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).
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Electrical Energy Conversion and Power Systems
Universidad
de Oviedo
Semester 1
Lecturer: Javier Sebastián
Power supply Systems
(Sistemas Electrónicos de Alimentación)
Javier Sebastián Dr. Electrical Engineer (Ingeniero Industrial)Full professor
Room 3.1.21Edificio nº 3, Campus Universitario de Viesques 33204 Gijón (Asturias). SpainPhone (direct): 985 18 20 85 Phone (secretary): 985 18 20 87Fax: 985 18 21 38
Email: [email protected]
Web: http://www.unioviedo.es/sebas/
Electrical Energy Conversion and Power Systems
Universidad
de Oviedo
Semester 1  Power Electronics Devices
At 0 K, empty


Energy of electrons








Interatomic spacing
At 0 K, filled
Actual spacing
Energy interatomic spacing of electrons




Concept of band diagramEmpty at 0 K
4 states/atom
Conduction band
Eg
Band gap
Valence band
4 electrons/atom
Filled at 0 K
Conduction band
Conduction band
Conduction band
Band gap Eg
Band gap Eg
Overlap
Valence band
Valence band
Valence band
Semiconductor
Eg=0.52 eV
Insulator
Eg= 510 eV
Metals
No Eg
Si 0 K
Si
Si
Si

















+




Band structure for semiconductors at room temperature. Concept of “hole”Conduction band
Eg
Valence band
+
Semiconductor
Eg=0.52 eV
Visualizationusingthebondingmodel
Si 0 K
Si



































Concepts of generation and recombination
Recombination
Generation
Si
Si
Eg
Eg


Si
Si
Si
Si
+
+




+
+
12
 0 K
Si
Si
Si
Si
Si
Si
Si
Si


































+ + + + + + +
+
+
+

Why both holes and electrons are electric charge carriers?
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)
 0 K























Sb
Al
Concept of extrinsic semiconductors: doping semiconductor materials

Si
Si

Si
Si

+
1
Donor
1
5
2
2

Si
Si


+
Acceptor
4

3

3
 0 K























5
Al
Sb+
Ntype and Ptype semiconductors

Si

Si
Si

Si
+
1
Donor
1
2
2
Si

Si

Acceptor
4


3
3
Al 0 K
Al
Al
Al
Al
Thermalgeneration
Al
Al
Al
Al
Al













+
+
+
+
+
+
+
+
+
+
+
+
+
Thermalgeneration
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Charges in Ntype and Ptype semiconductorsPtypesilicon
Acceptorions (negativeions)
hole
electron
Donorions(positive ions)
Ntypesilicon
Al 0 K
Al
Al
Al
Al
Al








+
+
+
+
+
+
+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Charge carries in Ntype and Ptype semiconductorsPtype
Very important equations!!!
Ntype
Al 0 K
Al
Al
Al
Al
Al








+
+
+
+
+
+
+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Static charges in Ntype and Ptype semiconductorsPtype
Ntype
Al 0 K
Al
Al
Al
Al
Al








+
+
+
+
+
+
+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Neutrality in Ntype and Ptype semiconductorsPtype
Very important equations!!!
Ntype
Al 0 K
Al
Al
Al
Al
Al








+
+
+
+
+
+
+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Calculating the concentration of electrons and holes (I)Ptype
Ntype
Al 0 K
Al
Al
Al
Al
Al








+
+
+
+
+
+
+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Calculating the concentration of electrons and holes (II)Ptype
Very useful equations!!!
Ntype
0 K
jn
E
jp








+
+
+
+
+
+
+
+
Mechanisms to conduct electric current: Drift (I)    
+ + + + +
q = magnitude of the electronic charge (1.6·1019 coulombs).
p = hole mobility.
n = electron mobility.
p = hole concentration.
n = electron concentration.
E = electric field.
0 K
jn_Diff


















1
2
n2 < n1
n1
Mechanisms to conduct electric current: Diffusion (I)Electrons have migrated due to “diffusion” (you can see the same phenomenon in gases)
0 K
jn_Diff
































1
2
n2 < n1
n1
n
Mechanisms to conduct electric current: Diffusion (II) 0 K
jn_Diff






















