Fundamentals of Semiconductor Devices

Introduction
Chapter 5 Section 1
Copyright © 2016
The McGraw-Hill Companies, Inc.
Permission required for
presentation or display

Introduction: Diodes
• Diodes: two-terminal devices
• In semiconductors, made by junctions between
two different materials
– Homojunctions: junctions between two differently
doped regions of the same semiconductor material
– Heterojunctions: junctions between two different
types of materials (usually means semiconductor
materials)
– Metal-semiconductor junctions
• But not all metal-semiconductor junctions are diodes

Diodes
• Diodes can act as switches
• For one polarity of applied voltage, current flows
(nearly a short circuit)
• For other polarity, current does not flow (nearly
an open circuit)
Silicon homojunction:
turns on at around 0.7V

Circuit equivalent
• Open circuit (for V<0.7
V)
• Looks like a power
supply: 0.7 V and
supplies current

Circuit symbol
• Ideal diode:
– For V<0, no current flows
– For V>0, short circuit
• More realistic silicon
diode:
– For V>≈0.7 V, current flows
(current increases
exponentially with voltage)
– For V<0.7 V, current flow is
neglible

Rectifier circuit example
• Let input be AC signal
of 60 Hz, with 5 volt
amplitude
• Take output voltage
across the resistor
– Let R=2KΩ
• Let the diode be
represented by
equivalent circuit
shown
• Plot the input and
output waveforms

Input
voltage
Voltage:
• Remains 0 until input reaches 0.7 V
• Then Vout= Vin-0.7V until Vin returns to
0 (0.7 V dropped across the diode)
• Vout stays at zero until Vin>0.7 V
Current:
• Is zero until input reaches 0.7
V
• Then
I =
Vout
R
=
5sin 377t( )- 0.7
2 ´103

How we’ll proceed
• Operation of semiconductor diode junctions will
be explained using energy band diagrams
– Draw diagrams in electrical neutrality first, lining up
vacuum levels
– Let regions come into contact (in your imagination);
charges flow until equilibrium is reached
– At equilibrium, Ef is a constant
– Redraw with Fermi levels lined up
• Then we’ll investigate influence of applied
voltages

Prototype pn homojunctions
• Junction between n-type and p-type of same
kind of semiconductor
• Example: start with a p-type substrate
• Implant phosphorus atoms (donors) in certain
region
• Junction occurs along B-B’ cross section

Doping profile
• Background acceptor concentration is
constant (NA)
• Typical donor concentration profile shown
• We’ll approximate as step function
• Junction occurs at x0

Step junction model
• Assume doping profile is a step function
– On the n-side, ND’=ND-NA (constant)
– On the p-side, NA’=NA-ND (constant)
• Assume all impurities are ionized
– Then n0 on the n side is n0=ND’
– Then p0 on the p side is p0=NA’
• Neglect impurity-induced band gap narrowing for
this model
– If either side of junction is degenerate, take Ef to be at
band edge EC0 or EV0 if ND’ or NA’ is greater than 1018
cm-3

Key points
• A pn junction will be a diode
– Two-terminal device
– Diode conducts current in
one direction but not the
other
• Will analyze using step
junction approximation
• Will understand using
energy band diagrams

Prototype pn Junctions
(Qualitative)- Energy
Band Diagrams Part A
Chapter 5 Section 2.1
Copyright © 2016

Introduction
• We’ll assemble the energy band diagram of a
pn junction
• There will be a built-in voltage
• There will be an internal electric field
• Two parts: Part A: Equilibrium, Part B: under
bias
• This section is all qualitative- understand the
physics

Step 2: Join sides
• When the sides are
“joined,” charges flow
from one side to the other
• As electrons flow, they
leave behind (positively
charged) ionized donors
– Donors cannot move
• Holes flow, leaving behind
(negatively charged)
ionized acceptors
• Separation of charges sets
up electric field

Currents
Jn
= qmn
nE + qDn
dn
dx
= qmn
n +
kT
q
dn
dx
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
Jp
= qmp
pE - qDp
dp
dx
= qmp
pE -
kT
q
dp
dx
é
ë
ê
ù
û
ú
Drift currents Diffusion currents

The built-in field
• Near interface, electrons from
n-side fill holes on p-side
– Results in negative charges on
p-side (un-neutralized
acceptors)
• Electrons annihilate holes
that diffused into n-side
– Results in positive charges on
p-side (un-neutralized donors)
• Results in a built-in electric
field near junction

The band edges are parallel
• Quantities γ, ϕ, and χ are
constants of the material
– Therefore EC is parallel to
Evac and EV
• An electron at EC on n-side
must acquire χ energy to
escape to vacuum level on
n-side- but to escape to
vacuum level on p-side
must also acquire
additional energy qVbi

The space charge region
• On each side is a region of
uncompensated charge:
space charge region
• Width on each side
depends on impurity
concentration on each side
and amount of charge
needed to equalize the
Fermi levels

The depletion region
• In the space charge region,
there is an electric field
– Will accelerate mobile
charges out of region
– Mobile charges are
depleted here
• Space charge region also
called the “depletion
region”

The built-in voltage
• Total space charge on
either side of the junction
is equal (but with opposite
sign)
• There exists a built-in
potential energy barrier
qVbi
– Vbi is called the built-in
voltage
– Always taken as positive (by
convention)
qVbi = Fp - Fn

The junction width
• Potential energy barrier is the
same in EC as for EV
• Thus barrier is the same for
electrons as for holes
• If doping concentration is
same on both sides (NA’=ND’),
width of junction is same on
both sides
– When doping is unequal,
width is wider on the more
lightly doped side

The band bending
• As doping increases,
built-in voltage increases
– Work functions increase,
so bands have to bend
more to align Fermi levels

The electric field
• Electric field
proportional to slope of
Evac
• Is steepest at
metallurgical junction
• Built-in voltage is mostly
on lightly-doped side
E =
1
q
dEvac
dx

One-sided junction
• When one side
is degenerately
doped, most of
junction
appears on
lightly doped
side
• Called “one-
sided junction”

Key points
• Constructed the energy band diagram under
equilibrium
• There is a built-in voltage created by uncovered
donors and acceptors
– Space charge region
• There is an electric field in space charge region
that sweeps out electrons and holes, depleting
them
– Space charge region= depletion region
• Next: what happens when you apply a voltage?

Prototype pn Junctions
(Qualitative)- Energy
Band Diagrams Part B
Copyright © 2016

Introduction
• We saw how to draw the energy band diagram
at equilibrium
• What happens when we apply a bias?

Apply a voltage Va
• Applied voltage by
convention measured
from p to n
• When VA is positive,
junction is forward-
biased
• When VA is negative,
junction is reverse-
biased

Reverse bias
• Let Va =-1 V
• Consider pn junction in three
regions: n, p, and junction
• On n-side number of donors is
equal to number of electrons (no
field)- hence called quasi-neutral
region
• There is another quasi-neutral
region on p-side
• Depletion region contains ions but
virtually no free carriers
Equilibrium
diagram
repeated here

Consider diode as series connection of
three resistances of these regions
• On the n-side: ρn is inversely proportional to
nn0 (subcript n means on n-side)
• On p-side: ρp inversely proportional to pp0
• In transition region, almost no free carriers
– Resistance therefore high
– Therefore applied voltage dropped almost entirely
across transition region

To adjust energy band diagram
• Use Evac at
metallurgical
junction as
reference
• Pivot bands
around this point
• The p-side is
electrically more
negative, so
represents a
higher potential
energy for
electrons
• Thus p-side moves
up on diagram
• The n-side moves
down

The junction voltage
• Junction
voltage Vj is
increased
• (increased
because
recall Va is
negative)
• Greater
field at
junction
Vj =Vbi -Va

The junction voltage
• Greater field
means more
charge on either
side
• Charges are un-
neutralized
donors and
acceptors
• Space charge
region gets wider
(to uncover more
of them)
• Mobile carriers
are swept out by
field

Under forward voltage
• Now Va is
positive
• Space charge
region gets
smaller
• Evac not
shown-no
new
information
there

Key points
• Learned how to draw energy band diagrams
under bias
• Applied fields change internal electric fields
• Applied fields change width of depletion
region
• Next: what about the current?

Description of Current
Flow in a pn Prototype
Homojunction
Copyright © 2016

Introduction
• We previously saw how to draw the energy
band diagram of a pn junction
• Learned how to modify it for the case of
applied bias
• Now we’ll see how to use the energy band
diagram to predict the behavior of the
electrons and holes
– Get a qualitative description of the current

Diffusion and drift
• Electrons diffuse to regions of
lower concentration (n to p)
• Holes diffuse to regions of
lower concentration (p to n)
• Electrons are accelerated by
built-in field (toward downhill
side)
• Holes accelerated by built-in
field (toward uphill side)
• At equilibrium, these current
cancel out
• Net current is zero

Generation and recombination
• In the quasi-neutral regions, generation
and recombination rates are equal
– No fields. no concentration gradients,
thus no current flow
• In depletion region, EHPS are
generated– and there IS a field
– Electrons accelerated to left (generation
current to right)
– Holes accelerated to right (generation
current to right)
• Electrons in depletion region can
recombine with holes in depletion
region
– Results in regeneration current to the
left
• In any region, net G-R current is zero at
equilibrium

Reverse bias- drift and diffusion
• Electrons trying to diffuse to
p-side are turned back by
the field
– Same for holes
– Negligible current
• Small number of electrons
on p-side that wander
(diffuse) near junction are
swept over
– Similarly for holes
– Result is a net current- but
small, few carriers available
– Called minority diffusion
current

