p-njunctions · limit) instead of the full Fermi-Dirac statistics. 3. The quasi-neutral regions are inﬁnitely long. 4. Finally,there’s no electric ﬁeld in the quasi-neutral

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 fix the ideas) and ‘diffusing’ 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 diffused (orimplanted) area there will be a ‘junction’ in which the p-type and the n-type semiconductor will be in directcontact.

– 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 first, 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 diffuse from the n-region to the p-region, ‘pushed’ by the diffusion term Dn∇nin the DDE. Similarly, holes will be free to diffuse to the n region. As these diffusion 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 field which will balance and stop the diffusiveflow 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 figure in the next page.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 diffusion 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 ≤ ln, holeshave left the region −lp ≤ x < 0, so that for lp ≤ x ≤ ln we have np < n2

i . This is called the‘transition region’ or, more often, the ‘depletion region’ of the junction.

ECE609 Spring 2010 105

3. The voltage ‘barrier’ built by the difusion of carriers upon putting the n and p regions in contact with eachother is called the ‘built-in’ potential, Vbi (denoted by Φ0 by Colinge and Colinge). It is easy to calculateit, since it will be given by the difference between the equilibrium Fermi levels in the the two regions:

Vbi = EFn0 −EFp0 = Ei + kBT ln

(ND

ni

)−Ei + kBT ln

(NA

ni

)= kBT ln

(NDNA

n2i

).

(358)


– What happens to a biased junction?Let’s now apply a bias Va to the junction. We consider Va positive when positive bias is applied to thep-region, as illustrated in the figure. If Va > 0 (forward bias), the field in the depletion region will bereduced, so that it will not balance anymore the diffusion current of electrons flowing 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 diffuse 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 field in the depletion region will increase and theassociated drift current will be larger than the diffusion current. However, the flow 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 flow.

• Equilibrium.Let’s consider the band-bending and carrier densities at equilibrium.

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

d2φ(x)

dx2= − e

εs(p − n + ND − NA) . (359)

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

n(x) = ni exp

[EF,n0−Ei,n0+eφ(x)

kBT

]= nn0 eeφ(x)/(kBT )

p(x) = ni exp

[Ei,p0−EF,n0−eφ(x)

kBT

]= pp0 e−eφ(x)/(kBT )

, (360)


where nn0 and pp0 are the electron and hole densities in the quasi-neutral n and p regions, respectively.– Depletion approximation. We can simplify Poisson equation, Eq. (359), by employing the ‘depletion

approximation’: Let’s assume that the electric field vanishes outside the depletion region and let’s also ignorethe charge due to free carriers in the depletion region lp ≤ x ≤ ln (indeed we will have N+

D>> n for

0 ≤ x ≤ ln and N−A >> p for −lp ≤ x < 0, since we have depletion of free carriers in these regions), so

that:d2φ(x)

dx2= − eND

εsfor 0 ≤ x ≤ ln , (361)

d2φ(x)

dx2=

eNA

εsfor − lp ≤ x < 0 , (362)

so that, using the boundary conditions

dφ

dx

∣∣∣∣x=−lp

=dφ

dx

∣∣∣∣x=ln

= 0 ,

(expressing the fact that the electric field vanishes at the edges of the depletion region) and φ(−lp) = φp0,φ(x = ln) = φn0, expressing the fact that at the edges of the depletion region the carrier concentrationapproaches the concentration in the quasi-neutral regions, we have:

φ(x) =

−eND2εs

(x − ln)2 + φn0 (0 ≤ x ≤ ln)

eNA2εs

(x + lp)2 + φp0 (−lp ≤ x ≤ 0) .

(363)

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

Fmax =eNA

εslp =

eND

εsln , (364)


where the last equality follows from charge neutrality which requires

NAlp = NDln . (365)

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

Vbi = eφn0 − eφp0 . (366)

Moreover, the continuity of the potential at x = 0 implies, from Eq. (363):

0 = φ(0−) − φ(0+) = φp0 − φn0 +eNA

2εsl2p +

eND

2εsl2n . (367)

Using now the charge-neutrality condition, Eq. (365), this becomes

0 = φp0 − φn0 +eNA

2εs

N2D

N2A

l2n +eND

2εsl2n . (368)

Inserting into this equation the expression for φp0 − φn0 from Eq. (366) and using Vbi from Eq. (358), wehave:

Vbi = kBT ln

(NAND

n2i

)=

e2

2εs

ND

NA(NA + ND) l

2n , (369)

so that:

ln =

(2εsNAVbi

e2ND(NA + ND)

)1/2

. (370)

Similarly:

lp =

(2εsNDVbi

e2NA(NA + ND)

)1/2

. (371)


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

cm−3 and ND = 1019 cm−3, we can ignore NA with respect to ND in Eqns. (370) and (371) above, sothat:

lp ≈(

2εsVbie2NA

)1/2

, ln ≈(

2εsNAVbi

e2N2D

)1/2

<< lp . (372)

So, the heavily-doped n-region exhibits essentially no depletion. This is a general property: The larger theconcentration of free carriers, the smaller the voltage which can drop in the region. In the limit of a metal, weknow that no electric field can be sustained, because any voltage drop will be effectively screened by the largeconcentration of free carriers.

• Off equilibrium: Shockley’s equation.Let’s apply a bias Va to the junction. We shall first consider the ideal case of no generation/recombination inthe depletion region (ideal diode). We shall later consider deviations from this ideal condition.

– Ideal diode: No generation-recombination in depletion region.Let’s 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-Boltznmannstatistics (that is, the high-T limit) instead of the full Fermi-Dirac statistics.

3. The quasi-neutral regions are infinitely long.4. Finally, there’s no electric field in the quasi-neutral regions, so that only diffusion controls the current-flow

in these regions.Also, the calculation of all of the components of the current density J1 through J4 in the figure (right frameat page 106) is difficult. However, we know that J2 = J1 and J3 = J4, so we need to calculate only thediffusion 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 = ln and J3 = Jn at x = −lp (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 modified by the applied bias,Vbi − Va, obtaining:

ln =

[2εsNA(Vbi − Va)

e2ND(NA + ND)

]1/2. (373)

Similarly:

lp =

[2εsND(Vbi − Va)

e2NA(NA + ND)

]1/2, (374)

Thus, the depletion region shrinks under forward bias (Va > 0) but grows under reverse bias (Va < 0).From Eq. (360) and from the right frame of the figure at page 106 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 then region) we have (using assumption 2 above):

n(x = −lp) = nn0 e−e(Vbi−Va)/(kBT ) = np0eeVa/(kBT )

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

(375)

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

δn(−lp) = np0

[eeVa/(kBT ) − 1

]

δp(ln) = pn0

[eeVa/(kBT ) − 1

],

(376)


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

Jp = −eDpdpdx (x > ln)

Jn = eDndndx (x < −lp) .

(377)

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

∂pn∂t = − 1

e∂Jp∂x − pn−pn0

τp(x > ln)

∂np∂t = 1

e∂Jn∂x − np−np0

τn(x < −lp) ,

(378)

where τp and τn are the (recombination) lifetimes of the minority holes and electrons in the quasi-neutralregions. Combining Eq. (377) and Eq. (378) we get:

∂pn∂t = Dp

∂2pn∂x2 − pn−pn0

τp(x > ln)

∂np∂t = Dn

∂2np

∂x2 − np−np0τn

(x < −lp) .

(379)

At steady state, the general solutions are:

pn(x) = pn0 + A e−x/Lp + B ex/Lp (x > ln)

np(x) = np0 + C e−x/Ln + D ex/Ln (x < −lp) ,

(380)

where Lp = (Dpτp)1/2 and Ln = (Dnτn)1/2 are the hole and electron diffusion lengths in the quasi-

neutral regions and A, B, C, and D are integration constants to be determined by the boundary conditions.


The first 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 = ln) and np(x = −lp) given by Eq. (376) above, so that:

pn(x) = pn0 + pn0

[eeVa/(kBT ) − 1

]e−(x−ln)/Lp (x > ln)

np(x) = np0 + np0

[eeVa/(kBT ) − 1

]e(x+lp)/Ln (x < −lp) .

(381)

Finally, by Eq. (377):

Jp(x) =eDppn0

Lp

[eeVa/(kBT ) − 1


Jn(x) =eDnnp0

Ln

[eeVa/(kBT ) − 1

]e(x+lp)/Ln (x < −lp) .

(382)

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 figure at page 106) is transformed – via recombinationprocesses – into an electron current (J2 in that figure). So, the total current will be independent of x. Thisconstant value can be obtained thanks to our assumption that there’s no generation/recombination in thedepletion region. In this case we can evaluate the currents in Eq. (382) at a position in which they have notyet started to ‘decay’ due to recombination, so Jp at x = ln and Jn at x = −lp. Thefore the total currentwill be:

Jtotal = Jp(ln) + Jn(−lp) =

[eDppn0

Lp+

eDnnp0

Ln

] [eeVa/(kBT ) − 1

]= Js

[eeVa/(kBT ) − 1

],

(383)where Js is the ‘saturation current density’ (the maximum current density we have under reverse bias, that is,for Va → −∞). Equation (383) is known as ‘Shockley’s equation’.