1
2
n2 < n1
n1
n
Mechanisms to conduct electric current: Diffusion (III)The current density is proportional to the electron concentration gradient:
jn_Diff = q·Dn· n
Dn = electron diffusion coefficient.
0 K
jp_Diff
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
1
2
p2 < p1
p1
p
Mechanisms to conduct electric current: Diffusion (IV)The current density is proportional to the hole concentration gradient:
jp_Diff = q·Dp· p
Dp = hole diffusion coefficient.
q = magnitude of the electronic charge (1.6·1019 coulombs).
Dp = hole diffusion coefficient.
Dn = electron diffusion coefficient.
Ñp = hole concentration gradient.
Ñn = electron concentration gradient.
jp_Diff= q·Dp· p
jn_Diff= q·Dn· n
jp_Diff = q·Dp· p
jn_Diff = q·Dn· n
A 0 K
B
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.
 0 K


Light

A
A
A
B
B
B
+
+
+
+
+
+
Continuity equations (II)jn_B
jp_B
jp_A
0 K
·jp/q
p/t = GL [p(t)p]/p

·jn/q
+
n/t = GL [n(t)n]/n
Continuity equations (III)
Taking into account the three effects, we obtain the continuity equation for holes:
Variation due to the excess of carriers over the equilibrium
Total time variation of holes
Variation due to the generation of electronhole pairs by light
Variation due to the different current density of holes across “A” and “B”
GL: rate of generation of electronhole pairs by light.
tp: hole minoritycarrier lifetime.
p: hole concentration in steadystate.
Similarly, we can obtain the continuity equation for electrons:
0 K
·jp/q
p/t = GL [p(t)p]/p

N
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Time evolution of an “excess” of minority carries (I)
Þ p0= GL·p + p
0
0
p
p0
N
N
p0
p(t)
p
Tangent line 0 K
p0
Samearea
·jp/q
p/t = GL [p(t)p]/p

p(t)
p
p
t
p
Time evolution of an “excess” of minority carries (II)
0
0
After integrating Þp(t) = p+(pp)·etp
N 0 K
·jp/q
p/t = GL [p(t, x)p]/p

xN
x
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Spatial evolution of an “excess” of minority carries (I)
p0
0
0
p
0
Þ 0 =  [p(x)p]/p + Dp·2[p(x)p]/x2
After integrating Þp(x) = p + C1·ex/Lp + C2·ex/Lp
where : Lp=(Dp· p)1/2 is the minority hole diffusion length
p 0 K0
p(x)
p
x
xN
Spatial evolution of an “excess” of minority carries (II)
p(x) = p + C1·ex/Lp + C2·ex/Lp
xN: length of the Ntype crystal
Lp: hole diffusion length
Lp >> xN (narrow)
p0
Lp << xN (wide)
p(x)
p
Tangent line
x
xN
Lp
P 0 Ktypesilicon
Ntypesilicon
Al
Al
Al
Al
Al
Al
Al
Al










Barrier to avoid carrier diffusion
+
+
+
+
+
+
+
+
+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Concept of PN junction (I)
What happens if we remove the barrier?
Pside 0 K
Nside
Al
Al
Al
Al
Al
Al
Al
Al










+
+
+
+
+
+
+
+
+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Concept of PN junction (II)
Holes begin to diffuse from the Pside to the Nside. Similarly, electrons diffuse from the Nside to the Pside
Are all the carriers to be diffused?
Al 0 K
Al
Al
Al
Al
Al
Al
Al










+
+
+
+
+
+
+
+
+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Concept of PN junction (III)
Are all the carriers to be diffused?
Pside
Nside
Nonneutral Ntype region, but positively charged
Nonneutral Ptype region, but negatively charged
Is this situation “the final situation”?
The answer is no
Pside 0 K
Nside
Al
Al
Al
Al
Al
Al
Al
Al











+
+
+
+
+
+
+
+
+
+
+
E
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Concept of PN junction (IV)
An electric field appears just in the boundary between both regions (we call this boundary metallurgical junction)
Al 0 K
Al
Al
Al
Al
Al
Al
Al




+
+
+
+
E
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Concept of PN junction (V)
Pside
Ntype
Due to diffusion (® ¬)
Due to drift (electric field) (¬®)
The electric field limits the carrier diffusion
Al 0 K
Al
Al
Al
Al
Al
Al
Al







+
Neutral Ntype region (electronsare balanced by positive ions )
Neutral Ptype region (holes are balanced by negative ions )
+
+
+
+
+
+
E
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Concept of PN junction (VI)
Depletion region, or space charge region, or transition region
Unbalanced charge exists because carriers barely exist
Nside
(neutral)
Pside
(neutral)
+ 
Concept of PN junction (VII)
E
Many electrons, but neutral
Many holes, 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
 + 0 K
 +
 +


