power electronic devices
DESCRIPTION
Power Supply Systems. Electrical Energy Conversion and Power Systems . Universidad de Oviedo. Power Electronic Devices. Semester 1 . Lecturer: Javier Sebastián. Research Group Power supply Systems (Sistemas Electrónicos de Alimentación). - PowerPoint PPT PresentationTRANSCRIPT
Power Electronic Devices
Semester 1
Lecturer: Javier Sebastián
Electrical Energy Conversion and Power Systems
Universidadde Oviedo
Power Supply Systems
2
Research GroupPower supply Systems
(Sistemas Electrónicos de Alimentación)
Javier Sebastián
Dr. Electrical Engineer (Ingeniero Industrial)Full professorRoom 3.1.21Edificio nº 3, Campus Universitario de Viesques 33204 Gijón (Asturias). Spain
Phone (direct): 985 18 20 85 Phone (secretary): 985 18 20 87Fax: 985 18 21 38E-mail: [email protected]: http://www.unioviedo.es/sebas/
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).
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Outline
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Outline
Review of physical principles of semiconductors
Power electronics devices
G
D
S
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 Metal-Oxide-Semiconductor Field Effect Transistors (MOSFETs), both in N-channel- and P-channel types.
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Previous requirements
Lesson 1 - Review of the physical principles of operation of semiconductor devices
Semester 1 - Power Electronics Devices
Electrical Energy Conversion and Power Systems
Universidadde Oviedo
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OutlineReview 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.
8
Energy level in a semiconductor as a function of inter-atomic spacing
Inter-atomic spacing
Ener
gy o
f ele
ctro
ns
--
- -
------
Actual spacing
At 0 K, empty
At 0 K, filled
9
Concept of band diagram
Band gap
Valence band
Conduction band
Eg
4 electrons/atom--
--
4 states/atom
Ener
gy o
f ele
ctro
ns
Empty at 0 K
Filled at 0 KMaterial Eg [eV]
Ge 0.66 Si 1.1
4H - SiC 3.26GaN 3.39
10
Band structure for insulators, semiconductors and metals at 0 K
Band gap Eg
Conduction band
Valence band
SemiconductorEg=0.5-2 eV
Band gap Eg
Conduction band
Valence band
InsulatorEg= 5-10 eV
MetalsNo Eg
Valence band
Conduction band
Overlap
11
Band structure for semiconductors at room temperature. Concept of “hole”
SemiconductorEg=0.5-2 eV
Eg
Conduction band
Valence band
-- -+-
• 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.
Visualization using the bonding model
Si Si
Si Si
- -
- -
- -
- -
- --
- -
- -
--
+
12
Si
-
- -
- -
- -
- --
- -Si
Si Si
- -
-
12
-
+
Recombination
Si
-
- -
- -
- -
- --
- -Si
Si Si
- -
-
-
+
-
Concepts of generation and recombination
Eg
-- -+-
Generation
Eg
-- -+
-
13+-
+ + + + + + + -------
-
Si Si Si Si
Si Si Si Si
- - - - -
- - - - -
- - -
- - -
--
- --
--
- - - -
- - - -
-
+
--
+
Why both holes and electrons are electric charge carriers?
• In general, there will be electric current due to both electrons and holes• Example: when there is an electric field in the semiconductor lattice
14
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)
15
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
- -
12
34
Sb
Si
Si
--
- -
Donor
- - -
- -
-
-
- -Si
- -
12
3
Al
Si
Si
--
-
Acceptor
5-
+ -
-+
16
N-type and P-type semiconductors
- - -
- -
--
-
- -
Si
- -1
2
34
Sb+
5-
Si
Si
--
- -Donor
- - -
- -
-
-
- -
Si
- -1
2
3
Al-
Si
Si
+
--
--Acceptor
N-type semiconductor: • Majority carriers are electrons.• Minority carriers are holes.• Positively-charged atoms of donor
(positive ions).
P-type semiconductor:• Majority carriers are holes.• Minority carriers are electrons.• Negatively-charged atoms of
acceptor (negative ions).
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hole
electron
+
-
P-type silicon Acceptor ions (negative ions)
Al- Al- Al- Al- Al-
Al- Al- Al- Al- Al-
-
+
Thermal generation
Donor ions (positive ions)
Sb+ Sb+ Sb+Sb+
Sb+
Sb+ Sb+ Sb+ Sb+ Sb+
-
+
--
-
- -
--
---
Thermal generation
N-type silicon
Charges in N-type and P-type semiconductors
++
+
+
+
+
+
++
+
18
Sb+ Sb+ Sb+
Sb+ Sb+ Sb+
-
+
--
- ---
N-type
Charge carries in N-type and P-type semiconductors
P-type
Al- Al- Al-
Al- Al- Al-
-
+
+ +
+
+
+
+
• Concentration of majority carriers: pP
• Concentration of minority carriers: nP
• 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)
Very important equations!!!
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Sb+ Sb+ Sb+
Sb+ Sb+ Sb+
-
+
--
- ---
N-type
Static charges in N-type and P-type semiconductors
P-type
Al- Al- Al-
Al- 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)
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Sb+ Sb+ Sb+
Sb+ Sb+ Sb+
-
+
--
- ---
N-type
Neutrality in N-type and P-type semiconductors
P-type
Al- Al- Al-
Al- Al- Al--
+
+ +
+
+
+
+
• Positive charges in volume V: pP·V
• Negative charges in volume V: nP·V + NA·V
• Neutrality: pP = nP + NA
• Silicon, aluminium and antimony were neutral before being used Þ The extrinsic semiconductor must be neutral, too.
