p-n junctions

Intuitive description. What are p-n junctions?p-n junctions are formed by starting with a Si wafer (or substrate) of a given type (say: B-doped p-type,to x the ideas) and diusing or implanting impurities of opposite type (say: n-type, as from a gas sourceof P such as phosphine or implanting As ions) in a region of the wafer. At the edge of the diused (orimplanted) area there will be a junction in which the p-type and the n-type semiconductor will be in directcontact. Refer to the Streetman-Banerjee text, section 5.1, for a description of semiconductor processing.We will review this topic later on, before dealing with metal-oxide-semiconducor (MOS) ed-eect transisitors(FET).

What happens to the junction at equilibrium?Consider the idealized situation in which we take an n-type Si crystal and a p-type Si crystal and bringthem together, while keeping them grounded, that is, attached to contacts at zero voltage. At rst, theconduction and valence band edges will line up, while the Fermi level will exhibit a discontinuity at the junction.But now electrons are free to diuse from the n-region to the p-region, pushed by the diusion term Dnnin the DDE. Similarly, holes will be free to diuse to the n region. As these diusion processes happen, theconcentration of extra electrons in the p-region will build up, as well as the density of extra holes in the nregion. These charges will grow until they will build an electric eld which will balance and stop the diusiveow of carriers. Statistical mechanics demands that at equilibrium the Fermi level of the system is unique andconstant. Therefore, the band-edges will bend acquiring a spatial dependence. This is illustrated in the leftframe of the gure on page 97. Note:

1. Deep in the n-type region to the right and in the p-type region to the left the semiconductor remainsalmost neutral: The contacts have provided the carriers lost during the diusion mentioned above, so thatn = ND in the quasi-neutral n region and p = NA in the quasi-neutral p region.

2. There is a central region which is depleted of carriers: Electrons have left the region 0 x xn0, holeshave left the region xp0 x < 0, so that for xp0 x xn0 we have np < n2i . This is called thetransition region or, more often, the depletion region of the junction. Its total width is W = xn0 + xp0.

ECE344 Fall 2009 95

3. The voltage barrier built by the difusion of carriers upon putting the n and p regions in contact witheach other is called the built-in potential, Vbi (denoted by eV0 in the textbook). Streetman and Banerjeepresent one possible way to calculate it. But an alternative, easier approach is based on the observation thatVbi will be given by the dierence between the equilibrium Fermi levels in the the two regions:

Vbi = EFn0EFp0 = Ei+ kBT ln(ND

ni

)Ei+ kBT ln

(NA

ni

)= kBT ln

(NDNA

n2i

).

(123)

Note that since the hole concentration in the p region, pp, is equal to NA, and the hole concentration in

the n region, pn is equal to n2i /ND, the equation above can be rewritten as:

Vbi = kBT ln

(pp

pn

), (124)

orpp

pn= e

eVbi/(kBT ) =nn

np, (125)

where the last step is based on the fact that at equilibrium ppnp = pnnn. Note: Vbi dened above is a

potential energy, measured in joules. If measured in eV, its expression will be (kBT/e) ln(NDNA/n2i ),

which is V0 in the text.

ECE344 Fall 2009 96

ECE344 Fall 2009 97

What happens to a biased junction?Lets now apply a bias Va to the junction. We consider Va positive when positive bias is applied to thep-region, as illustrated in the gure. If Va > 0 (forward bias), the eld in the depletion region will bereduced, so that it will not balance anymore the diusion current of electrons owing to the left. Thus, electronsupplied from the contact at the extreme right will replenish those electrons entering the p-type region. Thiswill result in a current density J3. Having entered the p region, electrons will eventually recombine with holes.The contact at left will provide the holes necessary for this recombination process, giving rise to a componentJ4 of the total current density. A similar sequence of events will happen to holes: Some will diuse to then region (yielding the component J1 of the current density) recombining there with electrons provided by acurrent density J2 from the right-contact.If, instead, we apply a negative Va (reverse bias), the eld in the depletion region will increase and theassociated drift current will be larger than the diusion current. However, the ow of electrons from the pregion will be negligibly small, since there are very few electrons in p-doped Si. Similarly for holes in the nregion. Therefore, the reverse current will be very small. This shows that p-n junctions behave like diodes,rectifying the current ow.

Lets now get back to the equilibrium condition and start to analyze the junction quantitatively.

Equilibrium.Lets consider the band-bending and carrier densities at equilibrium.

Poisson equation. First, the Poisson equation describing the band bending in the depletion region is:

d2(x)

dx2= e

s(p n+ND NA) . (126)

ECE344 Fall 2009 98

This is a nonlinear equation since the carrier densities, p and n, depend on the potential (x) itself:

n(x) = ni exp

[EF,n0Ei,n0+e(x)

kBT

]= nn e

e(x)/(kBT )

p(x) = ni exp

[Ei,p0EF,n0e(x)

kBT

]= pp e

e(x)/(kBT ), (127)

where nn and pp are the electron and hole densities in the quasi-neutral n and p regions, respectively. Depletion approximation. We can simplify Poisson equation, Eq. (126), by employing the depletion

approximation: Lets assume that the electric eld vanishes outside the depletion region and lets also ignorethe charge due to free carriers in the depletion region xp0 x xn0 (indeed we will have N+D >> n for0 x xn0 and NA >> p for xp0 x < 0, since we have depletion of free carriers in these regions),so that:

d2(x)

dx2= eND

sfor 0 x xn0 , (128)

d2(x)

dx2=

eNA

sfor xp0 x < 0 , (129)

so that, using the boundary conditions

d

dx

x=xp0

=d

dx

x=xn0

= 0 ,

(expressing the fact that the electric eld vanishes at the edges of the depletion region) and (xp0) = p0,(x = xn0) = n0, expressing the fact that at the edges of the depletion region the carrier concentration

ECE344 Fall 2009 99

approaches the concentration in the quasi-neutral regions, we have:

(x) =

eND2s (x xn0)2 + n0 (0 x xn0)

eNA2s

(x+ xp0)2 + p0 (xp0 x 0) .

(130)

Note that the maximum electric eld occurs at x = 0:

Fmax =eNA

sxp0 =

eND

sxn0 , (131)

where the last equality follows from charge neutrality which requires

NAxp0 = NDxn0 . (132)

Depletion width. To calculate the width of the depletion region, note that

Vbi = en0 ep0 . (133)

Moreover, the continuity of the potential at x = 0 implies, from Eq. (130):

0 = (0) (0+) = p0 n0 +

eNA

2sx2p0 +

eND

2sx2n0 . (134)

Using now the charge-neutrality condition, Eq. (132), this becomes

0 = p0 n0 +eNA

2s

N2DN2A

x2n0 +

eND

2sx2n0 . (135)

ECE344 Fall 2009 100

Inserting into this equation the expression for p0 n0 from Eq. (133) and using Vbi from Eq. (123), wehave:

Vbi = kBT ln

(NAND

n2i

)=

e2

2s

ND

NA(NA +ND) x

2n0 , (136)

so that:

xn0 =

(2sNAVbi

e2ND(NA +ND)

)1/2. (137)

Similarly:

xp0 =

(2sNDVbi

e2NA(NA +ND)

)1/2. (138)

Asymmetric junction. In the simpler case of a very highly asymmetric junction (for example: NA = 1015

cm3 and ND = 1019 cm3, we can ignore NA with respect to ND in Eqns. (137) and (138) above, sothat:

xp0 (2sVbi

e2NA

)1/2, xn0

(2sNAVbi

e2N2D

)1/2

Ideal diode: Lets rst consider the ideal diode; that is: An ideal junction in which there are no generation-recombination processes in depletion region. In addition, lets make the following simplifying assumptions:

1. The concentrations of free carriers injected into the quasi-neutral regions are small enough so that we canneglect their charge compared to the charge of the majority carriers when solving Poisson equation (low-levelinjection).

2. The concentration of free carriers everywhere is small enough so that we can use Maxwell-Boltzmannstatistics (that is, the high-T limit) instead of the full Fermi-Dirac statistics.

