Lecture27_ali.ppt

Department of EECS University of California, Berkeley
EECS 105 Spring 2005, Lecture 27
Lecture 27:
PN Junctions
Prof. Niknejad

EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad
Diffusion
 Diffusion occurs when there exists a concentration
gradient
 In the figure below, imagine that we fill the left
chamber with a gas at temperate T
 If we suddenly remove the divider, what happens?
 The gas will fill the entire volume of the new
chamber. How does this occur?

Diffusion (cont)
 The net motion of gas molecules to the right
chamber was due to the concentration gradient
 If each particle moves on average left or right then
eventually half will be in the right chamber
 If the molecules were charged (or electrons), then
there would be a net current flow
 The diffusion current flows from high
concentration to low concentration:

Diffusion Equations
 Assume that the mean free path is λ
 Find flux of carriers crossing x=0 plane
)
(
n
)
0
(
n
)
( 

n
0

 
th
v
n )
(
2
1

th
v
n )
(
2
1


 
)
(
)
(
2
1

 n
n
v
F th 


























dx
dn
n
dx
dn
n
v
F th 
 )
0
(
)
0
(
2
1
dx
dn
v
F th


dx
dn
qv
qF
J th




Einstein Relation
 The thermal velocity is given by kT
kT
v
m th
n 2
1
2
*
2
1

c
th
v 
 
Mean Free Time
dx
dn
q
kT
q
dx
dn
qv
J n
th 








 

n
n
q
kT
D 









*
*
2
n
c
n
c
c
th
th
m
q
q
kT
m
kT
v
v



 



Total Current and Boundary Conditions
 When both drift and diffusion are present, the total
current is given by the sum:
 In resistors, the carrier is approximately uniform
and the second term is nearly zero
 For currents flowing uniformly through an interface
(no charge accumulation), the field is discontinous
dx
dn
qD
nE
q
J
J
J n
n
diff
drift 


 
2
1 J
J 
2
2
1
1 E
E 
 
1
2
2
1



E
E
)
( 1
1 
J
)
( 2
2 
J

Carrier Concentration and Potential
 In thermal equilibrium, there are no external fields
and we thus expect the electron and hole current
densities to be zero:
dx
dn
qD
E
qn
J o
n
n
n 

 0
0
0 
dx
d
n
kT
q
E
n
D
dx
dn
o
o
n
n
o 0
0



















0
0
0
0
n
dn
V
n
dn
q
kT
d th
o












Carrier Concentration and Potential (2)
 We have an equation relating the potential to the
carrier concentration
 If we integrate the above equation we have
 We define the potential reference to be intrinsic Si:
)
(
)
(
ln
)
(
)
(
0
0
0
0
0
0
x
n
x
n
V
x
x th



i
n
x
n
x 
 )
(
0
)
( 0
0
0
0

0
0
0
0
n
dn
V
n
dn
q
kT
d th
o












Carrier Concentration Versus Potential
 The carrier concentration is thus a function of
potential
 Check that for zero potential, we have intrinsic
carrier concentration (reference).
 If we do a similar calculation for holes, we arrive at
a similar equation
 Note that the law of mass action is upheld
th
V
x
ie
n
x
n /
)
(
0
0
)
( 

th
V
x
ie
n
x
p /
)
(
0
0
)
( 


2
/
)
(
/
)
(
2
0
0
0
0
)
(
)
( i
V
x
V
x
i n
e
e
n
x
p
x
n th
th

  


The Doping Changes Potential
 Due to the log nature of the potential, the potential changes
linearly for exponential increase in doping:
 Quick calculation aid: For a p-type concentration of 1016
cm-3, the potential is -360 mV
 N-type materials have a positive potential with respect to
intrinsic Si
10
0
0
0
0
0
0
10
)
(
log
10
ln
mV
26
)
(
)
(
ln
mV
26
)
(
)
(
ln
)
(
x
n
x
n
x
n
x
n
x
n
V
x
i
i
th 



