pn junction and mos structure - gonzaga...
TRANSCRIPT
PN Junction and MOS structure
Basic electrostatic equations
• We will use simple one-dimensional electrostatic equations to develop insight and basic understanding of how semiconductor devices operate
– Gauss's Law
– Potential Equation
– Poisson's Equation
It puts together Gauss's Lawand the potential equation
Gauss's Law
dE
dx
charge density [C/m3]
permittivity [F/m]
E = electric field [V/m]
x a
x
d E xxE x
aE x
a
xa
x
x dxS
x
charge per area in theinterval from x
a to x [C/m2]
The possibility of a change in permittivity due to a material interface has been accounted for by keeping the permittivity together with the field
Potential Equation
x xR
xR
x
E x dx
reference point
E xd x
dx
Poisson's Equation
• It directly links the potential with the charge distribution (there is no need to go through the field)
E xd
dx
dE
dx d2
x
dx2
dE x
dx
Boundary Conditions
• Electronic devices are made of layers of different materials
• We need conditions for and E at the boundary between two materials
Potential at a boundary
• An abrupt jump of (“along x”) would lead to an infinite electric field at the boundary
• Infinite electric fields are not possible (they would tear the material apart)
• Therefore (x) must be continuous:
E xd
dx
0 0
Where the boundary is located at x = 0
Electric Field at a boundary
• The electric field usually jumps at a boundary
• By letting 0 :
d E x 2 E x 1 E x x dx
2 E x 0 1 E x 0 0
x dx 0
2 E x 0 1 E x 0 S
x dx S
There can be a sheet of charge at the boundary
A sheet of charge is an Infinite amount of charge all distributed on the boundary surface (that is a Dirac function)
Boundary between materials
Boundary conditions
E x 01
2
E x 0
x dx 0
E x 01
2
E x 0S
2
x dx S
x 0 x 0
electric field jump for charge free boundary
electric field jump for charged boundary
continuity of potential at a boundary
Oxide-Silicon interface• Example of very common interface in microelectronic devices
Esi 0ox
si
Eox 0Eox 0
3.0
si 11.70
ox 3.90
08.85 10
12F m
Permittivity of vacuum
Metal–Metal Capacitor
• In many IC processes there are two or more levels of metal separated by silicon oxide
Metal–Metal Capacitor
Metal–Metal Capacitor
• Since there is no charge present in the oxide
– the electric field in the oxide is constant
• Since the electric field in the oxide is constant and the voltage drop across the gap is V
– it follows that:
Eox
metal
ox
Emetal
S
ox
Eox
V
t d
dE
dx
= 0
No electric field inside metals
t d 0
0
td
Eox dx
V 0 t d
V
t d
S
ox
S
d S
dV
ox
t d
Capacitance per unit area [F/m2]
Example
• Find potential, electric field and charge distribution for a metal-metal capacitor with t
d =1 μm and an applied
voltage of 1V
M-O-S Capacitor
M-O-S charge distribution
0q N
A
QB
charge per area in the substrate
(C/cm2)
B 0X d G
G ox Vox
t ox
VX d
ox V
t ox 0
ox V
tox q N A
M-O-S charge density profile
M-O-S Electric Field
• In the metal the electric field is 0
• In the oxide (–tox
< x < 0) the charge density is zero ( (x)=0),
therefore the electric field is constant
SOM
for tox x 0:dE x
dx
x
ox
0 E x Eox
The electric field is confined in the region – t
ox< x < X
d
NOTE: the total excess charge in the region – t
ox< x < X
d is zero
0
M-O-S Electric Field
• Outside the charged region of the silicon (x > Xd) the electric field is 0
• In the charged region of the silicon (0 < x < Xd) the charge density is
constant (0) therefore the electric field is a linear function of x
SOM
dE x0
s
dx s E xd s E 00
0
X d
dx
0
E 00X d
s
NOTE: The electric field jumps only at the interface between two different materials
M-O-S Electric Field
• The boundary condition at the oxide/silicon interface is:
SOM
ox Eox s E 0 Eox
s
ox
E 0 E ox
0X d
ox
3
Potential plot through M-O-S
M SO
Poisson ' s Equation :
d2
x
dx2
x
potential at the metal gate :
m tox
