. Guess you can't just look at a chart without reading the text!
    Last edited by a moderator: May 4, 2017
  16. Jul 16, 2010 #15
    When I say that,I mean that I thought that "VB electrons are more associated with the holes, since the two are constantly being generated in pairs and recombining. "

    I think I have an enormous misconception here! I thought the CB electrons are the ones that flow similar to the metal electrons whereas the VB electrons having less energy (and more under the influence of the positive nucleus) are what combine and recombine with the holes!!
  17. Jul 16, 2010 #16
    Thanks for the diagram,johng23. I think I understand what you mean.Even when we did dative bonds in lower classes,our teachers told us that after the bond has been formed,you can't tell it apart from an ordinary coordinate bond.

    About the effective mass issue,I think it's better to not penetrate into the intricate details. I'll just have to remember that due to certain quantum mechanical phenomena,holes and electrons (CB electrons and the holes they form when they move out of a bond) have different effective masses.
  18. Jul 24, 2010 #17
    Hi everyone!

    I have a few more queries regarding the biasing of a P-N junction diode....please do help.

    1.In reverse bias,the holes in the p-side migrate (get attracted ) to the negative terminal on the p side itself.So,due to this,neutral atoms stay back in the bulk of the p-side.A similar situation occurs in the n side.

    If this is true,then the atoms near the junction are still neutral...how does the depletion region form?

    2.Also,in order to maintain the current,there needs to constant circulation of charges..how does that occur in the diode?

    3.In the revere bias,the reverse saturation current is due to breakage of bonds and thermal energy...does this energy come from the breakage of bonds itself?

    4.After hole-electron pair generation,in the p-side,does the electron generated cross the potential barrier and enter n-side?

    5.In forward bias, the holes and electrons (from p and n sides respectively) are repelled toward the junction and thereafter combine…..their neutralization should increase the barrier....but this is opposite to what we would epect in the forward bias,isn't it?
  19. Jul 26, 2010 #18
    Please come back to my thread!!
  20. Jul 27, 2010 #19
    First of all, you need to understand that the depletion region is already there regardless of the biasing condition. When you bring a P and an N type region in contact, the density of carriers is very different on the two sides, so electrons go to the p-side simply in response to the concentration gradient. The depletion region forms because of this.

    If it weren't for the fact that the dopant atoms have a charge relative to the intrinsic lattice, the charges would diffuse until the concentration profile was uniform. Once the electrons and holes start to migrate, the charges on the dopants are no longer compensated and the field restricts further charge diffusion. It's these competing factors that determine the width of the depletion region.

    Now, you're exactly right that in reverse bias the holes in the P-side are attracted to the terminal of the P-side and likewise for the N-side. This is why the depletion region increases width, because carriers are being pulled away from the junction. Don't think about neutral atoms staying near the junction. Of course there is always some atomic migration at ambient temperatures, but it's very small. For the most part, all the atomic cores are staying exactly where they are, regardless of their charge. The depletion region only refers to the depletion of carriers. Picture the lattice as a neutral array with a very small number (like 1 in a million) of charged atoms fixed in position.

    In forward bias, electrons and holes from close to the junction are migrating a short distance and recombining. Since this continually happens, the external circuit appears like a steady current is flowing, but in reality, no electron is moving across the width of the N- region, for example.

    Not sure what you're asking here. Some do, some don't.

    Electrons and holes recombining doesn't do anything to the barrier. I don't know if you're thinking that, after they recombine, the concentration of carriers on each side is different, but it's implicit in all of this that we are talking about steady state. Every time an electron and hole recombines, a new electron hole pair will be formed somewhere so that the concentrations retain their equilibrium values.

    I don't know your question about the saturation current in reverse bias.
  21. Jul 28, 2010 #20
    Okay,so it's basically that the dopant atoms get charged after they give away or accept electrons from the host element...so when the charge migration begins,suppose in the p side,(where the dopants are negatively charged stationary charges) the electrons,are repelled by these stationary negative charges....right?

    Thanks for the reply,johng23.

    When we say that in reverse bias,the holes in the P-side are attracted to the terminal of the P-side,we are basically saying that the minority electrons (that reside in the conduction band of the host atoms) move toward the junction...so it's exactly analogous to when the electrons from the n-side move in during depletion layer formation...that explains your statement,doesn't it?

    Wow,that's interesting! But,then I don't exactly understand what you mean by
    "electrons and holes from close to the junction are migrating a short distance ".
    Are you referring to the electrons in the CB(in o side) that move toward the junction?.....there'll be some electrons that cross over into the n side also,as the electrons in the n side migrate towards the positive terminal,leaving behind holes...:confused:

    Oh...actually I thought that if the electrons after being generated are accumulated in the p side,there might be a negative charge accumulation??

    I was thinking that,actually....thanks for pointing this out.
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