Plan for the quantitative solution
n p p
n
n x
D n
22
0
p n n
p
p x
D p
2 20
x ≤ -x p x ≥ x n
1) Solve the minority carrier diffusion equations employing boundary conditions
0 )
(
p
nx
1
)
(
/2
qV kTA i p
p
e
AN x n
n ( )
2
qV /kT 1
D i n
n
e
AN x n
p 0
)
(
n
px
p
p x L
L x
n
x Ae Be
p ( )
/
/
n
n x L
L x
p
x Ae Be
n ( )
/
/
Plan for the quantitative solution
2) Compute the minority carrier current densities in the quasi- neutral regions using
dx n qD d
J
n n
p
dx p qD d
J
p
p
nx ≤ -x p x ≥ x n
3) Evaluate the quasi-neutral region solutions for J n (x) and J p (x) at the edges of the depletion region and then sum the two edge current densities.
) ( )
(
p p nn
x J x
J
J
4) Finally, multiply the result by the cross-sectional area of the diode.
AJ
I
It is convenient to shift the origin of coordinates to the n-edge or the p-edge of the depletion region.
Quantitative solution for ideal diode equation
For the n-type quasi-neutral region,
For the p-type quasi-neutral region,
x n
x x '
x p
x
x "
In the new coordinates, the minority carrier diffusion equations are
Quantitative solution for ideal diode equation
n p p
n
n x
D n
22
0
p n n
p
p x
D p
2 20
x ≤ -x p
x ≥ x n x ' x x
nx
px x "
p n n
p
p x
D p
2 20 ' x’ ≥ 0
n p p
n
n x
D n
22
"
0 x” ≥ 0
In the new coordinates, the boundary conditions are
0 )
'
(
p
nx
1
) 0 '
(
/2
nn
ie
qVA kTx
p
side -
n
0
)
"
(
n
px
1
) 0
"
(
/2
qV kTA i p
e
AN x n
n
side -
p
Quantitative solution for ideal diode equation
p n n
p
p x
D p
2 20 ' x’ ≥ 0
n p p
n
n x
D n
22
"
0 x” ≥ 0
n
n x L
L x
p
x Ae Be
n ( )
/
/
p
p x L
L x
n
x Ae Be
p ( )
/
/
In the new coordinates, the general solutions are
n
n x L
L x
p
x Ae Be
n ( " )
"/
"/
p
p x L
L x
n
x Ae Be
p ( ' )
'/
'/
where A and B are solution constants, L
p D
p
pL
n D
n
nx’ ≥ 0
x” ≥ 0
Quantitative solution for ideal diode equation
p n n
p
p x
D p
2 2
0 ' x’ ≥ 0
As x ' , p
n( x ' ) Ae
x'/Lp Be
x'/Lp Be
x'/LpAccording to the boundary conditions,
thus B = 0 0
) '
(
p
nx
1
) 0 '
(
/2
qV kTD i n
e
AN x n
p
side -
n
/1
2
qV kTD
i
e
AN A n
At x ' 0 , p
n( x ' 0 ) Ae
x'/Lp Be
x'/Lp A
Thus,
p
p x L
L x
n
x Ae Be
p ( ' )
'/
'/
qVA kT
x LpD i
n
e e
N x n
p
/ '/2
1 )
'
(
Quantitative solution for ideal diode equation
qVA kT
x LpD i
n
e e
N x n
p
/ '/2
1 )
'
(
dx p qD d
x
J
p( )
p
nIn the new coordinates of x’, the hole current density is x
nx x '
) ' '
( dx
p qD d
x
J
p
p
n
qVA kT
x LpD i p n p
p
p
e e
N n L q D dx
p qD d
x
J
/ '/2
' 1 )
'
(
dx dx '
At the n-side edge (x’ = 0) of the depletion region,
1
) 0 '
(
/2
p i qV kTp
e
AN n L q D x
J
x’ ≥ 0
Quantitative solution for ideal diode equation
As x " ,
According to the boundary conditions,
thus B = 0
At x " 0 ,
Thus,
n p p
n
n x
D n
2
2
"
0 x” ≥ 0
side -
p
1
) 0
"
(
/2
qV kTA i p
e
AN x n
n
0 )
"
(
n
px
n
n x L
L x
p
x Ae Be
n ( " )
"/
"/
n n
n x L x L
L x
p
x Ae Be Be
n ( " )
"/
"/
"/
A Be
Ae x
n
p
x Ln
x Ln
