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)