– Deviation from ideality: Generation/recombination in depletion region.Let’s now consider the more general case in which GR processes are active in the depletion region. Recall thatthese processes are proportional to np−n2

i (see Eqns. (333)-(335), page 99 or Eq. (340), page 100), so let’sconsider the product np:

np = n2i e

(EFn−EFp)/(kBT )= n2

i eeVa/(kBT ) . (384)

From Eq. (340) at page 100 of the Lecture Notes, assuming σp = σn, and setting τ0 = σvthNTT , wehave:

Us =np − n2

i

τ0

[p + n + 2ni cosh

(ET−EikBT

)] ≈n2i

(eeVa/(kBT ) − 1

)τ0(p + n + 2ni)

, (385)

the last step depending on the assumption that only traps at energy ET = Ei contribute effectively to theGR processes.Now, from current continuity, at steady state:

dJn(x)

dx= − dJp(x)

dx= e Us(x) . (386)

We have calculated so far the current density Jn at x = −lp which we assumed was identical to Jp(x = ln)thanks to the fact that there were no GR processes in the depletion region. We can now account for theseprocesses and evaluate the correction to the current they cause by integrating the equation for Jn from −lpto ln:

Jn(ln) = Jn(−lp) + e

∫ ln

−lpdxUs(x) . (387)

Therefore for the total current we get:

J = Jp(ln)+Jn(ln) = Jp(ln) + Jn(−lp) + e

∫ ln

−lpdxUs(x) = Js

[eeVa/(kBT ) − 1

]+ JGR .

(388)


The term JGR – the ‘generation/recombination current density – acts as a correction to the Shockley term.In order to avaluate it in a simple approximation, let’s note that JGR will be dominated by processes occurringat a position where p + n is a minimum, as seen from Eq. (385). But since np is constant,

d(n + p) = 0 → dn = − dp , d(np) = 0 → pdn + ndp = 0 ,

and, combining these two equations, these conditions imply n = p. By Eq. (384),

n = p = ni exp

(eVa

2kBT

),

so that, from Eq. (385) we have:

Us ≈n2i

(eeVa/(kBT ) − 1

)2niτ0(1 + exp[eVa/(2kBT )]

=ni

2τ0

[eeVa/(2kBT ) − 1

]. (389)

Assuming that Us takes this constant value – let’s call it Umax – throughout the depletion region (not strictlycorrect, we are going to overestimate JGR), we get for JGR:

JGR ≈ e Umax (ln + lp) = e (ln + lp)ni

2τ0

[eeVa/(2kBT ) − 1

]. (390)

From Eqns. (383) and (388) we have:

Jtot ≈ Js[eeVa/(kBT ) − 1

]+ e (ln + lp)

ni

2τ0

[eeVa/(2kBT ) − 1

]. (391)

For a small Va the GR term dominates, while the Shockley term takes over at larger bias. It is customary tolump these two types of asymptotic behavior by writing:

Jtot ≈ Jsat

[eeVa/(nkBT ) − 1

], (392)


where the index n (called ‘non-ideality factor’) takes a value close to 2 al low Va and drops to unity as Vagrows larger.

• Short ‘base’ diodes.The name ‘short base’ comes from bipolar transistors. These are two p-n junctions butted against each otherto form either p-n-p or n-p-n structures. The ‘central’ region is called ‘base’. Consider for simplicity justthe first p-n junction of a p − n − p structure. When the n-type base is very long, we can study that p-n

junction as we have done above. But if the base is ‘short’ (compared to the diffusion length Lp = (Dpτp)1/2

of minority holes in the n region), then the boundary condition pn(x → ∞) = pn0 used before Eq. (381)must be replaced by pn(x = xR) = pn0, where xR is the edge of the ‘base’. In addition, the minority holesmay transit the base in a time so short that they will not be able to recombine. Thus, the current density of theminority holes in the base will be constant. Since the DDEs imply

Jp = eµppF − eDpdp

dx, (393)

assuming as before that the field vanishes in the quasi-neutral region (F = 0), we have:

p(x) = − Jpx

eDp+ B , (394)

where B is an integration constant which we can find by imposing p(x = ln) = pn0 exp[eVa/(kBT )]:

B = pn0 exp

(eVa

kBT

)+

Jpln

eDp. (395)

But the boundary condition at x = xR we’ve considered before implies pn(x = xR) = pn0, or:

p(x = xR) = pn0 = − JpxR

eDp+ pn0 exp

(eVa

kBT

)+

Jpln

eDp=

Jp(ln − xR)

eDp+ pn0 exp

(eVa

kBT

).

(396)


Solving for Jp we get:

Jp =eDppn0

xR − ln

[exp

(eVa

kBT

)− 1

]. (397)

Since the slope dp/dx is larger than in a long-base junction, the diffusion current will generally be larger.

• Junction capacitance.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. One way to estimate this ‘time response’ of the device isto consider the capacitances involved. Capacitance is a measure of the charge stored per unit change of voltage.A larger capacitance means that more charge must be moved in or out, so that – for a fixed current – moretime is needed to complete the process. Thus, this translates into a longer delay in responding to the new biascondition.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 ‘diffusioncapacitance’. Let’s now consider these two components separately.

– Depletion capacitance.When the bias Va applied to the junction is varied, the width of the depletion region changes according toEqns. (373) and (374). The charge present in each depletion region due to the ionized dopants will be:

Qdepl = eANDln = eANAlp , (398)

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 definition of capacitance,

Cdepl =

∣∣∣∣dQdepl

dVa

∣∣∣∣ = eAND

∣∣∣∣ dlndVa

∣∣∣∣ = A

[eεs

2

NAND

(NA + ND)

]1/2 1

(Vbi − VA)1/2, (399)


or, noticing that from Eqns. (373) and (374)

lp + ln =

[2εs

e

(Vbi − Va)(NA + ND)

NAND

]1/2, (400)

we can rewrite Eq. (399) as:

Cdepl =εsA

ln + lp, (401)

which is just the capacitance of a parallel-plate capacitor with a dielectric of permittivity εs, with plates ofarea A separated by a distance ln + lp.

– Diffusion capacitance.The concentration of excess carriers diffusing in the quasi-neutral regions can be obtained from Eq. (381):

δpn(x) = pn(x) − pn0 = pn0

[eeVa/(kBT ) − 1


δnp(x) = np(x) − np0 = np0

[eeVa/(kBT ) − 1

]e(x+lp)/Ln (x < −lp) .

(402)

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

Qdiff,p = e

∫ ∞

lnδpn(x) dx = eLppn0

[eeVa/(kBT ) − 1

]. (403)

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

Cdiff,p ≈ dQdiff,p

dVa=

e2Lppn0

kBTeeVa/(kBT ) ≈ e

kBT

L2p

DpJp(ln) =

e

kBTτpJp(ln) , (404)

where we have used Eq. (383) in the last step. Accounting now for the charge of the minority electrons


diffusing into the p quasi-neutral region:

Cdiff ≈ dQdiff,p

dVa+

dQdiff,n

dVa=

e

kBT[τpJp(ln) + τnJn(−lp)] . (405)

Note: This is the equation found in most textbooks. However, in Karl Hess’ book one finds instead:

Cdiff ≈ e

2kBT[τpJp(ln) + τnJn(−lp)] , (406)

where the additional factor of 1/2 is explicitly commented and the claim is made that Eq. (405) is in error.This factor can be justified 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 diffusion charge extendsalso inside the depletion region, so that the integration in Eq. (403) should extend from 0 to ∞, not from ln(and similarly for the expression for Qdiff,n). Since as Va changes charges will leave the depletion region, wewill obtain a lower estimate for the charge, and so for the capacitance. In a way, this argument is equivalent toour ‘hand-waving’ argument since both reduce to accounting for the charges throughout the entire junction,not just in the quasi-neutral regions.

– Switching time.Read the description of the ‘recovery time’ and ‘fall time’ in the text by Colinge and Colinge (pages 123-125).

• Hot electron effects and junction breakdown.The entire analysis we have carried out so far is based on the DDEs which, as we saw, assume that theelectrons and holes remain at thermal equilibrium. Let’s now consider a few issues which clearly go beyond thisassumption.

– Electron heating.The first issue is apparently a trivial one: At equilibrium, we may have a very large electric field in the


depletion region, especially if we consider heavy doping. We saw at page 86 of the Lecture Notes that theelectron temperature can grow as F 2 as the electric field increases. So, should we expect electron (and hole)heating in the depletion region? A rigorous solution of the Boltzmann equation, most likely via a Monte Carlotechnique, can provide an answer. But a simple application of the hydrodynamic model (the third moment ofthe BTE) is sufficient here: The relevant equation (Eq. (293) at the bottom of page page 85) was:

∂〈E〉∂t

≈ 1

nj · F − 〈E〉

τw, (407)

or, setting (3/2)kBTc = 〈E〉,∂Tc

∂t≈ 2

3kBnj · F − T − Tc

τw. (408)

At equilibrium no current flows (j = 0), we are also at steady-state (∂Tc/∂t = 0), so we must have Tc = T ,which expresses the fact that at equilibrium the electron temperature is equal to the lattice temperature. Inother words, the ‘average’ electron energy remains (3/2)kBT . Occasionally, an electron will ‘roll down’ thefield and gain kinetic energy above the equilibrium value. But there will also be electrons attempting to ‘climbthe potential hill’, so losing kinetic energy. On average, the two types of process will compensate each other.

– Impact ionization.The current-voltage characteristics of a p-n junction show an exponentially increasing current as Va growspositive, while the ‘leakage’ current density saturates to the value (from Eq. (383)):

Js =

[eDppn0

Lp+

eDnnp0

Ln

], (409)

as the applied bias takes very large negative values (reverse bias). In practice, as Va grow sufficiently large,the reverse current ‘snaps’ to very large values. This is called ‘junction breakdown’ and it is mainly due toavalanche multiplication initiated and sustained by impact ionization, and by Zener tunneling (which we shalldiscuss below).