3. The quasi-neutral regions are innitely long.4. Finally, theres no electric eld in the quasi-neutral regions, so that only diusion controls the current-ow

in these regions.Also, the calculation of all of the components of the current density J1 through J4 in the gure (right frameat page 97) is dicult. However, we know that J2 = J1 and J3 = J4, so we need to calculate only thediusion currents in the quasi-neutral regions J1 and J3. Moreover, it will be convenient to compute J1 (thatis, the hole current Jp before it starts decreasing (due to recombination with electrons in the n region) atx = xn0 and J3 = Jn at x = xp0 (for the same reason).First of all, we can follow again the same procedure we have followed above to obtain the width of thedepletion region simply replacing the built-in potential Vbi with its value modied by the applied bias,Vbi Va, obtaining:

xn0 =

[2sNA(Vbi Va)e2ND(NA +ND)

]1/2. (140)

Similarly:

xp0 =

[2sND(Vbi Va)e2NA(NA +ND)

]1/2, (141)

Thus, the depletion region shrinks under forward bias (Va > 0) but grows under reverse bias (Va < 0).From Eq. (127) and from the right frame of the gure at page 97 we see that for the concentrations of thefree minority carriers which spill-over (electrons spilling over into the p regions and holes spilling over into the

ECE344 Fall 2009 102

n region) we have (using assumption 2 above):

n(x = xp0) = nn0 ee(VbiVa)/(kBT ) = np0eeVa/(kBT )

p(x = xn0) = pn0 eeVa/(kBT ) ,

(142)

so that the excess carriers at the edges of the depletion region are:

n(xp0) = np0[eeVa/(kBT ) 1

]

p(xn0) = pn0

[eeVa/(kBT ) 1

],

(143)

Using now assumption 4, the current in the quasi-neutral regions will be:

Jp = eDpdpdx (x > xn0)

Jn = eDndndx (x < xp0) .

(144)

The continuity equations for the hole current in the quasi-neutral n region (the component J1 in the gurefor x > xn0) and for the electron current in the quasi-neutral p region will be:

pnt = 1e

Jpx

pnpn0p

(x > xn0)

npt =

1eJnx

npnp0n

(x < xp0) ,(145)

where p and n are the (recombination) lifetimes of the minority holes and electrons in the quasi-neutral

ECE344 Fall 2009 103

regions. Combining Eq. (144) and Eq. (145) we get:

pnt = Dp

2pnx2

pnpn0p (x > xn0)

npt = Dn

2np

x2 npnp0n (x < xp0) .

(146)

At steady state, the general solutions are:

pn(x) = pn0 + A ex/Lp + B ex/Lp (x > xn0)

np(x) = np0 + C ex/Ln + D ex/Ln (x < xp0) ,

(147)

where Lp = (Dpp)1/2 and Ln = (Dnn)

1/2 are the hole and electron diusion lengths in the quasi-neutral regions and A, B, C, and D are integration constants to be determined by the boundary conditions.The rst of these boundary conditions is determined by the concentrations of the minority carriers far away inthe quasi-neutral regions: pn(x ) = pn0 and np(x ) = np0, which implies B = C = 0.Another boundary condition results from the requirement that values of pn and np at the edges of thedepletion region match the values pn(x = xn0) and np(x = xp0) given by Eq. (143) above, so that:

pn(x) = pn0 + pn0

[eeVa/(kBT ) 1

]e(xxn0)/Lp (x > xn0)

np(x) = np0 + np0

[eeVa/(kBT ) 1

]e(x+xp0)/Ln (x < xp0) .

(148)

ECE344 Fall 2009 104

Finally, by Eq. (144):

Jp(x) =eDppn0Lp

[eeVa/(kBT ) 1

]e(xxn0)/Lp (x > xn0)

Jn(x) =eDnnp0Ln

[eeVa/(kBT ) 1

]e(x+xp0)/Ln (x < xp0) .

(149)

It is important to note that these current densities vary with x because, for example, as Jp(x) decreases asx increases, the hole current it represents (J1 in the gure at page 97) is transformed via recombinationprocesses into an electron current (J2 in that gure). So, the total current will be independent of x. Thisconstant value can be obtained thanks to our assumption that theres no generation/recombination in thedepletion region. In this case we can evaluate the currents in Eq. (149) at a position in which they have notyet started to decay due to recombination, so Jp at x = xn0 and Jn at x = xp0. Therefore the totalcurrent will be:

Jtotal = Jp(xn0) + Jn(xp0) =[eDppn0

Lp+

eDnnp0

Ln

] [eeVa/(kBT ) 1

]= Js

[eeVa/(kBT ) 1

](150)

where Js is the saturation current density (the maximum current density we have under reverse bias, thatis, for Va ). Equation (150) is known as Shockley (or diode) equation. The gure in the nextpage illustrates the qualitative behavior of Jtotal as a function of applied bias Va.

ECE344 Fall 2009 105

Va

I

ECE344 Fall 2009 106

A few considerations about Shockleys equation:

1. Under very large reverse bias (Va ) we see that Jtotal Js, the saturated reverse current.This current is due only to the drift of those few minority carriers (holes in the n-region, electrons in thep-region) which are thermally generated. We can increase the reverse current by increasing the density ofminority carriers by for example iluminating the sample. This is a photo-diode.

2. Under forward bias the current is dominated by the diusion component. Its easy to see: The driftcomponent does not change much with bias. This is because this component is limited by how manyminority carriers (pn0 and np0) are available to drift and these do not depend on bias. On the contrary,the diusion current is limited by how many majority carriers (pp0 and nn0) can surmount the barriers ofheighth e(Vbi Va). And this depends exponentially on the applied bias.

3. In an asymmetric junction the total current is dominated by the most heavily-doped side of the junction. Forexample, in a junction with NA >> ND, we will have np0

having used the fact that Lp/p =Dpp/p =

Dp/p = Dp/

Dpp = Dp/Lp. Therefore, in

order to replenish this charge, we must have:

Jp(xn0) =Qp

Ap= e

Dp

Lppn0

(eeVa/(kBT ) 1

), (154)

which is the same result we have obtained before (the rst of Eqns. (149)) from the diusion current. Thismethod of calculating the current is called charge control method.

Junction breakdown.From what we have seen so far we may conclude that under reverse bias the current will approach asymptoticallythe value Js as we apply an ever increasing voltage. However, this is not what happens: The Shockleyequation (150) describes correctly the behavior of the diode only as far as Va > VBD, some criticalbreakdown voltage VBD. As soon as the reverse bias exceeds this value the current increases dramatically.This is due to eects which we have not yet considered and which go beyond the simple assumptions we havemade so far:

1. Avalanche Breakdown.We have assumed that electrons and holes can be described by the DDEs, so that they always remain at(or near) thermal equilibrium. But we have seen before (pages 75-77 of the Lecture Notes, Part 1) thatunder a high electric eld charge carriers can gain kinetic energy in excess of the thermal value (3/2)kBT .Whenever carriers gain from the eld an amount of kinetic energy exceeding the vaue of the gap of thesemiconductor, Eg, impact ionization can occur: This is a process by which a conduction carrier (say, anelectron in the CB, just to x the ideas) hits (via the Coulomb force) another electron in the valence band.In this collision process energy is transfered from the conduction electron to the valence electron. The latteris excited into the CB, leaving a hole in the VB. The net eect is that the original electron loses an amount Eg of kinetic energy generating an electron-hole pair. Actually, for such a process to occur, it is notenough to have an intial electron with energy equal to Eg, since momentum conservation also holds and thisimplies that for each semiconductor there exists a minumum threshold kinetic energy Eth > Eg belowwhich impact ionization cannot occur.