( " 0 )
"/ "/
/1
2
qV kTA
i
e
AN A n
qVA kT
x LnA i
p
e e
N x n
n
/ "/2
1 )
"
(
Quantitative solution for ideal diode equation
In the new coordinates of x”, the electron current density is
qVA kT
x LnA i
p
e e
N x n
n
/ "/2
1 )
"
(
dx n qD d
x
J
n n
p )
(
px x x "
dx
dx " ( " ) "
dx n qD d
x
J
n n
p
qVA kT
x LnA i n p n
n
n
e e
N n L q D dx
n qD d
x
J
/ "/2
" 1 )
"
(
1
) 0
"
(
/2
qV kTA i n
n n
e
AN n L q D x
J
At the p-side edge (x” = 0) of the depletion region, x” ≥ 0
x” ≥ 0
1
) 0 '
(
/2
qV kTD i p
p p
e
AN n L q D x
1 J
) 0
"
(
/2
qV kTA i n
n n
e
AN n L q D x
J
Quantitative solution for ideal diode equation
The total current density (J) is given by
1 1
) 0 ' ( )
0
"
(
/2 /
2
qV kTD i p kT p
qV A
i n
n p
n
A
A
e
N n L q D N e
n L q D x
J x
J J
/1
2
2
qV kTD i p
p A
i n
n
e
AN n L D N
n L q D
The total current (I) is given by I = AJ
/1
2
2
qV kTD i p
p A
i n
n
e
AN n L D N
n L qA D I
) ( )
(
p p nn
x J x
J
J
The ideal diode equation )
1
( /
0
I e qV
AkT
I
D i p
p A
i n
n
N n L D N
n L qA D I
2 2
0
It is also referred to as the Shockley equation.
1
0
I e V A V ref I
q V
ref kT
At 300 K, kT/q = 26 mV
I I
I
Ideal diode I-V characteristics )
1
( /
0
I e qV
AkT
I if V
A few kt/q I I 0 e qV
A/ kT
V A
kT I q
I ln 0
ln
D i p
p A
i n
n
N n L D N
n L qA D I
2 2
)
01
( /
0
I e qV
AkT I
Ideal diode I-V: saturation current
The saturation current can vary by many orders of magnitude depending on semiconducting materials and doping
concentrations.
diode Ge
diode
Si 0 ,
,
0 I
I
Ex) Because of Si ( ) and Ge ( ) n
i,Si 10
10/ cm
3n
i,Ge 10
13/ cm
3For p + n junction,
D i p
p
N n L qA D I
2
0
For pn + junction,
n iN n L qA D I
2
0
determined I
0
side doped
- lightly the
by
Ideal diode I-V: saturation current
One can neglect the heavily doped side of asymmetrical junctions in computing the depletion width and other
electrostatic variables.
As a general rule, the heavily doped side of an asymmetrical junction can be ignored in determining the electrical
characteristics of the junction.
V A = 0
Majority hole diffusion I
diff,hMinority hole drift I
drift,eMajority electron diffusion I
diff,eMinority electron I
drift,eP
+n
E
0 0
/
0 ( e 1 ) I I
I
I qV
AkT
At V = 0,
Example: two ideal p + n diodes
Two ideal p + n step junction diodes at room temperature are identical except that N D,1 = 10 15 /cm 3 and N D,2 =10 16 cm 3 .
Compare their I-V characteristics.
D i p
p
N n L qA D I
2
0
) 1
( /
0
I e qV
AkT
I
Ideal diode I-V: carrier currents J n & J p
A reverse-bias plot is essentially identical except all current
qVA kT
x LnA i n p n
n
n
e e
N n L q D dx
n qD d
x
J
/ "/2
" 1 )
"
(
qVA kT
x Lp
D i p n p
p
p
e e
N n L q D dx
p qD d
x
J
/ '/2
' 1 )
'
(
) ( )
(
p p nn
x J x
J
J Forward-biased pn junction
x J x
J
n p
0
Ideal diode: minority carrier concentrations
qVA kT
x LpD i
n
e e
N x n
p
/ '/2
1 )
'
(
qVA kT
x Ln
A i
p
e e
N x n
n
/ "/2