Reverse bias,
generation/recombination and
tunneling• Generation: electron
accelerated to left, hole to
right- results in net current
• Neglect recombination in
depletion region- number of
carriers miniscule
• Tunneling from VB on p-side
directly into CB on n-side
can happen
– Net tunneling current to the
right
– Increases rapidly with
reverse bias voltage

Total reverse current
• Small minority carrier diffusion current
• Small generation current
• Tunneling current can be large
• Multiplication current
– EHPs generated in depletion region can collide
– Crashing knocks more electrons and holes loose
– They also get accelerated
– Will discuss more later

Forward bias
• Electrons trying to diffuse to p-side have lowered barrier-
lots of them get across
• Minority electrons on p-side that diffuse into space charge
region get swept across- but there aren’t many of them
• Result is large diffusion current

Forward bias, recombination
• Many electrons and holes crossing through
depletion region- opportunities to recombine are
plentiful
• Result is net current to the left

More on forward bias
• When electrons are injected into p-side, their charge is neutralized
by holes almost instantly (dielectric relaxation time)
• Holes are supplied ultimately by contact at p-terminal
• No net charge distribution (very small) so no electric field
– Hence p-side is quasi-neutral
– Similarly for n-side

Excess carriers
• Excess carriers (Δn and Δp) injected across junction
• They recombine as they move, so excess minority
carrier concentration decays exponentially with
distance away from junction
• Away from junction, E ≈0, so no drift- current flow is
by diffusion

Away from junction?
• Excess carriers have recombined, so no
diffusion (no concentration gradient)
• We said no field in quasi-neutral region
– But there is a small field
– Resistance is small but not zero, so some voltage
dropped across QNR’s
– Small field but low resistance means current can
flow
– Result: in QNRs current flows by drift

Tunneling?
• Not possible
– No empty states in
one band at same
energy as occupied
state in another
band

Forward current
• Primarily diffusion across
junction
• Recall carrier
concentration with energy
decreases as energy
increases
• As energy barrier lowered,
number of carriers that
can cross increases
exponentially
• Thus current increase
exponentially with applied
voltage

Forward current summary
• When forward voltage applied, electrons and
holes are injected across depletion region
• Become minority carriers on other side
• Diffuse away from junction, recombine
• Majority carriers provided for recombination
are resupplied from contacts
• Majority carrier current powered by drift
– Field small but carrier concentrations large

Key points
• Under reverse bias, there is a small
leakage current due to minority
carriers being swept across junction
– Small generation current
– Tunneling possible at higher voltages
• Under forward bias, minority
carriers diffuse across junction
(minority carrier diffusion current)
– Minority carriers replaced by
majority carriers
– Away from junction, current
propelled by drift of majority carriers

Tunnel Diodes
Copyright © 2016

Introduction
• Tunnel diodes (also known as Esaki diodes)
have both side degenerately doped
– Ef is generally inside the conduction and valence
band edges
• They demonstrate quantum-mechanical
tunneling really exists

Energy Band Diagram
≈10 nm
E »106
V /cm

Assumptions
• Current is primarily electrons from n going to
empty states on p side
• Electrons cross the forbidden region but
remain at same energy (conservation of
energy)
• Electrons occupy states at same K before and
after tunneling

I-V characteristic
I = Ce
qV
hkT

Figure of Merit
• Peak-to-Valley ratio
• Values of IP/IV ≈ 20 have been observed in
GaAs tunnel diodes

Can be used as an oscillator
• Negative resistance can be used to make an
oscillator
– Pretty stable
• Can be used as a switch
– Valley current tends to increase over time
(degradation)

We said K was conserved
• Implies direct-gap semiconductors
• In Si (indirect), this assumption is violated
– Negative resistance is still observed
– Peak-to-valley ratio is much lower (≈4)
– Probably simultaneous emission or absorption of
phonons
• Ge is indirect but has a relative minimum at
K=0
– Peak-to-Valley ratios ≈16

Key points
• When both sides of
pn junction are
heavily doped, can
get tunneling
• Results in negative
resistance region in
I-V

Prototype pn Homojunctions
(Quantitative): Energy Band
Diagram at Equilibrium-Step
Junction
Copyright © 2016

Introduction
• We examined qualitatively how current flows in
pn junction
– Reverse bias
– Forward bias
• Considered
– Majority and minority carrier currents
– Drift and diffusion
– Generation and recombination
– Tunneling
– Touched on multiplication current
• Next: quantitative analysis

Find Vbi
• Recall
• Express as
where on the n-side (non
degenerate)
• If n-side is degenerate
• Take Ef=EC and δn=0
qVbi = Fp - Fn
qVbi
= Eg
- dn
+dp( )
dn
= EC
- Ef
= kT ln
NC
nno
= kT ln
NC
ND
'

Similary for p-type
• Non-degenerate
• Degenerate:
δp=0
dp
= Ef
- EV
= kT ln
NV
ppo
= kT ln
NV
NA
'

Combine
• To get
• For one-sided junctions (n+-p or p+-n)
qVbi
= Eg
- dn
+dp( ) dn
= EC
- Ef
= kT ln
NC
nno
= kT ln
NC
ND
'
dp
= Ef
- EV
= kT ln
NV
ppo
= kT ln
NV
NA
'
qVbi
= Eg
- kT ln
NC
ND
'
+ ln
NV
NA
'
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
p-n junction
qVbi
= Eg
- kT ln
NV
NA
'
é
ë
ê
ù
û
ú n +
-p junction
qVbi
= Eg
- kT ln
NC
ND
'
é
ë
ê
ù
û
ú p +
-n junction

For non-degenerate semiconductor
• Recall
• Thus
ni
2
= NC
NV
e
- Eg
kT
Eg
= kT ln
NC
NV
ni
2
Vbi
=
kT
q
ln
ND
'
NA
'
ni
2
pn junction
Vbi
=
kT
q
ln
NV
ND
'
ni
2
p+
-n junction
Vbi
=
kT
q
ln
NC
NA
'
ni
2
n+
-p junction
And we had
qVbi
= Eg
- kT ln
NC
ND
'
+ ln
NV
NA
'
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú

For pn junction
• Can also express Vbi as
• This says that as ND’ and NA’ approach NC and
NV, Vbi approached the band gap
• If you include impurity-induced band gap
narrowing, could be slightly less than Eg
Vbi
=
Eg
q
-
kT
q
ln
NC
NV
ND
'
NA
'
p-n junction

Example
• Find the built-in voltage for a silicon
homojunction with ND’=1016 cm-3 and NV’ =
1015 cm-3
• Solution:
– The onset of degenerate doping in Si is ≈4x1018
cm-3, so neither side is degenerately doped
– Therefore use
Vbi
=
kT
q
ln
ND
'
NA
'
ni
2
pn junction

Solution
• If you have studied diodes in circuits class, you
will recognize that this is close to 0.7 V, the
typical turn-on voltage for silicon diodes
Vbi =
kT
q
ln
NA
'
ND
'
ni
2
é
ë
ê
ê
ù
û
ú
ú
=
0.026eV i 1.6 ´10-19
J / eV
1.6 ´10-19
C
ln
1016
1015
1.08 ´1010
( )
2
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
= 0.026V ln 8.57 ´1010é
ëê
ù
ûú = 0.655V

What if it were a one-sided junction?
• Then we’d use one of these
• Could happen a lot- let’s plot these to avoid
repeating the same calculations over and over
Vbi
=
kT
q
ln
NV
ND
'
ni
2
p+
-n junction
Vbi
=
kT
q
ln
NC
NA
'
ni
2
n+
-p junction

Vbi for one-sided junctions
Curves for n+-p and p+-n are
indistinguishable

Note
• If impurity-induced band gap narrowing is
considered, reduce result from previous plot
by amount of apparent band-gap narrowing
ΔEg*

Key points
• We derived an expression to find Vbi
• The built-in voltage depends on the doping on
both sides of the junction
• Found Vbi≈0.7 V for a typical silicon diode
• Next: quantitative energy band diagram with
applied voltage

Copyright © 2016
Energy band diagram with
applied voltage: Part A

Introduction
• Here we will quantitatively determine the
features and shapes of the energy band diagram
for a pn homojunction
• We’ll start with variation of charge with position
• Use that to find the electric field
• Use the field to calculate the variation of voltage
V(x) with position
• Potential energy is qV(x)– gives us the band
shapes

Charges in a step junction
• We approximate the
charges on either
side as constant
• The n-side is more
heavily doped
• Donors (ND’) are
positively charged
• Acceptors (NA’) are
negatively charged
• No uncompensated
charges outside
depletion region

Find the electric field E
• Use Poisson’s equation (works for equilibrium
and also under bias)
• Here Qv(x) is the charge per unit volume as a
function of position; ε is permittivity
• Need to know all the charges
– Electrons, holes, ionized donors, ionized acceptors
dE
dx
=
QV
(x)
e

Very few electrons or holes in
depletion region
• Depletion approximation: n=p=0
• Next, write down QV(x)
• On the n-side
• On the p-side
• Outside depletion region, QV =0
xn
< x < xo QV
= qND
'
x0
£ x £ xp QV
= -qNA
'

Integrate to solve
• On n-side
• On p-side
dE
o
E (x)
ò = q
ND
'
e
dx
xn
x
ò
dE
dx
=
QV
(x)
e
E (x) = q
ND
'
e
x - xn( )
d
E (x)
0
ò = - q
NA
'
ex
xp
ò dx
E (x) = q
NA
'
e
xp
- x( )
xn
£ x £ x0
x0
£ x £ xp

But electric field has to be continuous
• We had
• Set the solutions equal at x=x0
qND
'
(xo
- xn
) = qNA
'
(xp
- xo
)
E (x) = q
ND
'
e
x - xn( ) E (x) = q
NA
'
e
xp
- x( )
Charge:
Field:

Space charge widths
• Earlier we predicted that the junction would
extend further into lightly doped side?
• Let wn be the width of the depletion region on
the n-side, e.g. wn=x0-xn and wp be the width
on the p-side
• From
We can write
qND
'
(xo
- xn
) = qNA
'
(xp
- xo
)
wn
wp
=
xo
- xn( )
xp
- xo( )
=
NA
'
ND
'
Checks
out!

