ECE 444

Semiconductor Microelectronics Fundamentals Review

The information contained in this document is a brief sketch of the material that every student is expected to understand prior to enrolling in ECE 444. Each student is expected to review this material and complete the homework problems given at the end of the document.

The material and figures contained in the document are largely taken from Semiconductor Device Fundamentals by Robert F. Pierret (Addison-Wesley, 1996). Students are encouraged to refer to this book as they study this and other ECE 444 materials. Note that this document is intended only for the use of currently enrolled ECE 444 students and may include copyrighted materials. As such, it may not be copied, reproduced or distributed in whole or in part by any means.

1. Semiconductor Materials and Structures Elemental semiconductors. The portion of the periodic chart that is of most interest for semiconductor microelectronics is shown in the figure. The elements in Column IV comprise a family of elemental semiconductors. That is they are semi-conducting naturally as elements. This is in contrast to other materials, compound semiconductors, such as GaAs in which the individual constituent atoms Ga and As are not semiconductors as elements but the compound that is formed from these constituents is a semiconductor material. All of the elements in Column IV are semiconductors. Germanium was the material used in the first transistor and silicon is by far the most common and most important of all semiconductors.

Impurities. Trace amounts of impurities dramatically change the conductivity of a semiconductor – a phenomenon that is critical to electronic devices. So, to begin with we reduce the content of unintentional impurities to a remarkable level, typically less than 1 ppb (one impurity for every 109 Si atoms). The density of atoms in crystalline Si is approximately 5x1022 atoms/cm3 (written in units of cm-3) which means that the background level of unintentional impurities is less than 5x1013 cm-3. To this we will add controlled trace amounts of desirable (intentional) impurities called dopants at levels ranging from 1014 cm-3 to 1021 cm-3. The most desirable dopants are impurities taken from Column III (acceptors - one outer shell electron less than Column IV) and Column V (donors - one outer shell electron more than Column IV). Good choices are B, Al, and P. Crystal Structure. The spatial arrangement of atoms within a material plays an important role in its mechanical, chemical, and electronic properties. Materials can be categorized as amorphous (without structure), crystalline, or polycrystalline. Amorphous materials have no long range order and no fixed relationship between adjacent atoms. Crystalline materials are the other extreme and its atoms make up a very precisely ordered solid array with a clear and highly replicated fixed relationship between adjacent atoms. Polycrystalline materials are somewhere in between. In polycrystalline materials, there are small local regions of highly ordered crystalline material called grains but adjacent grains have no predictable relationship with each other.

tetrahedral covalent bonds

shared electrons

Silicon crystals are formed from the tetrahedral bonds that are natural for elements with four outer shell electrons and, hence, four unfilled bonds. A small number of these bonded atoms with the correct geometry can form a cubic unit cell which in turn can be replicated indefinitely to form as large a crystal as we like. The natural arrangement for silicon is called the diamond lattice. This is a cubic structure with atoms in the corners and on the faces but with additional atoms within the body of the unit cell. There are a total of eight atoms per unit cell and most of them are shared

with adjacent unit cells. The dimensions of the unit cell are given by the lattice constant, a which for silicon is 5.43 Å. The planes and directions in a crystal structure are defined by a set of vectors called the Miller index. Any arbitrary plane can be defined by its Miller index which is found by listing the intercepts in units of the lattice constant, finding their reciprocals, and multiplying by the least common

denominator. The Miller index for the example shown is (233). The directions normal to the planes are shown in square brackets. Low (Miller) index planes are important in the semiconductor industry. These are the (100), (110) and (111) planes.

2. Energy Bands and Carriers Semiconductor energy models. An isolated Si atom can be described nicely using the Bohr model for an atom. In this model, Niels Bohr theorized that the first H-atom electron was restricted to certain values of angular momentum resulting in energy quantization – limiting the electrons to certain well defined orbits. These energy states are given by

This model can be easily extended to a more complex atom like Si which is element number 14 in the periodic chart. Thus silicon has 14 electrons configured as 1s2 2s2 2p6 3s2 3p2. The model is developed by assuming that the nucleus and filled-shell electrons (n = 1, 2) are strongly bound and make up the core of the atom while the four weakly bound outer-shell valence electrons (n = 3) behave similarly to the Bohr hydrogen model. Actually, the outer shell is finely split into s- and p-levels containing up 2 and 6 allowed states respectively.

