Transcript Lesson 1 - Introduction

Electrical Energy
Conversion and
Power Systems
Universidad
de Oviedo
Power Electronic Devices
Semester 1
Power Supply
Systems
Lecturer: Javier Sebastián
Research Group
Power supply Systems
(Sistemas Electrónicos de Alimentación)
Javier Sebastián
Dr. Electrical Engineer (Ingeniero Industrial)
Full professor
Room 3.1.21
Edificio nº 3, Campus Universitario de Viesques
33204 Gijón (Asturias). Spain
Phone (direct): 985 18 20 85
Phone (secretary): 985 18 20 87
Fax: 985 18 21 38
E-mail: [email protected]
Web: http://www.unioviedo.es/sebas/
2
Outline

Review of the physical principles of operation of semiconductor
devices.

Thermal management in power semiconductor devices.

Power diodes.

Power MOSFETs.

The IGBT.

High-power, low-frequency semiconductor devices (thyristors).
3
Outline
Review of physical principles
of semiconductors
Power electronics devices
D
G
S
4
Previous requirements

Basic electromagnetic theory.

Basic circuit theory.

The operation of basic electronics devices in circuits. The
student must understand the behaviour of the following
electronics devices in simple circuits:

Diodes.

Bipolar Junction Transistors, both PNP and NPN types.

Field Effect Transistor, especially enhancement-mode MetalOxide-Semiconductor Field Effect Transistors (MOSFETs),
both in N-channel- and P-channel types.
5
Electrical Energy
Conversion and
Power Systems
Universidad
de Oviedo
Lesson 1 - Review of the physical principles
of operation of semiconductor devices
Semester 1 - Power Electronics Devices
6
Outline
Review of the physical principles of operation of semiconductor
devices:
 Basic concepts about semiconductor materials: band diagrams, intrinsic
and extrinsic semiconductors, mechanisms for electric current conduction
and continuity equation and its use in simple steady-state and transient
situations.
 Basic concepts about PN junctions: Equilibrium conditions, forward- and
reverse-biased operation and calculation of the current flow when biased.
 Reverse-biased voltage limits of PN junctions.
 PIN junctions.
 Conductivity modulation.
 Transient effects in PN junctions in switching-mode operation.
 Metal-semiconductor junctions.
7
Energy of electrons
Energy level in a semiconductor as a function
of inter-atomic spacing
At 0 K, empty
- -
- -
- - -
- Inter-atomic spacing
Actual spacing
At 0 K, filled
8
Concept of band diagram
Energy of electrons
Empty at 0 K
4 states/atom
Eg
- - -
Conduction band
Band gap
4 electrons/atom
Material
Eg [eV]
Ge
0.66
Si
1.1
4H - SiC
3.26
GaN
3.39
Valence band
Filled at 0 K
9
Band structure for insulators,
semiconductors and metals at 0 K
Conduction band
Conduction band
Band gap Eg
Band gap Eg
Overlap
Valence band
Valence band
Valence band
Semiconductor
Eg=0.5-2 eV
Insulator
Eg= 5-10 eV
Metals
No Eg
Conduction band
10
Band structure for semiconductors at room
temperature. Concept of “hole”
-
Si
-
-
-
-
-
-
Semiconductor
Eg=0.5-2 eV
Si
+
Si
-
Valence band
-
-
-
-
+
-
Si
-
-
Eg
-
-
Conduction band
Visualization using the bonding model
• Some electrons jump from the valence band to the conduction band. They are
charge carriers because they can move from one atom to another.
• The empty state in the valence band is referred to as a “hole”.
• The holes have positive charge. They are also charge carriers.
11
Concepts of generation and recombination
Recombination
Generation
Eg
-
Si
-
-
-
-
Si
+
Si
-
Si
-
-
-
-
-
-
-
-
Si
-
-
-
-
-
+
Si
-
+
-
Si
-
-
-
Si
-
-
-
-
-
+
-
-
-
Eg
12
-
-
Why both holes and electrons are electric charge carriers?
Si
-
-
-
-
-
-- -
-
Si
+
-
-
-
-
Si
-
-
-
-
-
-
+
Si
-
Si
-
-
-
-
-
-
-
-
Si
Si
-
Si
+++++++
-
-
• In general, there will be electric current due to both electrons and holes
• Example: when there is an electric field in the semiconductor lattice
13
How many electrons and holes are there in 1 cm3?
• The number of these electrons and holes strongly depend on both Eg and the
room temperature. It is called intrinsic concentration and it is represented as
“ni”.
• The concentration of electrons in the conduction band (negative charge
carriers) is represented as “n”. The concentration of holes in the valence band
(positive charge carriers) is represented as “p”.
• Obviously n = p = ni in this type of semiconductors (intrinsic semiconductors)
• Some examples of the value of ni at room temperature:
Material
Eg [eV]
ni [elect./cm3]
Ge
0.66
2.4·1013
Si
1.1
1.5·1010
GaAs
1.4
1.8·106
4H - SiC
3.26
8.2·10-9
GaN
3.39
1.9·10-10
Taking into account the number of bonds
of valence band electrons in 1cm3 of
silicon, only one bond is broken for each
amount of 1013 unbroken bonds (at room
temperature)
14
Concept of extrinsic semiconductors:
doping semiconductor materials
• Can we have different concentration of electrons and holes?
• The answer is yes. We need to introduce “special” impurities into the crystal:
 Donors: atoms from column V of the Periodic Table. We obtain an extra
electron for each atom of donor.
 Acceptors: atoms from column III of the Periodic Table. We obtain an
extra hole for each atom of acceptor.
Si
-
3
-
Al -
-
-
-
Acceptor
1
-
-
-
2
Si
+
-
-
-
-
-
-
- -
-
3
4
-
Sb
Si
-
+
-
2
5
Si
-
-
Donor
-
1
Si
-
Si
15
-
N-type and P-type semiconductors
-
-
3
Si
-
-
Al-
-
-
1
Si
+
-
-
-
2
-
Acceptor
-
-
-
-
-
-
- -
-
-
-
-
-
-
3
4
Si
-
Sb+
Si
-
-
2
-
5
1
Donor
Si
-
Si
N-type semiconductor:
P-type semiconductor:
• Majority carriers are electrons.
• Minority carriers are holes.
• Positively-charged atoms of donor
(positive ions).
• Minority carriers are electrons.
• Negatively-charged atoms of acceptor
(negative ions).
• Majority carriers are holes.
16
Charges in N-type and P-type semiconductors
+
Al
Al
-
-
Al
Al
+
-
+
+
Al
+
+
P-type silicon
Al
-
Al
Acceptor ions
(negative ions)
Donor ions
(positive ions)
-
N-type silicon
-
Sb+
Sb+
-
Sb+
Sb+
Sb+
-
-
Sb+
Sb+
-
+
Sb+
-
-
Thermal
generation
Al
+
+
-
+
-
-
electron
-
-
hole
Al
Al
+
-
+
-
Thermal
generation
-
+
Sb+
Sb+
17
Charge carries in N-type and P-type semiconductors
-
Al
-
Sb+
-
-
Sb+
-
Al
Sb+
-
Sb+
Sb+
• Concentration of majority carriers: pP
• Concentration of minority carriers: nP
+
P-type
Sb+
Al
+
+
Al
+
+
Al
+
-
-
-
Al
+
-
+
• Mass action law:
pP·nP = ni2
(only valid at equilibrium)
• Concentration of majority carriers: nN
• Concentration of minority carriers: pN
• Mass action law:
nN·pN = ni2
(only valid at equilibrium)
N-type
Very important equations!!!
18
-
Static charges in N-type and P-type semiconductors
-
Sb+
-
Sb+
Al
-
+
P-type
Sb+
-
+
Al
-
+
-
+
+
Al
Al
+
Sb+
Sb+
Sb+
-
-
-
Al
Al
+
-
+
• Concentration of acceptors: NA
(only negative static charges in a
P-type semiconductor)
• Concentration of donors: ND
(only positive static charges in a
N-type semiconductor)
N-type
19
-
Neutrality in N-type and P-type semiconductors
• Silicon, aluminium and antimony were neutral before being used
The extrinsic semiconductor must be neutral, too.
-
Sb+
-
Sb+
• Neutrality:
Sb+
Sb+
N-type
Al
pP = nP + NA
-
+
P-type
-
+
Al
-
+
-
+
+
Al
Al
+
Sb+
Sb+
-
-
-
Al
Al
+
-
+
• Positive charges in volume V: pP·V
• Negative charges in volume V: nP·V + NA·V
• Negative charges in volume V: nN·V
• Positive charges in volume V: pN·V + ND·V
• Neutrality:
nN = pN + ND
Very important equations!!!
20
-
Calculating the concentration of electrons and holes (I)
Al
-
Al
Sb+
-
-
Sb+
2 known (NA and ni) and 2 unkown (pP
and nP) variables can be solved
-
+
P-type
Sb+
-
+
+
Al
Al
+
Sb+
Sb+
Sb+
-
-
+
+
Al
-
Al
+
-
+
• Neutrality: pP = nP + NA
• Mass action law: pP·nP = ni2
• Neutrality: nN = pN + ND
• Mass action law: nN·pN = ni2
2 known (ND and ni) and 2 unkown (nN
and pN) variables can be solved
N-type
21
-
Calculating the concentration of electrons and holes (II)
• Frequent case: quite heavy doped semiconductors
Al
-
+
Al
-
-
Sb+
Sb+
-
Sb+
Al
Sb+
N-type
• NA >> ni
• Neutrality: pP  NA
• Mass action law: nP  ni2/NA
-
+
P-type
-
+
+
+
Sb+
Sb+
-
-
+
Al
-
Al
Al
-
+
+
• ND >> ni
• Neutrality: nN  ND
• Mass action law: pN  ni2/ND
Very useful equations!!!
22
-
Mechanisms to conduct electric current: Drift (I)
• Semiconductors can conduct electric current due to the presence of an
electric field E

