p- n diode rectifiers
p- n diode rectifiers
 qV

− 1
exp 
 kT

~
IS
2
Ir = 0 in an ideal
rectifier.
VD = VSource
Voltage
1
VF = 0 in an ideal
rectifier
Current
1
2
Time
Time
Lecture objective: p-n diode at reverse bias
Simulated Electron - Hole profile in Si p-n junction
Junction
0
V bi
E
c
E
F
-0.8
Electron-hole concentration
in a Si p-n junction.
Acceptor density Na=5×
×1015 cm-3
Donor density Nd=1×
×1015 cm-3.
T=300K.
Dashed line show the boundaries
of the depletion region
V bi
-1.6
p-type
-2
Ev
n-type
-1
0
1
2
Distance (µm)
p = Na
10 16
10 14
n = Nd
Electron
concentration
Hole
concentration
10 12
10 10
n -type
p -type
10 8
10 6
p = n i2 / N d
n = n i2 / N a
10 4
-2
-1
0
Distance (µm)
1
2
Concentration
Charge distribution in the depletion region.
Total charge density in the semiconductor: ρ = q×
×(Nd + p - Na - n);
NA
ND
p
n
ρ = 0;
ρ = 0;
x
pn ≈ 0;
np ≈ 0;
nn ≈ 0;
pn ≈ 0;
ρ ≈ -q NA;
ρ ≈ +q ND;
Charge density
(C/cm3)
Mobile Charge density distribution in the p-n junction
Depletion
region
Free
electrons
Holes
p-type
pn ≈ 0;
np ≈ 0;
n-type
nn ≈ 0;
pn ≈ 0;
x
Charge density
(C/cm3)
Charge distribution in the depletion region.
Total charge density in the semiconductor: ρ = q×
×(Nd + p - Na - n);
qND
x
ρ = 0;
ρ = 0;
qNA
pn ≈ 0;
np ≈ 0;
nn ≈ 0;
pn ≈ 0;
ρ ≈ -q NA;
ρ ≈ +q ND;
Electric field profile in the p-n junction
In the following material, the electric field is
denoted as “F”
(to distinguish from “E” = electron/hole energy).
The electric field distribution is defined by Poisson’s equation:
dF
ρ
ρ
=
= ;
dx ε ε 0 ε s
where ε s = ε ε 0 , ρ is the charge density
ε0 – the dielectric permittivity of vacuum;
ε0 = 8.85×10-12 F/m = 8.85×10-14 F/cm
ε − relative dielectric permittivity of the material
In Si, ε ≈ 11.7
2
Charge, Field,
Potential Profiles in the
p-n junction
0
-2
n-type
p-type
-4
dF ρ
= ;
dx ε s
-6
-2
-1
0
1
2
X
2
Charge, Field,
Potential Profiles in the
p-n junction
0
-xp
-2
xn
n-type
p-type
-4
dF ρ
= ;
dx ε s
Using the depletion approximation, we obtain
 qNa
− ε ,
dF
s
=
dx  qNd
,
 ε s
for
for
− xp < x < 0
0 < x < xn
-6
-2
-1
0
1
2
X
2
Charge, Field,
Potential Profiles in the
p-n junction
0
-xp
-2
xn
n-type
p-type
-4
dF ρ
= ;
dx ε s
Using the depletion approximation, we obtain
 qNa
− ε ,
dF
s
=
dx  qNd
,
 ε s
for
− xp < x < 0
-6
-2
0
0
-1
1
-xp
2
xn
-5
-10
for
0 < x < xn


x
−Fm  1+  , for − x p < x < 0

 xp 
F=


x
−Fm  1−  , for 0 < x < xn
 xn 

-15
-2
0.2
-1
0
1
2
1
2
0
-0.2
-0.4
-0.6
-0.8
-2
-1
0
Distance (µm)
Maximum Field in the p-n junction
0
-xp
-5
xn
dF ρ
= :
From
dx ε s
-10
-15
-2
-1
0
1
qN A
2
On the p-side of the junction,
Fm =
On the n-side of the junction,
Fm =
xp
εε 0
qN D
εε 0
xn
At the junction interface: Fm(p-side) = Fm (n-side)
Fm =
qN D
εε 0
xn =
qN A
εε 0
xp
xp
ND
=
xn N A
Voltage drop across the p-n junction
Fm =
qN D
εε 0
xn =
qN A
εε 0
xp
The voltage drop across the p-n junction,
V p − n = − ∫ F ( x ) dx
= -{AREA UNDER the F(x) curve}
W
0
-xp
-5
In equilibrium:
xn
1
Vbi = Fm W
2
-10
-15
-2
-1
0
1
2
At reverse bias:
1
Vbi −Vr = Fm W
2
Depletion region width in the p-n junction
The total width of the depletion (space-charge) region.
From
x p = xn ( N D / N A )
W = xn + xp = xn (1 + ND/NA);
The voltage (the area of the triangle): Vbi –Vr = Fm× W/2;
W
Fm
Fm =
qN D
εε 0
xn =
qN A
εε 0
xp
 qN D 
qN D 2
1
q
2 ND N A
Vbi −Vr = 
xn W / 2 =
W
=
W
N A + ND
2εε 0
1 + N D / N A 2εε 0
 εε 0

Depletion region of the p-n junction summary
Depletion region (a.k.a space-charge region) width
Vbi − Vr =
q
2εε 0
2
W N ';