surface potential : s 0
substrate potential :sub
X d
Potential is continuos at boundaries:x = t
ox metal/oxide boundary
x = 0 oxide/silicon boundary
Vm sub
Vox
VB
drop across the oxide
drop across the chargedregion of the silicon
Surface potentialfor tox x 0:
m s
0
t ox
E ox dx Eox tox V ox
s m Eox tox m
0X d tox
ox
s m
B
ox
and V ox
B
ox
G
ox
Linear
The drop across the oxide is proportional to the charge stored on each side of the oxide
m xx
tox
Eox dx =
= Eox t ox xV ox
t ox
t ox x
=QG= Q
B
=1/Cox
x is negative inthe region considered
Potential drop across the substrate
for 0 x X d :
d2
x
dx2
0
s
>0
The potential is “concave up” in the charged region
charged region of silicon
V B 0 X d
1
2
0
s
X d
2 0X d
2 s
X d
B
2 s
X d
Quadratic
PN Junction in Thermal Equilibrium
• If no external stimulus is applied (zero applied voltage, no external light source, etc) the device will eventually reach a steady state status of thermal equilibrium
• In this state (“open circuit” and steady state condition) the current density must be zero:
• Eventually, the populations of electrons and holes are each in equilibrium and therefore must have zero current densities
J tot ,0 J p ,0 J n ,0 0
J p ,0 0
J n ,0 0
Diffusion Mechanism
NA > N
D
The charge on the two sides of thejunction must be equal (charge neutrality)
We assumed the n-side is the more lightly doped
Under “open circuit” and steady state conditions the built in electric field opposes the diffusion of free carriers until there is no net charge movement
Diffusion Current
• It’s a manifestation of thermal random motion of particles (statistical phenomenon)
• In a material where the concentration of particles is uniform the random motion balances out and no net movement result (drunk sail-man walk Brownian walks)
• If there is a difference (gradient) in concentration between two parts of a material, statistically there will be more particles crossing from the side of higher concentration to the side of lower concentration than in the reverse direction.
• Then we expect a net flux of particles
Diffusion Equations
I n Aqe
dn
dx
The more non uniform the concentration the larger the current
I n Dn Aqe
dn
dxDn A q
dn
dx
J n Dn qdn
dx
J p D p qdp
dx
I p D p Aqh
dn
dxD p Aq
dn
dx
Assuming the charge concentration decreases with increasing x It means that dn/dx and dp/dx are negative, so to conform with conventions we must put a – sign in front of the equations.
Proportionality Constants
charge in the cross section
Drift Current
vn n
E
v p p E
Eventually v saturates • too many collisions • effective electrons' mass increases
Mobility (proportionality constant)
I n vn W h n qeI
pv
pW h p q
h
charge per unit of volume
cross section area
volume travelled per unit of time
J n n E n qe n E n q
J p p E p qh p E p q
h
L
W
Drift and Diffusion currents
J n ,drift n E n qe n E n q
J p ,drift p E p qh p E p q
q qh qe 1.6 1019
C
J n ,diff Dn qe
dn
dxDn q
dn
dx
J p ,diff D p qh
dp
dxD p q
dp
dx
Built in Voltage
• At equilibrium: (drift and diffusion balance out)
• Let's consider the second equation:
J n J n ,drift J n , diff n E n q Dn qdn
dx0
J p J p ,drift J p ,diff p E p q D p qdp
dx0
p E p q Dp qdp
dx p
dV
dxp D p
dp
dx p p dV D p dp
p
V xn
V x p
dV D p
p xn
p x p
dp
pV x p V xn
Dp
p
lnp x p
p xn
xp x
n
Built in Voltage
0V xn V x p
D p
p
lnp x p
p xn0
KT
qln
N A N D
ni
2
Einstein ' s Relation :
D p
p
Dn
n
KT
qV T
Mass Action Law :
n p ni
2
ND
e.g.1017
ni
2N D
N A e.g.1018
ni
2N A
xn
xp
1018
1017
103
104
Since both μ and D aremanifestations of thermalrandom motion (i.e. statistical thermodynamics phenomena) they are not independent
If n then p A larger number of electrons causesthe recombination rate of electrons with holes to increase
Applying KVL to the PN junction in equilibrium
I0
R
P N
+ + +
+ + +
0
xp x
n
R
We cannot have current. Something is wrong !