ECE344 Fall 2009 108

ECE344 Fall 2009 109

The top frames of the gure on the previous page show the inization rate (that is, the number of ionizationsper unit time) obtained by various quantum-mechanical calculations for electrons and holes in Si. A commonapproximation made for the ionization rate, 1/ii, as a function of electron or hole kinetic energy, E, in theCB and VB, repectively, is the so-called Keldysh formula:

1

ii(E)=

B

I(Eth)

(E EthEth

)p, (155)

for E > Eth, vanishing otherwise. The constant B is a dimensionless parameter and 1/(Eth) is thescattering rate at the threshold energy. The energy Eth is the minimum energy (dictated by energy andmomentum conservation) a carrier must have in order to be able to generate electron-hole pairs. Clearly,Eth > Eg by energy conservation. But, as we said above, momentum conservation may require thatEth be much higher, since the recoil carriers and the generated carriers will have to share the availablemomentum and so cannot in general have vanishing energy. The coecient p at the exponent takes thevalue of 2 for semiconductors with direct gap. In general, it takes values as high as 6 (as determinedby complicated calculations). Lower values for p imply a soft threshold (carriers will ionize with slowlyincreasing probability as E grows beyond Eth). Large values of p imply a hard threshold: Ionizationwill occur almost immediately as the energy of the carrier crosses the threshold energy. Phonon-assistedprocesses (in which one or more phonons are emitted or absorbed during the ionization process) can softenthe threshold, relaxing the restrictions posed by momentum conservation... As you see, the situation isparticularly complicated...The bottom frame of the gures on page 109, instead, show the ionization coecient as a function ofthe electric eld in a homogeneous situation (that is: innitely long crystal with position-independent electriceld). This parameter is most often of interest in the breakdown of p-n junctions. It is dened as thenumber of pairs generated per unit length, so that it also represents the rate at which the current densityincreases (per unit length): Considering for now only electrons:

dJn

dx= n Jn . (156)

ECE344 Fall 2009 110

Since each electron at energy E generates electron-hole pairs at a rate 1/ii(E), if f(E) is the electrondistribution function, then:

n =1

nvd

dE (E)

f(E)

ii(E), (157)

where n is the electron density and vd is the drift velocity. It can be seen (perhaps not too trivially) thatthe functional dependence of ii as given by Eq. (155) does not aect n too much. On the contrary, theshape of the distribution function f(E) as a function of the eld dramatically aects n. Therefore, thewhole game consists in estimating how f changes with F .Shockley had the basic idea that only those electrons which manage to be accelerated to Eth withoutlosing energy to phonons will contribute to the ionization process. Lets dene by Pph(E

, E)dE theprobability per unit time that an electron of kinetic energy E will scatter (via phonon collisions) into a stateat energy E. Then, the probability that an electron scatters (into any other energy state) per unit time is1/ph(E) =

dE Pph(E, E). If Pnoph(t) is the probability that the electrons has not scattered up

to time t, the probability that it will not have scattered up to t+ dt will be:

Pnoph(t+ dt) = Pnoph(t)[1 dt

dE

Pph(E

, E)

]. (158)

Solving this equation for a time interval [t1, t2], we get

Pnoph = exp{

t2t1

dt

dE

Pph(E

, E)

}= exp

{

t2t1

dt

ph(E)

}. (159)

(Note that E and E are functions of time due to their acceleration in the eld, so the integration is nottrivial). Now, the probability that an electron will not scatter with phonons while being accelerated fromenergy E = 0 to E = Eth can be obtained from Eq. (159) by using the chain-rule (lets consider only the

ECE344 Fall 2009 111

1D case for simplicity):

dE

dt=

dE

dk

dk

dt=

dE

dk

eF

h= eFv dt = dE

eFv,

since v = (dE/dk)/h. Then:

Pnoph = exp{ 1eF

Eth0

dE

ph(E)v(E)

}= e

F0/F , (160)

where F0 = Eth0 dE/[eph(E)v(E)] is a constant (that is, independent of the eld), so that:

(F ) = 0 eF0/F . (161)

Clearly, one can argue about the assumption that after every collision the electrons must start from zeroenergy. Nevertheless, this lucky electron model captures the basic feature of the ionization coecient atsmall elds F , exp(constant/F ).

Avalanche. As soon as impact ionization begins to take place in a p-n junction under reverse bias, thecurrent begins to increase. This is obvious from the denition of the ionization coecient, Eq. (156). Buta run-away phenomenon can actually occur: If the electric eld in the depletion region of the reversed-biasjunction is suciently large, one electron may impact-ionize, creating an electron-hole pair. The generatedelectron and hole may now be strongly accelerated by the eld, thus gaining kinetic energy above theionization threshold and, in turn, impact-ionize themselves. Clearly, this process resembles a chain reactionand it can lead to a fast and dramatic increase of the current. For obvious reasons it is called avalanchebreakdown.[Ultimately, as more and more carriers are generated, we reach whats called a high-level injection situation in which the depletion

region ceases to be depleted. Under these conditions the resistivity of the depletion region drops, and so the potential drop across

it and, nally, the electric eld. This will put a limit to the run-away phenomenon.]

ECE344 Fall 2009 112

In order to see under which conditions avalanche can be triggered, lets generalize Eq. (156) to the case of ap-n junction under strong reverse bias. Lets refer to the gure on page 116. As we saw above, the currentis due to the drift of those few minority carriers which enter the depletion region. Consider now the initialminority-electron current injected from the left. This will grow as:

dJn

dx= p Jp + n Jn , (162)

ordJn

dx (n p) Jn = pJ , (163)

where J = Jn + Jp is the total electron and hole current, constant at steady-state. To determine thebreakdown condition, lets denote with Jn(xp0) the electron current density incident from the left-edgeof the depletion region (at x = xp0) and lets assume that the electron current exiting the high-elddepletion region (at x = xn0) is a multiple Mn of Jn(xp0). Since most of the current at x = xn0 willbe carried by electrons, we can assume Jn(xn0) = MnJn(xp0) J . With this boundary condition thesolution of Eq. (163) can be written as:

Jn(x) = J

{1

Mn+

xxp0

dx p exp[ xxp0

dx (n p)] }

exp

[ xxp0

dx (n p)]

(164)

(Note: This comes from the fact that the general solution of the rst-order linear dierential equation

dy(x)

dx+ a(x)y(x) = b(x)

is

y(x) = C eA(x) + eA(x)

dx b(x) eA(x) ,

where C is an integration constant determined by the boundary condition and A(x) = x dxa(x)).

ECE344 Fall 2009 113

Evaluating Jn(x) at x = xp0 yields Jn(xp0) = J/Mn, which is our boundary condition. Wemust also require Jn(xn0) = J . With some algebra this implies:

1 1Mn

=

xn0xp0

dx n exp

[

xxp0

dx (n p)]. (165)

The avalanche breakdown voltage is dened as the voltage at which Mn . Thus, at breakdown theintegral above called ionization integral approaches unity:

xn0xp0

dx n exp

[

xxp0

dx (n p)]

= 1 . (166)

For a hole-initiated avalanche we get a symmetrical result:

xn0xp0

dx p exp

[

xn0x

dx(p n)

]= 1 . (167)

Equations (166) and (167) are equivalent: The condition determining the onset of breakdown depends onlyon what happens inside the depletion region, not on which type of carrier initiates the ionization process.Note that for semiconductors for which n p (as in GaP, for example), the breakdown conditionbecomes simply xn0

xp0dx = 1 , (168)

with obvious meaning: If the probability of ionizing over the depletion region reaches unity, avalanche willoccur.Since we have dened the breakdown voltage VBD as the voltage at which the multiplication coecientMn calculated above diverges to innity, an empirical expression for Mn capturing this asymptotic behavior

ECE344 Fall 2009 114

is the following:

Mn 1

1 (V/VBD)n, (169)

where n is a suitable exponent which usually varies from3 to 6, depending on the type of junction andsemiconductor. The breakdown voltage can be calculated from known doping prole, ionization rates, etc.from the equations above. In the case of an abrupt junction we have:

VBD =Fmax(xn0 + xp0)

2, (170)

where the maximum eld in the depletion region is given by Eq. (131). For linearly graded junctions(junctions with a doping prole varying linearly, rather than abruptly), the formula above must be correctedby a factor 2/3.

ECE344 Fall 2009 115

ECE344 Fall 2009 116

2. Zener breakdown.Another cause of breakdown is related to the quantum mechanical process of tunneling (see Lecture Notes,Part 1, page 10). At a very large electric eld in a reverse-biased p-n junction, electrons in the valenceband may tunnel across the band gap of the semiconductor, thus creating an electron-hole pair. The processmay lead to breakdown either directly (so many pairs will be crated that the leakage current will grow) orindirectly: The electrons and holes generated by the tunneling process will impact-ionize and an avalancheprocess will begin. This breakdown mechanism is called Zener breakdown. It aects mostly heavily-dopedjunctions in which the built-in voltage and the applied (reverse) bias fall over a narrow depletion region, thusgiving rise to large electric elds.A relatively simple estimate of the strength of the process may be obtained by assuming that the electronshave to tunnel through a triangular barrier of height EG and length EG/(eF ) (see the gure in the previouspage). Using a suitable approximation, the tunneling probability can be calculated and one nds that it isproportional to:

PZener(F ) exp4(2m)1/2E3/2G

3ehF

, (171)

wherem is the electron eective mass in the gap (usually approximated by the smaller between the electronand hole eective masses). More rigorous calculation show that the current is given by:

JZener (2m)1/2e3FVa

42h2E1/2G

PZener(F ) . (172)

This expression is independent of temperature and it is valid only for semiconductors with a direct gap (suchas most of the III-V compound semiconductors). On the contrary, for semiconductors with indirect gap (suchas Si and Ge), the calculations must take into account the fact that crystal momentum must be supplied(mainly by phonons) in order to allow a transition from the top of the valence band at the symmetry point to the bottom of the conduction band at other locations in the BZ (at the symmetry points L in Ge, nearthe symmetry points X for Si). The role played by phonons in the process renders Zener breakdown quite

ECE344 Fall 2009 117

strongly temperature-dependent.

