6.23 Semi­con­duc­tors

Semi­con­duc­tors are at the core of mod­ern tech­nol­ogy. This sec­tion dis­cusses some ba­sic prop­er­ties of semi­con­duc­tors that will be needed to ex­plain how the var­i­ous semi­con­duc­tor ap­pli­ca­tions work. The main semi­con­duc­tor ma­nip­u­la­tion that must be de­scribed in this sec­tion is dop­ing, adding a small amount of im­pu­rity atoms.

Fig­ure 6.31: Vicin­ity of the band gap in the spec­tra of in­trin­sic and doped semi­con­duc­tors. The amounts of con­duc­tion band elec­trons and va­lence band holes have been vastly ex­ag­ger­ated to make them vis­i­ble.
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(400,12...
...){\vector(0,1){30}}
\put(-94,90){\vector(0,-1){30}}
\end{picture}
\end{figure}

If semi­con­duc­tors did not con­duct elec­tric­ity, they would not be very use­ful. Con­sider first the pure, or “in­trin­sic,” semi­con­duc­tor. The vicin­ity of the band gap in its spec­trum is shown to the left in fig­ure 6.31. The ver­ti­cal co­or­di­nate shows the en­ergy ${\vphantom' E}^{\rm p}$ of the sin­gle-elec­tron quan­tum states. The hor­i­zon­tal co­or­di­nate shows the den­sity of states ${\cal D}$, the num­ber of quan­tum states per unit en­ergy range. Re­call that there are no quan­tum states in the band gap. States oc­cu­pied by elec­trons are shown in red. At room tem­per­a­ture there are some ther­mally ex­cited elec­trons in the con­duc­tion band. They left be­hind some holes in the va­lence band. Both the elec­trons and the holes can pro­vide elec­tri­cal con­duc­tion.

Time for a re­al­ity check. The num­ber of such elec­trons and holes is very much smaller than the fig­ure in­di­cates. The num­ber $\iota_{\rm {e}}$ of elec­trons per quan­tum state is given by the Fermi-Dirac dis­tri­b­u­tion (6.19). In the con­duc­tion band, that may be sim­pli­fied to the Maxwell-Boltz­mann one (6.21) be­cause the num­ber of elec­trons in the con­duc­tion band is small. The av­er­age num­ber of elec­trons per state in the con­duc­tion band is then:

\begin{displaymath}
\fbox{$\displaystyle
\iota_{\rm e} = e^{-({\vphantom' E}^{\rm p}- \mu)/k_{\rm B}T}
$} %
\end{displaymath} (6.33)

Here $T$ is the ab­solute tem­per­a­ture, $k_{\rm B}$ is the Boltz­mann con­stant, and $\mu$ is the chem­i­cal po­ten­tial, also known as the Fermi level. The Fermi level is shown by a red tick mark in fig­ure 6.31.

For an in­trin­sic semi­con­duc­tor, the Fermi level is about in the mid­dle of the band gap. There­fore the av­er­age num­ber of elec­trons per quan­tum state at the bot­tom of the con­duc­tion band is

\begin{displaymath}
\mbox{Bottom of the conduction band:}\quad
\iota_{\rm e} = e^{-{\vphantom' E}^{\rm p}_{\rm gap}/2k_{\rm B}T}
\end{displaymath}

At room tem­per­a­ture, ${k_{\rm B}}T$ is about 0.025 eV while for sil­i­con, the band gap en­ergy is about 1.12 eV. That makes $\iota_{\rm {e}}$ about 2 10$\POW9,{-10}$. In other words, only about 1 in 5 bil­lion quan­tum states in the lower part of the con­duc­tion band has an elec­tron in it. And it is even less higher up in the band. A fig­ure can­not show a frac­tion that small; there are just not enough atoms on a page.

So it is not sur­pris­ing that pure sil­i­con con­ducts elec­tric­ity poorly. It has a re­sis­tiv­ity of sev­eral thou­sand ohm-m where good met­als have on the or­der of 10$\POW9,{-8}$. Pure ger­ma­nium, with a smaller band gap of 0.66 eV, has a much larger $\iota_{\rm {e}}$ of about 3 10$\POW9,{-6}$ at the bot­tom of the con­duc­tion band. Its re­sis­tiv­ity is cor­re­spond­ingly lower at about half an ohm-m. That is still many or­ders of mag­ni­tude larger than for a metal.