Ionization rate. As shown by the top figures at page 94, the frequency at which a carrier – electron or hole– of kinetic energy E above a ‘threshold’ Eth generates an electron-hole pair can be approximated by theKeldysh formula:

1

τii(E)=

B

τI(Eth)

(E − Eth

Eth

)p

, (410)

for E > Eth, vanishing otherwise. The constant B is a dimensionless parameter and 1/τ(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 momentum conservation may require that Eth be much higher,since the recoil carriers and the generated carriers will have to share the available momentum and so cannot ingeneral have vanishing energy. The coefficient p at the exponent takes the value of 2 for semiconductors withdirect gap. In general, it takes values as high as 6 (as determined by complicated calculations). Lower valuesfor p imply a ‘soft threshold’ (carriers will ionize with slowly increasing probability as E grows beyond Eth).Large values of p imply a ‘hard threshold’: Ionization will occur almost immediately as the energy of thecarrier crosses the threshold energy. Phonon-assisted processes (in which one or more phonons are emitted orabsorbed during the ionization process) can soften the threshold, as well as the so-called Umklapp processes(in which a carrier ends up in the next Brillouin Zone and G vectors of the reciprocal lattice is needed to ‘mapit back’ into the first BZ) can ‘soften’ the requirements of momentum conservation.Ionization coefficient. The parameter which is of interest in the breakdown of p-n junctions is the ‘ionizationcoefficient’ α (see Eq. (307), page 93). This is defined as the number of pairs generated per unit length, sothat it also represents the rate at which the current density increases (per unit length): Considering for nowonly electrons:

dJn

dx= αn Jn . (411)

Since each electron at energy E = E(k) generates electron-hole pairs at a rate 1/τii(E), if f(k) is theelectron distribution function, then:

αn =1

n〈vd〉2

(2π)3

∫dk

f(k)

τii[E(k)], (412)


where n is the electron density n = 2(2π)−3 ∫ dkf(k) and 〈vd〉 = (2/n)(2π)−3 ∫ dkf(k)vF (k) isthe drift velocity (i.e., the group velocity along the direction of the field averaged over the electron distributionfunction). It can be seen (perhaps not too trivially) that the functional dependence of τii as given by Eq. (410)does not affect αn too much. On the contrary, the shape of the distribution function f(k) as a function ofthe field dramatically affects αn. Therefore, the whole 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 without losingenergy to phonons will contribute to the ionization process. The probability that an electron scatters per unittime is

∑k′ P (k′, k). If Pnp(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:

Pnp(t + dt) = Pnp(t)

1 − dt

∑k′

P (k′, k)

. (413)

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

Pnp = exp

−

∫ t2

t1

∑k′

P (k′, k) dt

. (414)

(Note that k and k′ are functions of time due to their acceleration in the field, 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. (414) by noticing that 1/τph(k) =

∑k′ P (k′, k),

by using the chain-rule (let’s consider only the 1D case for simplicity):

dE

dt=

dE

dk

dk

dt=

dE

dk

eF

h→ dt = dE

h

eF

(dE

dk

)−1

.


Then,

Pnp = exp

{− h

eF

∫ Eth

0

dE

τph(E)

(dE

dk

)−1}

. (415)

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 coefficient atsmall fields F , α ∝ exp(−constant/F ). However, despite its empirical success in this case, Shockley’slucky-electron model should not be taken too seriously (indeed it has caused more harm – in the form ofconfusion and bad physics in the literature – than good). This may be seen by noticing that at larger electricfield experimentally one finds instead that an expression of the form α ∝ exp(−constant/F 2) matches thedata more accurately. This can be explained by assuming that at large enough fields the electron ionize assoon as they reach the threshold energy Eth, that they lose all of their energy, so that the energy-loss term is∫dkE/τiif(k). It may be shown that adding this term to the third-moment equations will yield the desired

behavior.Avalanche. Let’s generalize Eq. (411) to the case of a p-n junction under strong reverse bias. Underthese bias conditions, the reverse ‘leakage current’ density Js is very small and it is mainly due to those fewelectrons present in the p-type quasi-neutral region which flow into the n-type region, and holes undergoing asimilar (but opposite) process. If the electric field in the depletion region is sufficiently large, impact ionizationwill occur. Considering one of those few electrons entering the depletion region from the left, it will have anonzero probability of creating an electron-hole pair. The generated hole will travel back towards the left andit may impact-ionize, thus creating another electron-hole pair. The generated electron, in turn, will travelback towards the right and the cycle will repeat itself. There will be a critical field at which the probabilitythat impact-ionization occurs will be unity. At this point, the process will run-away, more and more carriersbeing generated and the current will grow without limits. This is the avalanche breakdow.In order to quantify the onset of this condition, let’s note that the initial electron current injected from theleft will grow as:

dJn

dx= αp Jp + αn Jn , (416)


ordJn

dx− (αn − αp) Jn = αpJ , (417)

where J = Jn + Jp is the total electron and hole current, constant at steady-state. To determine thebreakdown condition, let’s denote with Jn(−lp) the electron current density incident from the left-edge ofthe depletion region (at x = −lp) and let’s assume that the electron current exiting the high-field depletionregion (at x = ln) is a multiple Mn of Jn(−lp). Since most of the current at x = ln will be carriedby electrons, we can assume Jn(ln) = MnJn(−lp) ≈ J . With this boundary condition the solution ofEq. (417) can be written as:

Jn(x) = J

{1

Mn+

∫ x

−lpdx′ αp exp

[−∫ x′

−lpdx′′ (αn − αp)

] }exp

[∫ x

−lpdx′ (αn − αp)

].

(418)

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

dy(x)

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

is

y(x) = C e−A(x) + e−A(x)∫

dx b(x) eA(x) ,

where C is an integration constant determined by the boundary condition and A(x) =∫ x dx′a(x′)).

Evaluating Jn(x) at x = −lp yields Jn(−lp) = J/Mn, which is our boundary condition. We mustalso require Jn(ln) = J . With some algebra this implies:

1 − 1

Mn=

∫ ln

−lpdx αn exp

[−∫ x

−lpdx′ (αn − αp)

]. (419)


The avalanche breakdown voltage is defined as the voltage at which Mn → ∞. Thus, at breakdown theionization integral approaches unity:

∫ ln

−lpdx αn exp

[−∫ x

−lpdx

′(αn − αp)

]= 1 . (420)

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

∫ ln

−lpdx αp exp

[−∫ ln

xdx

′(αp − αn)

]= 1 . (421)

Equations (420) and (421) 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 αn ≈ αp (as in GaP, for example), the breakdown condition becomessimply ∫ ln

−lpdx α = 1 , (422)

with obvious meaning: If the probability of ionizing over the depletion region reaches unity, avalanche willoccur.Breakdown voltage. The breakdown voltage can be calculated from known doping profile, ionization rates,etc. from the equations above. In the case of an abrupt junction we have

VBD =Fmax(ln + lp)

2, (423)

where the maximum field in the depletion region is given by Eq. (364). For ‘linearly graded’ junctions, the


formula above must be corrected by a factor 2/3.


– Zener breakdown.Another cause of breakdown is related to the quantum mechanical process of tunneling. At a very largeelectric field in a reverse-biased p-n junction, electrons in the valence band may tunnel across the band gap ofthe semiconductor, thus creating an electron-hole pair. The process may lead to breakdown either directly (somany pairs will be crated that the leakage current will grow) or indirectly: The electrons and holes generatedby the tunneling process will impact-ionize and an avalanche process will begin. This breakdown mechanismis called ‘Zener breakdown’. It affects mostly heavily-doped junctions in which the built-in voltage and theapplied (reverse) bias fall over a narrow depletion region, thus giving rise to large electric fields.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 figure in the previouspage). Using the WKB approximation (see Homework 1), the tunneling probability is proportional to:

PZener(F ) ∝ exp

−4(2m∗)1/2E3/2

G

3ehF

, (424)

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

JZener ≈ (2m∗)1/2e3FVa

4π2h2E1/2G

PZener(F ) . (425)

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 Γto 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 quitestrongly temperature-dependent.


Other ‘diodes’: Heterojunctions,Metal-Semiconductor junctions (Shottky 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 different semiconductors) make up heterojunctionbipolar transistors (HBTs), high electron-mobility transistors (HEMTs, also known as ‘modulation-doped field-effecttransistors’, MODFETs), and injection lasers; Shottky 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 field-effect transistor’ (MOSFET).We shall now discuss in turn each of these ’junctions’ (or ‘diodes’, as they broadly fall into the category oftwo-terminal devices).

• Heterojunctions.– Thermionic emission.

As shown in the left frame of the figure in the next page, let’s consider two different semiconductors broughtinto contact. In the figure we use the example of GaAs and the alloy AlxGa1−xAs (where x, known as the‘mole fraction’, is the molar fraction of Al ions to Ga ions). GaAs has a smaller band gap (≈ 1.51 eV)than AlxGa1−xAs (≈ 1.72 eV for x = 0.3, for example). Most of the discontinuity of the 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 figure.Let’s now assume that the bands will remain fixed in this situation (we’ll worry later on about the possibility ofa new equilibrium position) and let’s attempt to compute the current caused by electrons flowing from GaAsto AlxGa1−xAs.