Next find voltage distribution
• We use
• For n-side, integrate
• And get
• For p-side
E = -
dV
dx
dV
V (xn )
V (x)
ò = - E (x)dx
xn
x
ò = -
qND
'
e
(x - xn
)dx
xn
x
ò
V(x) -V(xn
) = -
qND
'
2e
x - xn( )
2
V(xp
)-V(x) = -
qNA
'
2e
xp
- x( )
2
xn
£ x £ xo
xo
£ x £ xp

Find the total voltage
• Voltage accumulated on n side:
• On p-side
• Total voltage is
V(xn
) -V(xo
) =Vj
n
=
qND
'
2e
xo
- xn( )
2
=
qND
'
2e
wn
2
V xo( )-V xp( )=Vj
p
=
qNA
'
2e
xp
- xo( )
2
=
qNA
'
2e
wp
2
Vj
=Vj
n
+Vj
p
=
q
2e
ND
'
(xo
- xn
)2
+ NA
'
(xp
- xo
)2
é
ë
ù
û

Voltage has to be continuous
• Set solutions equal at junction
• Take previous results and take ratio
• Recalling , we get
V(xo
) =V(xn
) -
qND
'
2e
xo
- xn( )
2
=V(xp
)+
qNA
'
2e
xp
- xo( )
2
V(xn
)-V(x0
)
V(x0
)-V(xp
)
=
Vj
n
Vj
p
=
ND
'
wn
2
NA
'
wp
2
Vj
n
Vj
p
=
NA
'
ND
'
wn
wp
=
NA
'
ND
'
Most of voltage dropped across more lightly doped side

Key points
• We started with the charge on either side of
the junction
• We integrated the charge to get the electric
field (Poisson’s Equation)
• We integrated the field to get the voltage
• Next episode:
– We’ll find the widths of the space charge region
– We’ll calculate the actual shape of the energy
band diagram

Copyright © 2016
Energy band diagram with
applied voltage: Part B

Introduction
• In Part A we found the charge, field and voltage
• Next, junction widths and energy band diagram

Find expressions for junction widths
• Combine:
• To find
V(xn
) -V(xo
) =Vj
n
=
qND
'
2e
xo
- xn( )
2
=
qND
'
2e
wn
2
V(xn
)-V(x0
)
V(x0
)-V(xp
)
=
Vj
n
Vj
p
=
ND
'
wn
2
NA
'
wp
2
V(xo
) =V(xn
) -
qND
'
2e
xo
- xn( )
2
=V(xp
)+
qNA
'
2e
xp
- xo( )
2
wn
wp
=
NA
'
ND
'
wn
= xo
- xn( )=
2eVj
n
qND
'
é
ë
ê
ê
ù
û
ú
ú
1
2
=
2eVj
qND
'
1+
ND
'
NA
'
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
1
2

Similarly for p, and total width
• For the p-side
• And the total junction width is
wp
= xp
- x0( )=
2eVj
p
qNA
'
é
ë
ê
ê
ù
û
ú
ú
1
2
=
2eVj
qNA
'
1+
NA
'
ND
'
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
1
2
w = wn
+ wp
=
2eVj
NA
'
+ ND
'
( )
qNA
'
ND
'
é
ë
ê
ê
ù
û
ú
ú
1
2
pn junction

We had
• Solve for junction voltage Vj
• For one-sided junction, e.g. n+-p, w≈wp since
in this case ND’>>NA’ and
w = wn
+ wp
=
2eVj
NA
'
+ ND
'
( )
qNA
'
ND
'
é
ë
ê
ê
ù
û
ú
ú
1
2
Vj
=
qND
'
NA
'
w2
2e ND
'
+ NA
'
( )
pn junction
w =
2eVj
qNA
'
é
ë
ê
ê
ù
û
ú
ú
1
2
(n+-p) w =
2eVj
qND
'
é
ë
ê
ê
ù
û
ú
ú
1
2
(p+-n)

Maximum electric field
• Combine
• To find
E (x) = q
ND
'
e
x - xn( )
w = wn
+ wp
=
2eVj
NA
'
+ ND
'
( )
qNA
'
ND
'
é
ë
ê
ê
ù
û
ú
ú
1
2
Vj
=
qND
'
NA
'
w2
2e ND
'
+ NA
'
( )
Emax
=
qND
'
e
wn
=
qNA
'
e
wp
=
2qVj
ND
'
NA
'
e ND
'
+ NA
'( )
é
ë
ê
ê
ù
û
ú
ú
1/2
=
2Vj
w

Potential energy related to electric
potential
• And EC is the potential energy for electrons in
the conduction band
• Thus, EC(x) looks like Vj(x) except inverted
• And Ev is parallel to EC
dEP
dx
= -q
dV
dx
=
dEC
dx
=
dEV
dx

So resulting energy band diagram is

Junction width for one-sided junction
as function of voltage

Junction width for one-sided junction
as function of doping

Key points
• We quantitatively derived expressions for the
charge, electric field, voltage, potential energy,
and energy band diagram for a step junction
– Charge was constant on either side (step junction)
– Integrate charge to get the field
– Integrate the field to get the voltage
– Potential energy is qV
– Potential energy is Ec (for electron in CB)
– Band gap is constant so EV is parallel to EC

Copyright © 2016
Current-Voltage
Characteristics of pn
Homojunctions (Excess
minority carrier
concentrations)
Chapter 5 Section 3.3 Part A

Introduction
• We’ll obtain I-V characteristics
– pn junctions first
– One-sided junctions later
• Assumptions:
– Minority carrier concentration is much less than
majority carrier concentration
– In bulk, majority carrier concentration ≈ equilibrium
value (space charge neutrality Δn=Δp and low
injection condition)
np
pp
@ NA
'
pn
nn
@ ND
'
nn @ nno = ND
'
pp @ ppo = NA
'

More assumptions
• For minority carriers, can
neglect drift in quasi-neutral
regions
• Semiconductor is non-
degenerate (can use
Boltzmann statistics)
• Current is defined as positive
if Va is positive
– For positive current, electrons
flow from n to p and holes
flow from p to n
Jnp
= qDn
dnp
dx
Jpn
= -qDp
dpn
dx
no
= NC
e
-
EC -Ef
kT
æ
è
ç
ö
ø
÷
po
= NV
e
-
Ef -EV
kT
æ
è
ç
ö
ø
÷

Consider long-base diode
• Both quasi-neutral regions are much longer
than minority carrier diffusion lengths Ln or Lp
– Minority carriers are recombining in QNR’s, not
reaching contacts
• We’ll do short-base diode later on
• Most important current mechanism is
minority carrier injection or extraction current
at junction

Diffusion current
• Begin with continuity equations
• Quantities Δn and Δp are excess carrier
concentrations
– Thus n=n0+Δn and p=p0+Δp
¶n
¶t
=
¶Dn
¶t
=
1
q
¶Jn
¶x
æ
èç
ö
ø÷ + Gop
-
Dn
tn
æ
è
ç
ö
ø
÷
¶p
¶x
=
¶Dp
¶x
= -
1
q
¶Jp
¶x
æ
è
ç
ö
ø
÷ + Gop
-
Dp
t p
æ
è
ç
ö
ø
÷

Electrons on p-side of junction
• Electrons are minority carriers
• From assumption that we can neglect their
drift in QNR
• Combine with
• To get
Jn
= Jndiff
= qDn
dn
dx
¶n
¶t
=
¶Dn
¶t
=
1
q
¶Jn
¶x
æ
èç
ö
ø÷ + Gop
-
Dn
tn
æ
è
ç
ö
ø
÷
¶n
¶t
=
¶Dn
¶t
= Dn
¶2
n
¶x2
-
Dn
tn

In steady state
• We had
• But in steady state
• Thus
• Where minority carrier diffusion length Ln is
¶n
¶t
=
¶Dn
¶t
= Dn
¶2
n
¶x2
-
Dn
tn
¶n/ ¶t = 0
¶2
n
¶x2
=
d2
n
dx2
=
Dn
Dn
tn
=
Dn
Ln
2
Ln
= Dn
tn

But np=np0+Δnp
• We had
• And np0 is constant, so
• Solution is
¶2
np
¶x2
=
d2
Dnp
dx2
=
Dnp
Ln
2
¶n
¶t
=
¶Dn
¶t
= Dn
¶2
n
¶x2
-
Dn
tn
Dnp
= Ae
x
Ln
+ Be
-x
Ln

Solution was
• Boundary conditions: There are no excess
minority carriers away from junction (they
have all recombined, so Δnp=0 at x=∞ gives
A=0
• At x=xp, B.C. is
giving
Dnp
= Ae
x
Ln
+ Be
-x
Ln
Dn(xp
) = Be
- x-xp( )
Ln
B = Dn(xp
)e
xp
Ln
And therefore Dn(x) = Dn(xp
)e
-(x-xp )
Ln

Substituting
• Substitute into
• And obtain the electron current
• So electron (diffusion) current is decreasing as you
move away form the junction, but total current has to
be constant (Kirchhoff current law)
– Difference is made p by increasing hole current
– Hole current is drift caused by the small electric field in the
quasi-neutral region
Dn(x) = Dn(xp
)e
-(x-xp )
Ln
Jn
= qDn
dn
dx
= qDn
dDn
dx
Jn
=
qDn
Ln
Dn(xp
)e
-(x-xp )
Ln