ND N A 
 where N ' =

N
N
+

A
D
Depletion region width sharing between n- and p-sides:
W = xn + xp
x p = xn ( N D / N A )
For strongly asymmetrical p-n junction:
1 q
N A >> N D , Vbi − Vr =
ND W 2
2 ε ε0
1 q
N D >> N A , Vbi − Vr =
NAW 2
2 ε ε0
W ≈ xn
W ≈ xp
The space-charge region is extended mainly to the low-doped side
of the p-n junction
Depletion region width as a function of voltage
From
Vbi − Vr =
q
2εε 0
W 2N '
The depletion region width W:
W=
2ε ε 0 (Vbi − Vr )
qN '
where
ND N A
N '=
N A + ND
For N D >> N A , N ' = N A
For N A >> N D , N ' = N D
Peak electric field in the p-n junction summary
Maximum electric field
1
Vbi − Vr = Fm W
2
Fm =
Vbi −Vr =
2q N ' (Vbi − Vr )
ε ε0
ε ε 0 Fm2
2q N '
Vmax =
2
ε ε 0 FBD
2q N '
P-n Junction Breakdown
Avalanche Breakdown mechanism
Mechanisms for breakdown
Two quantum processes give rise to breakdown:
(1) Impact ionization plus avalanche multiplication.
(2) Quantum tunneling (Zener effect).
Relative importance depends on doping level in the pand n- regions:
Avalanche Breakdown mechanism
• Electric field accelerates electrons (and holes);
• The kinetic energy of electron in the electric field
increases;
• Electron with high enough excessive energy can ionize
the atom at collision.
• As a result, collision produces an electron-hole pair.
Avalanche Breakdown mechanism
Minimum additional electron energy required for
ionization can be estimated using:
m v2
0
2
≈ EG
Additional electron energy comes from the electric field:
m v2
0
2
≈ q F LFP
where F is the electric field and LFP is the “mean free path” – the
average distance that the electron passes between collisions.
From this, the ionization condition:
q FBD LFP ≈ EG
FBD is the critical or “breakdown” field.
The ionization only occurs if electric field exceeds the critical value
Breakdown in p-n junction
0
-xp
xn
-5
-10
-15
-2
-1
0
1
2
Fm
In reverse-biased p-n junction, the electric field is concentrated
within a narrow depletion region.
The maximum electric field occurs exactly at the p-n interface.
As the field Fm exceeds the critical (breakdown) value:
e1
Atom
h2
e1
e2
Avalanche multiplication during the breakdown
Ionization coefficient αe ≈ αh
Ionization coefficient αe >> αh
Multiplication and Ionization Coefficients
Consider uniform electric field in the avalanche region
(for simplicity)
The avalanche region width is W
Assume αe = αh ≈ const (x);
Multiplication factor:
M=
The condition for avalanching:
Number of e − h pairs at theoutput
Number of e − h pairs at theinput
(every carrier has one ionizing collision)
Avalanche breakdown interpretation using band diagrams
(1) Electrons from the p-side and holes from the nside get accelerated in the depletion regions by
strong electric filed
np
EV
p
EC
pn
n
Avalanche breakdown interpretation using band diagrams
(2) At collision, electron produces new electron – hole
(e-h) pairs
EV
p
EC
n
Avalanche breakdown interpretation using band diagrams
(3) These secondary e-h pairs in also acquire high energy and
may in turn produce more e-h pairs.
EV
p
EC
n
Avalanche breakdown interpretation using band diagrams
(4) New electron hole pairs will add up to the reverse current
EV
p
EC
n
Avalanche breakdown interpretation using band diagrams
(5) New electron-hole pairs will also create more electron
hole pairs (avalanche multiplication)
EV
p
EC
n
Breakdown voltage of the p-n junction
For a p+ - n junction
(NA >> ND):
Vabd =
2
ε s Fbd
2qN d
Si p-n junction. Example 1:
Maximum Field vs. Reverse Voltage
1.0E+06
9.0E+05
-3
N = 1E15 cm
Fm
Peak Electric Field, V/m
8.0E+05
7.0E+05
6.0E+05
5.0E+05
4.0E+05
FBD
3.0E+05
2.0E+05
1.0E+05
0.0E+00
0
500
1000
1500
Reverse Voltage, V
Fm =
2q N ' (Vbi − Vr )
ε ε0
2000
Si p-n junction. Example 2
Maximum Field vs. Doping
1.60E+06
1.40E+06
V=1000 V
Fm, V/cm
1.20E+06
1.00E+06
8.00E+05
6.00E+05
4.00E+05
FBD
2.00E+05
0.00E+00
0.00E+ 1.00E+ 2.00E+ 3.00E+ 4.00E+ 5.00E+ 6.00E+ 7.00E+
00
15
15
15
15
15
15
15
Doping, cm-3
Fm =
2q N ' (Vbi − Vr )
ε ε0
Si p-n junction. Example 3
Maximum Voltage vs. Doping
2.50E+03
Vmax, V
2.00E+03
1.50E+03
1.00E+03
5.00E+02
0.00E+00
0.00E 1.00E 2.00E 3.00E 4.00E 5.00E 6.00E 7.00E 8.00E
+00
+15
+15
+15
+15
+15
+15
+15
+15
Doping, cm-3
Vmax =
2
ε ε 0 FBD
2q N '
For Silicon,
FBD ≈ 3 ×105 V/cm
Zener (tunneling) breakdown
If the junction is highly doped then the depletion region can be very thin:
W=
2ε ε 0 (Vbi − Vr )
qN '
At high reverse bias, the conduction and valence bands on the opposite sides
of the junction are very close to each other
Thin junction (heavily doped)
EV
EC
Zener (tunneling breakdown)
At certain reverse bias, the electrons from the valence band of p-material
can tunnel into the conductance band of n-material.
The reverse current increases abruptly at a certain voltage:
EC
EV
Typical voltage for the Zener breakdown, VZ ≈ 4 EG/q
Avalanche vs. Zener Breakdown comparison
I
Avalanche (100V – 5 kV)
V
Zener (5V – 10V)