-
-
jp
+
+

+
-
-
-
-
+
-
+
+
+

+
+++++
-
E
jn


jp_Drift = q·p·p·E is the current density of holes due to drift.


jn_Drift = q·n·n·E is the current density of electrons due to drift.
23
Mechanisms to conduct electric current: Drift (II)


jp_Drift = q·p·p·E


jn_Drift = q·n·n·E
q = magnitude of the electronic charge (1.6·10-19 coulombs).
p = hole mobility.
n = electron mobility.
p = hole concentration.
n = electron concentration.
Ge
Si
GaAs
(cm2/V·s)
(cm2/V·s)
(cm2/V·s)
n
3900
1350
8500
p
1900
480
400
E = electric field.
24
Mechanisms to conduct electric current: Diffusion (I)
-
-
n1
-
1
2

jn_Diff
n2 < n1
Electrons have migrated due to “diffusion” (you can see
the same phenomenon in gases)
25
Mechanisms to conduct electric current: Diffusion (II)
• If we maintain a different concentration of electrons, we
also maintain the motion of electrons in the lattice
-
-
-
-
-
-
-
-
2
-
-
-
n1

jn_Diff

n

-
1
n2 < n1
26
Mechanisms to conduct electric current: Diffusion (III)
-
-
-
-
-
-
2
-
-

jn_Diff

n

-
-
n1
-
-
-
-
1
n2 < n1
The current density is proportional to the electron concentration gradient:


jn_Diff = q·Dn· n
Dn = electron diffusion coefficient.
27

Mechanisms to conduct electric current: Diffusion (IV)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
2
+
+
p1

jp_Diff

p

+
+
+
+
+
1
p2 < p1
The current density is proportional to the hole concentration gradient:


jp_Diff = -q·Dp· p
Dp = hole diffusion coefficient.
28

Mechanisms to conduct electric current: Diffusion (V)


jp_Diff = -q·Dp· p


jn_Diff = q·Dn· n


q = magnitude of the electronic charge (1.6·10-19 coulombs).
Dp = hole diffusion coefficient.
Dn = electron diffusion coefficient.

p = hole concentration gradient.

n = electron concentration gradient.
Ge
Si
GaAs
(cm2/s)
(cm2/s)
(cm2/s)
Dn
100
35
220
Dp
50
12.5
10
29
Summary of conduction mechanisms


jp_Drift = q·p·p·E


jn_Drift = q·n·n·E


jp_Diff = -q·Dp· p


jn_Diff = q·Dn· n
• Drift currents depend on the carrier
concentration and on the electric field.
• Diffusion currents do not depend on
the carrier concentration, but on the
carrier concentration gradient.
30


Continuity equations (I)
There are some relationships between spatial and time variations of
carrier concentrations because electrons and holes cannot mysteriously
appear and disappear at a given point, but must be transported to or
created at the given point via some type of ongoing action.
The concentration of holes can be time-changing due to:
• Different current density of holes across “A” and “B”.
• Excess of carriers over the equilibrium (mass action law).
• Generation of electron-hole pairs by radiation (light) .
A
B
31
Continuity equations (II)
-
• Different current density
of holes across “A” and “B”.
+
+
jp_A
jp_B
A
• Excess of carriers over the
equilibrium (mass action law).
jn_B
B
+
A
• Generation of electron-hole
pairs by radiation (light) .
Light
A
B
+
B
32
Continuity equations (III)
Taking into account the three effects, we obtain the continuity equation for
holes:
Total time variation
of holes
·jp/q
-
p/t = GL- [p(t)-p]/p

Variation due to the excess of
carriers over the equilibrium
Variation due to the generation
of electron-hole pairs by light
GL: rate of generation of electron-hole pairs by light.
p: hole minority-carrier lifetime.
p: hole concentration in steady-state.
Variation due to the
different current density of
holes across “A” and “B”
Similarly, we can obtain the continuity equation for electrons:
·jn/q
+
n/t = GL- [n(t)-n]/n

33
Time evolution of an “excess” of minority carries (I)
• We generate an “excess” of electron-hole pairs by injecting light
into a piece of N-type silicon and we reach the steady-state.
0
0
+
N
+
+
+
+
+
+
·jp/q p0= GL·p + p
-
p/t = GL- [p(t)-p]/p

+
+
+
+
+
p0
+
• Now the light injected disappears. We want to compute the time evolution of
the hole concentration afterwards.
+
N
+
+
+
+
+
+
+
+
+
+
+
+
p0
p(t)
p
34
Time evolution of an “excess” of minority carries (II)
• We can also compute the time evolution of the hole concentration from
the continuity equation:
p/t = GL- [p(t)-p]/p