Built in Voltage
P N
+ + +
+ + +
+
0
xp x
n
E
If a free electron in the P region or a hole in the N region somehow reach the edge of the depletion region get swept by the electric field ( drift)
pm mn
KVL at equilibrium:
pm+
mn+
0 = 0
Metal-semiconductorcontact potentials
• There are 4 charged particles in silicon, two mobiles (holes and electrons) and two fixed (ionized donors and ionized acceptors)
• The total positive change density and the total negative charge must be equal
Neutrality of charge
positive ions concentration
holes concentration
negative ions concentration
electrons concentration
N D p N A n
Depletion region in equilibrium
• The doping concentrations NA on the p side and N
D on the n side
are assumed constants
+q ND
q NA
x
(x)
xn
xp
x
E(x)
E(0)=Emax
= qNAx
p/
s= qN
Dx
n/
s
Positive and negative excess charge in the depletion region must balance out (neutrality of charge):
q ND x
n = q N
A x
p
Gauss ' Law :
dE x
dx
x
s
x q p n N D N A
the depletion regionis freeof
electrons and holes
N D N A n p
x q N D N A
on the P side N D 0 :
x q N A
on the N side N A 0:
x q N D
Depletion region in equilibrium x
nx
p x
E(x)
Emax
= qNAx
p/
s= qN
Dx
n/
s
(x)
x
1
2E max xn
1
2
q N D xn
2
s
AreaTriangle 0 Emax xn
1
2E max x p
1
2
q N A x p
2
s
Area Triangle 0 Emax x p
0
0
1
2
q N D
s
xn
2 1
2
q N A
s
x p
2
N D xn N A x p
xn
N A
N D
x p
x p
N D
N A
xn
Potential Equation :
d E x dx
Depletion region in equilibrium
xn
N A
N D
x p
x p
N D
N A
xn
0
1
2
q N D
s
xn
2 1
2
q N A
s
x p
2 1
2
q N D
s
X dep
2N A
N A N D
2
1
2
q N A
s
X dep
2N D
N A N D
2
1
2
q
s
X dep
N A N D
2
N D N A
2N A N D
2q X dep
2
2 s
N A N D
N A N D
X dep xn
xp
1N
A
ND
xp
1N
D
NA
xn
x p X dep
N D
N A N D
and xn X dep
N A
N A N D
Width and max field of the depletion region in equilibrium
X dep
2s
q
NA
ND
NAN
D
0
E max
q N A
s
x p
q N A
s
X dep
N D
N A N D
q
s
N A N D
N A N D
X dep
Emax
2q
s
N A N D
N A N D
0
with :0
KT
qln
N A N D
ni
2
Biased PN Junction
vjpm mn
v
P N
xE(x)
v=0v<0
v>0
v pm mnv j 0
0
v j 0v
+ -
Biased PN Junction
• if we keep decreasing the voltage eventually we'll break the material. For silicon the breakdown point is reached for an electric field of approx. 107 V/m.NOTE: the depletion region can't get bigger than the length of the bar !
• if we keep increasing the voltage the depletion region will disappear (v=0).
As v becomes comparable with 0 the PN junction behave like a sort of resistor (the
current is determined by the ohmic contacts and the resistance of the semiconductor)
X dep
2s
q
NA
ND
NAN
D
0v
0
KT
qln
N A N D
ni
2
Emax
2 q
s
NAN
D
NA
ND
0v
v
i
BV
0
Reverse Biased PN Junction• Under reverse bias the depletion
region becomes wider
• Then, it gets harder for the majority carriers to cross (diffuse through) the junction and easier for the minority carrier to be swept (drifted) across the junction
• Since there are only a FEW minority carriers, the current carried under reverse bias is negligible
E
E0
Reverse Biased PN JunctionNOTE:
As soon as a minority carrier, let's say an electron on the P side, is swept across the junction, on the N side it becomes a majority carrier. Every time a minority carrier on the P side is swept toward the N side it leaves one less minority carrier on the edge of the depletion region in the P region. The same is true for holes swept from the N side to the P side.