Key points
• That was a general derivation of the excess
minority carrier concentration as a function of
position- when there are excess carriers
present (non-equilibrium)
– Excess carriers injected across junction decay
exponentially away from junction

Next up
• It would be useful to obtain expressions for
the minority carrier concentrations at either
end of the transition region
• We’ll do this in Part B of this section

Copyright © 2016
Current-Voltage
Homojunctions (Diffusion
current, forward bias)
Chapter 5 Section 3.3 Part B

Introduction
• We had derived expressions for the excess
minority carrier concentration Δn(xp) and the
functional form away from the junction
(exponential decay)
• How does Δn(xp) vary with applied voltage?
– Start with equilibrium case

Energy band diagram at equilibrium
• On the n side:
• On the p side
• Barrier for both electrons and hole is qVbi
nno = ND
' pno =
ni
2
ND
'
ppo = NA
' npo =
ni
2
NA
'

Let’s relate Vbi to doping
• Neutral p-region
• Multiply by to obtain
• But barrier qVbi=(Ecp-Ecn),
so
npo
= NC
e
-
ECp -Ef
kT
æ
è
ç
ö
ø
÷
e
-
ECn
kT e
+
ECn
kT
npo
= NC
e
-
ECn-Ef
kT
æ
è
ç
ö
ø
÷
e
-
ECp -ECn
kT
æ
è
ç
ö
ø
÷
npo
= NC
e
-
ECn-Ef
kT
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
e
-
qVbi
kT
æ
è
ç
ö
ø
÷

Result from last slide
• But part in square brackets is nn0, so
• And similarly the minority carrier density on p-side is
• Minority carrier concentrations at edges of depletion
region are functions of the doping on the other side
• Don’t forget Vbi contains information about doping on both
sides
npo
= NC
e
-
ECn-Ef
kT
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
e
-
qVbi
kT
æ
è
ç
ö
ø
÷
np
(xp
) = npo
= ND
'
e
-
qVbi
kT
= nn0
e
-
qVbi
kT
pn
(xn
) = pno
= NA
'
e
-
qVbi
kT
= pp0
e
-
qVbi
kT

Now consider forward bias
• Barrier has now changed to ECp-Ecn=q(Vbi-Va)
• More electrons have enough energy to diffuse to p-side (injection of minority carriers)
• More holes have enough energy to diffuse to n-side (injection of minority carriers)
• Result is net current flow

Now consider forward bias
• At edges of depletion region
np
(xp
) = ND
'
e
_
q Vbi -Va( )
kT
pn
(xn
) = NA
'
e
-
q Vbi -Va( )
kT
or
np
(xp
) = npo
e
qVa
kT
pn
(xn
) = pno
e
qVa
kT

Under forward bias
• Excess electrons are
injected into p side
– They become minority
carriers
– They diffuse to the right
– They recombine as they go
– The excess carrier
concentration decays
exponentially with distance
• Holes injected to the n-
side, diffuse to the left,
same thing

Excess carrier concentrations
• Let Δnp be excess
carrier concentration on
p-side
• Let Δpn be excess carrier
concentration on p-side
• Then from
• We can write
• And similarly
Dnp
(xp
) = np
(xp
)- npo
Dpn
(xn
) = pn
(xn
) - pno
np
(xp
) = npo
e
qVa
kT
Dnp
(xp
) = npo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
Dpn
(xn
) = pno
e
qVa
kT
-1
æ
è
ç
ö
ø
÷

The excess carriers diffuse
• They diffuse to regions
of lower concentration
(away from junction)
• We already solved for
the distribution
• And on n-side, for holes
• Note that varying
distribution means
diffusion currents
Dnp
(x) = Dnp
(xp
)e
-
(x-xp )
Ln
Dpn
(x) = Dpn
(xn
)e
-
(xn-x)
Lp

Diffusion currents
• On p-side:
• Now combine with
and
To obtain
Similarly
Jn
= qDn
dn
dx
= qDn
dDn
dx
Dnp
(x) = Dnp
(xp
)e
-
(x-xp )
Ln
Dnp
(xp
) = npo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
Jp
(xn
) = q
Dp
Lp
pno
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
Jn
(xp
) = q
Dn
Ln
npo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷

Electron injection
• Recall we’re neglecting generation,
recombination
• Thus all excess electrons on p-side had to
come from n-side
– They had to cross xn as well as xp
• Recall we also assumed negligible drift in the
junction
• Therefore, all minority current across junction
is due to diffusion

Similarly for holes
• Thus total current across junction is sum of
minority carrier diffusion currents

Total current picture (fwd. bias)
• Outside the junction, total current is still constant
– Minority carrier diffusion current decreasing
– Thus majority carrier current must increase
– Majority carrier current is by drift
– Electric field is small, but there are lots of majority
carriers

Total current density
• We know
• And we had
• Thus
• Or where
J = Jn(xp)+ Jp(xn)
Jp
(xn
) = q
Dp
Lp
pno
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
Jn
(xp
) = q
Dn
Ln
npo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
J = q
Dn
npo
Ln
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷ e
qVa
kT
-1
æ
è
ç
ö
ø
÷
J = Jo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷ Jo
= q
Dn
npo
Ln
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷

Repeating previous results
• These yield
• Point: J0 is proportional to ni
2
Jo
= q
Dn
npo
Ln
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷
Ln
= Dn
tn
, Lp
= Dp
t p
, np0
=
ni
2
NA
'
and pn0
=
ni
2
ND
'
J0
= qni
2 Dn
tn
×
1
NA
'
+
Dp
t p
×
1
ND
'
æ
è
ç
ç
ö
ø
÷
÷

Diode equation
• We derived
• This is diode current equation used in circuits
courses
• Current flow is due to lowering of potential
barriers
• Current flow is due to diffusion of minority
carriers(!)
J = Jo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷

Ratio of electron to hole current
• Ratio depends on ratio of doping on n-side
(ND’) to doping on p-side (NA’)
• This result important to bipolar junction
transistors
Jn
(xp
)
Jp
(xn
)
=
Dn
npo
Ln
i
Lp
Dp
pno
=
Dn
Dp
i
Lp
Ln
i
ND
'
NA
'

Diode characteristic, forward bias
• We had (current density J=I/A)
• Thus
• When Va>> kT/q,
• For current to change by factor of 10:
• Then ln(I)=ln(10)=2.3, so Va varies by factor of 2.3
J = Jo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
I = Io
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
I » I0
e
qVa
kT
Va
=
kT
q
ln
I
I0
=
kT
q
ln I - ln I0( )

Key points
• We found expressions for excess carrier
concentrations on either side of junction
under forward bias
• We derived the diode equation
• Next: look at reverse bias
I = Io
e
qVa
kT
-1
æ
è
ç
ö
ø
÷

Copyright © 2016
Current-Voltage
Homojunctions (Reverse
Bias)
Chapter 5 Section 3.3 Part C

Introduction
• We had derived expressions for current
through a diode for forward bias
• We found
• Now let’s look at reverse bias
J = Jo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷ Jo
= q
Dn
npo
Ln
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷

Energy band diagram
• Minority carriers
diffuse
• If they wander
to close to the
depletion
region, they feel
electric field
• Minority carriers
get swept over
• Extraction

At edge of depletion region, minority
carrier concentration ≈ 0
• Excess carrier
concentration:
• p-side:
• n-side:
• Now net diffusion is
toward junction
Dnp
(xp
) = np
(xp
)- np0
= 0- np0
= -np0
Dpn
(xn
) = pn
(xn
)- pn0
= -pn0

These equations still valid
• We derived (forward bias)
• For Va<-3kT, exponential term 0 and
• Quantity J0 called “reverse leakage current”
J = Jo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷ Jo
= q
Dn
npo
Ln
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷
J » -J0

But what if it’s a short-base diode?
• What if one side is
shorter than a diffusion
length?
– Happens in “base” of
transistor, hence the
name
• Forward bias: carrier
injection process is the
same
• But, on short side,
carriers don’t have a
chance to recombine
“naturally”
• Their excess
concentrations forced
to zero by presence of
contact

Short base diode, continued
• Diffusion current is still
• But now concentration gradient nearly linear,
or
• So, from (we had before)
• We now have
Jn
= qDn
dn
dx
= qDn
dDn
dx
dDn
dx
=
Dnp
(xp
)
WB
Jn
(xp
) = q
Dn
Ln
npo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
Jn
(xp
) = q
Dn
WB
npo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
Jo
= q
Dn
npo
WB
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷

Short-base diode result
• We had (last slide) (short
side)
• Total current is (long (n) side is normal, p-side
short:)
• If both sides are short,
Jn
(xp
) = q
Dn
WB
npo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
Jo
= q
Dn
npo
WB
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷
Jo
= q
Dn
npo
WB( p)
+
Dp
pno
W(n)
æ
è
ç
ö
ø
÷

Current ratios
• If both sides short, electron-to-hole current
ratio is
Jn
Jp
=
Dn
Dp
i
W(n)
WB( p)
i
ND
'
NA
'

Key points
• Under reverse bias, carriers are
extracted
• Resulting current is very small
• We also looked at short-base
diodes
– One side or both so short that
carriers don’t have time to recombine
(forward bias) or be generated
(reverse bias) between depletion
region and contact
All results so far are for non-degenerate materials!