• Negative charges in volume V: nN·V
• Positive charges in volume V: pN·V + ND·V
• Neutrality: nN = pN + ND
Very important equations!!!
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Sb+ Sb+ Sb+
Sb+ Sb+ Sb+
-
+
--
- ---
N-type
Calculating the concentration of electrons and holes (I)
P-type
Al- Al- Al-
Al- Al- Al-
-
+
+ +
+
+
+
+
• Neutrality: pP = nP + NA
• Mass action law: pP·nP = ni2
2 known (NA and ni) and 2 unkown (pP and nP) variables Þ can be solved
• Neutrality: nN = pN + ND
• Mass action law: nN·pN = ni2
2 known (ND and ni) and 2 unkown (nN and pN) variables Þ can be solved
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Sb+ Sb+ Sb+
Sb+ Sb+ Sb+
-
+
--
- ---
N-type
Calculating the concentration of electrons and holes (II)
P-type
Al- Al- Al-
Al- Al- Al-
-
+
+ +
+
+
+
+
• NA >> ni
• Neutrality: pP » NA
• Mass action law: nP » ni2/NA
• ND >> ni
• Neutrality: nN » ND
• Mass action law: pN » ni2/ND
• Frequent case: quite heavy doped semiconductors
Very useful equations!!!
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Mechanisms to conduct electric current: Drift (I)
• Semiconductors can conduct electric current due to the presence of an electric field E
jp
jn
E
+ + + + +
- - - - -
-
-
-+
+
- +
+
+
+
+
+-
-
--
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.
Mechanisms to conduct electric current: Drift (II)
Ge (cm2/V·s)
Si (cm2/V·s)
GaAs (cm2/V·s)
n 3900 1350 8500 p 1900 480 400
q = magnitude of the electronic charge (1.6·10-19 coulombs).
p = hole mobility.
n = electron mobility.
p = hole concentration.
n = electron concentration.
E = electric field.
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jp_Drift = q·p·p·E
jn_Drift = q·n·n·E
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Electrons have migrated due to “diffusion” (you can see the same phenomenon in gases)
jn_Diff
1 2
n1 n2 < n1
--
--
--
--
--
--
--
--
--
Mechanisms to conduct electric current: Diffusion (I)
26jn_Diff
1 2
n1 n2 < n1
• If we maintain a different concentration of electrons, we also maintain the motion of electrons in the lattice
n
-
--
--
---
-
-
-
-
--
-
-
--
-
-
-
-
--
---
-
-
-
--
Mechanisms to conduct electric current: Diffusion (II)
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jn_Diff
1 2
n1 n2 < n1- - ---
--
--
-- -
--
-
-
- - --
-
-
n
The current density is proportional to the electron concentration gradient:
jn_Diff = q·Dn· n
Mechanisms to conduct electric current: Diffusion (III)
Dn = electron diffusion coefficient.
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1 2
p1 p2 < p1
++
+
+
+
+
+
+
+
+
++
+
++
+
+ + +
+
+
+
p
The current density is proportional to the hole concentration gradient:
jp_Diff = -q·Dp· p
Dp = hole diffusion coefficient.
jp_Diff
Mechanisms to conduct electric current: Diffusion (IV)
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Ge (cm2/s)
Si (cm2/s)
GaAs (cm2/s)
Dn 100 35 220 Dp 50 12.5 10
jp_Diff = -q·Dp· p
jn_Diff = q·Dn· n
Mechanisms to conduct electric current: Diffusion (V)
q = magnitude of the electronic charge (1.6·10-19 coulombs).
Dp = hole diffusion coefficient.
Dn = electron diffusion coefficient.
Ñp = hole concentration gradient.
Ñn = electron concentration gradient.
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Summary of conduction mechanisms
jp_Diff = -q·Dp· p
jn_Diff = q·Dn· n
jp_Drift = q·p·p·E
jn_Drift = q·n·n·E
• 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.
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A B
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.
Continuity equations (I)
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) .
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A B
Continuity equations (II)
• Different current density of holes across “A” and “B”.
-
+
+
-
+
+
jp_Ajp_B
jn_B
A B
+ -• Excess of carriers over the
equilibrium (mass action law).
A B
Light
-
+
• Generation of electron-hole pairs by radiation (light) .
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·jp/q
-
p/t = GL- [p(t)-p ]/p
Taking into account the three effects, we obtain the continuity equation for holes:
Continuity equations (III)
Variation due to the excess of carriers over the equilibrium
Total time variation of holes
Variation due to the generation of electron-hole pairs by light Variation due to the
different current density of holes across “A” and “B”GL: rate of generation of electron-hole pairs by light.
p: hole minority-carrier lifetime.p: hole concentration in steady-state.
·jn/q
+
n/t = GL- [n(t)-n ]/n
Similarly, we can obtain the continuity equation for electrons:
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• We generate an “excess” of electron-hole pairs by injecting light into a piece of N-type silicon and we reach the steady-state.
Time evolution of an “excess” of minority carries (I)
·jp/q
-
p/t = GL- [p(t)-p ]/p
00
Þ p0= GL·p + p
NN+ + + + ++ + + + +
p0+
+
+
+
+
+
+
+
p
• Now the light injected disappears. We want to compute the time evolution of the hole concentration afterwards.