And the num­ber of con­duc­tion elec­trons be­comes much smaller still at cryo­genic tem­per­a­tures. If the tem­per­a­ture is a frigid 150 K in­stead of a 300 K room tem­per­a­ture, the num­ber of elec­trons per state in sil­i­con drops by an­other fac­tor of a bil­lion. That il­lus­trates one im­por­tant rule:

You can­not just for­get about tem­per­a­ture to un­der­stand semi­con­duc­tors.
Usu­ally, you like to an­a­lyze the ground state at ab­solute zero tem­per­a­ture of your sys­tem, be­cause it is eas­ier. But that sim­ply does not work for semi­con­duc­tors.

The num­ber of holes per state in the va­lence band may be writ­ten in a form sim­i­lar to that for the elec­trons in the con­duc­tion band:

\begin{displaymath}
\fbox{$\displaystyle
\iota_{\rm h} = e^{-(\mu - {\vphantom' E}^{\rm p})/k_{\rm B}T}
$} %
\end{displaymath} (6.34)

Note that in the va­lence band the en­ergy is less than the Fermi level $\mu$, so that the ex­po­nen­tial is again very small. The ex­pres­sion above may be checked by not­ing that what­ever states are not filled with elec­trons are holes, so $\iota_{\rm {h}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1-\iota_{\rm {e}}$. If you plug the Fermi-Dirac dis­tri­b­u­tion into that, you get the ex­pres­sion for $\iota_{\rm {h}}$ above as long as the num­ber of holes per state is small.

From a com­par­i­son of the ex­pres­sions for the num­ber of par­ti­cles per state $\iota_{\rm {e}}$ and $\iota_{\rm {h}}$ it may al­ready be un­der­stood why the Fermi level $\mu$ is ap­prox­i­mately in the mid­dle of the band gap. If the Fermi level is ex­actly in the mid­dle of the band gap, $\iota_{\rm {e}}$ at the bot­tom of the con­duc­tion band is the same as $\iota_{\rm {h}}$ at the top of the va­lence band. Then there is the same num­ber of elec­trons per state at the bot­tom of the con­duc­tion band as holes per state at the top of the va­lence band. That is about as it should be, since the to­tal num­ber of elec­trons in the con­duc­tion band must equal the to­tal num­ber of holes in the va­lence band. The holes in the va­lence band is where the elec­trons came from.

Note that fig­ure 6.31 is mis­lead­ing in the sense that it de­picts the same den­sity of states ${\cal D}$ in the con­duc­tion band as in the va­lence band. In re­al­ity, the num­ber of states per unit en­ergy range in the con­duc­tion band could eas­ily be twice that at the cor­re­spond­ing lo­ca­tion in the va­lence band. It seems that this should in­val­i­date the above ar­gu­ment that the Fermi level $\mu$ must be in the mid­dle of the band gap. But it does not. To change the ra­tio be­tween $\iota_{\rm {e}}$ and $\iota_{\rm {h}}$ by a fac­tor 2 re­quires a shift in $\mu$ of about 0.01 eV at room tem­per­a­ture. That is very small com­pared to the band gap. And the shift would be much smaller still closer to ab­solute zero tem­per­a­ture. At ab­solute zero tem­per­a­ture, the Fermi level must move to the ex­act mid­dle of the gap.

That il­lus­trates an­other im­por­tant rule of thumb for semi­con­duc­tors:

Keep your eyes on the ther­mal ex­po­nen­tials. Usu­ally, their vari­a­tions dwarf every­thing else.
If ${\vphantom' E}^{\rm p}$ or $\mu$ changes just a lit­tle bit, $e^{-({\vphantom' E}^{\rm p}-\mu)/{k_{\rm B}}T}$ changes dra­mat­i­cally.

(For gal­lium ar­senide, the dif­fer­ence be­tween the den­si­ties of states for holes and elec­trons is much larger than for sil­i­con or ger­ma­nium. That makes the shift in Fermi level at room tem­per­a­ture more sub­stan­tial.)