There are two different ways to visualize a silicon crystal – a bond model and a band model. The bond model is based a ball-and-stick view of the crystal showing the core of the atoms and indicating bonds with lines in a two dimensional schematic representation. This model is particularly useful for visualizing the electronic nature of defects and impurities – donors and acceptors. Defects result in incomplete bonds while donors and acceptors yield complete bonds but with a single additional or missing carrier – electrons and holes. These mobile carriers are much more weakly bound (only a few meV) and can range over a volume containing a few thousand atoms.

shared valence electronsshared valence electrons

silicon atomssilicon atoms

The energy band model is based on an understanding of how the bond relationship changes as atoms become closer together. When silicon atoms are far apart enough to be essentially isolated, there is no overlap between the energy levels which are sharp and discrete as in the Bohr model. But as the atoms are made closer together, the energy levels overlap, merge into a three dimensional shared electron probability distribution. They broaden and eventually split into two large separate bands with an energy band gap between them.

This leads to the well known energy band diagram with a conduction band at higher energy (EC) and a valence band (EV) separated by an energy gap (EG). The conduction band is largely empty (few electrons) and any electrons in that band are relatively free to move long distances in terms of the atom-atom spacing under an applied electric field resulting in conduction. The valence band is nearly full of electrons but by symmetry we can consider the absence of an electron as a virtual charged particle with positive charge – a hole. This hole can also

move under an applied field but in the opposite direction. The concept of a hole is perhaps not intuitive but is a critical concept in solid state electronics. If an electron is excited into the conduction band, forming a hole at the same time, we have created an electron-hole pair (EHP). Current in a semiconductor is the opposite motion of an EHP.

EC

EV

EG

EC

EV

EG

mostly empty (few electrons)

mostly full (few holes)

The more carriers we have in a band, the more conduction. The size of the energy gap is an indication of how easily an EHP can be formed. A wider energy band gap means that it takes more energy to create each EHP. This allows us to roughly classify different materials as conductors (EG ~0), insulators (EG > 4 eV), and semiconductors (EG ~0.5 – 2.0 eV). Remember that at room temperature, kT ~ 25.9 meV. Carrier properties. The motion of a free carrier (electron or hole) in a semiconductor is similar to the motion of a carrier in a vacuum provided that the nature of the solid is appropriately included. This is done by assuming an effective mass m* for electrons and holes and accounting for a different permittivity in the solid using a relative permittivity εr where ε = εrεo. Thus, the motion is defined by

If we can make the semiconductor as nearly pure as possible and, if the energy gap is large compared to kT, the free carriers generated from impurities are insignificant and the intrinsic concentration of EHPs generated thermally is very small. If all of the carriers are generated thermally, the densities of electrons and holes are the same and equal the intrinsic concentration, ni for silicon at 300K

−= = = 10 3in p n 10 cm

where n and p are the concentrations (cm-3) of free electrons and holes. Intrinsic semiconductors have so few EHPs that they aren’t very interesting so we make them extrinsic by adding useful intentional impurities – dopants. The dopant atoms are relatively weakly bound and behave like a fairly mobile charged particle bound to the fixed charge ionized impurity atom. This is essentially the same as Bohr’s hydrogen atom with corrections for the effective mass and permittivity. Note the inclusion of these corrections in the equation for the binding energy in silicon.

The binding energies EA and ED, of typical Column III and Column V dopants are ~ 50 meV. Since this energy is much smaller than the silicon band gap of ~ 1.1 eV, we refer to these impurities as shallow donors and acceptors. How they behave depends strongly on the temperature. For low temperatures (kT << EA or ED), there is a fixed

charge from the donor or acceptor and the carrier (electron or hole) is bound to the fixed charge. For high temperatures (kT >> EA or ED), there is a fixed charge from the donor or acceptor but the carrier (electron or hole) is no longer bound to the fixed charge and instead moves into the band becoming a mobile carrier as shown schematically in the figure. This process is called ionization. The sign of the fixed charge is opposite the charge on the mobile carrier. An important parameter for doping is the concentration (cm-3) of donors or acceptors, given by ND or NA. These are chemical quantities of fixed impurities in the silicon lattice. Upon ionization, each donor or acceptor generates a single mobile carrier. Thus, for silicon at room temperature (large kT), the concentration of free carriers is equal to the chemical concentrations of fixed donors and acceptors:

material me* mh* εr

Si 1.08 0.56 11.9Ge 0.55 0.37 16.2GaAs 0.067 0.48 13.18

material me* mh* εr

Si 1.08 0.56 11.9Ge 0.55 0.37 16.2GaAs 0.067 0.48 13.18

EC

EV

EG

EC

EV

EG

0.050 eVdonors

0.050 eVacceptors

low temperature high temperature

+

-

D DN N n+→ + and A AN N p−→ + where n and p are the concentrations of electrons and holes. Note the charge sign on the ionized donors and acceptors. Density of States. The simplified energy band structure clearly indicates that there are no states in the gap and “some” states within the bands. The density of states, different for electrons and holes, is given by