N+ + + + +
p0+
+
+
+
+
+
+
+p(t)p
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Time evolution of an “excess” of minority carries (II)
• We can also compute the time evolution of the hole concentration from the continuity equation:
·jp/q
-
p/t = GL- [p(t)-p ]/p
00
After integrating Þ p(t) = p + (p- p)·e-tp
p p
p0
p(t)t
p
Tangent lineSame area
• Physical interpretation: There is an appreciable increase of holes during 3-5 times p.
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• 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.
Spatial evolution of an “excess” of minority carries (I)
·jp/q
-
p/t = GL- [p(t, x)-p ]/p
00
Þ 0 = - [p(x)-p ]/p + Dp·2[p(x)-p ]/x2
x xN
+ + + ++
+
++
+ N
+
+
+
+
+
+ +
+ +
0
p0
p
After integrating Þ p(x) = p + C1·e-x/Lp + C2·ex/Lp
where : Lp=(Dp· p)1/2 is the minority hole diffusion length
37
• Cases:
a) Lp << xN (wide crystal): Þ p(x) = p + (p- p)·e-xLp (decay exponentially).
b) Lp >> xN (narrow crystal): Þ p(x) = p + (p- p)·(xN-x)/xN (decay linearly).
c) Other cases Þ hyperbolic sine evolution.
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 crystalLp: hole diffusion length
p(x)
p
p0
xxN
p(x)p
p0
xxN
Lp << xN (wide) Lp >> xN (narrow)
Tangent line
Lp
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What happens if we remove the barrier?
P-type silicon
-+
Al- Al- Al- Al-
Al- Al- Al- Al-
++
+
+
+
+
+
+
Barrier to avoid carrier diffusion
N-type silicon
Sb+ Sb+ Sb+ Sb+
Sb+ Sb+ Sb+ Sb+
-
+
--
-
-
---
-
Concept of PN junction (I)
39
Are all the carriers to be diffused?
Al-
Al-
P-side
-+
Al- Al- Al-
Al- Al- Al-
++
+
+
+
+
N-side
Sb+ Sb+ Sb+ Sb+
Sb+ Sb+ Sb+ Sb+
+
-- -
---
-+
+
-
-Holes begin to diffuse from the P-side to the N-side. Similarly, electrons diffuse from the
N-side to the P-side
Concept of PN junction (II)
40
Al-
Al-
P-side
-
+
Al- Al- Al-
Al- Al- Al-
N-side
Sb+ Sb+ Sb+ Sb+
Sb+ Sb+ Sb+ Sb+
+
-+
+-
-+
+
+
+
+
+
-
-
-
-
-
-
Are all the carriers to be diffused?
Is this situation “the final situation”? The answer is no
Non-neutral P-type region, but negatively charged
Non-neutral N-type region, but positively charged
Concept of PN junction (III)
41
An electric field appears just in the boundary between both regions (we call this boundary
metallurgical junction)
+-E
Al-
Al-
P-side
-+
Al- Al- Al-
Al- Al- Al-
++
+
+
+
+
N-side
Sb+ Sb+ Sb+ Sb+
Sb+ Sb+ Sb+ Sb+
+
-- -
---
-+
+
-
-
Concept of PN junction (IV)
42
The electric field limits the carrier diffusion
Al-
Al-
P-side
Al- Al- Al-
Al- Al- Al-
N-type
Sb+ Sb+ Sb+ Sb+
Sb+ Sb+ Sb+ Sb+
+
-+-
E
Due to diffusion ( ¬)
Due to drift (electric field) (¬)
+
+
-
-
Concept of PN junction (V)• Now, we do zoom over the metallurgical junction
43
Depletion region, or space charge region, or transition regionUnbalanced charge exists because carriers barely exist
Neutral P-type region (holes are balanced by
negative ions )
Al- Al- Al-
Al- Al- Al-
+
+
+
+
+
+
Al-
Al-
Sb+
Sb+
+-E Neutral N-type region
(electrons are balanced by positive ions )
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
-
--
- -
-
Concept of PN junction (VI)• Steady-state situation near the metallurgical junction
44
Many holes, but neutral
Many electrons, but neutral
P-side(neutral)
N-side(neutral)
+ -
Metallurgical junction
E
V0
Concept of PN junction (VII)• Summary and terminology
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
45
Computing the built-in voltage V0 (I)
- +- +- +- +- +- +
- +- +
P-side
N-side
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
jp_Drift
+Due to diffusion
jp_Diff
-Due to drift
jn_Drift
- Due to diffusion
jn_Diff Currents must cancel each other
Currents must cancel each other
46
Computing the built-in voltage V0 (II)
+ -+ -+ -+ -
- +N-side
Zona P
V0
jp_Drift = - jp_Diff
+
+
(hole concentration in N-side ) pN
+
pP (hole concentration in P-side )
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
+ + + ++ + + +
+
Due to drift
jp_Drift
+Due to diffusion
jp_Diff
47
jp_Diff = -q·Dp·dp/dx
jp_Drift = q·p·p·E
E = -dV/dxE
jp_Drift = - jp_Diff
Equations:
Therefore: dV = -(Dp/p)·dp/p
Computing the built-in voltage V0 (III)
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:Dp/p = Dn/n = kT/q = VT (Einstein relation)
k = Boltzmann constant.VT = 26 mV at 300 K.