 +
Nside
Due to drift
 +
Due to diffusion
 +
+
+
Due to drift
 +
Due to diffusion
Pside
 +
jp_Diff
jp_Drift
jn_Drift
jn_Diff
Computing the builtin 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.
Currents must cancel each other
Currents must cancel each other
p 0 KP(hole concentration in Pside )
(hole concentration in Nside )pN
 +
+ 
+ 
+ 
+ 
Nside
V0
Due to diffusion
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Zona P
Due to drift
jp_Drift
jp_Diff
Computing the builtin voltage V0 (II)
jp_Drift=  jp_Diff
Computing the builtin voltage 0 KV0 (III)
Equations:
jp_Drift=  jp_Diff
jp_Drift= q·p·p·E
jp_Diff= q·Dp·dp/dx
E = dV/dx
Therefore: dV = (Dp/mp)·dp/p
After integrating :
V0= VNside – VPside = (Dp/p)·ln(pN/pP) = (Dp/p)·ln(pP/pN)
Repeating the same process with electrons, we obtain:
E
V0= (Dn/n)·ln(nN/nP)
k = Boltzmann constant.
VT = 26 mV at 300 K.
It could be demonstrated:
Dp/p = Dn/n = kT/q = VT(Einstein relation)
 + 0 K
+ 
+ 
+ 
+ 



































+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Computing the builtin voltage V0 (IV)
Summary (I)
nN
pN
pP
nP
V0
Zona P
Nside: many electrons
Pside: many holes
Almost no holes
or electrons
Almost no holes
Almost no electrons
V0 = VT·ln(pP/pN) and also V0 = VT·ln(nN/nP)
Pside 0 K
Nside
+ 
ND, nN, pN
NA, pP, nP
V0
If ND >> ni(current case)
nN = NDpN = ni2/ND
If NA >> ni (current case)
pP = NA nP = ni2/NA
Computing the builtin voltage V0 (V)
Summary (II)
V0= VT·ln(pP/pN) and also V0= VT·ln(nN/nP)
V0 = VT·ln(NA·ND/ni2), VT = 26 mV at 300 K
Nside 0 K
Pside
Al
Al
Al
Al
Al
Al
Al
Al



WP0
WN0
NA
ND
W0
+
+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Sb+
Depletion width in Pside and in Nside
Charge neutrality implies: NA·WP0 = ND·WN0
The heavier doped a side, the narrower the depletion region in that side
E(x) 0 K
Calculating the electric field E and the total depletion width W0 (I)
V0
rNside
rPside

+
+

Pside
+

Nside
WP0
WN0
W0
We will use Gauss’ law and the relationship between electric field and voltage
q·N 0 KD
(x)
Charge density
x
q·NA
E(x)
E(x)
x
Electric field
Emax0
V(x)
V0
x
Voltage
Calculating the electric field E and the total depletion width W0 (II)
+

Pside
+

Nside
 V0 +
W0
E(x) 0 K
2·q·NA·ND·V0
Emax0 =
·(NA+ND)
Calculating the electric field E and the total depletion width W0 (III)
+

Pside
+

Nside
 V0 +
After applying Gauss’ law and the relationship between electric field and voltage, we obtain:
q·ND
W0
Charge density
x
2··(NA+ND)·V0
W0 =
q·NA
q·NA·ND
x
Electric field
Emax0
V0
x
Voltage
2·q·N 0 KA·ND·V0
Emax0 =
·(NA+ND)
Summary of the study of the PN junction with no external bias
Electric field at the metallurgical junction Emax0
 V0 +
Pside
Doped NA
Nside
Doped ND

+
WP0
WN0
W0
2··(NA+ND)·V0
Very important equations!!!
Almost no holes or electrons, but space charge, electric field and voltage
W0 =
q·NA·ND
V0= VT·ln(NA·ND/ni2)
W0 = WP0 + WN0
NA·WP0 = ND·WN0
Nside: many electrons
Pside: many holes
Almost no holes
Almost no electrons
 0 K

+
+
VNm
VmP
No energy can be dissipated here
i = 0
V = 0
Connecting external terminals to a PN junction
metalsemiconductor contacts
+ 
Nside
Pside