Time-transients.We have so far limited our attention to the steady-state behavior of the device. We have seen that when wemove from the equilibrium situation (Va = 0) to forward or reverse bias (Va = 0) we must move charges out(reverse bias) or in (forward bias) the depletion region. In practical applications it is important to know howquickly the device can adjust to a new bias condition.Unfortunately a detailed analysis of time-transient behavior is complicated: We would have to solve self-consistently the DDEs (including their full time-dependence) and Poisson equations. However, a couple ofexamples can give an idea of how a diode will switch.

Decay of stored charge.Consider a forward-biased p-n junction with a curent I = jA owing through it. Using the charge controlmodel discussed before, we may view the current at steady state as replenishing the charge of minority carriersrecombining. In formulae, using current continuity and assuming a one-sided p+-n junction so that the totalcurrent is almost completely due to the hole current:

jp(x, t)x

= ep(x, t)

t+ e

p(x, t)

p. (173)

Integrating both sides over x from xn0 to at time t, since the current jp vanishes innitely deep into then region, we get:

jp(xn0, t) = j(t) =dQp(t)

dt+

Qp(t)

p, (174)

where Qp(t) = e x0 p(x, t) is the total hole charge per unit area stored in the n-type side of the junction.

At steady state we recover the usual result j = Qp/p, which have we obtained before. But lets now assumethat at t = 0 we disconnect the current source so that j = 0 for t > 0. The sudden removal of the currentmeans that the hole charge Qp(t) will decay in time, since there will be no inux of holes from the p

+ regionto replenish those which recombine in the n region. Indeed, we can solve Eq. (174) with the initial condition

ECE344 Fall 2009 118

Qp(t = 0) = jp and setting j = 0 in that equation:

dQp(t)

dt= Qp(t)

p, (175)

or:Qp(t) = jpe

t/p, (176)showing that indeed the positive charge due to holes stored in the n side of the junction decays with a timeconstant p.Knowing how the hole charge decays with time, we can calculate how the voltage across the junction changeswith time: At steady state (for t < 0) the junction was forward biased, so that V (t < 0) = Va. But atmuch later times, after the hole charge will have disappeared, we must have the junction at equilibrium sothat V (t ) = 0. We can calculate this time dependence by assuming that the spatial dependence ofthe hole concentration is exponential during the entire transient (condition which, as explained in Sec. 5.5.1of the text, is not fully satised). Then we know that the voltage drop is correlated to the excess hole densityat xn0 via

pn(t) = pn0

[exp

(eV (t)

kBT

) 1

]. (177)

Assuming now, as we said above, that hole density is exponentially decaying at all times, we have:

p(x, t) = pn(t) e(xxn0)/Lp , (178)

so that:

Qp(t) = e

xn0

dx p(x, t) = pn(t) Lp . (179)

Using Eq. (176) and (177) we have, nally:

V (t) =kBT

eln

(jp

eLppn0et/p + 1

). (180)

ECE344 Fall 2009 119

This shows that the voltage across the junction decays to zero at rst linearly but exponentially at large timest.

Switching Diodes.In most cases, a diode is supposed to switch from the conducting (forward-bias) to the non conducting (reversebias) state. The speed at which the switching can occur is limited by the time it takes to remove or addcharge to/from the depletion region. This depends on two parameters: the lifetime of the carriers, n and p,and on the capacitance of the junction.The importance of these parameters is illustrated by considering a sudden forward-to-reverse bias transition.Consider a diode in series with a resistor R. Assume initially the p-n junction in forward bias with currentIf owing in the circuit. Since only a small voltage will drop across the junction under these conditions,most of the voltage, V = RIf , will drop across the external resistor. Assume now that we switch suddenlythe polarity of the external bias. In order for the junction to go under reverse bias, a large amount ofcharge has to be removed from the junction in order to form the large depletion region. This process andthe speed at which happens is controlled by the depletion capacitance which we shall discuss below: Alarger capacitance requires more charge in order to establish a given potential drop. But even before thiscan happen, the excess minority carriers (holes in the n region and electrons in the p region) which existedunder forward bias must recombine away. The speed of this process is controlled by the carrier lifetime.Thus, often impurities (Au, for example) are intentionally added to the junction in order to speed up the process.

Junction capacitance.Capacitance is a measure of the charge stored per unit change of voltage. A larger capacitance means that morecharge must be moved in or out, so that for a xed current more time is needed to complete the process.Thus, this translates into a longer delay in responding to the new bias condition.

In a p-n junction two major capacitances are at play: 1. The capacitance associated with the charge whichmust be moved in or out of the depletion region. This is called the depletion capacitance. 2. The capacitancepresent under forward bias due to charges spilling over into the quasi-neutral regions. This is called diusioncapacitance. Lets now consider these two components separately.

ECE344 Fall 2009 120

Depletion capacitance.When the bias Va applied to the junction is varied, the width of the depletion region changes according toEqns. (140) and (141). The charge present in each depletion region due to the ionized dopants will be:

Qdepl = eANDxn0 = eANAxp0 , (181)

where A is the cross-sectional area of the junction. There will also be a component of charge due to themotion of majority carriers. But this happens very quickly (in the time scale of picoseconds or less), so we canignore this delay. Thus, by denition of capacitance,

Cdepl =

dQdepldVa = eAND

dxn0dVa = A

[es

2

NAND

(NA +ND)

]1/2 1(Vbi VA)1/2

, (182)

or, noticing that from Eqns. (140) and (141)

W = xp0 + xn0 =

[2s

e

(Vbi Va)(NA +ND)NAND

]1/2, (183)

we can rewrite Eq. (182) as:

Cdepl =sA

xn0 + xp0=

sA

W, (184)

which is just the capacitance of a parallel-plate capacitor with a dielectric of permittivity s, with plates ofarea A separated by a distance W = xn0 + xp0.

ECE344 Fall 2009 121

Diusion capacitance.The concentration of excess carriers diusing in the quasi-neutral regions can be obtained from Eq. (148):

pn(x) = pn(x) pn0 = pn0[eeVa/(kBT ) 1

]e(xxn0)/Lp (x > xn0)

np(x) = np(x) np0 = np0[eeVa/(kBT ) 1

]e(x+xp0)/Ln (x < xp0) .

(185)The charge per unit area will be (considering only holes, a similar expression will hold for electrons):

Qdiff,p = e

xn0

pn(x) dx = eLppn0

[eeVa/(kBT ) 1

]. (186)

Under strong forward bias, eeVa/(kBT ) >> 1, so:

Cdiff,p dQdiff,p

dVa=

e2Lppn0

kBTeeVa/(kBT ) e

kBT

L2p

DpJp(xn0) =

e

kBTpJp(xn0) , (187)

where we have used Eq. (150) in the last step. Accounting now for the charge of the minority electronsdiusing into the p quasi-neutral region:

Cdiff dQdiff,p

dVa+

dQdiff,n

dVa=

e

kBT[pJp(xn0) + nJn(xp0)] . (188)

Note: This is the equation found in most textbooks. However, in a text by Karl Hess (and in a brief commenton the lastest edition of the Streetman-Banerjees book) one nds instead:

Cdiff e

2kBT[pJp(xn0) + nJn(xp0)] , (189)

ECE344 Fall 2009 122

where the additional factor of 1/2 is explicitly commented and the claim is made that Eq. (188) is in error.This factor can be justied in hand-waving fashion by noting that the charges Qdiff,p and Qdiff,n are likethe charges in the two opposite plates of a capacitor, so that the capacitance should be given by the changew.r.t. the applied bias of the average of the electron and hole charges. A more sophisticated and rigorousexplanation is given by S. E. Laux and K. Hess, IEEE Trans. Electron. Device vol. 46, no. 2 (February 1999),p. 396. Their argument is based on the observation that rigorously speaking the diusion charge extendsalso inside the depletion region, so that the integration in Eq. (186) should extend from 0 to , not fromxn0 (and similarly for the expression for Qdiff,n). Since as Va changes charges will leave the depletionregion, we will obtain a lower estimate for the charge, and so for the capacitance. In a way, this argumentis equivalent to our hand-waving argument since both reduce to accounting for the charges throughout theentire junction, not just in the quasi-neutral regions.