0
·jp/q
-
0
After integrating  p(t) = p + (p- p)·e-tp
Tangent line
p0
Same area
p(t)
p
p
p
• Physical interpretation: There is an appreciable
increase of holes during 3-5 times p.
t
35
Spatial evolution of an “excess” of minority carries (I)
• We constantly inject an “excess” of holes into a surface of a
piece of N-type silicon and we reach the steady-state. No electric
field exists and the hole current is due to diffusion.
+
+
+
0
p0
+
+
+
+
+
+
+
+
+
+
+
+
N
+
+
+
p
xN
x
0
0
p/t = GL- [p(t, x)-p]/p

·jp/q
-
0 = - [p(x)-p]/p + Dp·2[p(x)-p]/x2
After integrating  p(x)
= p + C1·e-x/Lp + C2·ex/Lp
where : Lp=(Dp· p)1/2 is the minority hole diffusion length
36
Spatial evolution of an “excess” of minority carries (II)
p(x) = p + C1·e-x/Lp + C2·ex/Lp
xN: length of the N-type crystal
Lp: hole diffusion length
• Cases:
a) Lp << xN (wide crystal):  p(x) = p + (p- p)·e-xLp (decay exponentially).
b) Lp >> xN (narrow crystal):  p(x) = p + (p- p)·(xN-x)/xN (decay linearly).
c) Other cases  hyperbolic sine evolution.
p0
Lp << xN (wide)
p(x)
p
Lp
p0
p(x)
Tangent line
x
xN
Lp >> xN (narrow)
p
x
xN
37
Concept of PN junction (I)
P-type silicon
+
+
Al
+
Al
-
Al
-
Sb+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
-
Al
-
-
-
Al
-
+
-
-
-
Al
Al
-
-
+
-
Al
+
+
-
+
N-type silicon
+
Sb+
Barrier to avoid carrier diffusion
What happens if we remove the barrier?
38
Concept of PN junction (II)
P-side
+
+
+
Al
+
Al
-
Al
-+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
-
Al
-
-
-
Al
-
+
-
+
-
-
-
Al
Al
-
Al
+
-
+
N-side
+
Sb+
Holes begin to diffuse from the P-side to the N-side. Similarly, electrons diffuse from the
N-side to the P-side
Are all the carriers to be diffused?
39
Concept of PN junction (III)
Are all the carriers to be diffused?
P-side
-+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Non-neutral P-type region, but
Non-neutral N-type region, but
negatively charged
positively charged
Is this situation “the final situation”?
The answer is no
+
-
Al
+
-
-
-
-
Al +
Al
-
+
-
+
-
-
+
Al
-
+
-
Al
+
-
Al
+
Al
-
Al
N-side
40
Concept of PN junction (IV)
P-side
Al
Sb+
+
+
Al
+
Al
-
Al
-
-
+
-+
Sb+

E
Sb+
Sb+
Sb+
Sb+
Sb+
-
-
-
Al
-
+
-
+
-
-
-
Al
Al
-
Al
+
-
+
N-side
+
Sb+
+
An electric field appears just in the boundary
between both regions (we call this boundary
metallurgical junction)
41
Concept of PN junction (V)
• Now, we do zoom over the metallurgical junction
P-side
Al
+
Al
-
Al
-
Al
-
Sb+
Sb+
Sb+
-
Sb+
Sb+
Sb+
+
-
Al
-
Al
-

Due to diffusion ()
E
-
Al
Sb+
-
+
-
Al
N-type
Sb+
+
Due to drift (electric field) ()
The electric field limits the carrier diffusion
42
Concept of PN junction (VI)
• Steady-state situation near the metallurgical junction
-
Al
+
-
Al
Al
-
-
Sb+
Sb+
-
Sb+
Sb+
Al
+
Neutral P-type region
(holes are balanced by
negative ions )
-

E
Sb+
+
-
Al
-
Sb+
Sb+
-
+
Al
+
-
-
-
Al
-
-
+
-
Al
+
Sb+
Neutral N-type region
(electrons are balanced by
positive ions )
Depletion region, or space charge region, or transition region
Unbalanced charge exists because carriers barely exist
43
Concept of PN junction (VII)
• Summary and terminology
Metallurgical junction
Many holes, but
neutral
+ -
P-side
(neutral)

E
N-side
(neutral)
Many electrons, but
neutral
V0
Depletion, or transition, or space charge region (non neutral)

There is space charge and, therefore, there are electric field E and voltage V0.
However, there are almost no charge carriers
44
Computing the built-in voltage V0 (I)
Net current passing through any section must be zero. As neither holes nor
electrons are being accumulated in any parts of the crystal, net current due
to holes is zero and net current due to electrons is zero.
Due to drift
P-side
jp_Diff
jn_Drift
-
Due to diffusion
-+
-+
-+
+- +
N-side
+ Due to drift
-+
- + - Due to diffusion
-+
-+
jp_Drift
jn_Diff
Currents must cancel each other
Currents must cancel each other
45
Computing the built-in voltage V0 (II)
pP (hole concentration in P-side )
+
+
+
+
+
+
+
Due
+ to diffusion
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Zona
P+
+
+
+
+
+
+
-
+
-
+
pN
+
+
V0
+
+
+
+
+
+
(hole concentration in N-side )
jp_Diff
N-side
+
Due to drift
jp_Drift
jp_Drift = - jp_Diff
46
Computing the built-in voltage V0 (III)
Equations:
jp_Drift = - jp_Diff
jp_Drift = q·p·p·E
jp_Diff = -q·Dp·dp/dx
E = -dV/dx

E
Therefore:
dV = -(Dp/p)·dp/p
After integrating :
V0 = VN-side – VP-side = -(Dp/p)·ln(pN/pP) = (Dp/p)·ln(pP/pN)
Repeating the same process with electrons, we obtain:
V0 = (Dn/n)·ln(nN/nP)
It could be demonstrated:
k = Boltzmann constant.
VT = 26 mV at 300 K.
Dp/p = Dn/n = kT/q = VT (Einstein relation)
47
+
+
+
+
+
+
+
+
-
+
Zona
P+
+
+
+
+
+
P-side: many holes
Almost no electrons
-
+
+ +
+
+-
+
+
V0
+
Almost no holes
or electrons
V0 = VT·ln(pP/pN)
and also
-
+
- - - - - - -
+
+
-
+
-
+
-
+
+
-
+
+
-
+
-
+
+
-
+
+
-
+
+
-
+
-
Computing the built-in voltage V0 (IV)
Summary (I)
nN pN
pP nP
N-side: many electrons
Almost no holes
V0 = VT·ln(nN/nP)
48
Computing the built-in voltage V0 (V)
Summary (II)
V0
+
-
P-side
NA, pP, nP
If NA >> ni (current case)
p P = NA
nP = ni2/NA
V0 = VT·ln(pP/pN)
N-side
ND, nN, pN
If ND >> ni (current case)
nN = ND pN = ni2/ND
and also
V0 = VT·ln(nN/nP)
V0 = VT·ln(NA·ND/ni2), VT = 26 mV at 300 K
49
Depletion width in P-side and in N-side
N-side
Al
Al
+
Al
+
NA
Al
-
Al
-
Al
-
WP0
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
-
Al
-
Al
-
P-side
WN0
W0
ND
Charge neutrality implies: NA·WP0 = ND·WN0
The heavier doped a side, the narrower
the depletion region in that side
50
Calculating the electric field E and
the total depletion width W0 (I)
We need to know:
• The electric field E.
• The total depletion width W0.
rP-side
P-side
V0
-
+
E(x)
+
-
-
+
WP0
WN0
rN-side
N-side
W0
We already know:
• The charge density in both sides inside the depletion region:
rP-side = NA·q and rN-side = ND·q
• The relative width of the depletion region in the P-side and in the N-side:
NA·WP0 = ND·WN0
• The built-in (contact) voltage: V0 = VT·ln(NA·ND/ni2)
We will use Gauss’ law and the relationship
between electric field and voltage
51
E(x)
+
P-side
-
+
-
Calculating the electric field E and
the total depletion width W0 (II)
N-side
W0
- V0 +
r(x)
q·ND
Charge
density
x
E(x)
V(x)
Voltage
law:
·E(x) = r(x)/e
x
-Emax0
V0
• Voltage
x