48
+ -+ -+ -+ -
- +
N-side: many electrons
Zona P
V0
pN
+
pP
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
+ + + ++ + + +
+
+-
-P-side: many holes
- - - -- - - -- - - -- - - -- - - -- - - -
- - - -- - - -
nN
-
nP
V0 = VT·ln(pP/pN) and also V0 = VT·ln(nN/nP)
Computing the built-in voltage V0 (IV)Summary (I)
Almost no holesor electronsAlmost no electrons Almost no holes
49
V0 = VT·ln(pP/pN) and also V0 = VT·ln(nN/nP)
Computing the built-in voltage V0 (V)
Summary (II)
P-side N-side+ -
V0
If NA >> ni (current case)
pP = NA nP = ni2/NA
NA, pP, nP
If ND >> ni (current case)
nN = ND pN = ni2/ND
ND, nN, pN
V0 = VT·ln(NA·ND/ni2), VT = 26 mV at 300 K
50
P-side N-side
W0
Charge neutrality implies: NA·WP0 = ND·WN0
WN0
Sb+ Sb+
Sb+ Sb+
Sb+ Sb+ Sb+
Sb+
Sb+
--
-
ND
WP0
Al- Al-
Al-Al-
Al-
Al-
Al-
Al-
+
NA
+
The heavier doped a side, the narrower the depletion region in that side
Depletion width in P-side and in N-side
51
Calculating the electric field E and the total depletion width W0 (I)
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 need to know: • The electric field E.
• The total depletion width W0.
E(x)+ -
P-side N-side- +
W0
WN0WP0
V0- +rN-side rP-side
We will use Gauss’ law and the relationship between electric field and voltage
52
Calculating the electric field E and the total depletion width W0 (II)E(x)
+ -
P-side N-side- +
- V0 +
E(x) -Emax0
Electric field x
r(x)Charge density x
q·ND
-q·NA
V(x) V0Voltage x
W0
• Gauss’ law: ·E(x) = r(x)/e
• Voltage and electric field: E(x) = - V
53
Calculating the electric field E and the total depletion width W0 (III)
-Emax0
Electric field x
Charge density x
q·ND
-q·NA
V0Voltage x
E(x)+ -
P-side N-side- +
- V0 +W0
After applying Gauss’ law and the relationship between electric field and voltage, we obtain:
2·e·(NA+ND)·V0W0 =q·NA·ND
e·(NA+ND)Emax0 =
2·q·NA·ND·V0
54
Summary of the study of the PN junction with no external bias
Almost no holes or electrons, but space charge, electric
field and voltage
P-side: many holesAlmost no electrons
N-side: many electronsAlmost no holes
P-sideDoped NA
N-sideDoped ND
- +
W0
WN0WP0
Electric field at the metallurgical junction Emax0- V0 +
V0 = VT·ln(NA·ND/ni2)
2·e·(NA+ND)·V0W0 =q·NA·ND
e·(NA+ND)Emax0 =
2·q·NA·ND·V0
NA·WP0 = ND·WN0W0 = WP0 + WN0
Very important equations!!!
Connecting external terminals to a PN junction
Therefore:V = 0, i = 0Hence:VmP – V0 + VNm = 0And:VmP + VNm = V0
V0- +
P-side N-side
+ -
VmP
-+VNm
-+
V = 0
i = 0
Conclusion:The built-in voltages across each metal-semiconductor contact cancel out the effect of V0 in such a way that V0 does not appear externally.
55
metal-semiconductor contacts
No energy can be dissipated here
56
Vj
- +VmP
-+VNm
-+i 0
P-side N-side
+ -Vext
-+
Low resistivity:VN=0
Low resistivity:VP=0
VmP and VNm do not change and, therefore VmP+VNm= V0
Biasing the PN junction: forward bias
V0 becomes Vj now
Conclusion:The built-in voltage across the junction has decreased Vext volts
Vext = VmP - Vj + VNm = V0 - Vj
Therefore: Vj = V0 - Vext
57
Vj
- +VmP
-+VNm
-+i 0
P-side N-side
+ - Vext
- +
Low resistivity:VN=0
Low resistivity:VP=0
Biasing the PN junction: reverse bias
Conclusion:The built-in voltage across the junction has increased Vext volts
Vext = -VmP +Vj - VNm = -V0 + Vj
Therefore: Vj = V0 + Vext
VmP and VNm do not change and, therefore VmP+VNm= V0
58
Vext-+
=Vj
- +P-side N-side
+ -
i
Biasing the PN junction: notation for a general case
Conclusion: • Always: Vj = V0 - Vext, being: 0 < Vext < V0 (forward biased) Vext < 0 (reverse biased)
59
Effects of the bias on the depletion region
We must replace V0 with Vj, that is, replace V0 with V0-Vext
Always: V0 = VT·ln(NA·ND/ni2)
Without bias
2·e·(NA+ND)·V0W0 =q·NA·ND
e·(NA+ND)Emax0 =
2·q·NA·ND·V0
Vj0 = V0
With bias
2·e·(NA+ND)·(V0-Vext)W(Vext) =q·NA·ND
e·(NA+ND)Emax(Vext) =
2·q·NA·ND·(V0-Vext)
Vj (Vext) = V0 - Vext
60
P-side - + N-side
V0
W0
r(x)
x
E(x)
-Emax0
x
Vj(x)V0 x
-Emax
V0-V1
• Less spatial charge
V0-V1
P-side - + N-side
V1
W
Effects of the forward bias on the depletion
region
• Lower electric field
• Lower built-in voltage
61
P-side - + N-side
V0
W0
r(x)x
E(x)
-Emax0
x
Vj(x)
V0x
V0+V2
-Emax
V0+V2
V2
P-side - + N-side
W
• More spatial charge
• Higher electric field
• Higher built-in voltage
Effects of the reverse bias on the depletion
region
62
nNnP
No bias: V0 = VT·ln(nN/nP)
- +Zona P
P-side
- - -- - -- - -- - -- - -
------ - - -
- - --- - - -
-
-
-
+ -+ -+ -+ -
V0
Forward bias: V0-Vext =VT·ln(nNV/nPV)
nNVnPV
-
------
--
V0-Vext
Effects of the bias on the neutral regions (I)
For holes with forward bias: V0-Vext =VT·ln(pPV/pNV)
nPV = nN·e -(V0-Vext)/ VT
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) Þ