If we take the z axis as the axis perpendicular to the plane of the interface, we can compute the current bymultiplying the distribution function f(k) in GaAs (i.e., in the left semiconductor) by the z-component of thevelocity and by summing over all states with right-directed velocity, vz(k) > 0 and with sufficient energy and‘correct’ angle of travel to overcome the barrier. This means, classically:

1

2m∗v2

z =1

2m∗(hkz

m∗

)2

≥ ∆Ec . (426)

Thus the minimum value of the z-component of k an electron in GaAs must have to make it over the barrier

of height ∆Ec is kz0 = (2m∗∆Ec)1/2/h. Thus, this ‘left-to-right’ current can be written as:

jLR =2e

(2π)3

∫kxky

dkx dky

∫kz>kz0

dkz vz(k) f(k) . (427)

Taking for f the non-degenerate Boltzmann limit and switching integration variables from wavevector to


velocity we have:

jLR =

(m∗

h

)3e

4π3eEF,L/(kBT )

∫ ∞

−∞dvx dvy

∫ ∞

vz0

dvz vz exp

[− m∗

2kBT(v2

x + v2y + v2

z)

],

(428)where EF,L is the Fermi level in GaAs measured from the conduction-band minimum of GaAs itself.

We can use cylindrical coordinates, so that dvxdvy = vdvdφ (with v = (v2x + v2

y)1/2) and, defining

E‖ = m∗v2/2, we have dE‖ = m∗vdv, so:

∫ ∞

−∞dvxdvy exp

[− m∗

2kBT(v2

x + v2y)

]=

∫ 2π

0dφ

∫ ∞

0dv v e−m∗v2/(kBT ) =

=2π

m∗

∫ ∞

0dE‖ e

−E‖/(kBT )= =

2π

m∗kBT , (429)

so that:

jLR =e

2π2

m∗2

h3kBT e

EF,L/(kBT )∫ ∞

vz0

dvz vz e−m∗v2z/(2kBT ) = A∗T 2e

(EF,L−|∆Ec|)/(kBT ).

(430)This is called ‘thermionic emission current’, and A∗ = em∗k2

B/(2π2h3) is the so-called Richardson constant.Following similar arguments, the current from AlxGa1−xAs to GaAs will be

jRL = A∗T 2eEF,R/(kBT )

, (431)

where EF,RL is now the Fermi level in AlxGa1−xAs measured from the conduction-band minimum ofAlxGa1−xAs.

Let’s now assume that the AlxGa1−xAs is doped with ND donors per cm3 and that the GaAs is undopedor weakly n-type doped. Similarly to what we saw for p-n junction, at equilibrium the Fermi levels must


‘line up’. Physically, this is caused by the electrons in the AlxGa1−xAs flowing into the GaAs-side of the‘heterojunction’ until a built-in potential will block the flow. We can estimate the time-dependence of thisflow from the continuity equation:

e∂n

∂t=

∂j

∂z, (432)

The current will be mainly caused by electrons flowing from right to left via thermionic emission. We canassume that only electrons within a distance Ln from the interface will participate in the conduction, sinceelectrons in the AlxGa1−xAs farther away will have essentially zero vz (since thay are at equilibrium, sufferingtoo many collisions). Then, from Eqns. (431) and (432) we have:

e∂n

∂t≈ −A∗T 2e

EF,R/(kBT )/Ln . (433)

Assuming a time-dependent Fermi level EF,R(t) such that:

n(t) =

∫ ∞

0dE ρR(E) e

(EF,R(t)−E)/(kBT ), (434)

(where ρR(E) is the density of states at energy E in AlxGa1−xAs), we can write:

eEF,R(t)/(kBT )

= n(t)4

(πh2

2m∗kBT

)3/2

= n(t)CR . (435)

From Eqns. (433) and (435) we get:

∂n

∂t= −A∗T 2

eLnn(t)CR , (436)

whose solution is:

n(t) = nc exp

[−(A∗T 2CR

eLnt

)], (437)


where nc is the concentration of electrons in the AlxGa1−xAs at t = 0. The time constant for the decay

of the electron concentration in the AlxGa1−xAs is thus τ ≈ eLn/(A∗T 2CR) which is of the order of

picoseconds at room temperature. Thus, the AlxGa1−xAs will lose its electrons very quickly, leaving behindpositively ionized donors. The build-up of the positive charge in the AlxGa1−xAs and of the (electron)negative charge in the GaAs will eventually build a barrier preventing further flow of electrons from theAlxGa1−xAs to the GaAs. When, finally, equilibrium will be reached, the band diagram will look like what isshown in the right frame of the figure at page 129.

– The depletion approximation (again).Let’s quantify this band diagram by solving Poisson equation. Denoting by F the z-component of the electricfield, we have

∂F

∂z=

e

εs(p − n + N

+D − N

−A ) , (438)

where εs is the dielectric constant of the semiconductor (a function which has a discontinuity at theGaAs/AlxGa1−xAs interface). Let’s take z = 0 at the interface. Let’s also embrace once again the‘depletion approximation’, by assuming that in the AlxGa1−xAs, at a distance W away from the interface thesemiconductor is charge-neutral, with zero electric field, since the free carriers (or the positive charges of thedopants) will have screened the electric field at this distance. Therefore, this is a valid approximation as longas W >> 2π/βs, where βs is the screening parameter (in our case β2

s = eN+D/(εskBT )). Then, the

‘built-in’ potential will be:

Vbi = −∫ W

0dz F = − zF |W0 +

∫ W

0dz z

∂F

∂z. (439)

From this equation, since we have assumed F (z) = 0 for z ≥ W , we have:

Vbi ≈∫ W

0dz z

∂F

∂z. (440)


Neglecting the charges of the free carriers in the depletion layer, from Eq. (438):

Vbi ≈ −∫ W

0dz z

eN+D

εs=

eW 2N+D

2εs. (441)

This equation relates Vbi to W , but it gives us neither value. These values can be obtained by solving Poissonequation on both sides of the interface and matching the solutions. Only when the GaAs-side of the junction isvery heavily doped, so that the bands distort significantly only in the AlxGa1−xAs-side of the heterojunction,then we have approximately:

|eVbi| ≈ ∆Ec + Ec,L(−∞) − Ec,R(∞) , (442)

(where Ec,L is the energy of the bottom of the CB of GaAs at the far left, and Ec,R(∞) the energy of thebotom of the CB of AlxGa1−xAs at the far right) from which W can be obtained via Eq. (441).

– Beyond the depletion approximaton.A better solution of our problem can be obtained by going beyond the depletion approximation and by solvingPoisson equation on both sides of the interface. First, let’s recall that at the interface between two dielectricsthe normal component of the field D = εsF and the ‘in-plane’ components of the field F must be continuous.Since F = −∇φ, these conditions imply:

εL∂φ

∂z

∣∣∣∣z=0−

= εR∂φ

∂z

∣∣∣∣z=0+

(443)

andφ(z = 0−) = φ(z = 0+) . (444)

Now let’ consider Eq. (438). As noted when discussing p-n junctions, this equation is nonlinear since thecarrier density n depends on the electrostatic potential itself:

n(z) = nc(−∞) exp

[eφ(z)

kBT

], (445)


where nc(−∞) is the equilibrium electron concentration in GaAs far away to the left. Let’s denote with ψ(z)the quantity −eφ(z)/(kBT ). Then, on the left side of the interface, in the absence of holes and acceptors,Poisson equation becomes:

∂2ψ

∂z2= − e2

εGaAskBT[N+

D − nc(−∞)eψ] . (446)

If the doping in GaAs is uniform, nc(−∞) = N+D

and

∂2ψ

∂z2=

e2nc(−∞)

εGaAskBT(e

ψ − 1) . (447)

This equation cannot be solved in closed form. However, we can get some information about the electronconcentration on the left-side of the heterojunction as a function of the value of the potential, φi, and of theelectric field, Fi, at the interface. In order to do this, let’s multiply Eq. (447) by −∂ψ/∂z and integrate.The left-hand-side becomes:

−∫

∂2ψ

∂z2

∂ψ

∂zdz = −1

2

∫∂

∂z

(∂ψ

∂z

)2

dz = −1

2

(∂ψ

∂z

)2

, (448)

while for the right-hand-side we get:

e2nc(−∞)

εGaAskBT

∫(eψ−1)

∂ψ

∂zdz =

e2nc(−∞)

εGaAskBT

∫(eψ−1) dψ =

e2nc(−∞)

εGaAskBT(eψ−ψ+C) , (449)

where C is an integration constant. Therefore, from Eqns. (448) and (449), fixing C = −1 by requiring∂ψ/∂z = 0 for z = −∞, we have the relation:

F 2 = −2nc(−∞)kBT

εGaAs(eψ − ψ − 1) . (450)


Equation (447) can be integrated from z = −∞ to the interface z = 0 obtaining:

εGaAs Fi = Qtot , (451)

where Qtot is the total excess charge (due to the electrons which have spilled over from AlxGa1−xAs-side ofthe heterojunction) at the left of the interface. Note that electrons accumulate exponentially – the Boltzmann

law - with increasing interface potential, since from Eqns. (450) and (451) Qtot ∝ eψi/2. Thus, one speaksof an ‘accumulation layer’ at the GaAs side of the interface. Had we considered, instead, p-doped GaAs,the derivation would have been very similar. However, in the presence of holes, for small values of ψi theelectrons would have been recombining with the holes – thus forming a depletion layer – before startingto form an ‘inversion layer’ (so called because the p-type GaAs would become n-type, so ’inverted’) in

which the electron density would grow, as before, as ∝ eψi/2. As the interface potential reaches the valueψi = 2kBT ln(NA/ni), then one talks of ‘strong inversion’, since the electron density at the interfaceexceeds the hole density in the bulk of the p-type GaAs.Equations (450) and (451), together with the continuity conditions Eqns. (443) and (444), can be used toobtain a complete solution of Poisson equation.