Electric field
• Gauss’

-q·NA
and electric field: E(x) = - V
52
+
P-side
E(x)
-
+
-
Calculating the electric field E and
the total depletion width W0 (III)
N-side
W0
- V0 +
q·ND
Charge
density
After applying Gauss’ law and the
relationship between electric field and
voltage, we obtain:
x
-q·NA
Electric field
W0 =
x
-Emax0
Emax0 =
Voltage
V0 x
2·e·(NA+ND)·V0
q·NA·ND
2·q·NA·ND·V0
e·(NA+ND)
53
Summary of the study of the PN junction with no
external bias
Electric field at the metallurgical junction Emax0
- V0 +
P-side
Doped NA
P-side: many holes
Almost no electrons
-
+
WP0
WN0
N-side
Doped ND
N-side: many electrons
W0
Almost no holes or electrons,
but space charge, electric
field and voltage
V0 = VT·ln(NA·ND/ni2)
W0 =
2·e·(NA+ND)·V0
q·NA·ND
W0 = WP0 + WN0
Emax0 =
Almost no holes
Very important
equations!!!
NA·WP0 = ND·WN0
2·q·NA·ND·V0
e·(NA+ND)
54
Connecting external terminals to a PN junction
metal-semiconductor contacts
+-
P-side
+ VmP
-
V0
N-side
+V -
+
Nm
i=0
V=0
Therefore:
V = 0, i = 0
Hence:
VmP – V0 + VNm = 0
And:
VmP + VNm = V0
No energy can be
dissipated here
Conclusion:
The built-in voltages across each metalsemiconductor contact cancel out the effect of V0
in such a way that V0 does not appear externally.
55
Biasing the PN junction: forward bias
Low resistivity:
VP=0
+ VmP
i0
+-
P-side
Low resistivity:
VN=0
VmP and VNm do not change
and, therefore VmP+VNm= V0
N-side
-V +
j
+V Nm
V0 becomes Vj now
+
-
Vext
Vext = VmP - Vj + VNm = V0 - Vj
Therefore: Vj = V0 - Vext
Conclusion:
The built-in voltage across the junction has decreased Vext volts
56
Biasing the PN junction: reverse bias
Low resistivity:
VP=0
+-
P-side
Low resistivity:
VN=0
N-side
- +
+ -
Vj
VmP
i0
+V Nm
VmP and VNm do not change
and, therefore VmP+VNm= V0
- +
Vext
Vext = -VmP +Vj - VNm = -V0 + Vj
Therefore: Vj = V0 + Vext
Conclusion:
The built-in voltage across the junction has increased Vext volts
57
Biasing the PN junction: notation for a general case
+-
P-side
N-side
- +
Vj
i
+
=
-
Vext
Conclusion:
• Always: Vj = V0 - Vext, being:
 0 < Vext < V0 (forward biased)
 Vext < 0 (reverse biased)
58
Effects of the bias on the depletion region
We must replace V0 with Vj, that is, replace V0 with V0-Vext
With bias
Without bias
Vj0 = V0
W0 =
Emax0 =
Vj (Vext) = V0 - Vext
2·e·(NA+ND)·V0
q·NA·ND
W(Vext) =
2·q·NA·ND·V0
e·(NA+ND)
Emax(Vext) =
Always: V0 = VT·ln(NA·ND/ni2)
2·e·(NA+ND)·(V0-Vext)
q·NA·ND
2·q·NA·ND·(V0-Vext)
e·(NA+ND)
59
W0
W
P-side
-- +
Effects of the forward
bias on the depletion
region
N-side
VV
0-V1
0
r(x)
V1
x
E(x)
• Less spatial charge
x
• Lower electric field
-Emax
-Emax0
Vj(x)
V0-V1
V0
• Lower built-in voltage
x
60
WW0
P-side
- - ++
N-side
0 2
V0V+V
V2
Effects of the reverse
bias on the depletion
region
r(x)
x
E(x)
x
• More spatial charge
• Higher electric field
-Emax0
-Emax
Vj(x)
V0
V0+V2
x
• Higher built-in voltage
61
Effects of the bias on the neutral regions (I)
nNV nN
-
-
-
-
-
-
-
-
-
- - - - - - -
+
+
+
+
- -
-
P-side
V0V-V0ext
Zona P
-
- +
-
nP nPV
No bias: V0 = VT·ln(nN/nP)
Forward bias: V0-Vext =VT·ln(nNV/nPV)
For holes with forward bias: V0-Vext =VT·ln(pPV/pNV)
62
Effects of the bias on the neutral regions (II)
• The quotients nNV/nPV and pPV/pNV strongly change with bias.
• In practice, nNV and pPV do not change appreciably (i.e., nNV 
nN and pPV pP) for charge neutrality reasons.
• Therefore the concentration of minority carriers (i.e., pNV and
nPV) strongly changes at the depletion region edges.
• The values of nPV and pNV can be easily obtained:
V0-Vext =VT·ln(nN/nPV)  nPV = nN·e -(V0-Vext)/ VT
-(V0-Vext)/ VT
p
=
p
·e
V0-Vext =VT·ln(pP/pNV)  NV P
• As nN ND and pP NA, then:
nPV = ND·e-(V0-Vext)/ VT
pNV = NA·e-(V0-Vext)/ VT
63
Effects of the bias on the neutral regions (III)
Non-biased junction
-V0/ VT
2
pN = ni /ND = NA·e
Non-biased junction
nP = ni2/NA = ND·e-V0/ VT
Biased junction
-Vj/ VT
-Vj/ VT
nPV = ND·e
+-
pNV = NA·e
Biased junction
Zona P
pP = NA
- Vj +
Vj = V0-Vext
+
Zona N
nN = ND
=Vext
64
Effects of the bias on the neutral regions (IV)
Concentration of minority carriers :
-Vj/ VT
pNV = NA·e-Vj/ VT nPV = ND·e
• Forward bias: The concentration of minority carriers at the
depletion region edges increases, because Vj < V0
• Reverse bias: The concentration of minority carriers at the
depletion region edges decreases, because Vj < V0
+-
P-side
- Vj +
N-side
Vj = V0-Vext
• Forward and reverse bias:
The concentration of majority carriers in neutral regions does not change
65
Effects of the bias on the neutral regions (V)
What happens with the minority
carriers along the neutral regions?
+-
P-side
N-side
- Vj +
• This is a case of injection of an “excess” of minority carriers (see slide #36).
• Cases of interest:
a) Lp << xN (wide N-side)
 decay exponentially
b) Lp >> xN (narrow N-side)
 decay linearly
66
Effects of the bias on the neutral regions (VI)
The concentration of minority carriers along the
neutral regions under forward biasing.
Wide P and N sides
Narrow P and N sides
Vext
P-side
Minority carrier
concentration
nPV
Vext
N-side
N-side
Minority carrier
concentration
0.1 mm
pN
Length
0
0.001 mm
pNV
nPV
pNV
nP
0
P-side
nP pN
Length
Excess of minority carriers
It plays a fundamental role evaluating the switching speed of electronic devices.
67
Effects of the bias on the neutral regions (VII)
The concentration of minority carriers along the
neutral regions under reverse biasing.
Wide P and N sides
Narrow P and N sides
Vext
P-side
Minority carrier
concentration
nP
0
nPV
Vext
N-side
P-side
N-side
Minority carrier
concentration
0.1 mm
pNV
nPV
pN
Length
0
0.001 mm
nP pN
pNV
Length
Deficit (negative excess) of minority carriers
68
Carriers along the overall device
Example of a silicon PN junction
Properties of Si at 300 K
Dp=12.5 cm2/s
Dn=35 cm2/s
p=480 cm2/V·s
n=1350 cm2/V·s
ni=1010 carriers/cm3
er=11.8
P -side
N-side
NA=1015 atm/cm3 ND=1015 atm/cm3
p=100 ns
n=100 ns
Lp=0.01 mm
Ln=0.02 mm
Carriers/cm3
V0=0.596 V
1016
Forward biased
with Vext = 0.48 V
1014
pP
nN
1012
1010
nPV
pNV
108
They decay exponentially
(log scale)
Log scale
106
104
-0.3
-0.2
-0.1
0
0.1
Length [mm]
0.2
0.3
69
Calculating the current passing through a PN junction (I)
• We have addressed a lot of important issues related the
PN junction:
 Charge, electric field and voltage across the
depletion region.
 Concentration of majority and minority carriers
along the total device.
• However, the most important issue has not been
addressed so far:
How can we compute the current passing through
the device?
• Fortunately, we already have the tools to answer this
question.
70
Calculating the current passing through a PN junction (II)
Case of wide P and N sides