+
V0
Therefore:
V = 0, i = 0
Hence:
VmP – V0 + VNm = 0
And:
VmP + VNm = V0
Conclusion:
The builtin voltages across each metalsemiconductor contact cancel out the effect of V0 in such a way that V0does not appear externally.
 0 K
+ 
+
Nside
Pside
Vj


+
+
VmP
VNm
i 0
V0 becomes Vj now

+
Vext
Low resistivity:
VP=0
Low resistivity:
VN=0
Biasing the PN junction: forward bias
VmP and VNmdo not change and, therefore VmP+VNm= V0
Vext= VmP  Vj + VNm = V0  Vj
Therefore: Vj = V0  Vext
Conclusion:
The builtin voltage across the junction has decreased Vext volts
 0 K
+ 
+
Nside
Pside
Vj


+
+
VmP
VNm
i 0

+
Vext
Low resistivity:
VP=0
Low resistivity:
VN=0
Biasing the PN junction: reverse bias
VmP and VNmdo not change and, therefore VmP+VNm= V0
Vext= VmP +Vj  VNm = V0 + Vj
Therefore: Vj = V0 + Vext
Conclusion:
The builtin voltage across the junction has increased Vext volts
+  0 K
Nside
Pside

+
Vj
i
=

+
Vext
Biasing the PN junction: notation for a general case
2·q·N 0 KA·ND·V0
Emax0 =
·(NA+ND)
Effects of the bias on the depletion region
We must replace V0 with Vj, that is, replace V0 with V0Vext
With bias
Without bias
Vj0 = V0
Vj(Vext) = V0  Vext
2··(NA+ND)·(V0Vext)
W(Vext) =
q·NA·ND
2··(NA+ND)·V0
W0 =
q·NA·ND
2·q·NA·ND·(V0Vext)
Emax(Vext) =
·(NA+ND)
Always: V0= VT·ln(NA·ND/ni2)
W 0 K0
W
Nside
Nside
Pside
Pside
V0
V0V1
V1
(x)


+
+
x
E(x)
x
Emax
Emax0
Vj(x)
V0
V0V1
x
Effects of the forward bias on the depletion region
W 0 K0
W
Pside
Pside
Nside
Nside
V0
V0+V2
V2


+
+
(x)
x
E(x)
x
Emax0
Emax
Vj(x)
V0+V2
V0
x
Effects of the reverse bias on the depletion region
 0 K











































Effects of the bias on the neutral regions (I)
nNV
nPV
nN
nP
 +
+ 
+ 
+ 
+ 
Zona P
V0
V0Vext
Pside
No bias: V0 = VT·ln(nN/nP)
Forward bias: V0Vext =VT·ln(nNV/nPV)
For holes with forward bias: V0Vext =VT·ln(pPV/pNV)
(V 0 K0Vext)/VT
(V0Vext)/VT
(V0Vext)/VT
(V0Vext)/VT
nPV = nN·e
pNV = pP·e
nPV = ND·e
pNV = NA·e
Effects of the bias on the neutral regions (II)
V0Vext =VT·ln(nN/nPV) Þ
V0Vext =VT·ln(pP/pNV) Þ
 0 K
+
Vj
Vj =V0Vext
=

+
+ 
Vext
Zona N
Zona P
pP = NA
nN = ND
Effects of the bias on the neutral regions (III)
V0/VT
pN = ni2/ND = NA·e
Nonbiased junction
Nonbiased junction
V0/VT
nP = ni2/NA = ND·e
Biased junction
Biased junction
Vj/VT
pNV = NA·e
Vj/VT
nPV = ND·e
Effects of the bias on the neutral regions (IV) 0 K
Concentration of minority carriers :
Vj/VT
pNV = NA·e
Vj/VT
nPV = ND·e
+ 
Nside
Pside
Vj =V0Vext

+
Vj
Effects of the bias on the neutral regions (V) 0 K
What happens with the minority carriers along the neutral regions?
+ 
Nside
Pside

