Download 1. Diffusion: the flow of charge due to density gradients. a. Diffusion

1. Diffusion: the flow of charge due to density gradients.
a. Diffusion Current
The diffusion currents are expressed in terms of the electron or hole diffusion
coefficient Dn or Dp, as shown below:
dn
dn
J n,diff = (q)Dn
= qDn
dx
dx
dn
dp
J p,diff = (+q)Dp
= qDp
dx
dx
b. Einstein Relation:
kT
Dn =
μn
q
kT
Dp =
μp
q
2. Total Current: J = J diff + J drift
EXAMPLE1: We have a piece of uniformly doped silicon with a acceptor concentration
of Na=2E16cm-3. By shedding light on one end (x=0) of the semiconductor, we creat an
electron concentration of n(x)=1E13*exp(-x/2um). Calculate the electron diffusion
current as a function of x.
SOLUTION1:
N a = 2E16 μn = 1050cm 2 / Vs
kT
μn = 0.026 1050 = 27.3cm 2 / s
Dn =
q
dn
d
x / ( 2 10 4 )
Jn, diff = qDn
= 1.6 10 19 27.3 1013 e
dx
dx
x /2 μ m
2
= 0.22e
A / cm
(
)
3. Integrated Passives
a. Resistors
-Diffusion Resistor: ion implantation/diffusion
100ohm/sq(unsilicided),10ohm/sq(silicided)
-Poly Film Resistor: deposit a thin film of heavily doped poly-Si
10-100ohm/sq(unsilicided),1ohm/sq(silicided)
b. Capacitors
-IC MIM Capacitor: by forming a Metal plate-Insulator(thin oxide)-Metal plate structure.
-Parallel Plate Capacitor:
A
C=
d
-Nonlinear capacitor: Q-V curve is not a straight line
Small signal capacitance: differential capacitance
df (V )
Q = Q0 + q f (Vs) +
vs
dV V =Vs
C
df (V )
dV V =Vs
4. PN Junction
a. Structure
b. Distribution of charge, electric field, and potential
-Depletion Approximation: The transition region is completely depleted of free carriers,
the only charges exist are immobile dopants. This region is called “Depletion Region”
qN a (x po < x < 0)
0 (x) +qN d (0 < x < x no )
-Electric Field in Depletion Region:
qN a
( x + x p 0 )(x po < x < 0)
s
E 0 (x) = qN
d ( x x )(0 < x < x )
n0
no
s
-Potential in Depletion Region
2
qN
p (x) = p + a ( x + x p 0 )
2s
qN
2
n (x) = n + d ( x x n 0 )
2s
-Boundary Conditions:
Potential continuous:
2
qN
qN
2
p (0) = p + a ( x p 0 ) = n (0) = n + d (x n 0 )
2s
2s
Electrical Field continuous:
qN a x p 0 = qN d x n 0
-Depletion Width & Built-in Potential
2s bi 1
1 X d 0 = xn 0 + x p 0 =
+
q Na Nd xn 0 =
2s bi N a qN d N a + N d x p0 =
2s bi N d qN a N a + N d N N bi n p = Vth ln d + ln a n i ni = Vth ln
Na Nd
>0
ni
c. PN Junction Capacitor
-PN Junction under Bias
x n (VD ) =
2s ( bi VD ) N a VD
= x n 0 1
qN d
bi
Na + Nd x p (VD ) =
2s ( bi VD ) N d VD
= x p 0 1
qN a
bi
Na + Nd Cj =
C j0
1
C j0 =
VD
bi
=
s
Xd 0
qs N a N d
= s
2 bi N a + N d X d
C j (VD ) =
s
X d (VD )
-EXAMPLE2:
Consider the following silicon P-N junction:
(a) If Na = 4E16cm-3 in the P-region and Nd = 1E17cm-3 in the N-region, under
increasing reverse bias, which region (N or P) will become completely depleted
first?
(b) What is the reverse bias at this condition?
(c) What is the small-signal capacitance (F/cm2) at the bias condition?
SOLUTION2:
(a) Under reverse bias, the depletion region will expand. The side which has fewer
dopants will fully deplete first.
Na*Wp=4E16*1.2>Nd*Wn=1E17*0.4
So N side will deplete first
N N
(b) bi = Vth ln a d = 0.81V
ni
2s bi N a xn 0 =
= 0.055um
qN d N a + N d V
0.4 = x n (VD ) = x n 0 1 D
bi
So VD=-42V
(c) The small signal capacitance is determined by the depletion region width.
Xn=Wn=0.4um, Xp=(Nd/Na)*Xn=(1E17/4E16)*0.4=1um
Xd=Xn+Xp=0.4+1=1.4um
Cd=s/Xd=8.85E-14*12/1.4E-4=7.6E-9F/cm2
5. PN Junction Diode
-Thermal Equilibrium:
small diffusion current: high barrier
small drift current: few minority carriers and wide depletion region
net current: none
-Reverse Bias:
drift current: remains small
diffusion current: reduced exponentially with bias
net current: small reverse current
-Forward Bias:
drift current: remains small
diffusion current: increase exponentially with bias
net current: large forward current
-IV curve:
-Minority Carrier Concentration
with recombination, long base diode
without recombination or short base diode:
-Diode Current Densities
qV
D
Dn kTA p
J diff = J ndiff + J pdiff = qn i2 +
e 1
N d W n N aW p -Small Signal Model
qVd qv d
q (Vd +v d ) kT
ID + iD = IS e
1 IS e kT e kT
qv
iD d = gd v d
kT
-Charge Storage
Cd =
1 qId
2 kT
-EXAMPLE3
Consider an ideal, long-base, silicon PN junction diode with uniform cross section and
constant doping on either side of the junction. P side is heavily doped and N side has an
doping concentration of 1E16cm-3 .
For the n-side of the junction:
(a) Calculate the density of the minority carriers as a function of x (distance from
the junction) when the applied voltage is 0.589V (which is 23kT/q).
(b) Find the minority carrier currents as functions of x (distance from the junction)
SOLUTION3
(
)
ni2 qVa / kT
x / L
x / L
(a)
p( x) =
e
1 e p = 1014 e p cm 3 .
Nd
(b) Minority current:
(
)
dp( x)
ni2
x / L
x / L
J p ( x) = qD p
=q
D p e qVa / kT 1 e p = 0.5e p A / cm 2 .
dx
N d Lp