Vext
jtotal
2
P-side
several mm
jtotal
Vj
P
3 -+ 3
1
N
N-side
2
0.3m
jtotal = jp_total(x) + jn_total(x) = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x)
• Two questions arise:
 What carrier must be evaluated to compute the overall current?
 Where?
• Places:
 Depletion region 1 .
 Neutral regions far from the depletion region 2 .
 Neutral regions, but near the depletion region edges 3 .
71
Calculating the current passing through a PN junction (III)
Computing the overall current from the current
density due to carriers in the depletion region
Vext
jtotal
P-side
jtotal
Vj
-+
P
N
N-side
0.3m
Carriers/cm3
several mm
1016 p
N
1014
nPV
pNV
nP
Log scale
1m
jtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x)
Currents due to drift (jp_Drift and jn_Drift) have opposite direction to currents due to
diffusion. Both currents have extremely high values (very high electric field and
carrier concentration gradient) and cannot be determined precisely enough to
guarantee that the difference (which is the total current) is properly computed.
Therefore, this is not the right place.
72
Calculating the current passing through a PN junction (IV)
Vext
Computing the overall current from the current density due to
carriers in the neutral regions far from the depletion region
jtotal
P-side
Vj
P
-+
N
jtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x)
High concentration Weak field
0
jp_Drift(x) = q·p·p(x)·E(x)
jp_Diff (x) = -q·Dp·dp(x)/dx
jn_Drift(x) = q·n·n(x)·E(x)
jn_Diff (x) = q·Dn·dn(x)/dx
0
0
Constant
concentration
Few electrons in P-side
Current is due to drift of majority carriers. However, it cannot be determined properly
because we do not know the value of the “weak” electric field. Therefore, these are
not the right places.
73
Calculating the current passing through a PN junction (V)
Computing the overall current from the current density due to carriers
in the neutral regions but near the depletion region edges (I)
Vext
jtotal
P-side
Vj
P
-+
N
jtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x)
High concentration Weak field
jp_Drift(x) = q·p·p(x)·E(x)
0
jn_Drift(x) = q·n·n(x)·E(x)
jp_Diff (x) = -q·Dp·dp(x)/dx
jn_Diff (x) = q·Dn·dn(x)/dx
Few electrons in P-side
We cannot compute the total current yet, but we can compute the current density
due to minority carriers:
jn_total (x) = jn_Drift(x) + jn_Diff(x) jn_Diff (x) = q·Dn·dn(x)/dx
74
Calculating the current passing through a PN junction (VI)
Computing the overall current from the current density due to carriers
in the neutral regions but near the depletion region edges (II)
We can do the same for the holes just in the opposite side of the depletion region
Vext
Vj
jn_total(x)
P-side
Minority
carrier
concentration
P
nPV
N-side
N
pNV
nP
pN
0
Length
jn_total(x) = q·Dn·dnPV(x)/dx
Current density
-+
jp_total(x)
Taking derivatives
jn_total(x)
0
jp_total (x)= -q·Dp·dpNV(x)/dx
jp_total(x)
Length
75
Calculating the current passing through a PN junction (VII)
Computing the overall current from the current density due to carriers
in the neutral regions but near the depletion region edges (III)
What happens with carriers in the depletion region?
Vext
Vj
jn_total(x)
P-side
Current density of
minority carriers
P
jn_total(x)
0
-+
jp_total(x)
N
N-side
jp_total(x)
Length
The carrier density currents passing through the depletion region are constant
because the probability of carrier recombination is very low, due to the low
carrier concentration in that region.
76
Calculating the current passing through a PN junction (VIII)
Computing the overall current from the current density due to carriers
in the neutral regions but near the depletion region edges (IV)
Vext
Now the total current density can be easily computed
Vj
jn_total(x)
P-side
P
-+
jp_total(x)
N
N-side
Current density
jtotal
jn_total(x)
0
jp_total(x)
Very important
conclusion!!!
Length
• The total current density passing through the device can be computed as the
addition of the two minority current densities at the edges of the depletion
region
77
Calculating the current passing through a PN junction (IX)
Summary of the computing of the overall
current density in a PN junction
jn_total(x)
P-side
Vj
-+
jp_total(x)
N-side
• We need to know the variation of the minority carrier
concentrations at the depletion region edges.
• We have to calculate the gradients of these concentrations (taking
derivatives).
• We have to calculate the current densities due to these minority
carriers, which are diffusion currents.
• We must add both current densities to obtain the total current
density, which is constant along all the device. This is the total current
density passing through the device.
78
Calculating the current passing through a PN junction (X)
• Once the total current density and the minority-carrier
current densities are known, the majority-carrier current
density can be easily calculated by difference.
jn_total(x)
Vj
jp_total(x)
-+
P-side
N-side
Total current
Current density
jtotal
Majority-carrier currents, due
to both drift and diffusion
jn_total(x)
0
jp_total(x)
Length
Minority-carrier currents, only due to diffusion
79
Current passing through an asymmetrical junction (P+N-)
P-side is heavily doped (P+) and wide
N-side is slightly doped (N-) and narrow
Vext
Vj
jn_total(x)
Concentration
P+-side (wide)
-+
N--side
pNV
pN
nPV
nP
Length
Current density
0
0
jtotal
jp_total(x)
This is a case of special
interest, because it is
directly related to the
operation of Bipolar
Junction Transistors (BJTs)
jn_total(x)
Length
80
Qualitative study of the current in a
forward-biased PN junction
Vext
Vj
jtotal
P-side
Minority carrier
concentration
P
nPV
-+
N-side
N
pNV
nP
pN
0
Length
High slope High current density due
to electrons in the depletion region
High slope High current density
due to holes in the depletion region
Current density
jtotal
jn_total(x)
0
jp_total(x)
High and positive
total current density
Length
81
Qualitative study of the current in a
reverse-biased PN junction
Vext
Vj
jtotal
P-side
Minority carrier
concentration
P
nP
N-side
N
pNV
nPV
0
pN
Length
0
Low slope Low current density due
to electrons in the depletion region
Current density
-+
Low slope Low current density
due to holes in the depletion region
jn_total(x)
jp_total(x)
Low and negative total
current density
Length
jtotal
82
Quantitative study of the current in a PN junction (I)
Procedure:
1- Compute the concentration of minority holes (electrons) in the proper edge of
the depletion region when a given voltage is externally applied.
2- Compute the excess minority hole (electron) concentration at the above