+
Vj
V 0 Kext
Vext
Pside
Nside
Pside
Nside
Minority carrier concentration
Minority carrier concentration
pNV
nPV
pNV
nPV
pN
nP
pN
nP
0
0
Length
Length
Effects of the bias on the neutral regions (VI)
Wide P and N sides
Narrow P and N sides
The concentration of minority carriers along the neutral regions under forward biasing.
0.001 mm
0.1 mm
Excess of minority carriers
It plays a fundamental role evaluating the switching speed of electronic devices.
V 0 Kext
Vext
Pside
Nside
Pside
Nside
Minority carrier concentration
Minority carrier concentration
pNV
nPV
pNV
pN
nPV
nP
pN
nP
0
0
Length
Length
Effects of the bias on the neutral regions (VII)
Wide P and N sides
Narrow P and N sides
The concentration of minority carriers along the neutral regions under reverse biasing.
0.001 mm
0.1 mm
Deficit (negative excess) of minority carriers
Example of a silicon PN junction 0 K
Properties of Si at 300 K
P side
Nside
Dp=12.5 cm2/s
Dn=35 cm2/s
p=480 cm2/V·s
n=1350 cm2/V·s
ni=1010 carriers/cm3
r=11.8
NA=1015 atm/cm3
p=100 ns
Lp=0.01 mm
ND=1015 atm/cm3
n=100 ns
Ln=0.02 mm
Carriers/cm3
1016
pP
nN
1014
1012
nPV
1010
pNV
108
106
104
0.3
0.3
0.2
0.2
0.1
0
0.1
Length [mm]
Carriers along the overall device
V0=0.596 V
Forward biased with Vext = 0.48 V
Log scale
They decay exponentially
(log scale)
Calculating the current passing through a 0 KPN junction (I)
V 0 Kext
Vj
 +
Nside
Pside
P
N
0.3m
several mm
Calculating the current passing through a PN junction (II)
Case of wide P and N sides
jtotal
jtotal
jtotal = jp_total(x) + jn_total(x) = jp_Drift(x) + jp_Diff(x) +jn_Drift(x) + jn_Diff(x)
3
1
2
2
3
V 0 Kext
Vj
 +
Nside
Pside
P
N
0.3m
several mm
Log scale
Carriers/cm3
1016
nP
pN
pNV
nPV
1mm
1014
Calculating the current passing through a PN junction (III)
Computing the overall current from the current density due to carriers in the depletion region
jtotal
jtotal
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.
Calculating the current passing through a PN junction (IV) 0 K
Computing the overall current from the current density due to carriers in the neutral regions far from the depletion region
Vext
jtotal
Vj
 +
Pside
P
N
jtotal = jp_Drift(x) + jp_Diff(x) +jn_Drift(x) + jn_Diff(x)
» 0
» 0
» 0
High concentration
Constant concentration
Weak field
jp_Diff (x) = q·Dp·dp(x)/dx
jn_Diff(x) = q·Dn·dn(x)/dx
Few electrons in Pside
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.
Calculating the current passing through a PN junction (V) 0 K
Computing the overall current from the current density due to carriers in the neutral regions but near the depletion region edges (I)
Vext
jtotal
Vj
 +
Pside
P
N
jtotal = jp_Drift(x) + jp_Diff(x) +jn_Drift(x) + jn_Diff(x)
» 0
High concentration
Weak field
jp_Diff (x) = q·Dp·dp(x)/dx
jn_Diff(x) = q·Dn·dn(x)/dx
Few electrons in Pside
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
Calculating the current passing through a PN junction (VI) 0 K
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
Nside
Pside
jn_total(x)
P
N
 +
jp_total(x)
Current density
jn_total(x)
nPV
jp_total(x)
pNV
Minority carrier concentration
pN
nP
Length
0
Length
0
jn_total(x) = q·Dn·dnPV(x)/dx
jp_total(x)= q·Dp·dpNV(x)/dx
Taking derivatives
Calculating the current passing through a PN junction (VII) 0 K
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
Nside
Pside
jn_total(x)
P
N
 +
Current density of minority carriers
jp_total(x)
jn_total(x)
jp_total(x)
Length
0
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.
Calculating the current passing through a PN junction (VIII) 0 K
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
Nside
Pside
jn_total(x)
P
N
 +
jtotal
jp_total(x)
Current density
jn_total(x)
jp_total(x)
Very important conclusion!!!
Length
0
Calculating the current passing through a PN junction (IX) 0 K
Summary of the computing of the overall current density in a PN junction
Vj
jn_total(x)
Pside
Nside
 +
jp_total(x)
Calculating the current passing through a PN junction (X) 0 K
Vj
jn_total(x)
Pside
Nside
 +
Total current
jtotal
jp_total(x)
Current density
jn_total(x)
jp_total(x)
Majoritycarriercurrents, duetobothdrift and diffusion
Length
0
Minoritycarrier currents, only due to diffusion
Current passing through an asymmetrical junction (P 0 K+N)
Pside is heavily doped (P+) and wide
Nside is slightly doped (N) and narrow
nPV
Vext
Vj
Nside
P+side (wide)
jn_total(x)
 +
jn_total(x)
pNV
Current density
Concentration
pN
nP
Length
0
Length
0
This is a case of special interest, because it is directly related to the operation of Bipolar Junction Transistors (BJTs)
jtotal
jp_total(x)
Qualitative study of the current in a forwardbiased PN junction
Vext
Vj
Nside
Pside
jtotal
P
N
 +
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
nPV
jn_total(x)
jp_total(x)
pNV
Minority carrier concentration
pN
nP
Length
jtotal
0
Length
0
High and positive total current density
j junctionp_total(x)
Qualitative study of the current in a reversebiased PN junction
Vext
Vj
Nside
Pside
jtotal
P
N
 +
Current density
0
Length
pNV
nPV
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
Minority carrier concentration
pN
nP
Low and negative total current density
Length
0
jn_total(x)
jtotal
Quantitative study of the current in a junctionPN 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).
6Add both current densities.
7 Compute the total current by multiplying the current density by the crosssectional area.
i junction
+
P
Vext
N