Under reverse bias the minority carrier concentration at the edges of the depletion regions is depleted below their equilibrium value. Since the number of minority carriers is small anyway this won't be a major difference
E
E0
Forward Biased PN Junction• Under forward bias the depletion
region shrinks
• Then, it gets easier for the majority carriers to cross (diffuse through) the junction and harder for the minority carrier to be swept (drifted) across the junction.
• Since there are a LOT of majority carriers we expect the current to be considerable
Forward Biased PN Junction
I/V characteristic of PN Junction
Idiode
q A Dn
Ln
ni
2
NA
q A D p
L p
ni
2
ND
e
V diode
V T1 I
se
V diode
V T1
I s
with :
A cross section area of the diode
ni
2
N D
holes' concentration in the N region (minority carriers)
ni
2
N A
electrons' concentration in the P region (minority carriers)
D p diffusion constant for the holes in the N region
L p diffusion lenght for the holes in the N region D p p
p average time it takes for a hole into the N region to recombine
with a majority electron
...
I/V characteristic of PN JunctionSince the only region where we have “net charge” is between -x
p and x
n
such region (a.k.a. space charge region) is the only one where there is electric field.
The regions from A to -xp and from x
n to K are quasi-neutral (it is like they
were perfect conductors and in perfect conductors there is no electric field inside)
I/V characteristic of PN Junction
The situation is similar to the one at equilibrium but now the “built in voltage” is
0 – V
diode instead of
0
Vdiode
gnd
-Xp
Xn
X0
pp
pn
np
nn
A K
I/V characteristic of PN Junction
P NIn the N region we have a lot of electronsthat will diffuse toward the P region
In the P region we have a lot of holesthat will diffuse toward the N region
• Under forward bias close to the depletion edges we have:
– a greater hole concentration than normal on the N side (minority carriers)
– a greater electrons concentration than normal on the P side (minority carriers)
I/V characteristic of PN Junction
Extending the result derived at equilibrium we can write the voltage across the space charge region (between -x
p and x
n) as:
V j V xn
V x p 0V
diodeV
Tln
p x p
p xn
VTln
n xn
n x p
0V diode
V T
lnp x p
p xn
0V diode
V T
lnn xn
n x p
p xn
p x p
e
0V
diode
VT
p x p
e
0
VT
e
Vdiode
VT
n xp
n xn
e
0V
diode
VT
n xn
e
0
VT
e
V diode
V T
I/V characteristic of PN Junction• And noting that:
– at the boundary of the quasi neutral P region at -Xp the hole density (majority carriers) is approximately equal at equilibrium as well as under bias,
– and the same is true for the electron density (majority carriers) at the
boundary of the quasi neutral N region (at Xn)
p x p
e
0
V T
p p0
e
0
V T
pn0
ni
2
ND
n xn
e
0
V T
nn0
e
0
V T
n p0
ni
2
NA
the concentration of the majority carriers in the quasi neutral regions is approximately the same as the concentration at equilibrium
I/V characteristic of PN Junction• Thus:
p xn
p x p
e
0V
diode
VT
p x p
e
0
VT
e
Vdiode
VT
n xp
n xn
e
0V
diode
VT
n xn
e
0
VT
e
V diode
V T
p xn
ni
2
ND
e
Vdiode
VT
n xp
ni
2
NA
e
V diode
V T
p x p
e
0
V T
p p0
e
0
V T
pn0
ni
2
ND
n xn
e
0
V T
nn0
e
0
V T
n p0
ni
2
NA
I/V characteristic of PN Junction
• If we consider the quasi neutral regions, since in the quasi neutral regions there is no field there will be no drift