Note the square root (parabolic) dependence of density of states on energy. The relationship between the energy band diagram described earlier and the density of states is shown in this figure. The number of conduction band states per cm3 lying between E and E+dE is given by ρC(E)dE and the number of valence band states per cm3 lying between E and E+dE is given by ρV(E)dE. The density of states function tells us how many states exist at a given energy but gives us no information about how many of those states are occupied. At finite temperatures, there is always some probability that there is an electron or hole occupying the states. The function that defines the probability that the states are occupied by electrons is the Fermi function which is given by

where EF is the Fermi level. The probability that a state is occupied by a hole is 1-f(E). The Fermi level is the energy where the probability is exactly 0.5. The Fermi function is shown at T = 0 and T > 0 in the figures. The region where E is more than 3kT above Ef is called the Boltzmann tail and can be approximated by

f(E E )/kTf(E) e− −≈ .

EFE

f(E)

0

½

1

T = 0 K

EFE

f(E)

0

½

1

T = 0 K

EFE

f(E)

0

½

1

T > 0 K

EFE

f(E)

0

½

1

T > 0 K(higher T)

The carrier distributions for electrons and holes in the conduction and valence bands is found by taking the density of states for the bands and multiplying them by the probability that the states are occupied (the appropriate Fermi function).

EC

EV

E E

EF

1 - f(E)

f(E)

E

ρC(E)f(E)

ρV(E)[1-f(E)]

energy band diagram

density of states

occupancy carrier distributions

EC

EV

EC

EV

ρC(E)

E

ρV(E)

ρC(E)

E

ρV(E)

The carrier distributions will look different depending on the position of the Fermi level. The total concentration of carriers – n and p – is found by integrating the carrier distributions over all available energies (the areas under the curves in red and blue).

These functions cannot be solved analytically in closed form – usually a computer solution is required. In the degenerate case – when the Fermi level is either in the band or within 3kT of the band – the exact solution is required. If, however, the Fermi level is more than 3kT from the appropriate band edge, the Boltzmann tail can be used as an approximation for the Fermi function. This is the non-degenerate case and is the usual case for silicon electronic devices. Then we can define an effective density of states for the conduction and valence bands

and the solutions become

There is a set of alternative expressions that can be useful in some circumstances. Remember intrinsic semiconductors have very few carriers. For the intrinsic case, n = p = ni and EF = Ei. From this we can get alternative solutions

Using the two forms for the solution, we can solve to find

GE /2kTi C Vn N N e−= and 2

inp n= . Charge neutrality. At reasonable temperatures, there are two kinds of charges – fixed charges from donors (ND

+) and acceptors (NA

-) and mobile charges (n and p, electrons and holes) from ionized carriers. For complete ionization, the usual case at room temperature, ND

+ = ND and NA- = NA, so, from charge neutrality

Case 1: Intrinsic semiconductors D AN 0 and N 0≈ ≈

in p n≈ ≈ Case 2: Extrinsic semiconductors D A iN N n− >>

n-type extrinsic doping 2i

D A DD

nN N n N pN

>> ≈ ≈

p-type extrinsic doping 2i

A D AA

nN N p N nN

>> ≈ ≈

Case 3: Compensated semiconductors D AN N 0− ≈

E

EF

E

EF

E

EF

E

EF

E

EF

n

p

In a compensated semiconductor, ND and NA may be large but, if their difference is small, the material still appears much like intrinsic material. Calculations of Ei and EF. In intrinsic materials, n = p and EF = Ei.