Quantitative study of the current in a PN junction (II)
The final results is:
i = IS·(eVext/VT 1), where:
IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]
(Is is called reversebias saturation current)
VT = kT/q
where:
A = crosssectional area.
q = magnitude of the electronic charge (1.6·1019 coulombs).
ni = intrinsic carrier concentration.
Dp = hole diffusion coefficient.
Dn = electron diffusion coefficient.
Lp= hole diffusion length in Nside.
Ln= electron diffusion length in Pside.
ND = donor concentration.
NA = acceptor concentration.
k = Boltzmann constant.
T = absolute temperature (in Kelvin).
(Shockley equation)
i [mA] junction
100
Vext [V]
0
 0.25
0.25
0.5
Vext
i »IS·e
i [nA]
VT
Vext [V]
0
0.5
10
Quantitative study of the current in a PN junction (III)
i = IS·(eVext/VT 1)
IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]
VT = kT/q
Þexponential dependence
i » IS
Þ constant
(reversebias saturation current)
V junctionext
Vext
Pside
Nside
Pside
Nside
Minority carrier concentration
Minority carrier concentration
pNV
nPV
pNV
nPV
pN
nP
pN
nP
0
0
Length
Length
Quantitative study of the current in a PN junction (IV)
Wide versus narrow P and N sides
Wide sides
Narrow sides
XP
XP << Ln
XN
XN << Lp
XP
XP >> Ln
XN
XN >> Lp
IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]
IS = A·q·ni2·[Dp/(ND·XN)+Dn/(NA·XP)]
Equation i = IS·(eVext/VT 1) is valid in both cases
 junction
+ 
+
Nside
Pside
Vj


+
+
i [A]
VmP
VNm
3
Vext
i 0
VP¹ 0
VN¹ 0
Vext [V]
0
1
4
Quantitative study of the current in a PN junction (V)
The IV characteristic in a real scale of use
Actual IV characteristic
According to Shockley equation
+ 
+ 
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).
Decreases with T junction
Increases with T
Temperature dependence of the IV characteristic (I)
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.
In practice, forward current increases when the temperature increases. For extremely high currents, the dependence can become just the opposite.
Forward bias junction
Reverse bias
i
i [mA]
+
i [A]
30
P
V
Vext [V]
370C
N
0.25

27 0C
27 0C
Vext [V]
37 0C
0
1
10
Temperature dependence of the IV characteristic (II)
In both cases, the current increases for a given external voltage.
+  junction
N
+ 
P
+ 



i
+ Vext 
i
Vext
+
+
+
0
Maximum reverse voltage that a PN junction can withstand
It will be explained later
This phenomenon does not take place in power devices (two heavily doped regions are needs).
Electric field in the depletion region with reverse bias junction
No bias
Reverse bias
2··(NA+ND)·(V0+Vrev)
W(Vrev) =
q·NA·ND
2··(NA+ND)·V0
W0 =
q·NA·ND
W0
W(Vrev)
2·q·NA·ND·(V0+Vrev)
Emax(Vrev) =