• Shottky contacts.– Metal-semiconductor junction.

When we bring a metal and a semiconductor together, as shown in the figure 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 figure), electrons will flowfrom 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 eφB between the bottom of the CB of the semiconductor and themetal Fermi level. Denoting by eφM the metal workfunction (that is, the energy required to excite an electronat the metal Fermi level to the level of the vacuum) and by eχ the electron affinity of the semiconductor, the


Shottky barrier height will be (looking at the figure):

eφBn = e (φM − χ) , (452)

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

eφBp = EG − e (φM − χ) . (453)


– 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, interdiffusion 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 Shottky barrier heightshould be regarded as an experimentally measured quantity.

– Image force effects.Another cause of deviation from the ideal picture we used to reach Eq. (452) is caused by ‘image force effects’.This is a basic idea of elementary electrostatics. Let’s consider an interface between two dielectrics (withpermittivities ε2 – the metal, assumed to be in the semi-infinite space z < 0, and ε1 – the semiconductor).Let’s consider the electron in the semiconductor, which fills the half-space z ≥ 0, at a distance z = d fromthe interface. By the method of images we can write the potential along the line (the z-axis) normal to theinterface and passing through the particle location as:

V (z) =e2

4πε1

{1

|d − z| − ε2 − ε1

ε2 + ε1

1

|d + z|}

. (454)

The first term is the ‘usual’ potential due to the electron iteself. The second term is the image potential,Vim(z), which we shall consider now. For the electron sitting at z = d the force due to its own image(located at z = −d) will be:

Fim(d) = − dVim(z)

dz

∣∣∣∣z=d

= − e2

16πε1d2

ε2 − ε1

ε2 + ε1. (455)

Now, the work performed against the image required to bring the electron from infinity to the position z > 0


will be:

∆φim(z) =

∫ ∞

zds Fim(s) = − e2

16πε1z

ε2 − ε1

ε2 + ε1. (456)

This can be viewed as a force which lowers (for ε2 > ε1) or raises (in the opposite case) the barrier height.Coming to our explicit case, assuming for the metal ε2 → ∞, and setting ε1 = εs, and adding the effect ofa uniform field Fz, we have a total potential energy

Φim(z) = − e2

16πε1z− eFzz . (457)

This potential exhibits a maximum at some distance from the interface and it yields a barrier whose height isreduced by the amount:

∆φB =

(eFz

4πεs

)1/2

. (458)

An additional barrier-reducing effect is purely quantum-mechanical and it’s caused by the penetration ofelectronic wavefunctions of the metal into the semiconductor band gap. We shall not discuss it further.


– Thermionic current.We can now compute the thermionic current. The current flowing from the left (the metal) to the right (thesemiconductor) can be computed following exactly the arguments used to derive Eq. (430). Thus:

jLR = A∗T

2e−eφB/(kBT )

. (459)

At equilibrium, the same current will flow from the semiconductor to the metal, so the net current will bezero. If we now apply an external bias, we shall for now assume that the (quasi) Fermi levels remain constantin the metal, EF,L and in the semiconductor, EF,R, and we assume that the difference between the Fermilevels is just the applied bias, eVa = EF,R − EF,L, taken positive when we apply positive bias to thesemiconductor. (The assumption that the applied voltage drops in the depletion region of the semiconductorwhile the quasi-Fermi levels remain constant will be discussed below.) For Va > 0 we can consider JLRgiven, unchanged, by Eq. (459), since the barrier heights will remain unchanged by the application of theexternal bias (approximately: the image force lowering will by slightly modified by the applied bias via the fieldFz in Eq. (458), but we shall ignore this small effect). On the contrary, the current flowing from the right willbe enhanced (for Va > 0) and will be:

jRL = A∗T

2e−eφB/(kBT )

eeVa/(kBT )

, (460)


so that the total current, j = jRL − jLR will be:

j = A∗ T 2 e−eφB/(kBT )[eeVa/(kBT ) − 1

]. (461)

Note the similarity with the Shockley equation, Eq. (383), for the current in p-n-junctions.The thickness of the depletion region can be obtained as we did before for p-n-junctions and we obtain

W =

[2εs(eVbi − Va)

eND

]1/2, (462)

where the built-in potential Vbi is (see the figure) eφB − (Ec − EF ).– Diffusion current.

The limitation of the derivation leading to Eq. (461) lies in the assumption that the quasi-Fermi level remainsconstant in the semiconductor. Consider electrons flowing from the right, ‘jumping’ over the barrier towardsthe metal. They will do so with a velocity vz roughly given by 2eφB/m∗, of the order of at least 107

cm/s (and up to 108 cm/s) for most barrier-heights and semiconductors. This velocity is quite large becauseonly electrons streaming from right to left contribute to its average, since, having entered the metal, it’s veryunlikely that they will be reflected back. In other words, only the left-going half of the electron distributioncontributes to this average velocity. Therefore, the time constant characterizing the out-flow of these carriersinto the metal will be of the order of λ/vz, where λ is the electron mean-free-path. This flow of electronsout of the depletion region towards the metal must be balanced by electrons coming into the depletion regionfrom the right at z = W . These electrons will not be ‘streaming’ in one direction (to the left), but bediffused in all directions by scattering with impurities, phonons, etc. Thus, their characteristic velocity (nowaveraged over left- and right-going carriers) will be of the order of the ‘drift’ velocity vd, typically of the orderof 105 cm/s, or even smaller, for the typical concentration gradients of interest. Therefore, the associatedtime constant, λ/vd, will be at the very least two orders of magnitude smaller. This is true unless the widthW of the depletion region is smaller than λ. In this case also the supply of electrons from the right will occur‘ballistically’ at the fast time scale λ/vz, since scattering will be negligible also for this ‘supply’ of electronsfrom the right. This is the situation in which the supply of electrons from the right is able to keep up with


the escape of electrons to the left and our assumption of constant quasi-Fermi levels is valid. If, however,the supply from the right cannot match the escape-rate to the left (which happens, as we have seen, whenW >> λ), then our picture is clearly inconsistent (since it leads to a situation away from steady-state, incontrast to our assumption). In this case a diffusion-dominated model better describes the current flow inShottky diodes.From current continuity

j = enµnFz + eDn∂n

∂z. (463)

Using Einstein’s relation to express µn in terms of Dn, we can re-write this equation as:

j = eDn

(− e

kBTn

∂φ

∂z+

∂n

∂z

). (464)

Multiplying by exp[−eφ(z)/(kBT )] this becomes:

j e−eφ(z)/(kBT ) = eDn∂ne−eφ(z)/(kBT )

∂z. (465)

In steady-state the current does not depend on z, so

j

∫ W

0e−eφ(z)/(kBT )

dz = eDn

[n(W ) e

−eφ(W )/(kBT ) − n(0) e−eφ(0)/(kBT )

]. (466)

Using the solution φ(z) of Poisson equation in the depletion approximation we could in principle compute thefunction

F (ND) = e−eφ(0)/(kBT )∫ W

0e−eφ(z)/(kBT ) dz , (467)

(which depends only on the doping density ND) so that from Eq. (466), noticing that under an applied externalbias and assuming the the entire external potential drops in the depletion region, Va = φ(W ) − φ(0), wehave

jF (ND) = eDn

[n(W ) e

−e(Vbi−Va)/(kBT ) − n(0)]

. (468)


Now, since in the depletion approximation n(W ) is equal to its equilibrium value and, at equilibrium,neq(W ) exp[−eVbi/(kBT )] = neq(0), we have:

jF (ND) = eDn

[neq(0) e

eVa/(kBT ) − n(0)]

. (469)

As last step, we must find the non-equilibrium concentration n(0). To do that, let’s observe that the currentflowing to the left at the junction (z = 0) is the thermionic emission current with concentration n(0) andzero barrier height (since we are considering the current at the top of the barrier). Therefore,

j0th = A

∗T

2[n(0)

Nc− neq(0)

Nc

], (470)

having used Eqns. (430), (431), and the known expression neq ≈ Nc exp[EF − Ec)/(kBT )]. Thesecond term in the equation above represents the current coming back, which is essentially an equilibriumcurrent, since no voltage drops in the left side of the junction. The term A∗T 2/(eNc) is called ‘interfacerecombination velocity’ and it is denoted by vR. Thus, using now the fact that j0th is equal to j (since this isthe current at the interface and j does not depend on z), combining Eq. (469) and (470) we finally get:

j =eDnneq(0)[e

eVa/(kBT ) − 1]

F (ND) + D/vR. (471)

Note, once again, that is of the form j ∝ exp[eVa/(kBT )] − 1, the typical form of the current in manyjunctions.


– Tunneling current.