The first term is mid-gap (0.55 eV for silicon) and the second term depends on the difference between electron and hole effective masses. In silicon, the difference is not large. For doped materials, solve these equations for EF – Ei.

n-type doping 2i D

D A D F iD i

n NN N n N p E E kTlnN n

>> ≈ ≈ − =

p-type doping 2i A

A D A i FA i

n NN N p N n E E kTlnN n

>> ≈ ≈ − =

3. Carrier Motion Drift current. If a semiconductor with mobile carriers is inserted in an electric field, carriers will drift according to their sign which results in a current density. For an arbitrary area A, the drift current is given by

drift dJ qnv= where q is the fundamental charge, n is the concentration of electrons, and vd is the drift velocity. Current density has units of A/cm2. The drift velocity as a function of electric field , is shown in the figure. In the low field region (< 103 V/cm), the relationship is linear. We can write this as

dv = µ where µ is the mobility (different for electrons and holes) and has units of cm2/V-sec. Thus we get

ndrift nJ q n= µ and p

drift pJ q p= µ The maximum mobility is at low doping levels while at higher levels scattering effects reduce the mobility. In silicon the average atomic spacing is about 2.7 Å

and the average range of a mobile electron or hole is ~ 200 Å. The average spacing between impurities for a doping level of 1015 cm-3 is ~ 1000 Å which means that mobile carriers will rarely encounter another fixed charge. However, the spacing between impurities at 1017 cm-3 is ~ 200 Å. Thus, for doping levels above 1017 cm-3, the mobile charges will encounter and be affected by nearby fixed charges. The effect is always to retard the motion and slow the carriers down – reduced mobility. Increased temperature also reduces carrier mobility. From electrostatics, we can use the relationship between drift current and electric field to find the resistivity of a semiconductor.

1J = σ =ρ

where σ is the conductivity and ρ is the resistivity and the total drift current is given by

n pdrift drift drift n pJ J J q( n p)= + = µ + µ

so

n p

1q( n p)

ρ =µ + µ

.

Recall that for n-type material, n ~ ND>>NA and for p-type material, p ~ NA>>ND.

Diffusion current. In addition to drift current, we need to account for uneven distributions of carriers. Any unbalance in distribution is driven to become balanced. The imbalance leads to a flux, F of particles (units of 1/cm2-sec) that depends on the concentration difference and is characterized by a diffusion coefficient D (units of cm2/sec). Adding the charge to make it an electrical current density gives

ndiffusion n

pdiffusion p

nJ qDxpJ qDx

∂=

∂∂

=∂

.

The total current density simply sums all the contributions – electrons and holes and drift and diffusion. Thus, n n

n drift diffusion n n

p pp drift diffusion p p

nJ J J q n qDxpJ J J q p qDx

∂= + = µ +

∂∂

= + = µ +∂

n pJ J J= + At equilibrium, you can derive from these equations, the Einstein relationships.

and pn

n p

DD kT kTq q

= =µ µ

Carrier generation. Where do carriers come from?

EC

EV

EC

EV

Impact ionization – in a large electric field. Carriers gain kinetic energy from the electric field and become hot carriers. At high enough energy, the hot carrier can collide with the lattice and give up enough energy to create another EHP. Then all three carriers continue to gain kinetic energy from the field.

EC

EV

EG

EC

EV

EG

Shallow impurities – binding energies < 3kT. These are usually “good” impurities introduced intentionally. Carriers are easily ionized into the conduction or valence band and only cooling to very low temperatures can freeze out the carriers. Deep levels – binding energies >>3kT. These are usually “bad” impurities introduced unintentionally. Carriers are not easily ionized into the conduction or valence band and cooling only a little can freeze out the carriers. EC

EV

EG

EC

EV

EGBand-to-band generation –takes a lot of thermal energy but is quite easy with incident light having energy above the band gap energy (photogeneration).

Minority carriers. If carriers are generated or injected into a semiconductor material, they are in an excited state and cannot remain that way. If the carrier concentrations at equilibrium are no, po and the carrier concentrations under excitation are n, p, the excess carrier concentrations are

o

o

n n np p p

δ = −

δ = −

and, since they are electron hole pairs (EHP, δn = δp. These carriers are subject to the statistics of a large number of particles and have a characteristic lifetime in the excited state. These lifetimes are τn and τp. Consider an n-type sample and recall that

2i

DD

nn N pN

≈ ≈

If ND = 1015 cm-3, we can calculate that p ≈ 105 cm-3. This is a very small value and is the reason we refer to these as minority carriers. At low injection levels, δn = δp<< no, which means that n ≈ no and δp>> po. This leads to the important conclusion that it is the minority carriers that matter. Once the source of excess carriers is shut off, the excess concentration decays according to

andp n

p p n nt t

∂ δ ∂ δ= − = −

∂ τ ∂ τ.

