Semiconductor pn junctions
semiconductor pn junction context
Figure 8.1-2 pn junction representations.
pn junction in forward bias: fJ = f0 – V
Forward bias: fJ = f0 – V.
Reduction of junction
potential lowers E-field
barrier.
pn junction forward bias: Thermal statistics
Equilibrium:
Forward bias:
(nn = nn0) low-level injection. Only the minority-carrier levels
are appreciably affected.
Low-level injection: Minority-carrier levels affected.
Figure 8.5-2: The quasi-neutral regions (QNR)
Low-level injection: Injected carrier profiles
Injected carriers and diffusion
Figure 7.7-1a. Concept of carrier injection with losses due to recombination
Carrier recombination: Recombination time constants
Recombination of p-type carriers
Recombination of n-type carriers
Carrier flux change (Fick’s laws)
Change in the total
count N within the slice
dN
 F  F  dF   (G  R p ) Adx
dt
G = generation rate
R = recombination rate
Figure 7.7-3 Carrier flow in/out for a one-dimensional slice
Diffusion and recombination (p-type example)
Flux F recast as flow/area
(Flux due to diffusion)
dp 
d 
0    Dp
  Rp
dx 
dx 
Steady-state flux balance of recombination
Since recombination
Then
d 2 p  p
0  Dp

2
dx
p
Steady-state flux balance of recombination
d 2 p  p
0  Dp

2
dx
p
Solution:
p( x)  p n ( x)  p n0  p(0)e
 x / Lp
For which Lp = Recombination length for p-type:
L p  D p p
Similarly Ln = Recombination length for n-type:
Ln  Dn n
EXAMPLE: Determine the diffusion length for electrons injected into a
p-type material doped with 5 × 1016 #/cm3 of Boron, assuming
recombination time for the electrons tn = 200 ns. Assume T = 300K.
SOLUTION: The mobility for n-type carriers in a material of ionized impurity
density 5 × 1016 #/cm3, according to equation (7.3-7a) is:
 n  88 


1  5  10 / 1.26  10
Then Dn =nVT
And
1252
16
17

= 905 × .02585

= 905 cm2/Vs
0.88
= 23.4 cm2/s
Ln  Dn n  23.4  200 10 9

= 21.6 m
Low-level injection
 n p 0 
d

nx x0  qDn 
J n  qDn
dx
 Ln 
 p n 0  
d

px x 0  qD p 
J p  qD p

dx
 Lp 
n p (0)  n p (0)  n p 0  n p 0 exp V VT   n p 0
p n (0)  p n (0)  p n 0  p n 0 exp V VT   p n 0
Low-level injection
J = Jn + Jp
 pn 0  
 n p 0  

  qD p 
 qDn 
 L 
L
n
p






n p0
p
n
0
expV VT   1
  qDn
 qD p


L
L
n
p


 J S expV VT   1
Low-level injection
J  J S expV VT   1

n p0
p
J S   qDn
 qD p n 0

Ln
Lp

 Dn
Dp
J S  qn 

L N
 n A Lp N D
2
i








EXAMPLE E8.5-1: An abrupt silicon pn junction is formed by an ion implant of NA =
1017 #/cm3 into an n-type substrate of impurity level ND = 1015 #/cm3. Determine:
(a) Built-in potential f0,
(b) reverse saturation current JS for recombination time constants n = p = 20ns
(c) Current density level J for V = 0.6V. Assume default temperature (= 300K).
(a)
 NAND
f0  VT ln 
2
n
i


 5  10 31 
  .02585  ln 

20 

 2.25  10 

= 0.693V
(b) reverse saturation current JS for recombination time constants n = p = 20ns
Both types of carriers exist on each
side of the junction
NA side: pp , np
ND side: nn , pn
∴ find (per heuristic formula) n and p
on both sides of junction
The Shockley equation refers to the carriers that are injected into the
other side. Hence the mobilites of interest are n in the NA side and p
on the ND side, which are n = 777cm2/Vs and p = 458cm2/Vs,
respectively.
From the mobilities the diffusion coefficients are
Dn = nVT = 777 × .0259 = 20.1 cm2/s
Dp = pVT = 458 × .0259 = 11.84 cm2/s
From which the recombination lengths are
Ln   n Dn  (20  10 9 )  20.1
= 4.86 × 10-4 cm = 4.86 m
L p   p D p  (20  10 )  11.84
= 6.34 × 10-4cm = 6.34 m
9
Then the reverse saturation current is
= (1.6 ×
10 2
10-7pC)(1.5  10 )
 Dn
Dp
J S  qn 

L N
 n A Lp N D
2
i






20.1
11.84




4
17
4
15 
(4.86  10 )  10 
 (6.34  10 )  10
= 36 × [(3.17 × 10-13) + (2.43 × 10-11)] = 8.88 × 10-10 A/cm2
= 888pA/cm2
it is times like these that a spreadsheet would be a friend.