mentioned place. It is also a function of the externally applied voltage.
3- Compute the decay of the excess minority hole (electron) concentration
(exponential, if the semiconductor side is wide, or linear, if it is narrow).
4- Compute the gradient of the decay of the excess minority hole (electron)
concentration just at the proper edge of the depletion region.
5- Compute the diffusion current density due to the above mentioned gradient.
5- Once the current due to minority holes (electrons) has been calculated,
repeat the same process with electrons (holes).
6- Add both current densities.
7- Compute the total current by multiplying the current density by the crosssectional area.
83
Quantitative study of the current in a PN junction (II)
The final results is:
i = IS·(eV
/VT
ext
- 1), where:
IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]
(Is is called reverse-bias saturation current)
VT = kT/q
(Shockley equation)
i
+
Vext
P
N
-
where:
A = cross-sectional area.
q = magnitude of the electronic charge (1.6·10-19 coulombs).
ni = intrinsic carrier concentration.
Dp = hole diffusion coefficient.
Dn = electron diffusion coefficient.
Lp = hole diffusion length in N-side.
Ln = electron diffusion length in P-side.
ND = donor concentration.
NA = acceptor concentration.
k = Boltzmann constant.
T = absolute temperature (in Kelvin).
84
Quantitative study of the current in a PN junction (III)
i = IS·(eV /V - 1)
IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]
VT = kT/q
ext
T
100
i [mA]
• Forward bias VO > Vext >> VT
i  IS
Vext
·e VT
 exponential dependence
- 0.25
0
0.25
0.5Vext [V]
i [nA]
• Reverse bias Vext << -VT
Vext [V]
0
-0.5
i  -IS constant
(reverse-bias saturation current)
-10
85
Quantitative study of the current in a PN junction (IV)
Wide versus narrow P and N sides
Wide sides
Vext
Narrow sides
P-side
N-side
P-side
N-side
XP
XP >> Ln
XN
XN >> Lp
XP
XP << Ln
XN
XN << Lp
Minority carrier
concentration
nPV
Minority carrier
concentration
pNV
pNV
nPV
pN
nP
0
Vext
Length
IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]
Equation i = IS·(eV
ext
/VT
0
nP pN
Length
IS = A·q·ni2·[Dp/(ND·XN)+Dn/(NA·XP)]
- 1) is valid in both cases
86
Quantitative study of the current in a PN junction (V)
The I-V characteristic in a real scale of use
i [A]
Actual I-V
characteristic
Equation i = IS·(eV /V - 1) describes the
operation in the range VO > Vext > -.
However, three questions arise:
ext
3
According to
Shockley equation
• What happens if Vext > VO?
Vext [V]
0
-4
VP 0
-
P-side
+ VmP
i 0
1
VN 0
+
+-
+
T
- +
V
j
Vext
N-side
+ -
VNm
• How does the temperature affect
this characteristic?
• What is the actual maximum voltage
that the junction can withstand?
When Vext appraches V0 (or it is even
higher), the current passing is so
high that the voltage drop in the
neutral regions is not zero. This
voltage drop is proportional to the
current (it behaves as a resistor).
87
Temperature dependence of the I-V characteristic (I)
 Reverse bias: i  -IS
where: IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)].
It should be taken into account that ni strongly depends on the temperature.
Therefore: Reverse current strongly increases when the
temperature increases. It doubles its value
when the temperature increases 10 oC.
 Forward bias:
i  IS·eV
/VT
ext
= IS·eq·V
ext
/kT
Decreases with T
Increases with T
In practice, forward current increases when the
temperature increases. For extremely high currents,
the dependence can become just the opposite.
88
Temperature dependence of the I-V characteristic (II)
Forward bias
30
Reverse bias
i
+
i [A]
V
370C
i [A]
P
N
Vext [V]
-0.25
27 0C
27 0C
Vext [V]
0
1
37 0C
-10
In both cases, the current increases for a given external voltage.
89
Maximum reverse voltage that a PN junction can withstand
• There are three different physical processes which limit
the reverse voltage that a given PN junction can withstand:
 Punch-through
It will be explained later
 Zener breakdown
This phenomenon does not take place in
power devices (two heavily doped
regions are needs).
 Avalanche breakdown
• Actual reverse current is higher than predicted due to the generation of
electron-hole pairs by collisions with the lattice.
• If the electric field is high enough, this phenomenon becomes degenerative.
i
+
-
+ Vext -
+
+-
-
--
-
i
+
+
+
P
Vext
N
0
90
Electric field in the depletion region with reverse bias
No bias
W0 =
Emax0=
Reverse bias
2·e·(N
V
A+ND)·V0
UV
p  e T q·NA·ND
PN
2·q·NA·ND·V0
e·(NA+ND)
W(Vrev) =
2·e·(N
V
A+ND)·(V0+Vrev)
UV
p  e T q·N ·N
A D
PN
2·q·NA·ND·(V0+Vrev)
Emax(Vrev) =
e·(NA+ND)
W0
P
- +
V0
W(Vrev)
N
P
-
+
N
V0+Vrev
-Emax0
-Emax(Vrev)
• As already known, both the electric field and depletion length increase.
• When the maximum electric field is high enough, the avalanche
breakdown starts.
91
Limits for the depletion region with reverse bias
Punch-through limit
WPN
Emax(Vrev) 
W(Vrev)
P
-
N
+
 |Vrev|
2·q·NA·ND·|Vrev|
e·(NA+ND)
• We must design the semiconductor
according to: Emax(Vrev) < EBR.
• The breakdown voltage is:
VBR = EBR2·e·(NA + ND)/(2q·NA·ND).
W(Vrev) 
-Emax
-EBR
Avalanche
breakdown limit
V
2·e·(N
A+ND)·|Vrev|
UV
p  e T q·NA·ND
PN
V
UV
p  e T
PN
• Moreover, W(Vrev) < WPN to avoid the
phenomenon called punch-through.
• Usually W(VBR) < WPN, which means that practical voltage limit
is not due to punch-through, but to avalanche breakdown.
92
What must we do to withstand high-voltage?
VBR =
EBR2·e·(NA + ND)
2·q·NA·ND
 VBR =
EBR2·e
2q
1 1
·( + )
ND N A
• A high value of VBR is obtained if one of the two regions has been slightly
doped (i.e., either NA or ND is relatively low).
• However, it should taken into account that low values of ND (NA) implies:
 Wide WN (WP), which also implies wide XN (XP) to avoid punch-through.
 Low nN (pP) and, therefore, low conductivity.
• If we have long length and low conductivity, then we have high resistivity.
• Hence, a trade-off between resistivity and breakdown voltage must be
established.
P+
NA
p P = NA
NA >> ND
-
+
N--side
ND
n N = ND
WN
XN
93
Maximum electric field Emax with reverse bias Vrev
-
P+
NA
NA >> ND
+
N--side
ND
r(x)
q·ND
-q·NA
x
x
-Emax
• Can we increase VBR for a given
EBR value?
- Yes, we can. We must modify
the electric field profile.
- The result is the PIN junctions.
-EBR
VBR
Vrev_N
Vrev_P
Vrev
x
The main part of the reverse
voltage is dropping in the
slightly doped region.
94
PIN junctions (I)
W(Vrev)
P
-
Vrev
+
N
• Main idea: the voltage across the device is
proportional to the dashed area (E(x) = - dV/dx).
• Can we have the same area (same voltage across
the device) with a lower value of Emax(Vrev)?
-Emax(Vrev)
• Yes, we can. We need another E(x) profile.