So what happens if we create a localized pulse of excess carriers? They will spread out by diffusion (Dp, Dn) and eventually recombine (τn, τp). The average distance a carrier will diffuse before it recombines is called the minority carrier diffusion length and is given by

andp p p n n nL D L D= τ = τ . 4. PN Junctions Qualitative behavior. Putting two pieces of semiconductor together of opposite type (doping) will create a pn junction. To understand how the energy bands look for a pn junction, recall that fixed charges (ionized donors and acceptors) cannot move physically while mobile carriers (electrons and holes) can move within the bands. In addition, the Fermi level must be flat if there is no applied voltage at the junction. Before the junction is formed, the bands look like

EC

EV

Ei

EF

and after junction formation the bands look like this

EC

EV

Ei

EF

qVbiqVbi

The Fermi level is flat so the other bands must bend to provide continuity. The shape of the bends depends on the doping levels on either side of the junction. A built-in voltage Vbi accounts for the step from the Fermi level

difference. In order to provide charge neutrality, mobile charges move out of the junction region leaving behind a depletion region. There are fixed charges in this region but no mobile carriers. Note that energy increases upward in the figure for electrons and downward for holes.

EC

EV

Ei

EF

p-depletion region n-depletion region In the n-depletion region, mobile majority carriers (electrons) move to the right. This “depletes” the region of mobile charges while positive fixed charges (ionized donors) remain behind. In the p-depletion region, mobile majority carriers (holes) move to the left which depletes the region of mobile charges while negative fixed charges (ionized acceptors) remain behind. Built-in potential. From electrostatics we know that the voltage profile V(x), electric field (x), and charge density ρ(x) are related.

r o

dV ddx dx

ρ= − =

ε ε

. n n ndnJ 0 q n qDdx

= = µ +

solving these together will give the built-in voltage

A Dbi 2

i

N NkTV lnq n

=

.

As a numerical example choose mid-range doping – i.e. NA = ND = 1016 cm-3. Then Vbi = 0.7 V. This is the classic value used in circuit designs for the silicon pn junction built-in voltage. Step junction. Consider the step junction doping profile shown in (a) in the figure. This doping profile yields the charge density profile shown in (b). We can integrate Poisson’s equation

r o

ddx

ρ=

ε ε

to solve for the linear electric field shown in (c). Integration of this profile gives the potential profile shown in (d). which is given by these equations

2Ap

r o

2Dbi n

r o

qNV(x) (x x) x 02

qNV(x) V (x x) x 02

= − + <ε ε

= − − >ε ε

from continuity at x = 0 and recognizing from charge neutrality that NAxp = NDxn, we can solve for xn and xp and the total depletion width w = xn + xp.

ND-NA

x

ND-NA

x

ρ

x

ρ

x

x

xV

x

V

x

ND

-NA

qND

-qNA

-xpxn

-xp xn

-xp xn

a] doping profile b] charge density

c] electric field d] potential

12

r o A Dn p bi

A D

2 N Nw x x Vq N N

ε ε += + =

In order to include an applied bias, we recognize that a forward bias works to reduce the built-in potential. This adds a simple voltage term by subtraction.

12

r o A Dn p bi

A D

2 N Nw x x (V V)q N N

ε ε += + = −

We can see the effect of applied voltages – forward and reverse biases – shown in the figure in red and blue. These changes have a pronounced effect on the energy band diagrams and introduce the concept of quasi-Fermi levels EFn and EFp which are non-equilibrium values of the Fermi level on each side of the junction. Forward bias reduces the voltage barrier and reverse bias increases the voltage barrier. Current-voltage characteristic. The current-voltage (I-V) characteristic of a real pn junction diode is shown in the figure. Note that the voltage scales are different in forward and reverse bias. In addition to rectification and the built-in voltage, there is a slope in the curves that arise from series resistance, there are leakage currents, and at large enough reverse bias there is breakdown. The primary shape of the curve is obtained from the ideal diode equation (the Shockley equation).