+
+
P
N
P
N
·(NA+ND)
V0
V0+Vrev
2·q·NA·ND·V0
Emax0=
·(NA+ND)
Emax0
Emax(Vrev)
Limits for the depletion region with reverse bias junction
Punchthrough limit
WPN
2·q·NA·ND·Vrev

N
W(Vrev)
+
P
Emax(Vrev) »
·(NA+ND)
» Vrev
2··(NA+ND)·Vrev
Emax
W(Vrev) »
q·NA·ND
EBR
Avalanche breakdown limit
What must we do to withstand highvoltage? junction
Þ
EBR2·e 1 1
VBR = ·( + )
NA
ND
2q
EBR2·e·(NA + ND)
VBR =
2·q·NA·ND

+
P+
Nside
NA
ND
pP = NA
nN = ND
WN
NA >> ND
XN
Maximum electric field junctionEmax with reverse bias Vrev
q·ND
(x)
NA >> ND
x
q·NA
x
Emax
EBR
VBR

+
P+
Nside
NA
ND
The main part of the reverse voltage is dropping in the slightly doped region.
Vrev_N
Vrev
x
Vrev_P
94
PIN junctions (I) junction
Emax(Vrev) new profile (ideal)
W(Vrev)
Emax(Vrev)
Emax(Vrev)

+
P
N
Vrev
Emax(Vrev) new profile (real)
95
Emax(Vrev)
PIN junctions (II) junction
A few holes and electrons
Negative space charge
Positive space charge
Many holes
Many electrons
Pside
+

Nside
Intrinsic
q·ND
(x)
x
q·NA
x
Emax
Vrev
x
96
Other structure to withstand high voltage: P junction+NN+
Heavily doped P
Lightly doped N
Heavily doped N
q·(ND2ND1)
(x)
N
q·ND1
NA
ND1
P+
N+
ND2

+
+
Partially depleted
q·ND2
x
q·NA
Low reverse voltage Vrev1
x
Reverse voltage Vrev2
Emax(Vrev1)
Vrev2 > Vrev1
Emax(Vrev2)
97
Forwardbias behaviour of structures to withstand high voltage
PIN
Undoped
In both cases, there is a highresistivity layer
(called drift region)
N+
N+
P+
P+
N
Intrinsic
Lightly doped
P+NN+
Carriers/cm voltage3
1016
1014
1012
1010
108
106
104
0 0+
0 0+
0.3
0.3
0.3
0.3
0.2
0.2
0.2
0.2
0.1
0.1
0.1
0.1
Length [mm]
Length [mm]
Injection levels
Lowlevel injection:
nN(0+) >> pNV(0+)
Highlevel injection:
nN(0+) »pNV(0+)
nN
pP
pP
nN
nPV
P+side
P+side
Nside
Nside
Not possible!
pNV
Log scale
Log scale
nPV
pNV
Conductivity modulation voltage
ND1 = 1014
NA = 1019
N+
ND2 = 1019
P+
N
Drift region
nN»pN
1016
P+NN+
1014
Holes are injected from the P+side
pN+
nP+
106
10
10
Electrons are injected from the N+side
Semiconductor junctions designed to withstand high voltage voltage
Summary
P+NN+
PIN
Transient and AC operation of a PN junction voltage
If we change the bias conditions instantaneusly, can the current change instantaneusly as well?
 voltage

+
+
Pside
Zona N
VO+Vext
VO+Vext+Vext
(x)
Vext
Vext + Vext
x
Nside
Carriers are pulled out from the depletion region when Vext is increased in Vext . Additional space charge has been generated.
Parasitic capacitances: depletion layer capacitance (I)
(also known as junction capacitance)
This is the dominant capacitance in reverse bias
V voltageext + Vext
Vext
Vext
+ + +
+ + + + +