The last component of the current we must consider is of quantum-mechanical origin. It is the tunnelingcurrent illustrated in the figure above.In the example illustrated in the figure, electrons in the metal can tunnel across the Shottky 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 in Homework 1. Let the z-axisbe normal to the plane of the interface, let z = 0 be the location of the interface and let vz(k) be thez-component of the velocity of the electrons hitting the interface and attempting the tunneling process. Theprobability Pt that an electron will tunnel across the barrier can be estimated using the WKB approximation:

Pt ≈ exp

{−2

∫ zt

0dz κz(z)

}, (472)


where zt is the tunneling distance (that is, the position at wich the electrons will emerge from the barrier

in the semiconductor, κ(z) = {2m ∗ ∗[eφ(z) − Ez]}1/2/h is the imaginary component of the electronwavevector in the barrier, φ(z) the electrostatic potential in the depletion region, and Ez = h2k2

z/(2m∗M)

the component of the electron kinetic energy in the metal due to the z-component of the electrons wavectorin the metal with effective mass m∗

M . (Note that this separation of kinetic energy is only valid for parabolicbands and for directions along the principal axes of ellipsoidal band-structures.)Tunneling processes, in principle, conserve the component of the momentum parallel to the interface. Indeed,the translation symmetry on the plane of the interface ensures the conservation of this quantity. In practice, atsufficiently high temperatures, phonons can provide the ‘parallel momentum’ required to relax this conservationlaw. Also, interface roughness and deviations fron ideality of the interface, often have the same effect. Sinceit is a realistic effect, and since it simplifies considerably the calculations, we shall assume that parallelmomentum is not conserved and we can replace Ez with the total electron kinetic energy E in the expressionfor κ(z) above.The tunneling current from the metal to the semiconductor will then be:

jMS = 2e

∫kz>0

dk

(2π)3vz(k) fM(k) [1 − fS(Ek)] exp

{−2

∫ zt

0dz κz(z)

}, (473)

where fM(k) is the Fermi function in the metal and the factor [1 − fS(Ek)] ensures that we tunnel onlyinto unoccupied states in the semiconductor . Performing the integral in polar coordinates, recalling thatvz(k) = hkz/m

∗M = hk cos θ/m∗

M , we have:

jMS =eh

2π2m∗M

∫ ∞

0dk k3 fM(k) [1−fS(k)] exp

{−2

∫ zt

0dz κz(z)

} ∫ π/2

0dθ sin θ cos θ ,

(474)the upper integration limit for θ being π/2 since we restrict the intergration only over right-directed electrons


(kz > 0 in Eq. (473)). Thus, changing integration variable k to E, as usual:

jMS =em∗

M

2π2h3

∫ ∞

0dE E fM(E) [1 − fS(E)] exp

{−2

∫ zt

0dz κ(z)

}. (475)

Similarly, the current due to electrons tunneling from the semiconductor to the metal will be:

jSM =em∗

S

2π2h3

∫ ∞

0dE E fS(E) [1 − fM(E)] exp

{−2

∫ zt

0dz κ(z)

}. (476)

The total current will be, obviously, jT = jSM − jMS.Let’s consider two cases in which the WKB integral,

∫ zt0 dzκ(z) can be handled analytically. In the first

case, quite a realistic one, we approximate the potential φ(z) with its expression valid in the depletionapproximation,

φ(z) = φ′B

(1 − z

W

)2

, (477)

where φ′B, as illustrated in the figure, is the total voltage drop in the semiconductor, φ′

B = Vbi − Va =

φB − Va + EF,R − Ec and W = [2εs(Vbi − Va)/(eND)]1/2 is the width of the depletion region. Inthis case, with some algebra, we find:

∫ zt

0dz κ(z) =

(2m∗S)1/2

h

∫ zt

0dz [eφ(z) − E]1/2 =

=(2m∗

S)1/2

h

∫ W [1−(E/eφ′B)1/2]

0dz

[eφ′

B

W 2(W − z)2 − E

]1/2

, (478)

having used the fact that zt = W [1 − (E/eφ′B)1/2].


Changing integration variable y = (eφ′B/W 2)(W − z)2 − E, we get:

∫ zt

0dz κ(z) =

(m∗

S

2eφ′B

)1/2W

h

∫ eφ′B−E

0dy

(y

y + E

)1/2

=

=

(m∗

S

2eφ′B

)1/2W

h2eφ

′B

(1 − E

eφ′B

)1/2

− 1

2

. (479)

Thus, from Eqns. (475) and (479), we get:

jMS =em∗

M

2π2h3

∫ ∞

0dE E fM(E) [1−fS(E)] exp

{−2(2m∗

Seφ′B)1/2

hW

[(1 − E/(eφ

′B))1/2 − 1/2

](480)

At this point the integration must be performed numerically.Another example consists in approximating the Shottky barrier as a triangular barrier

φ(z) = φ′B − Fzz , (481)

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

depletion region. In this case we get:

jMS =em∗

M

2π2h3

∫ ∞

0dE E fM(E) [1 − fS(E)] exp

{−4(2m∗

S)1/2

3ehFz(eφ′

B − E)3/2}

. (482)

Given the complexity of these calculation, one may wonder how ohmic contacts could ever be realized. Onepossible explanation consists in assuming that height of the Shottky barrier, eφB, may be zero. This mayhappen in a few cases. But more likely is the scenario in which heavy amorphization of the interface mayresult in a very heavily doped – almost metallic – thin layer of semiconductor. The very high doping will result


in such a thin depletion layer, that tunneling across this barrier may be a dominant process, almost killing the

resistance of the contact. This can be seen easily from Eq. (480): For W << h/(2m∗Seφ

′B)1/2, the WKB

exponential approaches unity, thus resulting in a very large current. Similarly from Eq. (482), noticing thatthe field Fz is inversely proportional to W .

• 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 figures below.

eφBeφM eχ

eψBeVFB < 0EF,M

EC

EV

EiEF,S

METAL INSULATOR SEMICONDUCTOR

FLAT BANDS

EF,M

EC

EV

EiEF,S


ZERO BIAS


eVG < 0

EF,M

EC

EV

EiEF,S


ACCUMULATION

eVG > 0

EF,M

EC

EV

EiEF,S


INVERSION

Starting from a semiconducting substrate (assumed to be p-type in the figures), 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 figures a metal is considered.At ‘flat band’ the alignement of the bands is illustrated in the first figure. In the ideal case, charge would flowacross the insulator, so that the Fermi levels in the metal, EF,M , and in the semiconductor, EF,S, wouldline-up and the difference, φMS, between the metal and the semiconductor work-functions would vanish:

φMS = φM −(χ +

EG

2e+ ψB

)= 0 for p − type semiconductor , (483)


and

φMS = φM −(χ +

EG

2e− ψB

)= 0 for n − type semiconductor . (484)

In practice, the time required for the Fermi level to line-up is extremely long and this ideal situation is neverachieved. The figure shows that the application of a small bias, the flat-band voltage VFB , is required toline-up the Fermi levels. Its value will be given by the nonvanishing φMS of Eq. (483) or (484) above.The application of a negative bias, VG, to the metal (usually called the ‘gate’) drives the MOS diode into‘accumulation’: As seen in the figure, 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. Let’s now consider these processes in some detail.

– Interface space-charge region.

EC

Ei

EF

EV

eφBeψeψs

SiO2 Semiconductor

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


coordinate along the normal to the interace, z = 0 be the location of the interftace, and let’s define by ψ(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

[eψ(z)

kBT

]= np0 exp(βψ) , (485)

pp(z) = pp0 exp(−βψ) , (486)

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

ns = np0 exp(βψs) , ps = pp0 exp(−βψs) (487)

are the surface concentrations of the carriers.We can deal with the ‘exact’ Poisson equation as we have done above (see pages 134 and 135), extractingsome information before embracing the ‘usual’ depletion approximation. We have:

∂2ψ

∂z2= − e

εs[N+

D + p(z) − N−A − n(z)] , (488)

where εs is the (static) dielectric constant of the semiconductor. From Eqns. (485) and (486) we have:

pp − np = pp0 e−βψ − np0 eβψ , (489)

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

−N−A

= np0 − pp0, we have

∂2ψ

∂z2= − e

εs[pp0 (e−βψ − 1) − np0 (eβψ − 1)] . (490)


Let’s now integrate Eq. (490) from an arbitray location z towards the bulk (z → ∞, as we have done onpage 134-135, Eqns (448)-(451): Let’s multiply Eq. (490) by −∂ψ/∂z and integrate. The left-hand-sidebecomes:

−∫ ∞

z

∂2ψ

∂z2

∂ψ

∂zdz = −1

2

∫ ∞

z

∂

∂z

(∂ψ

∂z

)2

dz =1

2

(∂ψ(z)

∂z

)2

, (491)

having used the fact that the field vanishes as z → ∞. For the right-hand-side we get:

e

εs

∫ ∞

z[pp0(e

−βψ−1) − np0(eβψ−1)]

∂ψ

∂zdz =

e

εs

∫ 0

ψ[pp0(e

−βψ−1) − np0(eβψ−1)] dψ =

=e

εsβ[pp0(e

−βψ+ βψ − 1) − np0(e

βψ − βψ − 1)]. (492)

Therefore we have the following relationship between field and potential at any location z:

F 2 =

(2kBT

e

)2 ( epp0β

2εs

) [(e−βψ + βψ − 1) +

np0

pp0(eβψ − βψ − 1)

]. (493)

Introducing the Debye length, LD = [kBT/(e2pp0)]1/2 (the dielectric screening length in the p-type

non-degenerate bulk Si), and denoting as G(βψ, np0/pp0)2 the term in square brackets in Eq. (493), we

have:

F =∂ψ

∂z= ± 21/2kBT

eLDG

(βψ,

np0

pp0

), (494)

the plus (minus) sign valid for ψ > 0 (ψ < 0).The charge at the interface can now be expressed using Gauss law and the value of the field at the interface(obtained by setting ψ = ψs in Eq. (494)):

Qs = −εs Fs = ∓ 21/2εskBT

eLDG

(βψs,

np0

pp0

). (495)