The Fermi level may be di­rectly com­puted. Ex­pres­sions for the to­tal num­ber of con­duc­tion elec­trons per unit vol­ume and the to­tal num­ber of holes per unit vol­ume are, {D.30}:

\begin{displaymath}
\fbox{$\displaystyle
i_{\rm e} =
2 \left(\frac{m_{\rm eff...
...
e^{-(\mu - {\vphantom' E}^{\rm p}_{\rm v})/k_{\rm B}T}
$} %
\end{displaymath} (6.35)

Here ${\vphantom' E}^{\rm p}_{\rm {c}}$ and ${\vphantom' E}^{\rm p}_{\rm {v}}$ are the en­er­gies at the bot­tom of the con­duc­tion band, re­spec­tively the top of the va­lence band. The ap­pro­pri­ate ef­fec­tive masses for elec­trons and holes to use in these ex­pres­sions are com­pa­ra­ble to the true elec­tron masses for sil­i­con and ger­ma­nium. Set­ting the two ex­pres­sions above equal al­lows $\mu$ to be com­puted.

The first ex­po­nen­tial in (6.35) is the value of the num­ber of elec­trons per state $\iota_{\rm {e}}$ at the bot­tom of the con­duc­tion band, and the sec­ond ex­po­nen­tial is the num­ber of holes per state $\iota_{\rm {h}}$ at the top of the va­lence band. The bot­tom line re­mains that semi­con­duc­tors have much too few cur­rent car­ri­ers to have good con­duc­tiv­ity.

That can be greatly im­proved by what is called dop­ing the ma­te­r­ial. Sup­pose you have a semi­con­duc­tor like ger­ma­nium, that has 4 va­lence elec­trons per atom. If you re­place a ger­ma­nium atom in the crys­tal by a stray atom of a dif­fer­ent el­e­ment that has 5 va­lence elec­trons, then that ad­di­tional elec­tron is mis­matched in the crys­tal struc­ture. It can eas­ily be­come dis­lo­cated and start rov­ing through the con­duc­tion band. That al­lows ad­di­tional con­duc­tion to oc­cur. Even at very small con­cen­tra­tions, such im­pu­rity atoms can make a big dif­fer­ence. For ex­am­ple, you can in­crease the con­duc­tiv­ity of ger­ma­nium by a fac­tor of a thou­sand by re­plac­ing 1 in a mil­lion ger­ma­nium atoms by an ar­senic one.

Be­cause such va­lence-5 im­pu­rity atoms add elec­trons to the con­duc­tion band, they are called “donors.” Be­cause elec­tri­cal con­duc­tion oc­curs by the neg­a­tively charged ad­di­tional elec­trons pro­vided by the dop­ing, the doped semi­con­duc­tor is called “n-type.”

Al­ter­na­tively, you can re­place ger­ma­nium atoms by im­pu­rity atoms that have only 3 va­lence elec­trons. That cre­ates holes that can ac­cept va­lence band elec­trons with a bit of ther­mal en­ergy. There­fore such im­pu­rity atoms are called “ac­cep­tors.” The holes in the va­lence band from which the elec­trons were taken al­low elec­tri­cal con­duc­tion to oc­cur. Be­cause the holes act like pos­i­tively charged par­ti­cles, the doped semi­con­duc­tor is called “p-type.”

Sil­i­con has 4 va­lence band elec­trons just like ger­ma­nium. It can be doped sim­i­larly.

Now con­sider an n-type semi­con­duc­tor in more de­tail. As the cen­ter of fig­ure 6.31 in­di­cates, the ef­fect of the donor atoms is to add a spike of en­ergy states just be­low the con­duc­tion band. At ab­solute zero tem­per­a­ture, these states are filled with elec­trons and the con­duc­tion band is empty. And at ab­solute zero, the Fermi level is al­ways in be­tween filled and empty states. So the Fermi level is now in the nar­row gap be­tween the spike and the con­duc­tion band. It il­lus­trates that the Fermi level of a semi­con­duc­tor can jump around wildly at ab­solute zero.

But what hap­pens at ab­solute zero is ir­rel­e­vant to a room tem­per­a­ture semi­con­duc­tor any­way. At room tem­per­a­ture the Fermi level is typ­i­cally as shown by the tick mark in fig­ure 6.31. The Fermi level has moved up a lot com­pared to the in­trin­sic semi­con­duc­tor, but it still stays well be­low the donor states.