+
N
P
    
  
Vext + Vext

+
N
P

+
Parasitic capacitances: depletion layer capacitance (II)
PN junction
Capacitor
dQ voltage
dQ
W(Vext)
2··(NA+ND)·(V0Vext)
W(Vext) =
q·NA·ND
·q·NA·ND
Cj = A·
2·(NA+ND)·(V0Vext)
Cj
Vext
0
Parasitic capacitances: depletion layer capacitance (III)
Cj=dQ/dV=·A/W(Vext)
As:
Then:
In an “abrupt” PN junction (as we have considered so far), this capacitance is a K·(V0Vext)1/2 type function
C voltagej
Vext
0
Reverse bias
Forward bias
Parasitic capacitances: diffusion capacitance (I)
This capacitance is the one dominant in forward bias
10 voltage16
pP
nN
1014
V=240mV
nPV
Carriers/cm3
pNV
Increase of minority carriers due to a increase of 60mV in forward bias
1012
V=180mV
1010
Longitud [mm]
3
3
2
2
1
0
1
Parasitic capacitances: diffusion capacitance (II)
R voltage
i
a
b
+
V2
v
V1

i
V1/R
t
t
v
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.
Transition from “a” to “b” (switching off) in a wide time scale (ms or s).
The diode behaviour seems to be ideal in this time scale.
V2
R voltage
i
a
b
+
V2
v
V1
V1/R
i

trr
t
V2/R
ts
tf(i= 0.1·V2/R)
v
t
V2
Switching times in PN junctions (II)
Transition from “a” to “b” (switching off) in a narrow time scale (ms or ns).
ts = storage time.
tf = fall time.
trr = reverse recovery time.
Reverse recovery peak
R voltage
i
a
b
+
V2
v
V1
Carriers/cm3

i
8·1013
V1/R
t0
t0
t1
pNV
nPV
t3
t4
t
4·1013
t0
t0
t1
t2
t3
t2
t4
V2/R
0
v
t
0.1
0.1
0
Length [mm]
V2
Switching times in PN junctions (III)
R voltage
i
a
b
+
V2
v
V1

i
Carriers/cm3
8·1013
0,9·V1/R
t4
t1
t3
pNV
nPV
t4
0,1·V1/R
4·1013
td
t2
t3
t1
t2
t0
tr
tfr
0
t0
0. 1
0.1
0
Length [mm]
Switching times in PN junctions (IV)
Transition from “b” to “a” (switching on) in a narrow time scale (ms or ns).
td = delay time.
tr = rise time.
tfr = td + tr = forward recovery time.
Tradeoff between static and dynamic voltagebehaviourin PN junctions
ND1 = 1014
NA = 1019
N+
ND2 = 1019
P+
N
Excess of electrons in N
nN»pN
Excess of holes in N
1016
nP+
pN+
1014
Log scale
106
Excess of electrons in P+
Excess of holes in N+
10
10
Ntype semiconductor voltage
Metal


+
+



+
+
N

Ntype

+
+

+
+
Electrons (thin sheet)
Donor ions
Introduction to the metalsemiconductor junctions (I)
Case #1: an Ntype semiconductor transfers electrons to a metal
Ptype semiconductor voltage
Metal
Lack of electrons (thin sheet )
P
Acceptor ions


+
+
+

+

+
Ptype

+

+
+


Introduction to the metalsemiconductor junctions (II)
Case #2: a metal transfers electrons to a Ptype semiconductor
Recombination between the transferred electrons and the Pside holes takes place in this edge.
Ntype semiconductor voltage
Metal


+
+


+
+
Lack of electrons
(thin sheet)
Electrons
(thin sheet)


+
+
Ntype


+
+


+
+


+
+


+
+


+
+
Ptype semiconductor
Metal
Holes
(thin sheet)
Electrons
(thin sheet)
Ptype
Introduction to the metalsemiconductor junctions (III)
Case #3: a metal transfers electrons to an Ntype semiconductor
Case #4: a Ptype semiconductor transfers electrons to a metal
W voltage0
2··V0


W0 =
+
+


q·ND

+
+
N

Ntype

Metal
+
+

+
+
2·q·ND·V0
Emax0 =
·q·ND
Cj0 = A·
2·V0
Rectifying contacts (I)
Case #1: an Ntype semiconductor transfers electrons to a metal
ND
However, the builtin voltage and the IV characteristic depend on the work function of both the semiconductor and the metal.
Rectifying contacts (II) voltage
To define these concepts properly, we should introduce others. This task, however, is beyond the scope of this course.
Schematic Symbol voltage
Rectifying contacts (III)
Ohmic contacts voltage
P
N+
P+
N
NA1 = 1019
NA2 = 1016
ND1 = 1016
ND2 = 1019
Beyond the scope of this course, as well.
Ohmic contacts
Ohmic contacts
N+
N
PN diode
ND1 = 1016
ND2 = 1019
Rectifying contacts
Ohmic contacts
Schottky diode