-Emax(Vrev) new profile (ideal)
-Emax(Vrev)
• To obtain this profile, we need a region
without space charge (undoped) inside
the PN junction.
-Emax(Vrev) new profile (real)
-Emax(Vrev)
95
PIN junctions (II)
Negative
space charge
Many holes
P-side
-
• It means P-Intrinsic-N
Positive
space charge
A few holes
and electrons
Intrinsic
r(x)
Many electrons
+
N-side
q·ND
Characteristics:
x
-q·NA
x
- Good forward operation due
to conductivity modulation.
- Low depletion capacitance.
- Slow switching operation.
-Emax
Vrev
All these characteristics will be
explained later.
x
96
Other structure to withstand high voltage: P+N-N+
Heavily doped P
P+
NA
-
q·ND1
Lightly doped N
+
r(x)
N-
Heavily doped N
ND1
q·(ND2-ND1)
+
N
+ N
D2
q·ND2
Partially
depleted
x
-q·NA
Low reverse voltage Vrev1
x
-Emax(Vrev1)
-Emax(Vrev2)
Reverse voltage Vrev2
Vrev2 > Vrev1
97
Forward-bias behaviour of structures to
withstand high voltage
P+
Intrinsic
Undoped
Lightly doped
P+
N-
N+
PIN
In both cases, there is a high-resistivity layer
(called drift region)
N+
P+N-N+
• This means that, when forward biased, bad behaviour might be expected.
• However, a new phenomenon arises and the result is quite better than
expected.
• This phenomenon is called conductivity modulation. In this case,
high-level injection takes place.
98
Injection levels
Carriers/cm3
Low-level injection:
nN(0+) >> pNV(0+)
High-level injection:
nN(0+)  pNV(0+)
1016
1014
pP
pP
nN
1012
1010
108
106
104
-0.3
P+-side
nPV
-0.2
P+-side
N--side
Log scale
0.2
N--side
nPV
Not possible!
pNV
-0.1 0- 0+
0.1
Length [mm]
nN
pNV
Log scale
0.3-0.3
-0.2
-0.1 0- 0+
0.1
Length [mm]
0.2
0.3
• Low-level injection has been assumed so far, for PN and P+N- junctions.
• In the case of a P+N- junction, this assumption is only valid if the forward
bias is not very intense. Else, high-level injection starts.
• If the forward voltage is high enough, pNV(0+) approaches nN(0+). In this
case, nN does not remain constant any more, but it notably increases.
99
Conductivity modulation
Drift region
P+
NA = 1019
Holes are injected
from the P+-side
nP+
10
N-
N+
ND1 = 1014
ND2 = 1019
nN-  pN-
1016
P+N-N+
1014
Electrons are injected
from the N+-side
106
pN+
10
• There is carrier injection from both highly doped regions to the drift
region. This is called double injection.
• This phenomenon substantially increases the carrier concentration in
the drift region, thus dramatically reducing the device resistivity.
100
Semiconductor junctions designed to
withstand high voltage
PIN
P+N-N+
Summary
• A high-resistivity region (drift region) is needed to withstand high
voltage when the junction is reverse biased.
• Fortunately, this high-resistivity “magically” disappears when the
junction is forward biased if the device is properly designed to have
conductivity modulation.
• Due to this, devices where the current is passing through P-type and
N-type regions (bipolar devices) have superior performances in on-state
than devices where the current always passes through the same type
(either P or N) of extrinsic semiconductor (unipolar devices).
• Unfortunately, bipolar devices have inferior switching characteristics
than unipolar devices.
• Due to this, a trade-off between conduction losses and switching losses
has to be established frequently selecting power semiconductor devices.
101
Transient and AC operation of a PN junction
If we change the bias conditions instantaneusly,
can the current change instantaneusly as well?
• The answer is “no, it cannot”.
• This is due to the fact that the current conducted by a PN
junction depends on the minority carrier concentration just at the
edges of the depletion region and the voltage withstood by a PN
junction depends on the depletion region width.
• In both cases, carriers have to be either generated or
recombined or moved, which always takes time.
These non-idealities can characterize as:
• Parasitic capacitances (useful for linear applications)
• Switching times (useful for switching applications)
102
Parasitic capacitances: depletion layer capacitance (I)
(also known as junction capacitance)
This is the dominant capacitance in reverse bias
P-side
-
+
N-side
Zona N
VO+Vext+Vext
Vext + Vext
r(x)
x
Carriers are pulled out from the depletion region when Vext is
increased in Vext . Additional space charge has been generated.
103
Parasitic capacitances: depletion layer capacitance (II)
PN junction
Capacitor
Vext
-
P
N
+
Vext + Vext
P
-
- +
+
Vext
Vext + Vext
+ + +
+++++
- - -
-----
N
• Capacitor: new charges are located at the same
distance  constant capacitance.
• PN junction: new charges are located farther away
from each other  non-constant capacitance.
104
Parasitic capacitances: depletion layer capacitance (III)
dQ
-dQ
• As it is a non-constant capacitance,
static and dynamic capacitances could be
defined. The latter is defined as:
Cj=dQ/dV=e·A/W(Vext)
W(Vext)
As:
Cj
V
2·e·(N
A+ND)·(V0-Vext)
UV
p  e T
q·NA·ND
PN
W(Vext) =
Then:
Vext
0
V
Cj = A·
UV
e·q·NA·ND
p  e T
PN
2·(N
A+ND)·(V0-Vext)
In an “abrupt” PN junction (as we have considered so far),
this capacitance is a K·(V0-Vext)-1/2 -type function
105
Parasitic capacitances: diffusion capacitance (I)
This capacitance is the one dominant in forward bias
Cj
0
Reverse bias
Vext
Forward bias
• Cj increases when the PN junction is forward biased.
• However, depletion layer capacitance only dominates the reactance of a
PN junction under reverse bias.
• For forward bias, the diffusion capacitance (due to the charge
stored in the neutral regions) becomes dominant.
106
Parasitic capacitances: diffusion capacitance (II)
Carriers/cm3
1016
pP
1014
nN
V=240mV
nPV
1012
Increase of minority carriers due
to a increase of 60mV in forward
bias
pNV
V=180mV
1010
-3
-2
-1
0
1
2
3
Longitud [mm]
• This increase in electric charge is a function of the forward bias voltage.
• This means that a capacitive effect takes place in these conditions.
• The dynamic capacitance thus obtained is called diffusion capacitance.
107
Switching times in PN junctions (I)
Let us consider a PN diode as PN junction. The results obtained can be
generalized to PN junctions in other semiconductor devices.
R
a
V1
b
V2
i
V1/R
v
-V2
i
+
Transition from “a” to “b” (switching off)
in a wide time scale (ms or s).
v
t
t
The diode behaviour seems to
be ideal in this time scale.
108
Switching times in PN junctions (II)
Transition from “a” to “b” (switching off) in a narrow time scale (s or ns).
R
a
V1
b
V2
i
+
i
V1/R
trr
v
-
ts = storage time.
tf = fall time.
trr = reverse recovery time.
Reverse recovery peak
t
ts
tf (i= -0.1·V2/R)
-V2/R
v
t
-V2
109
Switching times in PN junctions (III)
R