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

and weak inversion the term βψ in the function G dominates, so that he charge grows as ψ1/2s . Finally,

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

strong inversion begins at ψs = 2ψB, 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(–βψs/2)accumulation

~ exp(βψs/2)strong inversion

flatband

depletionweak

inversion

ψB EC

~ ψs1/2

p–type Si 300K

NA = 4x1015 cm–3

ψs (Volt)

Qs

(cm

–2)


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−βψ − 1) dz =

epp0LD

21/2kBT

∫ 0

ψs

e−βψ − 1

G(βψ, np0/pp0)dψ , (496)

and

∆n = np0

∫ ∞

0(e

βψ − 1) dz =epp0LD

21/2kBT

∫ 0

ψs

eβψ − 1

G(βψ, np0/pp0)dψ . (497)

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

CD =∂Qs

∂ψs=

εs

21/2LD

1 − e−βψs + (np0/pp0)(eβψs − 1)

G(βψs, np0/pp0). (498)

At flat-band condition, ψs = 0, so:

CD,FB =εs

LD. (499)

– 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 field-effect transistor. Let’s recall that we are nowinterested in the ‘differential 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, ψs, drops in the semiconductor and part, Vox, in the insulator. The latter will be given obviously by:

Vox = Foxtox =Qstox

εox, (500)


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

Cox =dQs

dVox=

εox

tox. (501)

But the total capacitance will also depend on the charge induced by the voltage drop in the semiconductor,Eq. (498). Therefore the total capacitance will be the series-capacitance of the insulator, Cox and of thedepletion region, CD:

Ctot =CoxCD

Cox + CD. (502)

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

GATE VOLTAGE

GA

TE C

AP

AC

ITA

NC

E

high–frequency

lowfrequency

Cox Cox

CFB

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


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:

ψ(z) ≈ ψs

(1 − z

W

)2

, (503)

where the depletion width W can be obtained in the usual way (see Eq. (441) above):

W ≈(

2εsψs

eNA

)1/2

. (504)

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

CD =εs

W, (505)

andCtot =

εox

tox + (εox/εs)W. (506)

At flat 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 MOSfield-effect transistors), we have

W = Wmax ≈(

2εs2ψB

eNA

)1/2

=

[4εskBT ln(NA/ni)

eNA

]1/2

. (507)

When the surface potential reaches the strong-inversion value, ψs = 2ψB, 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 field and it will prevent anyfurther widening of the depletion region. Indeed, as we can see in the plot Qs vs. ψs at page 152, a smallincrement of surface potential will result in a huge increase of the electron charge. Thus, one can assume


that in strong inversion most of the additional VG will drop in the insulator and no significant increas ofψs will occur. Therefore, Eq. (507) 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 ψ will exceed2ψB, W will grow beyond Wmax and the capacitance will drop below its minimum value Cmin given belowby E. (508).As the gate bias becomes even more positive, we must distinguish two different 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 differential 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 valueεs/Wmax, and so the total capacitance will remain at the minimum value

Cmin =εox

tox + (εox/εs)Wmax, (508)

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 variedsufficiently 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 first 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


the insulator. These defects may be broadly classified into two categories: Fixed charges in the oxide andelectrically-active interface states. (The word ‘fixed’ 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 manyefforts), 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 effect of shifting the threshold voltage

VT = Vox,si + 2ψB =Qs(2ψB)

Cox+ 2ψB , (509)

or the flat-band voltage. In order to derive a general expression for the flat-band voltage shift due toan arbitrary distribution of charges ρ(z) inside the insulator (where z now denotes the distance from thegate-insulator interface) let’s consider first an ideal (free of charge) MOS capacitor originally at flat-bandcondition and now add a sheet of charge λ (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 flat-band. Let’s now apply an additional bias to the metal, soto bring the semiconductor at flat band. In this situation there will be no field in the metal (by definition),no field in the semiconductor (as we are at flat band). There will only be a constant electric field λ/εoxfor −tox < z < 0. Thus, the metal potential will have moved by an amount zλ/εox with respect to theinitial situation. This will be the shift of the flat-band voltage caused by the sheet of charge. Therefore, for adistribution ρ(z) of charges inside the insulator, the flat-band shift will be given by:

∆VFB =

∫ tox

0zρ(z)

εoxdz =

1

Cox

∫ tox

0

z

toxρ(z) dz . (510)

Note that charges close to the gate-insulator interface have no effect on VFB (the polarization charges at themetal surface screen completely the oxide charges), while charges near the semiconductor-insulator interfacehave maximum effect.


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 effect: Dependingon their occupation, they shift the flat-band voltage, as in Eq. (510). Since for these states z = tox, theirelectrostatic effect 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. (511)

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 ψs) will be ∝ Cit(ψs): Havingobtained Cit from the total capacitance via Eq. (511), at a given gate bias we have:

Cit =dQit

dψs= e

Nit

dE= e Dit , (512)

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 + ψs, we have dVG = dψs + dVox. SincedQ = Cox dVox = Ctot dVG, we have dVG = dψs + (Ctot/Cox) dVG, and so we can obtaindψs/dVG from the relation

dψs

dVG= 1 − Ctot

Cox. (513)

Thus, from Eqns. (506), (512), and (513) we can extract dQit/dψs:

Dit(E) =dQit

dψs=

Ctot

e

(dψs

dVG

)−1

− Cd

e. (514)


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

τ =1

vthσniexp

[−e(ψB − ψs)

kBT

], (515)

(for a p-type substrate), either a low-temperature measurement (so that at a given ac frequency ω the responsetime τ 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 first 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 field/potential orsurface-field/total-charge given by Eqns. (494) and (495). 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 affected by quantum mechanics when the chargecarriers are confined 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. Let’sconsider this situation in more detail, limiting our attention to inversion layers.


– Quantization effects in inversion layers.The thermal wavelength, λth, of an electron is given by the deBroglie relation λth = h/pth = 2π/kth,where pth = hkth is the average momentum of electrons at thermal equilibrium. For the average carrierenergy at thermal equilibrium we have Eth = (3/2)kBT = h2k2

th/(2m∗), so that, at 300K,

λth =2πh

(3m∗kBT )1/2≈ 6.2

(m0

m∗

)1/2

nm (516)

Consider the Si-SiO2 interface in inversion/strong-inversion. The electron density ns (the ‘sheet density’ perunit area, see the figure at page 152) is of the order of 1011-to-1013 cm−2, corresponding to an interfacialfield Fs = ens/εs of the order of 104-to-106 V/cm. Thus, an electron of thermal energy ((3/2)kBT ≈ 40meV) will be ‘squeezed’ by the field against the interface over a confining distance ∆z ≈ 3kBT/(2Fs),very much like a particle in a box considered before. This distance is of the order of 40-to-0.4 nm, comparableor even smaller than the electron thermal wavelength. We are not allowed to ignor the wave-like natureof electrons when we confine them so tightly: We expect that discrete energy levels will emerge fromthe confinement. If we confine a particle in a region of width ∆z, by Heisenberg’s principle the particlemomentum will suffer an uncertainty ∆k ∼ 1/∆z, so that the confined particle will have a minimum energyE0 ∼ h2∆k2/(2m∗) ∼ h2/(2∆z2m∗), called the ‘zero-point energy‘. In strong inversion this energymay be comparable to (or even larger than) the thermal energy, and quantum effects due to the confinementshould not be ignored.A simplified case exemplifies the situation. Consider a ‘triangular well’: Electrons are confined at the left(z = 0) by an infinitely high potential wall (mimicking the Si-SiO2 barrier) and at the right by a potentialeφ(z) = Fs z, where Fs is constant field given for example, by Gauss law, Fs = ens/εs, as we just saw.Assuming either spherical bands or ellipsoidal valleys with principal axes aligned with the plane of the interfaceand its normal, since the potential is a function of the coordinate z alone, the ‘effective mass’ (or ‘envelope’)wave equation (see Eq. (104), page 32 of the Lecture Notes) can be solved by separating variables (seeEq. (113), page 35 of the Lecture Notes): Writing ψ(x, y, z) = ψx(x)ψy(y)ψz(z), the factors ψx(x)and ψy(y) will be just right-travelling or left-travelling plane waves of the form ∼ exp(±ikxx) and


∼ exp(±ikxx), respectively. The factor ψz(z), instead, obeys the Schrodinger-like effective mass equation:

− h2

2mz

d2

dz2ψz(z) + eFzzψz(z) = E ψz(z) , (517)

where mz is the effective mass along the direction normal to the interface. The boundary conditions on ψz

are, obviously, ψz(0) = 0 and ψz(z → ∞) = 0.Equation (517) can be written as

d2

dz2ψz(z) − 2mzeFs

h2

(z − E

eFs

)ψz(z) = 0 . (518)

Let’s now change independent the variable to t = (2mzeFs/h2)1/3[z−E/(eFs)] and set u(t) = ψz(z),

so that Eq. (518) becomes:

u′′ − t u = 0 , (519)

where the prime denotes derivative with respect to t and the boundary conditions are u(t0) = 0 andu(t → ∞) = 0, having indicated with t0 the value of t corresponding to z = 0:

t0 = − E

eFs

(2mzeFs

h2

)1/3

(520)

As trivial as this equation may appear, it is actually one of the possible forms in which the Bessel equationmay be cast. Its solutions – shown in the figure together with the similar function Bi(t) we are not interestedin now – are functions called ‘Airy functions’:


u(t) = Ai(t) =1

3t1/2[I−1/3

(2

3t2/3)

