Post on 31-Mar-2015
transcript
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Chapter 5-1. PN-junction electrostatics
You will also learn about:
Poisson’s Equation
Built-In Potential
Depletion Approximation
Step-Junction Solution
In this chapter you will learn about pn junction electrostatics:
Charge density, electric field and electrostatic potential existing
inside the diode under equilibrium and steady state conditions.
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PN-junction fabrication
PN-junctions are created by several processes including:
1. Diffusion
2. Ion-implantation
3. Epitaxial deposition
Each process results in different doping profiles
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Ideal step-junction doping profile
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Equilibrium energy band diagram for the pn junction
kT
EEnp Fi
i exp
EF = same everywhere under equilibrium
Join the two sides of the band by a smooth curve.
kT
EEnn iF
i exp
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Electrostatic variables for the equilibrium pn junction
x
E
qx
E
q d
d1
d
d1 iC E
Potential, V = (1/q) (EC–Eref). So,potential difference between the two sides (also called built-in voltage, Vbi) is equal to (1/q)(EC).
xd
dE
refC1
EEq
V
= charge density = Ks o
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Conceptual pn-junction formationConceptual pn-junction formation
Holes and electrons will diffuse towards opposite directions, uncovering ionized dopant atoms. This will build up an electric field which will prevent further movement of carriers.
p and n type regionsbefore junction formation
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The built-in potential, Vbi
When the junction is formed, electrons from the n-side and holes from the p-side will diffuse leaving behind charged dopant atoms. Remember that the dopant atoms cannot move! Electrons will leave behind positively charged donor atoms and holes will leave behind negatively charged acceptor atoms.
The net result is the build up of an electric field from the positively charged atoms to the negatively charged atoms, i.e., from the n-side to p-side. When steady state condition is reached after the formation of junction (how long this takes?) the net electric field (or the built in potential) will prevent further diffusion of electrons and holes. In other words, there will be drift and diffusion currents such that net electron and hole currents will be zero.
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Equilibrium conditions
Under equilibrium conditions, the net electron current and hole current will be zero.
electron drift currentopposite to electron flux
NA = 1017 cm3 ND = 1016 cm3
electron diffusion currentopposite to electron flux
hole diffusion current
hole drift current
net current = 0
net current = 0
E-field
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The built-in potential, Vbi
p-side n-side
EC
Ei
EV
iFi ln
n
pkTEE
q Vbi = (Ei EF)p-side + (EF Ei)n-side
iiF ln
n
nkTEE
EC
Ei
EV
EF
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The built-in potential, Vbi
The built-in potential, Vbi, measured in Volts, is numerically equal to the “shift” in the bands expressed in eV.
Vbi = (1/q) {(Ei EF)p-side + (EF Ei)n-side }
sidenonionconcentratelectronand
sideponionconcentratholewhere
ln
lnln
n
p
2i
np
ii
n
p
n
np
q
kT
n
n
q
kT
n
p
q
kT
kT
Vq
n
n
p
p biexpp
n
n
pAn interesting fact:
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Majority and minority carrier concentrations
pp
np
nn
pn
xp xn
x
p-side NA ND n-side
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Built-in potential as a function of doping concentration for an abrupt p+n or n+p junction
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Depletion approximation
elseeverywhere0
for)(0
0d
d
npADs
s
xxxNNK
q
Kx
EPoisson equation
We assume that the free carrier concentration inside the depletion region is zero.
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Example 1
A p-n junction is formed in Si with the following parameters. Calculate the built-in voltage, Vbi.
ND = 1016 cm–3 NA = 1017 cm–3
Calculate majority carrier concentration in n-side and p-side. Assume nn = ND = 1016 cm3 and pp = NA= 1017 cm3.
2i
DA2i
npbi lnln
n
NN
q
kT
n
np
q
kTV
Plug in the numerical values to calculate Vbi.
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Example 2
A pn junction is formed in Si with the following parameters. Calculate the built-in voltage, Vbi.
ND = 2 1016 cm–3 NA = 3 1017 cm–3
NA = 1 1016 cm–3 ND = 2 1017 cm–3
Calculate majority carrier concentration in n-side and p-side. nn = “effective ND” = 1016 cm-3; pp = “effective NA” = 1017 cm–3
2i
DA2i
npbi lnln
n
NN
q
kT
n
np
q
kTV
Plug in the numerical values to calculate Vbi.
Here NA and ND
are “effective” ornet values.