Next up
• We still need to consider effects of
regeneration and recombination: Parts D&E
• There are also more considerations about
reverse bias, like tunneling and carrier
multiplication: Part F

Copyright © 2016
Current-Voltage
Homojunctions (Generation
and recombination currents,
reverse bias)
Chapter 5 Section 3.3 Part D

Introduction
• We had derived expressions for current
through a diode
• We considered diffusion of minority carriers
across junction
– Forward bias: injection
– Reverse bias: extraction
• Now we consider effects of generation and
recombination

Recall G-R
• Primarily happens via interband (trap) states
ET near Ei
• Suppose there are trap states in our diode
• For simplicity, assume lifetimes
• Then net recombination rate is
• At equilibrium, np=ni
2 and R=G=0 as expected✔
tn
= t p
= to
R - G =
np - ni
2
to
(n+ ni
+ p + ni
)

Net recombination rate
• Depends on numbers of available electrons and
holes
• At position x inside the transition region
• Thus n decreases rapidly as you approach the p-
side, and p decreases rapidly as you approach the
n-side
n(x) = nnoe
-[EC (x)-EC (xn)]
kT
p(x) = ppoe
[EV (x)-EV (xp )]
kT

Energy band diagram under reverse
bias
• Recall transition
region widens
(compared to
equilibrium)
• Means more
opportunities for G-R
in depletion region
• But, not many
carriers there
because electric field
sweeps them out

The depletion approximation
• Depletion approximation:
assume n and p can be
neglected for xn<x<xp
• If no carriers, no
recombination, so neglect
that (reverse bias only)
• Implies generation rate
constant in depletion region
R = 0, G =
ni
2to

Current density
• Generation current density is
• And depletion region width w for step
junction is
• Thus (reverse bias)
JG
= -qGw = -
qni
w
2to
w =
2e ND
'
+ NA
'
( ) Vbi
-Va( )
qND
'
NA
'
é
ë
ê
ê
ù
û
ú
ú
1
2
JG
= -
ni
to
qe ND
'
+ NA
'
( ) Vbi
-Va( )
2ND
'
NA
'
é
ë
ê
ê
ù
û
ú
ú
1
2

Example: Estimate the value of generation current
relative to the diffusion current for a typical Si pn
junction under reverse bias conditions.
• Solution: assume a protype junction with
• From graph find D’s
NA
'
= 1017
cm-3
ND
'
= 1017
cm-3
tn
@ t p
= to
@ 6 ´10-6
s
Vj
= Vbi
-Va( )= 5V
Dn
= 20 cm2
/s
Dp
= 11 cm2
/s

Example, continued
• Also find L’s from graph
• Minority carrier concentrations are
np0=pn0=ni
2/1017=1.16 x103 cm-3
• Built-in voltage is
Lp
Ln
Lp
= 102 mm
Ln
= 73mm
Vbi
=
kT
q
ln
ND
'
NA
'
ni
2
= 0.026V ln
1017
´1017
1.08´1010
( )
2
æ
è
ç
ç
ö
ø
÷
÷
= 0.83V

We said we’d assume Vj=5V
• Vj=5V=Vbi-Va=0.83-Va or Va=-4.17V
• Next, find diffusion current- find leakage
current first
Jo
= q
Dn
npo
Ln
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷
= 1.6 ´10-19
C
20
cm2
s
æ
èç
ö
ø÷ 1.16 ´103
cm-3
( )
73´10-4
cm
+
11
cm2
s
æ
èç
ö
ø÷ 1.16 ´103
cm-3
( )
102 ´10-4
cm
é
ë
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
= 7.1´10-13
A/ cm2

Plug result into
• For generation current, use
J = Jo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
Jdiff
= J0
e
qVa
kT
-1
æ
è
ç
ö
ø
÷ = 7.1´10-13
A/ cm2
e
1.6´10-19
C´(-4.17V )
0.026×1.6´10-19
J /eV
-1
æ
è
ç
ö
ø
÷ @ -J0
= -7.1´10-13
A/ cm2
= -7.1´10-21
A/ mm2
JG
= -
qni
w
2to
= -
1.6 ´10-19
C 1.08 ´1010
cm-3
( ) 0.36 ´10-4
cm( )
2 6´10=6
s( )
= -5.2 ´10-9
A/ cm2
= 5.2 ´10-17
A/ mm2

Ratio
• We find Jdiff<<JG so under reverse bias, we can
safely neglect diffusion current
JG
Jdiff
@
5.2 ´10-17
7.1´10-21
= 7.3´103

Key points
• Under reverse bias, there are essentially no
free carriers in depletion region (they are
swept out by the field)
– Therefore negligible recombination
– There is generation current in depletion region
• Diffusion current negligible under reverse bias
• Next: what happens under forward bias?

Copyright © 2016
Current-Voltage
Homojunctions (Generation
and recombination currents,
forward bias)
Chapter 5 Section 3.3 Part E

Introduction
• We are examining generation and
recombination currents in pn homojunctions
• Under reverse bias, found that diffusion
current was negliglible
• Negligible recombination current
• There was some generation current (small)
• Here we will consider forward bias

GR current under forward bias
• Here, electrons in transition region are in thermal
equilibrium with n-side
• Holes in transition region in thermal equilibrium with p-side
• Can show
n(x) = nn0
e
-
EC (x)-ECn( )
kT
and p(x) = pp0
e
-
EVp -EV (x)( )
kT

Then the np product
• We had
• And we recall
• Can show that
n(x) = nn0
e
-
EC (x)-ECn( )
kT
and p(x) = pp0
e
-
EVp -EV (x)( )
kT
np = ni
2
e
qVa
kT
Vbi
=
kT
q
ln
ND
'
NA
'
ni
2

Then the np product
• We had
• And we recall
n(x) = nn0
e
-
EC (x)-ECn( )
kT
and p(x) = pp0
e
-
EVp -EV (x)( )
kT
np = ni
2
e
qVa
kT
Vbi
=
kT
q
ln
ND
'
NA
'
ni
2

For reasonable forward bias…
• We had
• When Va≥3kT/q, recombination current is
large compared to generation current
– Lots of electrons crossing the transition regions
– Lots of holes crossing the other way
– Same place, same time, recombination likely
np = ni
2
e
qVa
kT

Forward bias
• Since under forward bias np>>ni
2 , from
• We can write
• And max recombination rate is where dR/dn=0
– More convenient: d(1/R)/dn=0
R - G =
np - ni
2
to
(n+ ni
+ p + ni
)
R =
np
to
n+ p( )
=
ni
2
e
qVa
kT
to
n+ p( )

Where is recombination rate greatest?
• We had
• Set slope=0, use
• Result is where or
R =
ni
2
e
qVa
kT
to
n + p( )
p =
ni
2
n
e
qVa
kT
n = p = ni
e
qVa
2kT
Rmax
=
ni
e
qVa
2kT
2to

More on recombination in junction
• Most of the recombination occurs where
R≈Rmax
– Therefore “barrier” to recombination current
looks like half the potential barrier for diffusion
– G-R current often approximated by
where JGR0 is slightly voltage
dependent
Vbi
-Va
2
æ
èç
ö
ø÷
JGR
= JGR0
e
qVa
2kT
-1
æ
è
ç
ö
ø
÷

Leakage currents
• Compare JGR0 to J0(diff) (and recall that under
reverse bias J=-J0
• From reverse bias discussions we had
• So we see that while
JGR0
= -
ni
to
qe ND
'
+ NA
'
( ) Vbi
-Va( )
2ND
'
NA
'
é
ë
ê
ê
ù
û
ú
ú
1
2
J0(diff )
= qni
2 Dn
tn
×
1
NA
'
+
Dp
t p
×
1
ND
'
æ
è
ç
ç
ö
ø
÷
÷
JGR0 µ ni J0(diff ) µ ni
2

Total current density
• And usually JGR0>>J0
• For forward bias (Va>3kT/q),
• At small Va, recombination current
predominates because JGR0>>J0 but at larger Va
diffusion dominates
J = JGR
+ Jdiff
= JGR0
e
qVa
2kT
-1
æ
è
ç
ö
ø
÷ + J0
e
qVa
kT
-1
æ
è
ç
ö
ø
÷
J = JGRo
e
qVa
2kT
+ Jo
e
qVa
kT

Compare
• Over some range of
current, generally
approximated as
• Where n is the
“quality factor” and
usually varies
between 1 and 2.
J = Js
(e
qVa
nkT
-1)

Key points
• Under forward bias, recombination current
occurs in junction due to high numbers of
carriers crossing the depletion region
– Dominates under small forward bias
• Diffusion current also significant
– Dominates under “normal” forward bias

Next up
• Tunneling under reverse bias
• Current amplification under reverse bias
(avalanche)

Copyright © 2016
Current-Voltage
Homojunctions (Tunneling and
current amplification, reverse
bias)
Chapter 5 Section 3.3 Part F

Introduction
• We have considered diffusion and drift
• We have examined generation and
recombination
• Two more effects: tunneling and current
amplification under reverse bias

Reverse bias tunneling
• Remember
electrons tunnel
to empty states at
the same energy
• If forbidden region
is not too thick
• Here “barrier” is
the forbidden gap

Tunneling
• There is a finite probability that electrons will
cross the forbidden gap
• Also called “Zener tunneling”
• Tunneling probability for electron at energy E
normally incident on forbidden region is
T = e
-2 a dxò
a =
2m*
2
EP
*
(x)- E( )é
ë
ê
ù
û
ú
1
2
M* is some average of effective masses for
both bands
EP* is effective potential energy

Effective potential energy
• Recall potential energy for electron in
conduction band is EC
• Potential energy for hole in valence band is EV
• Both are varying with position in this case

While electron is tunneling..
• While electron is in forbidden gap, it’s
affected by both potential energies
• Effective potential energy analogous
to resistances in parallel
EP
*
(x)- E( )=
EC
(x) - E( ) E - EV
(x)( )
EC
(x)- E( )+ E - EV
(x)( )
=
EC
(x)- E( ) E - EV
(x)( )
Eg

Can show
• Tunneling probability is
• Where WT is the tunneling distance
• Tunneling probability depends on band gap
and reverse bias voltage, and doping and
effective mass
T = e
-
pWT m*
Eg
2
3
2
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
WT
=
Eg
qE

Current-voltage characteristics
• We’ll discuss
avalanche
multiplication,
a different
effect, next

Carrier multiplication
1. EHP generated
thermally
2. Electron accelerated
to left (high field
under reverse bias)
3. Electron collides
1. Loses energy
2. Knocks another
electron loose,
generating a another
EHP (impact
ionization)
3. Now there are three
carriers

Carrier multiplication, continued
4. Both electrons
accelerated to
left, new
collision
• Two electrons
arrived at left side
of junction while
only one entered-
multiplication of 2Note: for multiplication, electron must gain enough energy to exceed the band
gap to generate a new EHP. This doesn’t happen at steps 4 and 5 here so there is
no further electron multiplication.