+ I1/3

(2

3t2/3)]

=

(t

3π

)1/2

K1/3

(2

3t2/3),

(521)where I±1/3(x) are ‘modified Bessel functions of the first kind’ and K1/3(x) is the ‘modified Bessel functionof the second kind’, all of order 1/3. An integral representation of the Airy function (useful for numericalevaluation) is:

Ai(t) =1

2π

∫ ∞

−∞ei(tz+z3/3)

dz . (522)

This shows that our solution u(t) = Ai(t) vanishes as t → ∞, so the boundary condition at the right isalways satisfied. Not so for the other boundary conditions: Only particular values of E such that Ai(t0) = 0(where t0 is given by Eq. (520) above) are acceptable. The first few roots of Ai(t) are at t=-2.33811, -4.08795, -5.52056, -6.7867144, ..., so that only discrete values of E (let’s call them En) are acceptable. An


approximated formula for the roots of the Airy functions yields:

En =

(h2

2mz

)1/3

(eFs)2/3 |tn| ≈

(h2

2mz

)1/3 [3πeFs

2

(n +

3

4

)]2/3

, (523)

for n = 0, 1, 2, ..., having approximated the roots of the Airy function by tn ≈ −[(3π/2)(n + 0.75)]2/3.Thus, the solutions of Eq. (517) are associated to a discrete spectrum of eigenvalues En and can be writtenas:

ψz,n(z) = Ai

[(2mzeFs

h2

)1/3 (z − En

eFs

)]. (524)

These functions are shown in the figure in the next page, together with the confining potential eFsz. Notethat there are two families (called ‘ladders’) of solutions: Indeed we are illustrating the case of Si. Wehave two minima of the conduction band sitting at the center of ellipsoidal equi-energy surfaces with thelongitudinal axis along z. For this two-fold degenerate family, mz = mL ≈ 0.9m0. Their energy levels andwavefunctions are shown as solid lines. In addition, there are 4 minima such that mz = mT ≈ 0.19m0. ByEq. (523), for a given n their energy is higher, thanks to the smaller mass. This second four-fold degenerateladder of eigenlevels is called the ‘primed’ ladder (with eigenvalues denonetd by En′) and the correspondingeigenvalues and (squared) wavefunctions are shown as dashed lines in the figure.


0 5 10 15 20 250.00

0.05

0.10

0.15

DISTANCE FROM INTERFACE ( nm )

SQ

UA

RE

D W

AV

EFU

NC

TIO

NS

(ar

b. u

nits

) an

d P

OTE

NTI

AL

( V

)


0 20 40 60 80–0.1

0.00.10.20.30.40.50.60.70.80.91.01.1

z (nm)

pote

ntia

l (V

)

0 5 10 15–0.20

–0.10

–0.00

0.10

0.20

0.30

0.40

0.50

0.60

z (nm)

pote

ntia

l (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 (

1020

cm

–3)

The ‘energy levels’ are actually ‘bands’ (properly called ‘subbands’) since only the motion along the z axis isquantized. The motion of the electrons on the plane of the interface is still ‘free’, so each index n labels aband of energy En(kx, ky) = (h2/2)(k2

x/mx + k2y/my) + En.

The ‘triangular well approximation’ just considered is ‘instructive’ but not generally accurate. To properlyaccount for confinement effects we must solve the envelope Schrodinger-like equation

− h2

2mz

d2

dz2ψz(z) + eV (z)ψz(z) = E ψz(z) , (525)

where the potential eV (z) is the solution of the Poisson equation

d2V (z)

dz2=

e

εs[N

+A − n(z)] , (526)

where N+A

is the density of ionized acceptors in the p-type substrate and the electron density n(z) is given


by:

n(z) =∞∑i=0

ni(EF ) |ψz,i(z)|2 , (527)

where

ni(EF ) =mdkBT

πh2ln

[1 + exp

(FF − Ei

kBT

)], (528)

is the occupation of the subband i (see homework assignement 5). Thus, we must solve simultaneously (‘self-consistently’) Eqns. (525) and (526) (note that the electron charge density depends on the wavefunctionswhich, in turn, depend on the potential), while computing the Fermi level so that

∑i ni(EF ) yields the

desired total electron sheet density. The problem is usually solved iteratively, by starting with an appropriate‘guess’ for the potential (the semiclassical solution or even the depletion approximation), by solving theSchrodinger equation, Eq. (525), computing the Fermi level EF at the desired density for the subbandenergies Ei obtained from the solution of the Schrodinger equation, computing the charge density en(z)and solving the Poisson equation Eq. (526). The cycle is then repeated, by solving Eq. (525) with the newpotential, etc. When some convergence criterion is met (the potential or the charge density change from oneiteration to the next by no more than some preset amount), the procedure is halted. The figure above showsa typical solution (subbands, wavefunctions, potential), as well as a comparison between the classical andquantum-mechanical charge distribution.There are three major effects due to carrier confinement in inversion layers:

1. Threshold voltage shift. The ‘zero-point energy’ E0 (the energy of the ground-state subband) can benoticeably above the bottom of the conduction band at the interface. Moreover, the density of states in theCB will be reduced with respect to the bulk, 3D value. Therefore, a negative gate bias of a larger magnitudewill be required to induce a given charge density. This results in a shift of the threshold voltage, VT , whichis approximately given by E0/e.

2. Reduction of the gate capacitance. As shown in the right-most frame of the previous figure, the electroncharge density in the inversion layer peaks at the interface in the classical model, but reaches a maximumat z = 〈z〉, a few tenths of nm away in the quantum-mechanical case (up to a full nm for ns ∼ 1013


cm−3). For an insulator thickness of the order of 1 or 2 nm, this can be a significant effect. Indeed theoxide capacitance Cox = εox/tox in inversion will be become

Cox → εox

tox + (εox/εs)〈z〉<

εox

tox. (529)

This reduction of gate capacitance is indeed a major concern in VLSI technology.3. Modification of the transport properties. From the discussion above, it should be clear that electrons in

inversion layers are confined in two dimensions, since they are not free to move along the direction normalto the interface. They constitute what’s known as a ‘two dimensional electron gas’ (2DEG). Only electronsoccupying subbands at very high energy can be described once more as bulk, 3D particles, since the energeticseparation of the subbands is so small that a continuum representation is suitable. Under very strongconfinement, only a few subbands will be populated. Furthermore, the energetic separation of the subbandsgrows with confining field (see Eq. (523), so the density of states (DOS) is very different from the bulk, 3DDOS. Finally, the wavefunctions associated to the electronic states are not plane-waves any longer, so thatthe matrix elements of the various collision processes will be significantly different from their 3D counterparts.The net effect is that the transport properties (e.g., mobility and saturated velocity) are significantly affectedby the quantum confinement. Note that we can still employ the semiclassical Boltzmann Transport Equationto describe transport (under the same approximations), but now the BTE describes a 2D system with onlytwo degrees of freedom, those on the plane of the interface. We shall see in detail some of these effectswhen dealing with MOS Field-Effect Transistors.

– Tunneling effects.The final effect we must consider is electron tunneling across the insulator. This is becoming an increasinglyimportant process. Let’s see why. As illustrated in the figure, electrons can tunnel either via ‘direct tunneling’(at left, for thin insulators and/or small insulator fields) or via ‘Fowler-Nordheim tunneling’ (at right, for thickinsulators and/or strong fields). Using the usual WKB approximation, the tunneling probabilities in the two


cases are: For direct tunneling:

Pd ≈ exp

{− 4(2emox)

1/2

3hFoxφ′3/2B

[1 − (1 − toxFox/φ′B)]3/2

}∼ exp

{− 2(2moxeφ

′B)1/2

htox

}

= e−2κ tox , (530)

where Fox is the field in the insulator, mox the effective mass in the gap of the insulator, φ′B = φB − E0

is the effective barrier height, reduced by the energy E0 of the bottom subband in the inversion layer, tox isthe insulator thickness, the last step has been made assuming a thin insulator (tox << φ′

B/Fox), and we

have defined an ‘average’ decay constant κ = (2moxeφ′B)1/2/h. For FN-tunneling, instead:

PFN ≈ exp

{− 4(2emox)

1/2

3hFoxφ′3/2B

}= = e−(4/3)κ zt , (531)

where zt = φ′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 nmthick and the applied bias was of the order of 5 V, FN tunneling was the only concern. This possibly causedonly minor ‘leakage’ when the devices were turned-on very strongly, so only during the ‘on’ state. Today,instead, insulators are as thin as 2 nm (or even less). Even if the applied bias has been reduced to less than1 V, from Eq. (530) it is clear that direct tunneling is becoming increasingly important. The major causeof concern is its independence of bias: Electrons can tunnel across the ‘trapezoidal’ (or almost ‘rectangular’barrier) under any bias condition. This leakage in the ‘off-state’ causes unwanted power dissipation and itconstitutes one of the problems (if not ‘the’ problem) we must face attempting to scale devices to even smallerdimensions.


–20 0 20 40 60 80–1.0

–0.5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

eφB – E0

direct tunneling

z (nm)

pote

ntia

l (V

)

–80 –60 –40 –20 0 20 40 60 80–1.0

–0.5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

eφB – E0

Fowler–Nordheim tunneling

z (nm)

pote

ntia

l (V

)


Documents

p-njunctions · limit) instead of the full Fermi-Dirac statistics. 3. The quasi-neutral regions are inﬁnitely long. 4. Finally,there’s no electric ﬁeld in the quasi-neutral