V0-Vext =VT·ln(pP/pNV) Þ pNV = pP·e-(V0-Vext)/ VT
63
• As nN » ND and pP » NA, then:
nPV = ND·e -(V0-Vext)/ VT pNV = NA·e-(V0-Vext)/ VT
Vext
-+ =
Vj- +
Vj = V0-Vext
Zona P Zona N
+ -
pP = NA nN = ND
Effects of the bias on the neutral regions (III)
64
pNV = NA·e -Vj/ VT
Biased junction
pN = ni2/ND = NA·e -V0/ VT
Non-biased junction
nPV = ND·e -Vj/ VT
Biased junction
nP = ni2/NA = ND·e-V0/ VT
Non-biased junction
Vj = V0-Vext
Vj- +
P-side N-side
+ -
Effects of the bias on the neutral regions (IV)
65
pNV = NA·e -Vj/ VT nPV = ND·e -Vj/ VT
• 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
• Forward and reverse bias: The concentration of majority carriers in neutral regions does not change
Concentration of minority carriers :
Vj- +
P-side N-side
+ -
Effects of the bias on the neutral regions (V)
66
What happens with the minority carriers along the neutral regions?
• This is a case of injection of an “excess” of minority carriers (see slide #36).
• Cases of interest:
a) Lp << xN (wide N-side)Þ decay exponentially
b) Lp >> xN (narrow N-side) Þ decay linearly
67
Length
pNVnPV
0
Minority carrier concentration
pNnP
Length
pNVnPV
0
Minority carrier concentration
nP pN
Vext
P-side N-side
Wide P and N sides
Vext
P-side N-side
Narrow P and N sides
Effects of the bias on the neutral regions (VI)
The concentration of minority carriers along the neutral regions under forward biasing.
0.1 mm 0.001 mm
Excess of minority carriersIt plays a fundamental role evaluating the switching speed of electronic
devices.
68
Length
pNVnPV
0
Minority carrier concentration
pNnP
Length
pNVnPV
0
Minority carrier concentration
nP pN
Vext
P-side N-side
Wide P and N sides
Vext
P-side N-side
Narrow P and N sides
Effects of the bias on the neutral regions (VII)
The concentration of minority carriers along the neutral regions under reverse biasing.
0.1 mm 0.001 mm
Deficit (negative excess) of minority carriers
69
Carriers along the overall device
Properties of Si at 300 KDp=12.5 cm2/sDn=35 cm2/s p=480 cm2/V·sn=1350 cm2/V·sni=1010 carriers/cm3
er=11.8
Example of a silicon PN junction
V0=0.596 V
NA=1015 atm/cm3
p=100 ns
Lp=0.01 mm
ND=1015 atm/cm3
n=100 ns
Ln=0.02 mm
P -side N-side
Forward biased with Vext = 0.48 V
pNV
pP
nPV
nN
Carriers/cm3
104
1012
1014
1016
-0.3 -0.2 -0.1 0 0.1 0.2 0.3Length [mm]
1010
108
106 Log scaleThey decay exponentially
(log scale)
70
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.
71
several mm
Vext
Vj
0.3m
P N- +P-side N-side
• Two questions arise: What carrier must be evaluated to compute the overall current? Where?
Calculating the current passing through a PN junction (II)
jtotal = jp_total(x) + jn_total(x) = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x)
jtotal jtotal
Case of wide P and N sides
• Places: Depletion region 1 . Neutral regions far from the depletion region 2 . Neutral regions, but near the depletion region edges 3 .
122 3 3
72
Carr
iers
/cm
3
nPpN
1014
1016
pNVnPVLog scale
1m
Calculating the current passing through a PN junction (III)
several mm
Vext
Vj
0.3mP N- +P-side N-side
jtotal jtotal
Computing the overall current from the current density due to carriers in the depletion region
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.
73
Calculating the current passing through a PN junction (IV)
Vext
Vj
P N- +P-side
jtotal
Computing the overall current from the current density due to carriers in the neutral regions far from the depletion region
jtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x)
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.
jp_Drift(x) = q·p·p(x)·E(x)
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
» 0
» 0» 0
Few electrons in P-side
Constant concentration
Weak field High concentration
74
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
Vj
P N- +P-side
jtotal
jtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x)
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
jp_Drift(x) = q·p·p(x)·E(x)
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» 0
Few electrons in P-side
Weak field High concentration
jp_total(x)
Length0Min
ority
ca
rrie
r co
ncen
trati
on
pNnP
75
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)
jn_total(x) = q·Dn·dnPV(x)/dx
We can do the same for the holes just in the opposite side of the depletion region
jp_total (x)= -q·Dp·dpNV(x)/dx
Vext Vj
P N- +P-side N-side
jn_total(x) jp_total(x)
pNVnPV
Length0Curr
ent d
ensit
y
jn_total(x)
Taking derivatives
76
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)
Vext Vj
P N- +P-side N-side
jn_total(x) jp_total(x)
jp_total(x)
Length0Curr
ent d
ensit
y of
m
inor
ity ca
rrie
rs
jn_total(x)
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.