Carrier multiplication, holes
• Holes can also
create impact
ionization (5 here)
• In this case total
multiplication is
three (if no more
impacts)
Above some critical field, process can avalanche

Derivation of multiplication
• Let P be probability of either electron or hole
creating EHP
• Let nin be number of electrons entering
depletion region from p-side
• There will be Pnin ionizing collisions
– Result is nin(1+P) electrons reaching n-side
– Also generates Pnin holes
– Result is P(Pnin)=P2nin pairs
– Total number of carriers crossing junction is
nin
(1+ P+ P2
+ P3
+ )

Multiplication, continued
• We had
• Which can be expressed as
• Thus multiplication factor M is
• If P=1, M=∞, and avalanche occurs
nin
(1+ P+ P2
+ P3
+ )
nin
(1- P)
M =
1
(1- P)

Breakdown
• When current
exceeds some
value, diode is in
“breakdown”
– Means the I-V
curve has broken
downward on
the graph,
– Device is not
harmed

Reverse breakdown in silicon
• For Si, if Vbr is greater than
about 8 V, breakdown
mechanism is primarily
avalanche
• If Vbr is less than around 6V,
is Zener tunneling
• When you buy a “Zener
diode,” not necessarily
breaking down via tunneling
mechanism- in fact, probably
not

Key points
• Under reverse bias, diodes can break down
– Not destructive, means the curve breaks down
• Two mechanisms
– Tunneling
– Carrier multiplication
• If multiplication large, can have avalanche

Copyright © 2016
Reverse Bias Breakdown

Introduction
• We saw that a diode doesn’t conduct much
for reverse voltages- up to a point
• At some voltage, current starts to flow and the
I-V characteristics turns downward
– Called reverse breakdown
– Does not mean the device is damaged
• Breakdown mechanism can be via:
– Tunneling (Zener breakdown)
– Avalanche

Explanation: avalanche
• For avalanche to occur, need a big enough field that electrons gain Eg or more to
create ionizing impacts
• Takes a larger field (reverse voltage) to get there
• With increasing doping, junction width gets smaller so field increases for a given
voltage

Avalanche breakdown
• For one-sided junctions
Vbr avalanche( )= CN
-
3
4
Semiconductor Bandgap (eV) C
Ge 0.67 2.4 x 1013
Si 1.12 5.3 x 1013
GaAs 1.43 7.0 x 1013
4H-SiC 3.26 3.0 x 1015
GaN 3.44 6.1 x 1015

Explanation: tunneling
• Tunneling probability increases with decreasing width
• Gap narrows with decreasing band gap, so probability increases and
tunneling happens at lower voltages
• Increased doping means increased field; higher slope decreases WT

Example: Estimate the tunneling distance for
appreciable tunnel current. Consider a p+-n Si junction
with ND’=8.0 x 1017 cm-3 (8.0 x 1023 m-3).
• Tunneling begins at breakdown, so find the
junction voltage at breakdown.
• One-sided junction, so use plot to find Vbi
Vbi=1.04 V

Find
breakdown
voltage
• Vbr =4 V
(Va=4V)
4V

Find maximum field and junction
width
• Junction width: ( we used Vj=Vbi-Va=1.04-(-
4)=5.04V)
• Maximum electric field:
w =
2eVj
qND
'
é
ë
ê
ê
ù
û
ú
ú
1
2
=
2 ´11.8 ´ 8.85´10-12
F / m( )´ 5.04V( )
1.6 ´10-19
C( )´ 8´1023
m-3
( )
é
ë
ê
ê
ù
û
ú
ú
1
2
= 9.1´10-8
m = 91nm
Emax
=
2Vj
w
=
2´5.04V
9.1´10-8
m
=1.1´108
V / m

Now to find the tunneling width
WT
=
Eg
(eV )
qEmax
=
1.12eV
1.6 ´10-19
C( ) 1.1´108
V / m( ) 1eV
1.6 ´10-19
V
æ
èç
ö
ø÷
= 1.0 ´10-8
m = 10nm.
Thus, for silicon, an appreciable
tunneling current flows for WT<10
nm

Key points
• Breakdown voltage increases
with increasing band gap
– Important for power devices
that must withstand large
voltages- wide band gap
materials advantageous
• Breakdown voltage
decreases with increasing
doping
– Provides a way to “engineer”
reverse breakdown devices

Introduction
• We have looked at I-V characteristics in detail
• Now look at small-signal ac response
• Bias at some DC voltage
• Vary voltage by small amount around that
point

Small signal equivalent
• Series resistance RS (constant with voltage,
contact resistance plus resistance of QNR)
• Differential resistance RP varies with voltage
• Two capacitances:
– Junction capacitance Cj
– Stored charge capacitance Csc
– Both associated with the junction

Junction conductance
• Small signal conductance
GP:
• Is slope of I-V curve at a
given voltage
• For small variation of input
voltage, can determine
output current
– Small so slope is a constant
in the vicinity
GP
=
dI
dVa

Junction resistance
• Resistance is reciprocal of conductance, or
RP
=
1
GP
=
dI
dVa
é
ë
ê
ù
û
ú
-1

Example: Find the junction resistance of a diode at
forward currents of 1 mA and 1 μA. Assume the ideality
factor is unity and RS = 0.
• We had
• And we know for reasonable forward bias
• “Reasonable” means
• Then
RP
=
dI
dVa
é
ë
ê
ù
û
ú
-1
I = I0
e
qVa
kT
-1
æ
è
ç
ö
ø
÷ = I0
e
qVa
kT
- I0
@ I0
e
qVa
kT
e
qVa
kT
>>1
dI
dVa
=
q
kT
I0
e
qVa
kT
=
qI
kT
=
1
RP

Example, continued
• We had
• Thus
• Since kT/q=0.026V:
– At I=1 mA, RP=26 Ω
– At I=1 μA, RP=26 KΩ
dI
dVa
=
q
kT
I0
e
qVa
kT
=
qI
kT
RP
=
kT
qI
.

Key points
• Small signal resistance is reciprocal of the local
slope at the bias point
• There is also a series resistance due to
contacts and quasi-neutral regions
• Next up: junction capacitance

Introduction
• We were looking at the small-signal model of
a prototype homojunction
• We looked at the junction resistance (and
mentioned the series resistance)
• Two capacitances to take into account
– Junction capacitance
– Stored charge capacitance

Recall there are charges on either side
of the transition region
• Charges are due to ionized impurities
• Number of charges depends on depletion
width
– Varies with voltage
– Change in charge on either side of a dielectric
region with voltage is capacitance

As voltage changes…
• Suppose applied voltage changes by dVA
• Charge on one side changes by dQ
• Charge on other side changes by –dQ
• As applied voltage changes, mobile charges move
– They move out of junction if reverse bias
– They move into junction if forward bias
• Electrons and holes moving=current
• Must be an equal displacement current flowing
across the junction

• Also called small-signal junction capacitance
• Already some charge on either side of junction at
equilibrium
• When voltage changes, charge is added or
subtracted in sheets at edges of depletion region
• Looks like parallel plates
Differential junction capacitance
Cj
º
dQ
dVa
Cj
=
e A
w
A=junction area
ε=permittivity of material
w=junction width

But, w depends on √(voltage)
• Thus Cj is not linear with voltage
• Expression for w (pn junction):
• So (pn junction)
w = wn
+ wp
=
2eVj
NA
'
+ ND
'
( )
qNA
'
ND
'
é
ë
ê
ê
ù
û
ú
ú
1
2
Cj
=
e A
w
= A
qeND
'
NA
'
2 ND
'
+ NA
'
( ) Vbi
-Va( )
é
ë
ê
ê
ù
û
ú
ú
1
2
= A
qeND
'
NA
'
2 ND
'
+ NA
'
( )Vj
é
ë
ê
ê
ù
û
ú
ú
1
2

Results
• For pn junction (repeated)
• For one-sided step junction
Cj
= A
qeN'
2 Vbi
-Va( )
é
ë
ê
ê
ù
û
ú
ú
1
2
= A
qeN'
2Vj
é
ë
ê
ê
ù
û
ú
ú
1
2
Cj
= A
qeND
'
NA
'
2 ND
'
+ NA
'
( )Vj
é
ë
ê
ê
ù
û
ú
ú
1
2
N’ is the net doping concentration on the lightly doped side

Junction capacitance per unit area for
one-sided junction
• Reverse bias shown

What if Va approaches Vbi?
• Cj appears to go infinite
– But as Va increases, current increases
– Have IRs drop across series resistance
– Junction voltage is
– So in practice Vj is always greater than zero
Cj
= A
qeN'
2 Vbi
-Va( )
é
ë
ê
ê
ù
û
ú
ú
1
2
= A
qeN'
2Vj
é
ë
ê
ê
ù
û
ú
ú
1
2
Vj
=Vbi
- Va
- IRS( )