• Then, the most suitable traverse sections for the evaluation of the total current I
diode are those at the boundary of the depletion layer (x=–x
p or x=x
n)
Idiode
In
I p I n ,drift I n , diff I p , drift I p , diff
I n I n , diff neutral regionW N
I p I p ,diff neutral regionW P
Idiode
In
xp
Ip
xp
I n.diff xp
I p ,diff xp
I/V characteristic of PN Junction If we make the simplifying assumption that the flow of the carriers in the
depletion region is approximately constant (in other words we assume the recombination in the depletion region is negligible)
The currents due to diffusing carriers moving away from the junction are given by the well know diffusion equations:
I p ,diff xp
I p ,diff xn
Idiode
I n.diff xp
I p ,diff xp
I n.diff xp
I p , diff xn
I n , diff x q A Dn
dnp
x
dx
I p ,diff x q A D p
dpn
x
dx
I/V characteristic of PN Junction• If we assume that the carriers distribution is linear (SHORT DIODE)
dn p x
dx
|
| x p
n p x p n p A
W p
n p0 e
Vdiode
VT
n p0
W p
n p0 e
Vdiode
VT
1
W p
dpn
x
dx
|
| xn
pn
xn
pn
K
Wn
pn0
e
Vdiode
VT p
n0
Wn
pn0
e
Vdiode
VT
1
Wn
PN Junction: I/V characteristic
• Recalling that:
dn p x p
dx
n p0 e
Vdiode
VT
1
W p
ni
2
NA
e
Vdiode
VT
1
dpn
xn
dx
pn0
e
V diode
V T
1
Wn
ni
2
ND
e
V diode
VT
1
n p0
ni
2
N A
(electrons are minority carriers in P)
pn0
ni
2
N D
(holes are minority carriers in N)
I n ,diff x p q A Dn
dn p x p
dxq An
i
2D
n
NAW p
e
Vdiode
VT
1
I p , diff xn
q A Dp
dpn
xn
dxq An
i
2D p
NDW
n
e
V diode
V T
1
PN Junction: I/V characteristic • And finally:
• In the case of a LONG DIODE the minority carriers will recombine before reaching the diode terminals
I n ,diff x p q Ani
2D
n
NAW p
e
Vdiode
VT
1
I p ,diff xn
q A ni
2D p
NDW
n
e
V diode
V T
1
I diode I n.diff x p I p ,diff x p q Ani
2Dn
N AW p
N D W n e
V diode
V T1
Is
Idiode
I n.diff x p I p ,diff x p q Ani
2D
n
NA
Ln
D p
ND
L p
e
V diode
V T1
PN Junction: I/V characteristic SHORT DIODE:
LONG DIODE:
Where Ln is a constant known as the diffusion length of electrons in
the P side and Lp is a constant known as the diffusion length for
holes in the N side. The constants Ln and L
p are dependent on the
doping concentrations NA and N
D respectively.
Idiode
I n.diff x p I p , diff x p q Ani
2D
n
NAW p
D p
NDW
n
e
V diode
V T1
Is
Idiode
I n.diff x p I p ,diff x p q Ani
2D
n
NA
Ln
D p
ND
L p
e
V diode
V T1
Is
Diode Capacitances
• Depletion Capacitance (= Junction Capacitance)
• Diffusion Capacitance
• Reverse Biased Diode
– Depletion Capacitance
• Forward Biased Diode
– Diffusion Capacitance + Depletion Capacitance
C j
Cd
Depletion Charge • The depletion region stores an immobile charge of equal amount
on each side of the junction ( it forms a capacitance !!)
qJ
qP
q NAx
pA q
Nq N
Dx
nA
x p X dep
N D
N A N D
and xn X dep
N A
N A N D
qJ
qN
AN
D
NA
ND
A X dep
X dep
2 s
q
N A N D
N A N D0 vD
qJ
vD
A 2qs
NA
ND
NA
ND
0v
D
The charge of the depletion region is a function of the voltage v
D applied to the diode
P side has “ ” ions
N side has “+” ions
The decision to take qJ negative is
totally arbitrary. (But it turns out tobe a good one if we prefer to work with positive capacitances)
Depletion Capacitance • Since the depletion charge does not change linearly with the applied
voltage the resulting capacitor is non linear !!