What happens with carriers in the depletion region?
• 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 (VIII)Computing the overall current from the current density due to carriers in
the neutral regions but near the depletion region edges (IV)
Vext
Vj
P N- +P-side N-side
jn_total(x) jp_total(x)
jp_total(x)
Length0Curr
ent d
ensit
y
jn_total(x)
Now the total current density can be easily computed
jtotal
Very important conclusion!!!
• 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 (IX)
Vj
P-side N-side- +
jn_total(x) jp_total(x)
Summary of the computing of the overall current density in a PN junction
79
Calculating the current passing through a PN junction (X)
Vj
P-side N-side- +
jn_total(x) jp_total(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.
jp_total(x)
Length0Curr
ent d
ensit
y
jn_total(x)
jtotal
Majority-carrier currents, due to both drift and diffusion
Minority-carrier currents, only due to diffusion
Total current
nP
80
Current passing through an asymmetrical junction (P+N-)
Vext Vj
- +P+-side (wide) N--side
jn_total(x)
Length0
Curr
ent d
ensit
y
pNV
nPV
Length0
pN
Conc
entr
ation
jp_total(x)
jn_total(x)
jtotal
P-side is heavily doped (P+) and wideN-side is slightly doped (N-) and narrow
This is a case of special interest, because it is directly related to the operation of Bipolar Junction Transistors (BJTs)
Length0Min
ority
carr
ier
conc
entr
ation
pNnP
81
Qualitative study of the current in a forward-biased PN junction
Vext Vj
P N- +P-side N-side
jtotal
pNVnPV
jp_total(x)
Length0Curr
ent d
ensit
y
jn_total(x)
jtotal
High and positive total current density
High slope Þ High current density due to electrons in the depletion region
High slope Þ High current density due to holes in the depletion region
Length0Min
ority
carr
ier
conc
entr
ation
pNnP
82
Qualitative study of the current in a reverse-biased PN junction
Vext Vj
P N- +P-side N-side
jtotal
nPV
Low and negative total current density
pNV
Low slope Þ Low current density due to electrons in the depletion region
Low slope Þ Low current density due to holes in the depletion region
Length0
Curr
ent d
ensit
y
jn_total(x)
jtotal
jp_total(x)
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 cross-sectional area.
83
Quantitative study of the current in a PN junction (I)
The final results is:
i = IS·(eVext/VT - 1), where:
IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]
(Is is called reverse-bias saturation current)
VT = kT/q
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 (II)
PN
+
-
i
Vext(Shockley equation)
i = IS·(eVext/VT - 1)
IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]
VT = kT/q
85
Quantitative study of the current in a PN junction (III)
• Forward bias VO > Vext >> VT
• Reverse bias Vext << -VT
Þ exponential dependencei » IS·eVext
VT
i » -IS Þ constant (reverse-bias saturation current)
Vext [V]0
100
0.25- 0.25
i [mA]
0.5
-10
-0.5 0
i [nA]
Vext [V]
86
Quantitative study of the current in a PN junction (IV)
Length
pNVnPV
0
Minority carrier concentration
pNnP
Length
pNVnPV
0
Minority carrier concentration
nP pN
Vext
P-side N-side
Wide sides Vext
P-side N-side
Narrow sides
Wide versus narrow P and N sides
XN XN >> Lp
XN XN << Lp
IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)] IS = A·q·ni
2·[Dp/(ND·XN)+Dn/(NA·XP)]
XP XP >> Ln
XP XP << Ln
Equation i = IS·(eVext/VT - 1) is valid in both cases
Equation i = IS·(eVext/VT - 1) describes the operation in the range VO > Vext > -. However, three questions arise:
• What happens if Vext > VO?
• How does the temperature affect this characteristic?
• What is the actual maximum voltage that the junction can withstand?
87
Quantitative study of the current in a PN junction (V)The I-V characteristic in a real scale of use
0 1-4
3i [A]
Vext [V]
Vj
- +VmP
-+VNm
-+i 0
P-side N-side
+ -
VN 0VP 0
Vext
+ - + - 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).
According to Shockley equation
Actual I-V characteristic
88
Decreases with T
Increases with T
Forward bias: i » IS·eVext/VT = IS·eq·Vext/kT
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:
Temperature dependence of the I-V characteristic (I)
Reverse current strongly increases when the temperature increases. It doubles its value when the temperature increases 10 oC.
In practice, forward current increases when the temperature increases. For extremely high currents, the dependence can become just the opposite.
89
30
10
i [A]
Vext [V]
Forward bias
PN
+
-
i
V
37 0C
27 0C
-0.25
-10
Vext [V]
i [A]
Reverse bias
27 0C
370C
Temperature dependence of the I-V characteristic (II)
In both cases, the current increases for a given external voltage.