ECE344 Fall 2009 123

Other diodes: Heterojunctions,Metal-Semiconductor junctions (Schottky contacts), MOS capacitors

p-n junctions are ubiquitous in semiconductor devices: We had to analyze their characteristics in some detailbecause this understanding is required to analyze, in turn, the characteristics of many types of devices (bipolarjunction transistors BJTs in particular). However, other types of junctions play a major role in several types oftransistors: Heterojunctions (that is, the junction between two dierent semiconductors) make up heterojunctionbipolar transistors (HBTs), high electron-mobility transistors (HEMTs, also known as modulation-doped eld-eecttransistors, MODFETs), and injection lasers; Schottky contacts (that is, the junction between a semiconductorand a metal) enter heavily in the operation of metal-semiconductor FETs (MESFETs), among other devices;MOS - metal-oxide-semiconductor- capacitors are at the heart of what is arguably the most important type oftransistor in VLSI technology, the metal-oxide-semiconductor eld-eect transistor (MOSFET).We shall now discuss in turn each of these junctions (or diodes, as they broadly fall into the category oftwo-terminal devices). We shall not go into too much detail as far as heterojunction and Shottttky contacts areconcerned, but we shall analyze carefully MOS diodes (or capacitors), since the operation of MOSFETs requiressome basic understanding of the accumulation, depletion, and inversion of the surface of semiconductors in contactwith an insulating layer.

Heterojunctions. Band-alignment.

As we have seen in the midterm exam 1, the workfunction of a solid is the energy required to excitean electron from the bottom of the conduction band and bring it to the energy of the vacuum, so thatthe electron becomes free from the attractive potential of the crystal. Therefore, if we consider twosemiconductors, simply labeled 1 and 2 in the gure below, each isolated from the other, they will have aband alignment as shown in the gure below (left), since the energy level of free electrons (the vacuum level)is the same for both materials. The edges of the conduction and valence bands will exhibit discontinuitiesEC and EV respectively, as indicated in the gure. Figure on page 29 of the Notes, Part 1, shown

ECE344 Fall 2009 124

how the valence and conduction of several semiconductors (characterized by their lattice constants) bands align.

EC

EVEV1EF1

EC1

Evac

EV2

EF2

EC2e1

e2

EC

EVEV1EF1

EC1

EV2

EF2

EC2

As we bring the two semiconductors in electrical contact, we know that their Fermi levels must line-up underequilibrium conditions. On the other hand, the discontinuitiesEC andEV (which depend on the presenceof microscopic dipoles at the interface) must be present in order to account for the dierence of the bandgaps.In order to accommodate these two requirements, upon bringing the materials in contact, electrons ow fromone material to the other (from 2 to 1 in the gure) and holes in the opposite direction (1 to 2 in the gure),much as it happens in a conventional junction (homojunction, same junction, from Greek). As electronsdeplete region 2, positive charge is formed, thus forming a depletion layer with the potential energy showingpositive second-derivative (upwards curvature). Simililarly, hole leave behind negative charge, resulting in thediagram shown above (right). This is the equilibrium, zero-bias band-diagram of a heterojunction (that is, ajunction made up of two dierent materials).

ECE344 Fall 2009 125

Thermionic emission.In the gure below we use the example of GaAs and the alloy AlxGa1xAs (where x, known as the molefraction, is the molar fraction of Al ions to Ga ions). This type of heterojunction can be obtained by startingfrom a GaAs substrate and growing successive layers of AlxGa1xAs via Molecular Beam Epitaxy, in whichbeams of Ga, Al, and As ions (beams whose intensities are in carefully monitored proportions) are shot onthe GaAs surface. If the correct ratio of beam intensties is used, the process will result in the deposition ofthe alloy Al0.3Ga0.7As on GaAs. As seen from the gure on page 29, at this mole fraction Al0.3Ga0.7As hasthe same lattice constant of GaAs, so that we will end up with a crystallographically perfect heterojunction. Aslight mismatch of lattice constants can be tolerated for thin lms. The top layer will be under compressive ortensile biaxial strain (depending on whether its lattic constant is larger or smaller that that of the substrate).If the lm thickness is small enough, the energetic cost of mainting the strain will be smaller than thatof creating crystal defects (broken arrays of ions, known as stacking faults, crystal planes misaligned withone-another, known as dislocations etc.). But is the lm is too thick, these kind of defects will be eventuallyenergetically favorable and the crystal structure of the top layer will shown a non-ideal (and usually unwanted)crystalline structure.Coming back to our example, GaAs has a smaller band gap ( 1.51 eV) than AlxGa1xAs ( 1.72 eV for x= 0.3, for example). Experimentally it is found that most of the discontinuity of the energy gap falls in theconduction band. Therefore, as soon as the two materials are brought together, the conduction bands willline-up as shown in the left frame of the gure.The resulting structure behaves like a diode, but in order to evaluate the magnitude of the current owingacross the heterointerface, we must modify the analysis we have done in the context of p-n homojunctions,since the presence of the band discontinuity presents some new physical aspects.

ECE344 Fall 2009 126

In principle, the qualitative aspects of the junction are similar. We saw that one possible way to look atthe electron ux from the n side of a p-n junction to the p side consisted in looking at all electrons whichhave sucient kinetic energy to overcome the barrier of height eVbi eVa. Their number is exactlynp0[exp(eVa/(kBT ) 1], the very factor which controls the electron diode current, since it xes thegradient of the electron concentration, and so the diusion current. At zero bias, the drift term opposes exactlythis ow, thus yielding zero current. But under forward bias, exponentially more electrons can overcome thebarrier, the drift term remaining essentially unchanged, and a large current would result. The same happens ina heterojunction. There are two main dierences. First, since the band discontinuitues EC and EV arein general very dierent, unlike what happens in homojunctions, the type of carriers facing the smaller barrierwill usually dominate. Lets assume that in our example EC < EV (unlike what shown in the gure), sothat we can deal only with electrons. The second dierence lies in the fact that the barrier is now caused notby the built-in potential, but by the conduction-band discontinuity EC . Only electrons which are thermallyexcited to energies above the barrier can make it across. This is called thermionic current, in analogy withthe electron current emitted by a heated lament (as the W lament in an incandescent light bulb).The derivation of an accurate formula describing the current through the heterojunction is somewhat

ECE344 Fall 2009 127

complicated. An approximate expression is given by:

jhet(Va) = AB T

2e|Ec|)/(kBT )

[eeVa/(nkBT ) 1

], (190)

where A = emk2B/(22h3) is the so-called Richardson constant, B a slowly-varying function of the bias,

depending on the nature of the junction, and n an ideality index varying between 1 and 2. Note the similarityof this equation with the diode equation, Eq. (150).

Schottky contacts. Schottky contacts are ideal junctions between semiconductors and metals. In theiry andin their ideal songuration, they should be rectifying contacts. In practice, whenever we need to make goodcontact to a semiconductor (meaning: we need a perfect ohmic contact, not rectifying), we resort to metals,typically Al. The nature of ohmic contacts and why systems which should behave like rectifying contacts inpractice end up showing ohmic behavior is still relatively poorly understood. So, we start by discussing the ideal

ECE344 Fall 2009 128

rectifying Schottky contacts. But at some point we shall stop and discuss briey how one can obtain ohmicbehavior from these structures.

Metal-semiconductor junction.When we bring a metal and a semiconductor together, as shown in the gure in the next page, we have asituation not too dissimilar from what we have seen regarding heterojunctions: If the semicondutor (assumedto be n-type) Fermi level is larger than Fermi level in the metal (as shown in the gure), electrons will owfrom the semiconductor untill a new equilibrium will be reached. This will result in the creation of a depletionregion in the semiconductor side of the contact. Similarly to what we have sen in the case of heterojunctions,there will be an energy barrier of height eB between the bottom of the CB of the semiconductor and themetal Fermi level. Denoting by eM the metal workfunction (that is, the energy required to excite an electronat the metal Fermi level to the level of the vacuum) and by e the electron anity of the semiconductor, theSchottky barrier height will be (looking at the gure):

eBn = e (M ) , (191)

where the subscript n reminds us that we have considered an n-type semiconductor. For a p-typesemiconductor, instead, we would have obtained

eBp = EG e (M ) . (192)

The thickness of the depletion region can be obtained as we did before for p-n-junctions and we obtain

W =

[2s(eVbi Va)

eND

]1/2, (193)

where the built-in potential Vbi is (see the gure) eB (Ec EF ).