Key points
• Junction capacitance arises because of ionized donors
and acceptors in transition region
• As voltage is applied, depletion region gets wider (or
narrower)
• Sheets of charge are added or removed
• Sheets of charge look like a parallel plate capacitor
• Junction capacitance is smallish for large reverse bias,
increases nonlinearly as voltage becomes more
positive
• Next: stored charge capacitance

Introduction
• There are two kinds of capacitance associated
with pn junctions
• We already looked at junction capacitance
– Caused by ionized donors and acceptors on each
side of junction
• Now we’ll investigate stored-charge
capacitance
– Caused by change in minority carrier density

Consider an n+p junction
• Take the case of
forward bias
• For n+p, have injection
of electrons primarily
• When you change the
bias, the amount of
injected carriers
changes
• Change in charge
caused capacitance

First, steady state
• We already know the density of excess electrons
on p-side is
• In steady state, this concentration remains
constant in time (still varies with distance)
– Electrons are constantly diffusing across junction
– Electrons are constantly diffusing into p-region and
recombining
– There is a constant “pile” of charge
on p-side
Dnp
(x) = Dnp
(xp
)e
-
x-xp
Ln

When the voltage changes
• The amount of injection changes
– The size of the “pile” changes
– Change in charge “stored” in the pile is
capacitance
– Total charge stored is integral of injection pile
– Will relate this to current
Qs
= -qA Dnp
(x)dx
xp
¥
ò = qADnp
(xp
)Ln

Current is by diffusion of minority
carriers (primarily)
• So current across x=xp is
• Combine with
• Result
I xp
= qADn
dDnp
(x)
dx
xp
= -qADn
Dnp
(xp
)
Ln
Qs
= -qA Dnp
(x)dx
xp
¥
ò = qADnp
(xp
)Ln
Qs
=
ILn
2
Dn
= Itn
Ln
= Dn
tn

Now change voltage by dVa abruptly
• Size of pile changes
• Since
then Δn(xp) changes abruptly
• But the rest of the
distribution can’t change
instantaneously
• Now peak concentration is
not at xp but to right of it
• Since
• the stored electrons will
diffuse to left and right of peak
• The ones that diffuse to the
right recombine and disappear
• Only the ones that diffuse to
the left flow into external
circuit and contribute to the
capacitance
Dn xp( )µ e
qVa
kT
Jn
= qDn
dn
dx
Referred to as reclaimable stored charge

Let δ be the fraction of charge that is
reclaimable
• Then Qsr is reclaimable stored charge and recalling that
• and
• We obtain
• For prototype step junction, δ=0.5
– For graded doping, δ depends on the profile
– We’ll visit in more detail when we talk about bipolar junction
transistors (Chapter 10)
Csc
=
dQsr
dVa
= d
dQs
dVa
=
q
kT
d Itn
I xp
= -qADn
Dnp
(xp
)
Ln
= -qADn
np0
e
qVa
kT
Ln
Qs
= Itn

Stored-charge capacitance diffusion current
• Thus Csc increases exponentially with Va
• But junction capacitance goes as square root of
Va
• Therefore, for reverse bias and small forward
bias, junction capacitance dominates
• For large forward bias, stored-charge capacitance
dominates
µ
µ
Csc
=
q
kT
d Itn
=
q
kT
dnp0
e
qVa
kT
tn
Cj
= A
qeNA
2 Vbi
-Va( )

Example: Compare the junction capacitance and stored
charge capacitance under reverse (Va=-5V) and forward
bias (Va =+0.75V).
• Consider a prototype n+-p junction with
NA’=NA=1017 cm-3, with junction area A=100
μm2 and fractional reclaimable charge δ =0.5
• Junction capacitance:
Cj
= A
qeNA
2 Vbi
-Va( )
é
ë
ê
ê
ù
û
ú
ú
1
2

First find Vbi
• Find Vbi from chart for one-sided junction
Vbi=0.98V

Calculate Cj’s
• For Va=-5V
• and for Va=0.75V
Cj
(-5) = 100mm2 10-8
cm2
1mm2
æ
èç
ö
ø÷ i
i
(1.6 ´10-19
C) 11.8( ) 8.85´10-14 F
cm
æ
èç
ö
ø÷ 1017
cm-3
( )
2 0.98+5( )
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
1
2
= 0.053pF
Cj
(0.75) = 0.27 pF
x
x
Cj
= A
qeNA
2 Vbi
-Va( )

For stored-charge capacitance
• Need to find I
– Use the diffusion current
– Generation/recombination current does not
contributed to stored-charge capacitance
• We find
I = I0
e
qVa
kT
-1
æ
è
ç
ö
ø
÷ = qA
Dn
np0
Ln
æ
è
ç
ö
ø
÷ e
qVa
kT
-1
æ
è
ç
ö
ø
÷
np0
=
ni
2
NA
'
=
(1.08´1010
cm-3
)2
1017
cm-3
=1.17 ´103
cm-3
Csc
=
dQsr
dVa
= d
dQs
dVa
=
q
kT
d Itn

• Look up Dn (minority carriers on p-side)
I = I0
e
qVa
kT
-1
æ
è
ç
ö
ø
÷ = qA
Dn
np0
Ln
æ
è
ç
ö
ø
÷ e
qVa
kT
-1
æ
è
ç
ö
ø
÷
Dn=20 cm2/s

So I0 becomes
• and I at Va=-5V is -5.3 x 10-19 A
• At Va=+0.75V,
I0
= qA
Dn
np0
Ln
æ
è
ç
ö
ø
÷
= 1.6 ´10-19
C ×100 ´10-8
cm2
´
20
cm2
s
×1.17 ´103
cm-3
70 ´10-4
cm
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
= 5.3´10-19
A
I = I0
e
qVa
kT
-1
æ
è
ç
ö
ø
÷ = 5.3´10-19
A e
0.75
0.026
-1
æ
èç
ö
ø÷ =1.8mA

Stored-charge capacitances
• At Va=-5V:
• At Va=+0.75V
Csc
(-5) =
q
kT
d Itn
=
1
0.026V
æ
èç
ö
ø÷ (0.5)(5.3´10-19
A)(3´10-6
s) = 3.1´10-23
F @ 0
Csc
(0.75) =
q
kT
d Itn
=
1
0.026V
æ
èç
ö
ø÷ (0.5)(1.8´10-6
A)(3´10-6
s) = 100 pF

Compare results
Cj Csc
Va= -5 V 0.053 pF ≈0
Va= +0.75 V 0.27 pF 100 pF

Key points
• Under forward bias, as voltage changes, the
amount of charge “stored” in the injection
changes, leading to capacitance
• When voltage is changed, not all the “stored”
charge is recovered- reclaimable fraction is δ=1/2
for step junction
• Under reverse bias, junction capacitance
dominates
• Under large forward bias, stored-charge
capacitance dominates
• Next: we’ll look at transient behavior of diodes

Introduction
• A pn junction is often used as a switch
– Change state from “on” (forward bias, VF) to “off”
(reverse bias, VR)
• Junction has capacitance
– Cannot change state instantly
– There will be turn-on and turn-off transients
• Investigate turn-off transient first

Here is the situation
• We’ll assume that both VF and VR are larger
than Vbi
• Assume R1>>Rs (so we can neglect internal
series resistance of diode)

Take an n+-p junction
• Then can ignore stored hole charge on n-side
• At time t=0, the diode is on
Dnp
(x,t = 0) = Dnp
(xp
,0)e
-
x-xp
Ln
• Vj(on)<Vbi
• VF>>Vbi>Vj
• Thus forward
current IF is (up
to t=0)
IF
(0) @
VF
R1

At t=0 switch voltage to VR
• Excess carrier
concentration goes
from injection to
extraction
• Some carriers diffuse
to the right (not all
are reclaimed)
• Notice slope of Δnp is
constant near
junction during
transition

Find the current
• Recall current crossing the plane xp is same as the
current anywhere else in the device
– But it’s easy to evaluate here- it’s entirely due to diffusion
• Since slope is constant during the
switch, current is constant until
t=ts (ts is storage time)
• After that, slope flattens to
steady state
• Current decays to steady state
value (reverse current ≈0)
IR
@
VR
R1

The derivation is involved
• But approximate solution is
• For VF=VR, |IF|≈|IR| and
• Note IR is zero in steady state, only reaches
|IF| during transition
ts
@ tn
ln 1+
IF
IR
æ
è
ç
ö
ø
÷
ts
= tn
ln2 = 0.68tn

To shorten turn-off time, reduce
charge in p-region
• Can do this by decreasing (adjust voltages in
circuit)
• Can do this by reducing electron lifetime
– By doping with traps such as Au or Cu
– Traps, however, increase G current in the off state
• Increases power consumption
– Shortening lifetime also decreases diffusion length Ln
• Can also reduce thickness of more lightly doped
side
• Can also grade doping- more in Chapter 9
IF
IR

Key points
• When switching from “on” to “off”, initial
condition is injected charge stored on lightly
doped side of one-sided junction
• When voltage is reversed, takes time to deplete
the excess charge (source of capacitance)
• On turn-off, current is constant for storage time
ts, then decays
• To speed up turn-off: shorten lifetime via traps
states, adjust on/off voltages

Introduction
• We looked at the turn-off transient for a step-
junction diode
• Now we’ll consider turn-on
• Transient associated with going from VR back
to VF

Now must build up stored charge
• Slopes constant again
• Current IF constant
• Reach steady state when
injection rate equals
recombination rate