90
• There are three different physical processes which limit the reverse voltage that a given PN junction can withstand:
Punch-through Zener breakdown Avalanche breakdown
Maximum reverse voltage that a PN junction can withstand
0
i
Vext
i+ Vext -
PN
+ -+ -
+ -
+
-
+
-+
-This phenomenon does not take place in power devices (two heavily doped regions are needs).
• 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.
It will be explained later
• As already known, both the electric field and depletion length increase.• When the maximum electric field is high enough, the avalanche
breakdown starts. 91
W0 = p P p N e
VU V
T
2·e·(NA+ND)·V0
q·NA·ND
Emax0=e·(NA+ND)
2·q·NA·ND·V0
No bias Reverse bias
W(Vrev) = p P p N e
VU V
T
2·e·(NA+ND)·(V0+Vrev)
q·NA·ND
Emax(Vrev) =e·(NA+ND)
2·q·NA·ND·(V0+Vrev)
W(Vrev)NP - +
V0+Vrev
-Emax(Vrev)
V0
P N- +
W0
-Emax0
Electric field in the depletion region with reverse bias
Punch-through limit
92
Emax(Vrev) »e·(NA+ND)
2·q·NA·ND·|Vrev|
W(Vrev)
NP - +» |Vrev|
-Emax
WPN
W(Vrev) » p P p N e
VU V
T 2·e·(NA+ND)·|Vrev|
q·NA·ND p P p N e
VU V
T
Avalanche breakdown limit
-EBR
• We must design the semiconductor according to: Emax(Vrev) < EBR.
• The breakdown voltage is:
VBR = EBR2·e·(NA + ND)/(2q·NA·ND).
• 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.
Limits for the depletion region with reverse bias
93
• 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.
What must we do to withstand high-voltage?
2·q·NA·ND
VBR =EBR
2·e·(NA + ND)NA2qVBR = ·( + )
EBR2·e 1 1
NDÞ
WN
XN
pP = NA nN = ND
P+ - N--side+ NDNA
NA >> ND
9494
-Emax
x
Vrev
x
r(x) xq·ND
-q·NA
NA >> ND
P+ - N--side+ NDNA
Maximum electric field Emax with reverse bias Vrev
-EBR
The main part of the reverse voltage is dropping in the slightly doped region.
Vrev_P
Vrev_N
• 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.
VBR
9595
PIN junctions (I)W(Vrev)
NP - +Vrev
-Emax(Vrev)
• 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)?
• Yes, we can. We need another E(x) profile.
-Emax(Vrev) new profile (ideal)
-Emax(Vrev)
-Emax(Vrev) new profile (real)
-Emax(Vrev)
• To obtain this profile, we need a region without space charge (undoped) inside the PN junction.
9696
P-side - N-side+
PIN junctions (II)
q·NDr(x)x
-q·NA
-Emax
x
Vrev x
Intrinsic
Many holes Many electronsA few holes
and electrons
• It means P-Intrinsic-N
Negative space charge
Positive space charge
Characteristics:
- Good forward operation due to conductivity modulation.
- Low depletion capacitance.
- Slow switching operation.
All these characteristics will be explained later.
9797
Other structure to withstand high voltage: P+N-N+
P+ - + N- + N+
-q·NA
q·ND1
r(x)
x
q·(ND2-ND1)
x
-Emax(Vrev1)
-Emax(Vrev2)
Heavily doped P Heavily doped NLightly doped N
Vrev2 > Vrev1
Low reverse voltage Vrev1
Reverse voltage Vrev2
ND1NA ND2
q·ND2Partially depleted
98
Forward-bias behaviour of structures to withstand high voltage
In both cases, there is a high-resistivity layer(called drift region)
• 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.
P+ N+Intrinsic
Undoped
PIN
P+ N+N-
Lightly doped
P+N-N+
99
Injection levels
Log scale
Carriers/cm3
104
1012
1014
1016
-0.3 -0.2 -0.1 0- 0+ 0.1 0.2 0.3Length [mm]
1010
108
106
P+-side N--side
nPV
nNpP
pNV
Low-level injection:nN(0+) >> pNV(0+)
Log scale
-0.3 -0.2 -0.1 0- 0+ 0.1 0.2 0.3Length [mm]
P+-sideN--sidenPV
nNpP
pNV
High-level injection:nN(0+) » pNV(0+)
• 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.
Not possible!
100
Conductivity modulation
1016
106
10
1014
10
P+ N+N-
Drift region
P+N-N+
NA = 1019 ND2 = 1019ND1 = 1014
nP+ pN+
nN- » pN-Holes are injected from the P+-side
Electrons are injected from the N+-side
• 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.
101
Semiconductor junctions designed to withstand high voltage
PIN P+N-N+
• 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.
Summary
102
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)
Transient and AC operation of a PN junction
103
Parasitic capacitances: depletion layer capacitance (I) (also known as junction capacitance)
This is the dominant capacitance in reverse bias
x
r(x)Vext
P-sideVO+Vext
- + Zona NVO+Vext+Vext
- + N-side
Vext + Vext
Carriers are pulled out from the depletion region when Vext is increased in Vext . Additional space charge has been generated.
104
- +P N
Vext
PN junction
Vext
+ + +
- - -+ + + + +- - - - -
Vext + Vext
Capacitor
• Capacitor: new charges are located at the same distance Þ constant capacitance.• PN junction: new charges are located farther away
from each other Þ non-constant capacitance.