qV/kToI I (e 1)= −

2 2

pn i io

n A p D

DD n nI qAL N L N

= +

EC

EV

Ei

EF

zero bias

EC

EV

Ei

EFn

reverse bias

EFp

EC

EV

Ei

EFn

reverse bias

EFp

EC

EV

Ei

EFn

forward bias

EFp

EC

EV

Ei

EFn

forward bias

EFp

x

V

x

-xp xn

-xp xn

ρ

x

qND

-qNA

-xpxn

x

x

V

x

V

x

-xp xn

-xp xn

ρ

x

ρ

x

qND

-qNA

-xpxn

V > 0V < 0

Junction breakdown. Reverse breakdown takes place in different ways in a pn junction diode. One of these is avalanche breakdown. If the reverse bias is large enough there is a very large electric field in the depletion region of the diode. Carriers that enter into this region are accelerated by the electric field and can gain significant kinetic energy and undergo impact ionization described earlier. This effectively multiplies the current with a multiplication factor M, which can get large quickly resulting in a rapidly growing reverse current under reverse bias and blows up when the reverse bias equals the breakdown voltage.

mo

BR

I 1MI V

1V

= =

−

Zener breakdown is based on quantum mechanical tunneling. In this process, an energy barrier that is thin enough (~ 100 Å) will allow a significant probability that a carrier can cross the barrier. This can happen under reverse bias when the doping on both sides of the junction is heavy and the depletion region is thin. Junction capacitance. A pn junction under zero or reverse bias behaves like a parallel plate capacitor with the dielectric constant of the semiconductor.

EC

EV

Ei

EF

depletion region

EC

EV

Ei

EF

depletion region

metal plates

semiconductor dielectric One important difference is that the depletion width varies with applied bias, so the capacitance for a pn junction diode is a voltage-dependent capacitance. The capacitance of a parallel plate capacitor is given by the simple formula

r oACw

ε ε=

where

12

r o A Dbi

A D

2 N Nw (V V)q N N

ε ε += −

.

1 The surface of a Si wafer is a (100) plane.

(a) Determine the number of atoms per cm2 at the surface of the wafer. (b) Repeat, this time taking the surface of the Si Wafer to be a (111) plane.

2 A crystalline plane has intercept of 1a, 3a, and 1a on the x, y, and z axes, respectively. a is the cubic cell

side length. (a) What is the Miller index notation for the plane? (b) What is the Miller index notation for the direction normal to the plane?

3 Semiconductor A has a band gap of 1 eV, while semiconductor B has a band gap of 2 eV. What is the

ratio of the intrinsic carrier concentrations in the two materials (niAlniB) at 300 K. Assume any differences in the carrier effective masses may be neglected.

E

Lz

E

Lz

EC

EV

Ei

EFn

EFp

EC

EV

Ei

EFn

EFp

4 Determine the equilibrium electron and hole concentrations inside a uniformly doped sample of Si under

the following conditions: (a) T = 300 K, NA << ND = 1015/cm3. (b) T = 300 K, NA = 1016/cm3, ND << NA. (c) T = 300 K, NA = 9 x 1015/cm3, ND = 1016/cm3.

For each of these conditions, determine the position of Ei, compute EF – Ei, and draw a carefully dimensioned energy band diagram for the Si sample.

5 The equilibrium and steady state conditions before and after illumination of a semi-conductor are

characterized by the energy band diagrams shown in Fig. P3.24. T = 300 K, ni = 1010/cm3, µn = 1345 cm2/V-sec, µp = 458 cm2/V-sec. From the information provided, determine

(a) n0 and p0, the equilibrium carrier concentrations. (b) n and p under steady state conditions (c) ND (d) What is the resistivity of the semiconductor before and after illumination?

6 More resistivity Questions

(a) A silicon sample maintained at room temperature is uniformly doped with ND = 1016/cm3 donors. Calculate the resistivity of the sample. Compare your calculated result with the p deduced from GT-1.

(b) The silicon sample of part (a) is “compensated” by adding NA = 1016/cm3 acceptors. Calculate the resistivity of the compensated sample. Exercise caution in choosing the mobility values for this part of the problem.

(c) Compute the resistivity of intrinsic (NA = 0, ND = 0) silicon at room temperature. How does your result here compare with that for part (b)?

7 A Si step junction maintained at room temperature under equilibrium conditions has a p-side doping of

NA = 2x1015/cm3 and an n-side doping of ND = 1015/cm3. Compute (a) Vbi. (b) xp, xn, and W. (c) at x = 0. (d) V at x = 0. (e) Make sketches that are roughly to scale of the charge density, electric field, and electrostatic

potential as a function of position. (f) What is the junction capacitance at a bias of -5 V?