Effects of the two capacitances
• For turn-on, beginning state is reverse bias
– Junction capacitance important there
– Amount of charge needed to discharge the
junction capacitance small compared to amount
needed to fill stored charge pile
– Stored charge established by diffusion (slow)
• Thus turn-on time determined mostly by time
needed to set up the minority carrier steady-
state distribution
small
large

When switching from high to low
• Current switches
from VF/R1 to VR/R1
• During storage time,
diode voltage
constant (current is
constant)
• Real diode: series
resistance Rs not
really 0
VOLTAGE
IN
DIODE
VOLTAGE
DIODE
CURRENT

Voltage drop across series resistance
of diode
• Diode voltage is
junction voltage plus
voltage across Rs
• Voltage across diode
terminals goes from
(Vj(on)+IFRS) to
(Vj(on)+IRRS)
– Remember IF is
positive, IR is negative
Diode
model
DIODE
VOLTAGE
DIODE
CURRENT

After storage time…
• Current flow decays to
“off” value
• Voltage across
terminals of diode
decays to VR
• Notice turn-on time
very short compared
to turn-off
• Maximum switching
frequency limited by
turn-off
VOLTAGE
IN
DIODE
VOLTAGE
DIODE
CURRENT

Real life
• Usually use V=0 for off voltage (not a negative
voltage)
• Junction capacitance and stored-charge
capacitance vary with voltage
• Messy
• Simulate with SPICE or equivalent

For short-base diode
• WB<<Ln
• Recombination mostly happens at contact
(neglect recombination in p-QNR)
• and
Dnp
(x) = Dnp
(0) 1-
x
WB
æ
è
ç
ö
ø
÷
In
= qADn
dDn
dx
= -
qADn
Dn(0)
WB

Excess charge in short base is
• where tT is transit time across the short QNR
(“base transit time”)
• Base transit time given by
Qs
= -qADn(0)i
WB
2
= In
tT
tT
=
WB( )
2
2Dn
x

Example: Compare the amount of minority carrier
stored charge in a forward-biased short-base diode
with that in a long-base diode.
• Solution: We had
• So we can write
Qs
= -qADn(0)i
WB
2
= In
tT
tT
=
WB( )
2
2Dn
Qs
=
ILn
2
Dn
= Itn Ln
2
= Dn
tn
Qs
(short - base)
Qs
(long - base)
=
1
2
i
WB
Ln
æ
è
ç
ö
ø
÷
2
x
x
short
long

Let WB≈0.1 μm, Ln=31 μm
• This value of WB typical for the base of a
transistor
• Can speed up switching time by about 5
million by using a short-based diode
Qs
(Short-Base)
Qs
(Long Base)
@
1
2
i
0.1
31
æ
èç
ö
ø÷
2
@ 5´10-6x

Key points
• Turn-off time is much slower than
turn-on time
• Switching frequency limited by
turn-off
• Reduce time, increase frequency by
shortening the width of the lightly
doped side
• Can also introduce trap states to
speed up recombination but that
incurs a power penalty

Introduction
• When designing a diode (or any component)
have to consider the temperature range over
which it has to operate
– Commercial grade products might have to operate
over 0 C to 70 C
– Military grade: -55 C up to 125 C

Current in an n+-p diode
• Quantities Dn , τn , NC , NV , and Eg all vary with
temperature
• Setting those aside, have kT in exponent
• Compare fractional change in J0 to J0
J0
=
q
NA
'
Dn
tn
ni
2
=
q
NA
'
Dn
tn
NC
NV
e
- Eg
kT
dJ0
dT
æ
èç
ö
ø÷
J0
=
de
- Eg
kT
dT
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
e
- Eg
kT
=
Eg
kT 2
æ
è
ç
ö
ø
÷ =
Eg
kT
×
1
T

Fractional change for Si
• For Si,
• or J0 varies by 14% per °C!
dJ0
dT
æ
èç
ö
ø÷
J0
=
de
- Eg
kT
dT
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
e
- Eg
kT
=
Eg
kT 2
æ
è
ç
ö
ø
÷ =
Eg
kT
×
1
T
1.12
0.026
×
1
300
= 0.14

What about JGR0?
• That depends on ni rather than ni
2 (J0 depends
on ni
2)
• Recall
• So JGR0 changes with temperature at about
half the rate of J0
ni
2
= NC
NV
e
- Eg
kT

And the voltage
• Suppose a diode is biased at a constant
forward current in typical range 0.1 to 1 mA
• Diode voltage increases about 2 mV per
degree C

Key points
• Diode current and voltages vary with
temperature
• Have to take into account when designing
devices

Introduction
• We considered prototype pn junctions
– Real diodes have complex doping gradients
– Have to solve equations numerically- provides little
insight
• We chose step junction to illustrate physical
processes
• We considered
– pn junctions (neither side degenerate)
– one-sided junctions (one side is degenerate)
• n+-p
• p+-n

Fermi levels
• Non-degenerate n and p regions:
• Degenerate: can often assume that
Ef
= EC
- kT ln
NC
ND
'
= Ei
+ kT ln
ND
'
ni
n-region
Ef
= EV
+ kT ln
NV
NA
'
= Ei
- kT ln
NA
'
ni
p-region
Ef
@ EC
n+
-region
Ef
@ EV
p+
-region

Built-in voltage
Vbi =
1
q
Fp - Fn( )
Vbi
=
1
q
Eg
- kT ln
NC
NV
ND
'
NA
'
é
ë
ê
ù
û
ú =
kT
q
ln
ND
'
NA
'
ni
2
prototype pn junction
Vbi
=
1
q
Eg
- kT ln
NV
NA
'
é
ë
ê
ù
û
ú =
kT
q
ln
NV
NA
'
ni
2
prototype n+
-p junction
Vbi
=
1
q
Eg
- kT ln
NC
ND
'
é
ë
ê
ù
û
ú =
kT
q
ln
NC
ND
'
ni
2
prototype p+
-n junction
in general:

Junction width
• Width of region where there are uncompensated
ions
• No free carriers
– Depletion region
– Transition region
– Space charge region

Junction width equations
• pn
• n+-p
• p+-n
w = wn
+ wp
=
2eVj
NA
'
+ ND
'
( )
qNA
'
ND
'
é
ë
ê
ê
ù
û
ú
ú
1
2
w =
2eVj
qNA
'
é
ë
ê
ê
ù
û
ú
ú
1
2
w =
2eVj
qND
'
é
ë
ê
ê
ù
û
ú
ú
1
2
Vj I the junction voltage
Vj=Vbi-Va
Va is the applied voltage

Junction width
• Most of the junction width appears on the
lightly doped side
• Most of the voltage is dropped across the
lightly doped side
wn
wp
=
xo
- xn( )
xp
- xo( )
=
NA
'
ND
'
Vj
n
Vj
p
=
NA
'
ND
'

Three current mechanisms
• Drift (caused by electric field)
• Diffusion (caused by gradients in carrier
concentrations)
• Generation/recombination
• Total current density is
• if evaluated at edge of transition region where
you can neglect drift
J = JGR
+ Jdiff
@ JGRo
e
qVa
2kT
-1
æ
è
ç
ö
ø
÷ + Jo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷

Diode current density
• J0=J0diff+JGR0
• J0 increases by about 14% per °C
• J increases by factor of 10 for Va increase of 60
mV (room temperature)
• Total current is I=AJ where A is the cross-
sectional area of the junction
Jdiff
= Jo
e
qVa
kT
-1
æ
è
ç
ö
ø
÷

Leakage currents
Transition width w goes as √VjJGRo
@
qni
w
2to
Jo
= q
Dn
npo
Ln
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷
Long base diode
Short base diode (p-side short)Jo
= q
Dn
npo
WB
+
Dp
pno
Lp
æ
è
ç
ö
ø
÷
J0
= q
Dn
np0
WB( p)
+
Dp
pn0
W(n)
æ
è
ç
ö
ø
÷
Both sides short

Leakage current comments
• Usually JGR0>>J0
– JGR0 important for reverse bias and small forward
bias
– J0 important for larger forward bias

Junction breakdown
• Two mechanisms
– Tunneling (Zener breakdown)
– Avalanche multiplication
– Zener is softer breakdown, occurs in heavily doped
junctions
– Avalanche sharper breakdown, comes with lightly
doped junctions
J e
-
pwT m*
Eg
2
3
2
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
M =
1
(1- P)

Capacitance
• Two kinds
– Junction capacitance (variation in ionized charges
in depletion region)
– Stored charge capacitance (variation in charge in
excess carriers near junction)

Junction Capacitance
• Junction capacitance (variation in ionized
charges in depletion region)
• Goes as Va
-1/2
• Predominates under reverse bias and low
forward bias
Cj
= A
qeND
'
NA
'
2 ND
'
+ NA
'
( ) Vbi
-Va( )
é
ë
ê
ê
ù
û
ú
ú
1
2

Stored Charge Capacitance
• Stored charge capacitance (variation in charge
in excess carriers near junction)
• Dominates under large forward bias
Csc
=
q
kT
d Itn
(longbase) Csc
=
q
kT
d ItT
=
q
kT
d I
WB
2
2Dn
æ
è
ç
ö
ø
÷ (shortbase)

Transient effects
• Switching time determined by how quickly
you can charge/discharge internal
capacitances
• Turn-on time small compared to turn-off
• Operating frequency limited by turn-off
– Use short-base diode to improve

Key points
• Prototype (step) junction greatly simplified
• Does provide great physical insight
• Will refine the model in Chapter 6
• See also Supplement to Part II which has
additional information

Fundamentals of Semiconductor Devices

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Fundamentals of Semiconductor Devices