ECE344 Fall 2009 129

eB

METAL SEMICONDUCTOR

EV

EF

EC

Vbi = e(BEC+EF)

Non-ideal junctions.This is what we expect in an ideal case. In practice, it often happens that the Fermi level of the semiconductoris pinned at some particular energy in the gap due to the presence of electron traps at the interface (interfacestates). Years ago J. Bardeen had suggested that electronic states associated to unterminated bonds presentat the semiconductor interface may be an intrinsic cause of this pinning. Now it seems that things are morecomplicated: For example, bare surfaces may reconstruct and, when exposed to metals, may deconstruct incomplicated ways, leading to amorphization, interdiusion and clustering in the interfacial region. It is fair tosay that we do not know exactly what pins the Fermi level, so that in many cases the Schottky barrier heightshould be regarded as an experimentally measured quantity.

Thermionic and tunneling current.The current owing through the Schottky diode is due two major processes: The thermionic ow of electrons

ECE344 Fall 2009 130

over the barrier, and quantum-mechanical tunneling of electrons from the metal into the semiconductor. Thethermionic component can be calculated in a way very similar to the one described dealing with heterojunctions.The net result is:

j = AT2eeB/(kBT )

[eeVa/(kBT ) 1

]. (194)

Note, once more, the similarity with the Shockley equation, Eq. (150), for the current in p-n-junctions.

The second component of the current we must consider is of quantum-mechanical origin. It is the tunnelingcurrent illustrated in the gure above.In the example illustrated in the gure, electrons in the metal can tunnel across the Schottky barrier and enterthe semiconductor. Similarly, electrons can tunnel from the semiconductor into the metal.Such a current can be calculated using the WKB approximation we have seen when dealing with Zenerbreakdown. We can approximate the shape of the potential barrier by assuming a constant eld in the

ECE344 Fall 2009 131

depletion layer, so that we must deal with a simple triangular barrier of the type:

(z) = B Fzz , (195)

where Fz may be chosen as the interfacial eld, 2B/W , or some suitable average of the eld over

the depletion region. The quantity B is the total voltage drop in the semiconductor (see the gure),B = Vbi Va = B Va EC + EF,S. In this case we get:

PMS = exp

{4(2m

S)

1/2

3ehFz(eB E)3/2

}, (196)

and so the tunneling current will be proportional to this expression integrated over the electron energydistribution M(E)fM(E) in the metal. Given the complexity of these calculation, one may wonder howohmic contacts could ever be realized. One possible explanation consists in assuming that height of theSchottky barrier, eB, may be zero. This may happen in a few cases. But more likely is the scenario inwhich heavy amorphization of the interface may result in a very heavily doped almost metallic thin layerof semiconductor. The very high doping will result in such a thin depletion layer, that tunneling across thisbarrier may be a dominant process, almost killing the resistance of the contact. This can be seen easily fromEq. (196), noticing that the eld Fz is inversely proportional to the width of the depletion layer, W .

ECE344 Fall 2009 132

MOS capacitors.The Metal-Oxide-Semiconductor (MOS) or Metal-Insulator-Semiconductor (MIS) diode is arguably themost used and useful device in VLSI technlogy. Its ideal structure is shown in the gures below.

eBeM e

eBeVFB < 0EF,M

EC

EV

EiEF,S

METAL INSULATOR SEMICONDUCTOR

FLAT BANDS

EF,M

EC

EV

EiEF,S

METAL INSULATOR SEMICONDUCTOR

ZERO BIAS

ECE344 Fall 2009 133

eVG < 0

EF,M

EC

EV

EiEF,S

METAL INSULATOR SEMICONDUCTOR

ACCUMULATION

eVG > 0

EF,M

EC

EV

EiEF,S

METAL INSULATOR SEMICONDUCTOR

INVERSIONStarting from a semiconducting substrate (assumed to be p-type in the gures), an insulator is grown ordeposited on the substrate. Typically, the natural oxide of Si, SiO2, is thermally grown by heating the Si waferto temperatures in the range 850-1000oC in oxygen-rich ambient. The relative simplicity of this process (which,however, must be extremely clean) and the unsurpassed electronic properties of SiO2 are probably the reasonswhy Si has been the dominant material in microelectronics. After the growth or deposition of the insulator ametal (or highly-doped polycrystalline Si) is deposited over it. In the gures a metal is considered.At at band the alignement of the bands is illustrated in the rst gure. In the ideal case, charge would owacross the insulator, so that the Fermi levels in the metal, EF,M , and in the semiconductor, EF,S, wouldline-up and the dierence, MS, between the metal and the semiconductor work-functions would vanish:

MS = M ( +

EG

2e+ B

)= 0 for p type semiconductor , (197)

ECE344 Fall 2009 134

and

MS = M (+

EG

2e B

)= 0 for n type semiconductor . (198)

In practice, the time required for the Fermi level to line-up is extremely long and this ideal situation is neverachieved. The gure shows that the application of a small bias, the at-band voltage VFB , is required toline-up the Fermi levels. Its value will be given by the nonvanishing MS of Eq. (197) or (198) above.The application of a negative bias, VG, to the metal (usually called the gate) drives the MOS diode intoaccumulation: As seen in the gure, the bias causes an accumulation of holes at the Si-SiO2 interface. Theapplication of a positive gate bias, instead, results in the inversion of the semiconductor surface, electrons nowpiling up at the interface. Lets now consider these processes in some detail.

Interface space-charge region.

EC

EiEF

EV

eBees

SiO2 Semiconductor

The gure above shows in more detail the band-diagram near the semiconductor-SiO2 interface. Let z be the

ECE344 Fall 2009 135

coordinate along the normal to the interace, z = 0 be the location of the interftace, and lets dene by (z)the potential, taken as zero in the bulk and measured from the intrinsic Fermi level Ei. Then, assuming thenon-degenerate limit, the electron and hole concentrations will be:

np(z) = np0 exp

[e(z)

kBT

]= np0 exp() , (199)

pp(z) = pp0 exp() , (200)

where is positive downward (as in the gure), = e/(kBT ), and np0 and pp0 are the equilibriumelectron and hole concentrations in the bulk.Let s be the surface potential so that

ns = np0 exp(s) , ps = pp0 exp(s) (201)

are the surface concentrations of the carriers.We can deal with the exact Poisson equation as follows, extracting some information before embracing theusual depletion approximation. We have:

2

z2= e

s[N+D + p(z)NA n(z)] , (202)

where s is the (static) dielectric constant of the semiconductor. From Eqns. (199) and (200) we have:

pp np = pp0 e np0 e , (203)

so that, since by charge neutrality in the bulk N+DN

A= np0 pp0, we have

2

z2= e

s[pp0 (e

1) np0 (e 1)] . (204)

ECE344 Fall 2009 136

Lets now multiply Eq. (204) by /z and integrate Eq. (204) from an arbitray location z towards thebulk (z ). The left-hand-side becomes:

z

2

z2

zdz = 1

2

z

z

(

z

)2dz =

1

2

((z)

z

)2, (205)

having used the fact that the eld vanishes as z . For the right-hand-side we get:e

s

z

[pp0(e1) np0(e1)]

zdz =

e

s

0

[pp0(e1) np0(e

1)] d =

=e

s[pp0(e

+ 1) np0(e 1)]. (206)Therefore we have the following relationship between eld and potential at any location z:

F 2 =

(2kBTpp0

s

) [(e + 1) + np0

pp0(e 1)

]. (207)

Introducing the Debye length, LD = [skBT/(e2pp0)]

1/2 (the dielectric screening length in the p-type

non-degenerate bulk Si), and denoting as G(, np0/pp0)2 the term in square brackets in Eq. (207), we

have:

F =

z= 2

1/2kBT

eLDG

(,

np0

pp0

), (208)

the plus (minus) sign valid for > 0 ( < 0).The charge at the interface can now be expressed using Gauss law and the value of the eld at the interface(obtained by setting = s in Eq. (208)):

Qs = s Fs = 21/2skBT

eLDG

(s,

np0

pp0

). (209)

ECE344 Fall 2009 137

This represents the total charge per unit area, shown in the gure below. For negative s the charge ispositive (holes), corresponding to accumulation. At at band the total charge is obviously zero. In depletion

and weak inversion the term in the function G dominates, so that he charge grows as 1/2s . Finally,

in strong inversion the charge is negative, the term (np0/pp0)es being the dominant one. By denition,

strong inversion begins at s = 2B, the value of the surface potential at which the electron concentrationat the interface equals the hole concentration in the bulk.