• An important physical consideration: we are dealing with a capacitor that no matter where I put the + of the applied voltage it always accumulate the positive charge on the N side of the junction, and the negative charge on the P side of the junction.
qJ
vD
A 2qs
NA
ND
NA
ND
0v
D
qJ
vD
0
Small Signal Depletion Capacitance • For small changes of the applied voltage about a specified DC
voltage VD we can derive an equivalent linear capacitor
approximation
qJ
vD
since qJ vs. D relationship is
non linear the capacitor is non linear
small “quantities”
Common conventions:v
DV
Dv
d
Vd
VD
vd
total “quantities”(applied)
DC “quantities”
D
qJ
0VD D
= VD +
d
QJ
d
qJ= Q
J + q
j
slopeq
j
dqJ
dvD
|V
D
vd
slope
qJ
vD
qJ
VD
dqJ
dvD
|V
D
vD
VD
Small Signal Depletion Capacitance
C j C j VD
dqJ
dvD
|V
D
d
dvD
A 2 qs
NA
ND
NA
ND
0v
D
1 2
VD
A 2 qs
NA
ND
NA
ND
1 2
d
dvD
0v
D
1 2
VD
A 2qs
NA
ND
NA
ND
1 2
1
20
vD
1 2
VD
Aq
s
2
NA
ND
NA
ND
1 2
0V
D
1 2A
2
qs
NA
ND
NA
ND
0V
D
1 2
A
2
qs
NA
ND
NA
ND
0V
D
A
2
qs
NA
ND
NA
ND
01
VD
0
= Cj0
Small Signal Depletion Capacitance
C j C j VD
q j
vd
dqJ
dvD
|V
D
C j0
1V
D
0
As
X dep ,0 1V
D
0
As
X dep
Zero Bias Capacitance = junction capacitance in thermal equilibrium (V
D=0)
C j0 C j VD
0A
2
qs
NA
ND
NA
ND
0
As
X dep ,0
X dep ,0
2 s
q
N A N D
N A N D0 s
2
q s
N A N D
N A N D0
Small Signal Depletion Capacitance:Physical Interpretation
C j
q j
vd
As
X dep
Capacitance of a parallel plate capacitor with its plates separated by the depletion width X
dep(V
D) at the particular DC voltage V
D.
The charges separated by Xdep
are the small signal charge layers ±qj.
For vd 0, the small signal charges become sheets that are separated by a gap width of exactly Xdep
Physical Interpretation
C j
q j
vd
As
X dep
Capacitance of a parallel plate capacitor with its plates separated by the depletion width X
dep(V
D) at the particular DC voltage V
D. The charges
separated by Xdep
are the small signal charge layers ±qj. For vd 0, the
small signal charges become sheets that are separated by a gap width of Xdep
Small Signal Depletion Capacitance
• In the practice the depletion capacitance is usually provided per cross-section area:
j0 j VD
0s
X dep ,0
1
2
qs
NA
ND
NA
ND
0
j j VD
dqJ
dvD
|V
D
s
X dep
s
X dep ,0 1V
D
0
j0
1V
D
0
NOTE: when the diode is forward biased with vD = 0
the equation for Cj “blows up” (i.e., is equal to infinity).
As vD approaches 0 the assumption that the depletion
region is free of charged carriers is no longer true.
Graded Junctions
• All the equations derived for the depletion capacitance are based on the assumption that the doping concentration change abruptly at the junction. Although this is a good approximation for many integrated circuits is not always true.
• More in general:
• Mj is a constant called grading coefficient and its value ranges from 1/3 to depending on the way the concentration changes from the P to the N side of the junction
j
j0
1V
D
0
M j
Large Signal Depletion Capacitance
• The equations for the depletion capacitance given before are valid only for small changes in the applied voltage
• It is extremely difficult and time consuming to accurately take this non linear capacitance into account when calculating the time to charge or discharge a junction over a large voltage change
• A commonly used approximation is to calculate the charge stored in the junction for the two extreme values of applied voltage, and then through the use of Q = C V, calculate the average capacitance accordingly
j av
V2
V1
V2
V1
The approximation ispessimistic
Large Signal Depletion Capacitance
j av
V2
V1
V2
V1
20 j0
1V
2
0
1V
1
0
V2
V1
V 2 qs
NA
ND
NA
ND
0V 2 q
s
NA
ND
NA
ND
01
V
0
20 j0 1
V
0
j0
qs
20
NA
ND
NA
ND
qs
NA
ND
NA
ND
j0 20
Example
Find a rough approximation for the junction capacitance to be used to estimate the charging time of a reverse biased junction from 0V to 5V (or vice versa). Assume
0 0.9 V
j av2
0 j0
1V
2
0
1V
1
0
V2
V1
2 0.9 j0
15
0.91
5 0
j0
2
0V
5V
VD