a
i
+
b
V2
V1
i
• This is because the junction cannot
withstand voltage until all the excess of
minoritary carriers disappears from the
neutral regions.
v
-
Carriers/cm3
8·1013
V1/R
t3
t0
Why does this evolution occur?
t4
t
t1 t2
pNV
nPV
4·1013
t1
-V2/R
0
v
t
t2
-0.1
-V2
t0
t3 t4
0
Length [mm]
0.1
110
Switching times in PN junctions (IV)
R
a
i
+
b
V2
V1
Transition from “b” to “a” (switching on)
in a narrow time scale (s or ns).
v
-
i
0,9·V1/R
t0
td
td = delay time.
8·1013
t4
tr
t2 t3
pNV
nPV
4·1013
t1
0,1·V1/R
Carriers/cm3
t3
0
t2
t4
t1 t0
tfr
tr = rise time.
tfr = td + tr = forward recovery time.
-0. 1
0
Length [mm]
0.1
111
Trade-off between static and dynamic behaviour in
PN junctions
P+
NA = 1019
1016
Excess of
electrons
in N-
nP+
N-
N+
ND1 = 1014
ND2 = 1019
nN-  pN1014
106
Excess of
holes in N-
pN+
Log scale
10
Excess of
electrons in P+
10
Excess of
holes in N+
• P+N-N+ and PIN structures allow us to
combine high reverse voltage (due to a wide
drift region) and low forward resistivity (due
to conductivity modulation).
• However, these structures imply large
excess of minority carriers (even majority
carries due to conductivity modulation). This
excess of carriers must be eliminated when
the device switches off to allow the device to
withstand voltage.
• The time to remove this excess of carriers
depends on the width of the drift layer. If the
drift layer is shorter than a hole diffusion
length, then very little charge is stored and
the device switches off fast. In this case,
however, the device cannot withstand high
reverse voltages.
• The switching process can be made still faster by purposely adding “recombination
centers”, such as Au atoms in Si, to increase the recombination rate. However, this fact
112
can deteriorate the conductivity modulation.
Introduction to the metal-semiconductor junctions (I)
• Metals have many more electrons than semiconductors. However, metals
and semiconductors are different materials. This is not the case of a PN
junction, where the two sides (P and N) are made up of the same material.
• In a PN junction made up of a given semiconductor, electrons (holes) move
from the N-side (P-side) to the P-side (N-side) due to diffusion, until the builtin voltage establishes an equilibrium between diffusion and drift currents.
• In a metal-semiconductor junction, the electron movement when the
junction is being built strongly depends on the work function of both
materials. The higher the work function, the more difficult for the electrons to
eject the material.
• 4 possibilities exist when you build a Metal-Semiconductor (MS) junction:
N-type semiconductor
Metal
Case #1: an N-type
semiconductor transfers
electrons to a metal
Electrons (thin sheet)
-- + +
-- +
+
-- + +
- + +
N
N-type
Donor ions
113
Introduction to the metal-semiconductor junctions (II)
Case #2: a metal transfers electrons to a P-type semiconductor
Lack of electrons (thin sheet )
Recombination between the
transferred electrons and the
P-side holes takes place in this
edge.
Metal
P-type semiconductor
+ - +
+
+ - P
+
P-type
+ - +
+ - -
Acceptor ions
• In Case #1 and Case #2, a depletion region in the semiconductor side has
been generated.
• This depletion region has a built-in voltage that stops the electron diffusion.
• This built-in voltage can be decreased by external forward bias (thus
allowing massive electron diffusion) or increased by external reverse bias
(avoiding electron diffusion).
• The final result is that it works like a rectifying contact (similar to a PN
114
junction).
Introduction to the metal-semiconductor junctions (III)
Case #3: a metal transfers electrons to an N-type semiconductor
N-type semiconductor
Metal
Lack of electrons
(thin sheet)
+ + + + + + + + -
N-type
Electrons
(thin sheet)
Case #4: a P-type semiconductor transfers electrons to a metal
P-type semiconductor
Metal
Electrons
(thin sheet)
-
- +
- +
+
+
+
+
+
+
P-type
Holes
(thin sheet)
• We have an ohmic contact (non-rectifying contact) in both cases.
115
Rectifying contacts (I)
W0
Case #1: an N-type
semiconductor transfers
electrons to a metal
-- + +
-- +
+
Metal
-- + +
- + +
N
N-type
ND
• The width of the depletion region, the maximum electric field and the
depletion layer capacitance can be calculated as in the case of a PN junction
with an extremely-doped P side (i.e., NA  ).
• Therefore:
W0 =
2·e·V0
q·ND
Emax0 =
2·q·ND·V0
e
Cj0 = A·
However, the built-in voltage and the I-V characteristic depend on
the work function of both the semiconductor and the metal.
V
e·q·N
D
U
V
T
p  e2·V
PN
0
116
Rectifying contacts (II)
• Built-in voltage:
V0 = (Fm - Fs_N)/q, where:
Fm = metal work function.
FS_n = N-type semiconductor work function.
• Barrier voltage to avoid electron diffusion without bias:
VB = (Fm - cS_n)/q, where:
cS_n = N-type semiconductor electron affinity.
To define these concepts properly,
• I-V characteristic:
we should introduce others. This
task, however, is beyond the scope
i = IS·(eVext/VT - 1), as in a PN junction.
of this course.
However, the value of Is has a very different value:
IS = A*·A·T2·e-V /V , where:
B
T
A* = Richardson constant (120 amps/(cm2·K2)).
117
Rectifying contacts (III)
• There is a type of diode based on the operation of a rectifying
contact. It is the Schottky diode. Schottky diodes are widely
used in many applications, including RF (telecom) circuits and
low-voltage, medium-power power converters.
• Main features:
 Lower forward voltage drop than a similar-range, PNjunction diode.
 They are faster than PN diodes because minority carriers
hardly play any role in the current conduction process. They
are majority carrier devices.
 However, they always have a higher reverse current (this
is not a big problem).
 When they are made up of silicon, the maximum reverse
voltage (compatible with reasonable drop voltage in forward
bias) is about 200 V.
 However, Schottky diodes made up of wide-bandgap
materials (such as silicon carbide and gallium nitride) reach
breakdown voltages as high as 600-1200 V.
Schematic
Symbol
118
Ohmic contacts
• There are two different possibilities to obtain ohmic contacts:
a) According to the previous slides, to have an MS junction
corresponding to Case #3 or to Case #4.
b) To have MS junctions corresponding to Cases #1 or #2, but
with an extremely-doped semiconductor side (1019 atoms/cm3).
In this situation, electrons can flow in both directions by a
tunneling process.
Beyond the scope of this course, as well.
Ohmic contacts
Ohmic contacts
P+
P
NA1 = 1019 NA2 = 1016
Rectifying contacts
N
ND1 = 1016
N
ND1 = 1016
N+
PN diode
ND2 = 1019
Ohmic contacts
N+
ND2 = 1019
Schottky diode
119