Vext + Vext
- +P N- +
Parasitic capacitances: depletion layer capacitance (II)
105
In an “abrupt” PN junction (as we have considered so far), this capacitance is a K·(V0-Vext)-1/2 -type function
Cj=dQ/dV=e·A/W(Vext)
W(Vext) = p P p N e
VU V
T 2·e·(NA+ND)·(V0-Vext)
q·NA·ND
Cj = A· p P p N e
VU V
T 2·(NA+ND)·(V0-Vext)
e·q·NA·ND
W(Vext)-dQ
dQ• As it is a non-constant capacitance,
static and dynamic capacitances could be defined. The latter is defined as:
Then:
0
Vext
Cj
Parasitic capacitances: depletion layer capacitance (III)
As:
106
This capacitance is the one dominant in forward bias
Reverse bias 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.
0 Vext
Cj
Parasitic capacitances: diffusion capacitance (I)
107
Increase of minority carriers due to a increase of 60mV in forward bias
1010
1012
1014
1016
Carr
iers
/cm
3
-3 -2 -1 0 1 2 3
Longitud [mm]
pP
pNV
nN
nPV
V=180mV
• 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.
V=240mV
Parasitic capacitances: diffusion capacitance (II)
108
Switching times in PN junctions (I)
The diode behaviour seems to be ideal in this time scale.
Transition from “a” to “b” (switching off) in a wide time scale (ms or s).
a b
V1
V2
R i
v+
-i
v
t
t
V1/R
-V2
Let us consider a PN diode as PN junction. The results obtained can be generalized to PN junctions in other semiconductor devices.
109
a b
V1
V2
R i
v+
-
Transition from “a” to “b” (switching off) in a narrow time scale (s or ns).
i
v
t
t
trr
V1/R
-V2/R
ts
tf (i= -0.1·V2/R)
-V2
ts = storage time. tf = fall time. trr = reverse recovery time.
Reverse recovery peak
Switching times in PN junctions (II)
110
a b
V1
V2
R i
v+
-
Why does this evolution occur? • This is because the junction cannot
withstand voltage until all the excess of minoritary carriers disappears from the neutral regions.
V1/R
vt
i
t pNVnPV
Carriers/cm3
8·1013
4·1013
0
-0.1 0 0.1Length [mm]
t0t0
t3
t3
t1
t1
t2
t2
-V2
-V2/R
t4
t4
t0t0
Switching times in PN junctions (III)
111
a b
V1
V2
R i
v+
-
pNVnPV
Carriers/cm3
8·1013
4·1013
0
-0. 1 0 0.1Length [mm]
i
td = delay time.
tr = rise time.
tfr = td + tr = forward recovery time.
tr
0,9·V1/R
td
0,1·V1/R
tfr
t0
t0
t1
t1
t2
t2
t3
t3
t4t4
Switching times in PN junctions (IV)
Transition from “b” to “a” (switching on) in a narrow time scale (s or ns).
112
Trade-off between static and dynamic behaviour in PN junctions
• 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.
1016
106
10
1014
10
P+ N+N-
NA = 1019 ND2 = 1019ND1 = 1014
nP+ pN+
nN- » pN-
Log scale
Excess of holes in N-
Excess of electrons in N-
Excess of electrons in P+
Excess of holes in N+
• 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 can deteriorate the conductivity modulation.
113
Introduction to the metal-semiconductor junctions (I)
Case #1: an N-type semiconductor transfers
electrons to a metal
N-type semiconductorMetal
N+++
+++
+ +-------
-N-type
Donor ions Electrons (thin sheet)
• 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 built-in 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:
• 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 junction).
114
P-type semiconductorMetal
P---
---
- -+++++++
+
P-type
Acceptor ions
Lack of electrons (thin sheet )
Case #2: a metal transfers electrons to a P-type semiconductor
Introduction to the metal-semiconductor junctions (II)
Recombination between the transferred electrons and the P-side holes takes place in this edge.
115
N-type semiconductorMetal
N-type+
++++
+
++
-----
-
--
Electrons (thin sheet)
Lack of electrons(thin sheet)
Case #4: a P-type semiconductor transfers electrons to a metal P-type semiconductorMetal
P-type+
++++
+
++
-----
-
--
Electrons (thin sheet)
Holes(thin sheet)
Introduction to the metal-semiconductor junctions (III)
Case #3: a metal transfers electrons to an N-type semiconductor
• We have an ohmic contact (non-rectifying contact) in both cases.
116
Rectifying contacts (I)
• 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:
eEmax0 =
2·q·ND·V02·e·V0W0 = q·ND
Cj0 = A· p P p N e
VU V
T 2·V0
e·q·ND
However, the built-in voltage and the I-V characteristic depend on the work function of both the semiconductor and the metal.
Case #1: an N-type semiconductor transfers
electrons to a metal
W0
MetalN+
++
+++
+ +-------
-N-type
ND
117
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.
• I-V characteristic:
i = IS·(eVext/VT - 1), as in a PN junction.
However, the value of Is has a very different value:
IS = A*·A·T2·e-VB/VT , where:
A* = Richardson constant (120 amps/(cm2·K2)).
To define these concepts properly, we should introduce others. This task, however, is beyond the scope of this course.
118
Schematic Symbol
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, PN-
junction 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.
119
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.
P+ N+NNA1 = 1019 ND2 = 1019ND1 = 1016NA2 = 1016
P
N+NND2 = 1019ND1 = 1016
Ohmic contactsOhmic contacts
Rectifying contacts Ohmic contacts
PN diode
Schottky diode
Beyond the scope of this course, as well.