0.4 0.0 0.4 0.81010

1011

1012

1013

1014

1015

~ exp(s/2)accumulation

~ exp(s/2)strong inversion

flatband

depletionweak

inversion

B EC

~ s1/2

ptype Si 300KNA = 4x1015 cm3

s (Volt)

Q s (cm

2 )

ECE344 Fall 2009 138

In order to obtain separately the electron and hole charges, n and p, we must integrate p(z) and n(z)from the surface to the bulk:

p = pp0

0

(e 1) dz = epp0LD21/2kBT

0s

e 1G(, np0/pp0)

d , (210)

and

n = np0

0

(e 1) dz = epp0LD

21/2kBT

0s

e 1G(, np0/pp0)

d . (211)

The dierential capacitance of the semiconductor depletion layer is given by:

CD =Qs

s=

s

21/2LD

1 es + (np0/pp0)(es 1)G(s, np0/pp0)

. (212)

At at-band condition, s = 0, so:

CD,FB =s

LD. (213)

Ideal capacitance-voltage characteristics.It is very important to understand the capacitance-voltage characteristics of an MOS diode (or capacitor),in view of their relevance to the operation of an MOS eld-eect transistor. Lets recall that we are nowinterested in the dierential capacitance. That is, we apply a dc (or slowly-varying) gate bias to the diode,but, in addition, we apply a small (of the order of kBT/e or less) ac bias at a given frequency.If we apply a gate bias VG to the gate (while keeping the semiconductor substrate grounded), part of thevoltage, s, drops in the semiconductor and part, Vox, in the insulator. The latter will be given obviously by:

Vox = Foxtox =Qstox

ox, (214)

ECE344 Fall 2009 139

having used the fact that sFs = oxFox, and having denoted with tox the thickness of the insulator. Thus,the oxide capacitance will be

Cox =dQs

dVox=ox

tox. (215)

But the total capacitance will also depend on the charge induced by the voltage drop in the semiconductor,Eq. (212). Therefore the total capacitance will be the series-capacitance of the insulator, Cox and of thedepletion region, CD:

Ctot =CoxCD

Cox + CD. (216)

For a given insulator thickness, the oxide capacitance is thus the maximum capacitance.

GATE VOLTAGE

GATE

CAP

ACIT

ANCE

highfrequency

lowfrequency

Cox Cox

CFB

Looking at the gure above, in accumulation (VG < 0) holes pile-up very close to the semiconductor-

ECE344 Fall 2009 140

insulator interface. As the gate bias is reduced, the depletion capacitance begins to matter, depressing thetotal capacitance. To estimate its value, in the depletion approximation we can write the potential in thesemiconductor as:

(z) s(1 z

W

)2, (217)

where the depletion width W can be obtained in the usual way (see the rst of the Eqns. (139) above):

W (2ss

eNA

)1/2. (218)

Thus, the depletion capacitance will be the result of charges responding to the ac-bias at the edge of thedepletion region, so that

CD =s

W, (219)

andCtot =

ox

tox + (ox/s)W. (220)

At at band we should replace W with LD, while the onset of strong inversion (also called the turn-onvoltage or, more commonly, the threshold voltage, as this marks the onset of strong conduction in MOSeld-eect transistors), we have

W = Wmax (2s2B

eNA

)1/2=

[4skBT ln(NA/ni)

eNA

]1/2. (221)

When the surface potential reaches the strong-inversion value, s = 2B, the electron concentration at theinsulator-semiconductor interface is so large (provided enough time is given to the minority carriers so thatan inversion layer is indeed formed, as we shall see below) that it will screen the eld and it will prevent anyfurther widening of the depletion region. Indeed, as we can see in the plot Qs vs. s at page 138, a smallincrement of surface potential will result in a huge increase of the electron charge. Thus, one can assume

ECE344 Fall 2009 141

that in strong inversion most of the additional VG will drop in the insulator and no signicant increase ofs will occur. Therefore, Eq. (221) gives the maximum width of the depletion region. The only exceptionto this is given by the application of a very quickly-varying dc bias: If the gate bias VG is increased veryquickly to positive values, there will be little or no time for generation/recombination processes in the bulkto provide/absorb enough minority carriers (electrons) to feed the inversion layer. In this case will exceed2B, W will grow beyond Wmax and the capacitance will drop below its minimum value Cmin given belowby E. (222).As the gate bias becomes even more positive, we must distinguish two dierent situations. If the applied biasis varied slowly and the applied ac bias is of low frequency, generation and recombination processes in thebulk may be able to follow the ac signal. Thus, electrons in the inversion layer will respond, the response ofthe depletion layer will be screened by the ac-varying inversion charge and the dierential capacitance will riseback to the value given by Cox. If, on the contrary, the frequency of the ac signal is large, than the inversioncharge will not be able to follow the signal. The depletion capacitance will be clamped at the minimum values/Wmax, and so the total capacitance will remain at the minimum value

Cmin =ox

tox + (ox/s)Wmax, (222)

independent of gate bias.Experimentally the transition between the low-frequency and the high-frequency behavior happens around1-50 Hz. However, MOS diodes built on substrates of excellent quality (small density of SRH centers) mayexhibit the low-frequency behavior only when there is no applied ac signal and the dc gate bias is variedsuciently slowly (even a fast varying dc VG say, a C-V sweep in a few seconds, may push the diode intoits high-frequency response).

Deviations from ideality: Oxide charges and interface traps.So far we have considered an ideal MOS structure in which both the insulator and the interfaces are free fromdefects and impurities. This was not the case early on when MOS devices were rst fabricated. Even today,after decades of research and development has enabled the realization of almost-ideal structures, during theoperation of a device defects may be created by energetic carriers hitting the interface or being injected into

ECE344 Fall 2009 142

the insulator. These defects may be broadly classied into two categories: Fixed charges in the oxide andelectrically-active interface states. (The word xed labelling the charges in the oxide refers to the fact thattheir charge-state or electronic population does not depend on the applied bias, while their location asin the case of mobile Na or K ions may depend on the bias and thermal history of the device.)Insulator charges are typicaly either impurities (historically mobile Na and K ions were the subject of manyeorts), or defects, such as dangling bonds, local stress in the SiO2 ionic network, oxygen vacancies inducedby growth, processing such as irradiation. Such charges have the main eect of shifting the threshold voltage

VT = Vox,si + 2B =Qs(2B)

Cox+ 2B , (223)

or the at-band voltage. In order to derive a general expression for the at-band voltage shift due toan arbitrary distribution of charges (z) inside the insulator (where z now denotes the distance from thegate-insulator interface) lets consider rst an ideal (free of charge) MOS capacitor originally at at-bandcondition and now add a sheet of charge (charge per unit area) at z. This charge will induce polarization(image) charges both in the metal and in the semiconductor. The latter charge will bend the bands and somodify the gate bias we must apply to recover at-band. Lets now apply an additional bias to the metal, soto bring the semiconductor at at band. In this situation there will be no eld in the metal (by denition),no eld in the semiconductor (as we are at at band). There will only be a constant electric eld /ox fortox < z < 0. Thus, the metal potential will have moved by an amount (tox + z)/ox with respect tothe initial situation. This will be the shift of the at-band voltage caused by the sheet of charge. Therefore,for a distribution (z) of charges inside the insulator, the at-band shift will be given by:

VFB =

tox0

(tox + z)(z)

oxdz =

1

Cox

tox0

(1 +

z

tox

)(z) dz . (224)

Note that charges close to the gate-insulator interface, z = tox, have no eect on VFB (the polarizationcharges at the metal surface screen completely the oxide charges), while charges near the semiconductor-insulator interface have maximum eect.