If the Fermi level would not move up, then the to­tal num­ber of elec­trons in the con­duc­tion band would not change. And there would be ex­tremely few elec­trons in the donor states for that Fermi level. That is not pos­si­ble, be­cause all the other donor elec­trons can­not just dis­ap­pear. In fact, the amount of elec­trons con­tributed by the donor states is dra­matic; that is be­cause there are so ex­tremely few con­duc­tion elec­trons in the in­trin­sic case. The Fermi level has to move up sig­nif­i­cantly to ex­plain the in­crease. An in­crease in Fermi level $\mu$ in­creases the num­ber of elec­trons per quan­tum state (6.33) in the con­duc­tion band. The Fermi level has to go up far enough that the com­bined num­ber of elec­trons in the con­duc­tion band and the donor states is just a bit more than the num­ber of donor elec­trons.

But the Fermi level can­not move too close to the donor states ei­ther. For as­sume the con­trary, that the Fermi level is re­ally close to the donor states. Then the donor states will be largely filled with elec­trons. But at room tem­per­a­ture the gap be­tween the donor states and the con­duc­tion band is com­pa­ra­ble to ${k_{\rm B}}T$. There­fore, if the donor states are largely filled with elec­trons, then the states at the bot­tom of the con­duc­tion band con­tain sig­nif­i­cant num­bers of elec­trons too. Since there are so many of these states com­pared to a typ­i­cal num­ber of donors, the num­ber of elec­trons in them would dwarf the num­ber of elec­trons miss­ing from the donor states. And that is not pos­si­ble since the to­tal num­ber of elec­trons can­not ex­ceed the num­ber of donor states by any no­tice­able amount. So the Fermi level has to stay low enough that the num­ber of elec­trons $\iota_{\rm {e}}$ stays small in both the donor states and con­duc­tion band. That is as sketched in fig­ure 6.31.

If more donors are added, the Fermi level will move up more. Light dop­ing may be on the or­der of 1 im­pu­rity atom in a 100 mil­lion, heavy dop­ing 1 in 10,000. If the donor atoms get too close to­gether, their elec­trons start to in­ter­act. If that hap­pens the spike of donor states broad­ens into a band, and you end up with a metal­lic “de­gen­er­ate” semi­con­duc­tor. For ex­am­ple, low tem­per­a­ture mea­sure­ments show that phos­phor donors turn sil­i­con metal­lic at about 1 phos­phor atom per 15 000 sil­i­con ones. It may seem strange that im­pu­rity elec­trons at such a small con­cen­tra­tion could in­ter­act at all. But note that 1 im­pu­rity in 15 000 atoms means that each $25\times25\times25$ cube of sil­i­con atoms has one phos­phor atom. On av­er­age the phos­phor atoms are only about 25 atom spac­ings apart. In ad­di­tion, the or­bit of the very loosely bound donor elec­tron is re­ally far from the pos­i­tively charged donor atom com­pared to the crys­tal spac­ing.

The up­ward shift in the Fermi level in n-type ma­te­r­ial has an­other ef­fect. It dec­i­mates the al­ready mis­er­ably small num­ber of holes in the va­lence band, (6.34). There­fore the num­ber of con­duc­tion band elec­trons pro­vided by the va­lence band be­comes many times smaller still than it was al­ready for the in­trin­sic semi­con­duc­tor. Es­sen­tially all con­duc­tion band elec­trons are pro­vided by the donors. Also, al­most all elec­tri­cal con­duc­tion will now be per­formed by elec­trons, not holes. The elec­trons in n-type ma­te­r­ial are there­fore called the “ma­jor­ity car­ri­ers” and the holes the “mi­nor­ity car­ri­ers.”

Fig­ure 6.32: Re­la­tion­ship be­tween con­duc­tion elec­tron den­sity and hole den­sity. In­trin­sic semi­con­duc­tors have nei­ther much con­duc­tion elec­trons nor holes.
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(300,16...
...17.2){\vector(0,-1){15}}
\put(53,22){\emph{p}-type}
\end{picture}
\end{figure}

The fact that rais­ing the amount of con­duc­tion band elec­trons low­ers the amount of va­lence band holes may be ver­i­fied math­e­mat­i­cally from (6.35). That equa­tion im­plies that the prod­uct of the elec­tron and hole den­si­ties is con­stant at a given tem­per­a­ture:

\begin{displaymath}
\fbox{$\displaystyle
i_{\rm e} i_{\rm h} =
4 \left(\frac{...
...ght)^3
e^{-{\vphantom' E}^{\rm p}_{\rm gap}/k_{\rm B}T}
$} %
\end{displaymath} (6.36)

This re­la­tion­ship is called the “law of mass ac­tion” since non­ex­perts would be able to make sense out of elec­tron-hole den­sity re­la­tion. And if you come to think of it, what is wrong with the name? Doesn’t pretty much every­thing in physics come down to masses per­form­ing ac­tions? That in­cludes semi­con­duc­tors too!