ECE344 Fall 2009 143

Interface traps (or states) are defects typically Si dangling bonds whose occupation depends on theposition of the Fermi level at the insulator-semiconductor interface. They have a twofold eect: Dependingon their occupation, they shift the at-band voltage, as in Eq. (224). Since for these states z = tox, theirelectrostatic eect is strong. More importantly, becaue of the intrinsic delay in responding to an ac signal (thetime constants for emission and capture are those of SRH centers), they store charge, thus contributing to thetotal capacitance of the MOS diode. Their capacitance, Cit, is in series with Cox and in parallel to CD, sothat

Ctot =

(1

Cox+

1

CD + Cit

)1=

Cox(CD + Cit)

Cox + CD + Cit. (225)

Several methods have been devised to measure the density, Dit, of the interface traps. They all rely on ameasurement of their capacitance. Ideally, if one knew very accurately the theoretical C-V characteristics ofthe diode, a comparison of the theoretically computed and experimentally measured characteristics will yieldthe desired capacitance, Cit, and so the density Dit, since the density of the interface-trap charge at energyE in the gap (the position of the Fermi level in the gap at a surface potential s) will be Cit(s): Havingobtained Cit from the total capacitance via Eq. (225), at a given gate bias we have:

Cit =dQit

ds= e

dNit

dE= e Dit , (226)

where Nit is the total number of traps up to energy E in the gap and Dit is the trap density per unitarea and energy in the gap. Recalling that VG = Vox + s, we have dVG = ds + dVox. SincedQ = Cox dVox = Ctot dVG, we have dVG = ds + (Ctot/Cox) dVG, and so we can obtainds/dVG from the relation

ds

dVG= 1 Ctot

Cox. (227)

Thus, from Eqns. (220), (226), and (227) we can extract dQit/ds:

Dit(E) =dQit

ds=

Ctot

e

(ds

dVG

)1 Cd

e. (228)

ECE344 Fall 2009 144

This is the density of interface traps per unit energy in the gap at the energy E which, as we said above,indicates the position of the Fermi level inside the semiconductor gap at the interface.In practice, detailed theoretical curves C V are hard to compute. Therefore, typically one replaces thetheoretical curve with a C-V curve obtained under conditions such that the interface traps do not respond.Since the characteristic time of the response of the trap is of the order of

=1

vthniexp

[e(B s)

kBT

], (229)

(for a p-type substrate), either a low-temperature measurement (so that at a given ac frequency the responsetime becomes so long that the trap occupation does not vary) or a high-frequency measurement will providealmost ideal C-V characteristics. A comparison between high-frequency and low-frequency measurements (orhigh-T and low-T measurements) will provide Cit.

There are two major corrections we should make to the analysis followed so far: We have used the non-degeneratelimit (using Maxwell-Boltzmann instead of Fermi-Dirac statistics in relating carrier density to (quasi)Fermi levels)and we have ignored completely all quantum mechanical properties of the electrons.The rst correction important at large densities such as strong inversion and accumulation complicates themathematical analysis so that it becomes impossible to derive analytically even the relations eld/potential orsurface-eld/total-charge given by Eqns. (208) and (209). Only numerical work can give us reliable answers.Yet, the analysis followed so far is qualitatively correct and gives a quantitatively correct picture in the importantregion covering weak-accumulation to weak inversion.Quantum mechanical properties are usualy important in transport. So far, we have limited our attention toelectrostatics. Yet, even the electrostatic poperties can be aected by quantum mechanics when the chargecarriers are conned within regions of size comparable to (or, a fortiori, smaller than) the thermal wavelengthof the electrons. Accumulation and inversion layers are such reasons and they do present this problem. Indeed,the wavelength of an electron at equilibrium (called its thermal wavelength) is given by the de Broglie relationth = h/pth = 2/kth, where pth = hkth is the average momentum of electrons at thermal equilibrium.For the average carrier energy at thermal equilibrium we have Eth = (3/2)kBT = h

2k2th/(2m), so that,

ECE344 Fall 2009 145

at 300K,

th =2h

(3mkBT )1/2 6.2

(m0

m

)1/2nm (230)

Consider the Si-SiO2 interface in inversion/strong-inversion. The electron density ns (the sheet density perunit area, see the gure at page 138) is of the order of 1011-to-1013 cm2, corresponding to an interfacialeld Fs = ens/s of the order of 10

4-to-106 V/cm. Thus, an electron of thermal energy ((3/2)kBT 40meV) will be squeezed by the eld against the interface over a conning distance z 3kBT/(2Fs), verymuch like a particle in a box considered at the beginning of the course. This distance is of the order of 40-to-0.4nm, comparable or even smaller than the electron thermal wavelength. We are not allowed to ignore thewave-like nature of electrons when we conne them so tightly: We expect that discrete energy levels will emergefrom the connement. If we conne a particle in a region of width z, by Heisenbergs principle the particlemomentum will suer an uncertainty k 1/z, so that the conned particle will have a minimum energyE0 h2k2/(2m) h2/(2z2m), called the zero-point energy. In strong inversion this energy maybe comparable to (or even larger than) the thermal energy, and quantum eects due to the connement shouldnot be ignored. The major changes to the classical picture that these quantum-mechanical correction providesare:

1. The electron charge in inversion layeres is removed somewhat (typically by a length t of the order of 1.0nm) from the interface. This causes a reduction of the gate capacitance in accumulation, Cox, since theeective thinckness of the oxide is increased by the amount ox/Sit.

2. The threshold voltage shifts to higher values, since we must move the Fermi level in the inversion layer to anenergy higher by an amount E0 (the zero-point energy mentioned above.

3. The properties of electron transport (and so, in particular, the electron mobility) are modied, since both theshape of the wavefunctions and the density of states are modied.

Detailed calculations of electron transport in these subbands (as they are called) is beyond the scope of thiscourse. We simply show in the gure below a typical example of energy levels, wavefunctions, and chargedistribution in a Si inversion layer.

ECE344 Fall 2009 146

0 20 40 60 800.1

0.00.10.20.30.40.50.60.70.80.91.01.1

z (nm)

pote

ntial

(V)

0 5 10 150.200.100.00

0.100.200.300.400.500.60

z (nm)

pote

ntial

(V)

0 1 2 3 4 5 60.0

0.3

0.6

0.9

1.2

1.5

quantumclassical

z (nm)

elec

tron

dens

ity (1

020 cm

3 )

A nal phenomenon we must mention is the possibility of quantum-mechanical tunneling of electrons in theinversion layer across the gate insulator. This becomes important for thin oxides and/or at very large elds. Therst situation (illustrated by the left frame of the gure below) is called direct tunneling, the second situationFowler-Nordheim tunneling. Using the usual WKB approximation, the tunneling probabilities in the two casesare: For direct tunneling:

Pd exp{ 4(2emox)

1/2

3hFox3/2B

[1 (1 toxFox/B)]3/2} exp

{ 2(2moxe

B)

1/2

htox

}

= e2 tox , (231)

where Fox is the eld in the insulator, mox the eective mass in the gap of the insulator, B = B E0 is

the eective barrier height, reduced by the energy E0 of the bottom subband in the inversion layer, tox is theinsulator thickness, the last step has been made assuming a thin insulator (tox

dened an average decay constant = (2moxeB)

1/2/h. For FN-tunneling, instead:

PFN exp{ 4(2emox)

1/2

3hFox3/2B

}= = e

(4/3) zt , (232)

where zt = B/Fox is the tunneling distance across the triangular barrier. As device scaling progresses, thinner

and thinner insulators are used, together with a reduced applied bias. When insulators were 10 nm thick andthe applied bias was of the order of 5 V, FN tunneling was the only concern. This possibly caused only minorleakage when the devices were turned-on very strongly, so only during the on state. Today, instead, insulatorsare as thin as 2 nm (or even less). Even if the applied bias has been reduced to less than 1 V, from Eq. (231) itis clear that direct tunneling is becoming increasingly important. The major cause of concern is its independenceof bias: Electrons can tunnel across the trapezoidal (or almost rectangular barrier) under any bias condition.This leakage in the o-state causes unwanted power dissipation and it constitutes one of the problems (if notthe problem) we must face attempting to scale devices to even smaller dimensions.

ECE344 Fall 2009 148

20 0 20 40 60 801.0

0.5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

eB E0

direct tunneling

z (nm)

pote

ntial

(V)

80 60 40 20 0 20 40 60 801.0

0.5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

eB E0

FowlerNordheim tunneling

z (nm)

pote

ntial

(V)

ECE344 Fall 2009 149