The re­la­tion­ship is plot­ted in fig­ure 6.32. It shows that a high num­ber of con­duc­tion elec­trons im­plies a very low num­ber of holes. Sim­i­larly a p-type ma­te­r­ial with a high num­ber of holes will have very few con­duc­tion elec­trons. The p-type ma­te­r­ial is an­a­lyzed pretty much the same as n-type, with holes tak­ing the place of elec­trons and ac­cep­tors the place of donors.

The law of mass ac­tion can also be un­der­stood from more clas­si­cal ar­gu­ments. That is use­ful since band the­ory has its lim­its. In ther­mal equi­lib­rium, the semi­con­duc­tor is bathed in black­body ra­di­a­tion. A very small but nonzero frac­tion of the pho­tons of the ra­di­a­tion have en­er­gies above the band gap. These will move va­lence band elec­trons to the con­duc­tion band, thus cre­at­ing elec­tron-hole pairs. In equi­lib­rium, this cre­ation of elec­tron-hole pairs must be bal­anced by the re­moval of an iden­ti­cal amount of elec­tron-hole pairs. The re­moval of a pair oc­curs through “re­com­bi­na­tion,” in which an con­duc­tion band elec­tron drops back into a va­lence band hole, elim­i­nat­ing both. The rate of re­com­bi­na­tions will be pro­por­tional to the prod­uct of the den­si­ties of elec­trons and holes. In­deed, for a given num­ber of holes, the more elec­trons there are, the more will be able to find holes un­der suit­able con­di­tions for re­com­bi­na­tion. And vice-versa for holes. Equat­ing a cre­ation rate $A$ of elec­tron-hole pairs by pho­tons to a re­moval rate of the form $Bi_{\rm {e}}i_{\rm {h}}$ shows that the prod­uct $i_{\rm {e}}i_{\rm {h}}$ is con­stant. The con­stant $A$$\raisebox{.5pt}{$/$}$$B$ will de­pend pri­mar­ily on the Maxwell-Boltz­mann fac­tor $e^{-{\vphantom' E}^{\rm p}_{\rm {gap}}/{k_{\rm B}}T}$ that lim­its the num­ber of pho­tons that have suf­fi­cient en­ergy to cre­ate pairs.

This pic­ture also pro­vides an in­tu­itive ex­pla­na­tion why adding both donors and ac­cep­tors to a semi­con­duc­tor does not dou­ble the amount of cur­rent car­ri­ers over just one type of dop­ing alone. Quite the op­po­site. As fig­ure 6.32 shows, if the num­ber of holes be­comes com­pa­ra­ble to the num­ber of elec­trons, there are not many of ei­ther one. The semi­con­duc­tor be­haves again like an in­trin­sic one. The rea­son is that adding, say, some ac­cep­tors to an n-type ma­te­r­ial has the pri­mary ef­fect of mak­ing it much eas­ier for the con­duc­tion band elec­trons to find va­lence band holes to re­com­bine with. It is said that the added ac­cep­tors com­pen­sate for the donors.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Dop­ing a semi­con­duc­tor with donor atoms greatly in­creases the num­ber of elec­trons in the con­duc­tion band. It pro­duces an n-type semi­con­duc­tor.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Dop­ing a semi­con­duc­tor with ac­cep­tor atoms greatly in­creases the num­ber of holes in the va­lence band. It pro­duces an p-type semi­con­duc­tor.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The mi­nor­ity car­rier gets dec­i­mated.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The Fermi level is in the band gap, and to­wards the side of the ma­jor­ity car­rier.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
There is com­pen­sa­tion in dop­ing. In par­tic­u­lar, if there are about the same num­bers of elec­trons and holes, then there are not many of ei­ther.