10
0
0
10
)
(
log
mV
60
)
(
x
n
x 

10
0
0
10
)
(
log
mV
60
)
(
x
p
x 



n-type
p-type
ND
NA
PN Junctions: Overview
 The most important device is a junction
between a p-type region and an n-type region
 When the junction is first formed, due to the
concentration gradient, mobile charges
transfer near junction
 Electrons leave n-type region and holes leave
p-type region
 These mobile carriers become minority
carriers in new region (can’t penetrate far due
to recombination)
 Due to charge transfer, a voltage difference
occurs between regions
 This creates a field at the junction that causes
drift currents to oppose the diffusion current
 In thermal equilibrium, drift current and
diffusion must balance
− − − − − −
+ + + + +
+ + + + +
+ + + + +
− − − − − −
− − − − − −
−
V
+

PN Junction Currents
 Consider the PN junction in thermal equilibrium
 Again, the currents have to be zero, so we have
dx
dn
qD
E
qn
J o
n
n
n 

 0
0
0 
dx
dn
qD
E
qn o
n
n 

0
0
dx
dn
n
q
kT
n
dx
dn
D
E
n
o
n
0
0
0
0
1





dx
dp
p
q
kT
n
dx
dp
D
E
p
o
p
0
0
0
0
1





PN Junction Fields
n-type
p-type
ND
NA
)
(
0 x
p
a
N
p 
0
d
i
N
n
p
2
0 
diff
J
0
E
a
i
N
n
n
2
0 
Transition Region
diff
J
d
N
n 
0
– – + +
0
E
0
p
x
 0
n
x

Total Charge in Transition Region
 To solve for the electric fields, we need to write
down the charge density in the transition region:
 In the p-side of the junction, there are very few
electrons and only acceptors:
 Since the hole concentration is decreasing on the p-
side, the net charge is negative:
)
(
)
( 0
0
0 a
d N
N
n
p
q
x 




)
(
)
( 0
0 a
N
p
q
x 


0
)
(
0 
x

0
p
Na 
0
0 

 x
xp

Charge on N-Side
 Analogous to the p-side, the charge on the n-side is
given by:
 The net charge here is positive since:
)
(
)
( 0
0 d
N
n
q
x 


 0
0 n
x
x 

0
)
(
0 
x

0
n
Nd 
a
i
N
n
n
2
0 
Transition Region
diff
J
d
N
n 
0
– – + +
0
E

“Exact” Solution for Fields
 Given the above approximations, we now have an
expression for the charge density
 We also have the following result from
electrostatics
 Notice that the potential appears on both sides of
the equation… difficult problem to solve
 A much simpler way to solve the problem…












0
/
)
(
/
)
(
0
0
)
(
0
)
(
)
(
0
0
n
V
x
i
d
po
a
V
x
i
x
x
e
n
N
q
x
x
N
e
n
q
x
th
th



s
x
dx
d
dx
dE


 )
(
0
2
2
0




Depletion Approximation
 Let’s assume that the transition region is
completely depleted of free carriers (only immobile
dopants exist)
 Then the charge density is given by
 The solution for electric field is now easy











0
0
0
0
)
(
n
d
po
a
x
x
qN
x
x
qN
x

s
x
dx
dE

 )
(
0
0

)
(
'
)
'
(
)
( 0
0
0
0
0
p
x
x
s
x
E
dx
x
x
E
p


  

Field zero outside
transition region

Depletion Approximation (2)
 Since charge density is a constant
 If we start from the n-side we get the following
result
)
(
'
)
'
(
)
(
0
0
0 po
s
a
x
x
s
x
x
qN
dx
x
x
E
p



  


)
(
)
(
)
(
'
)
'
(
)
( 0
0
0
0
0
0
0
x
E
x
x
qN
x
E
dx
x
x
E n
s
d
x
x
s
n
n




  


)
(
)
( 0
0 x
x
qN
x
E n
s
d




Field zero outside
transition region

Plot of Fields In Depletion Region
 E-Field zero outside of depletion region
 Note the asymmetrical depletion widths
 Which region has higher doping?
 Slope of E-Field larger in n-region. Why?
 Peak E-Field at junction. Why continuous?
n-type
p-type
ND
NA
– – – – –
– – – – –
– – – – –
– – – – –
+ + + + +
+ + + + +
+ + + + +
+ + + + +
Depletion
Region
)
(
)
( 0
0 x
x
qN
x
E n
s
d




)
(
)
(
0 po
s
a
x
x
qN
x
E 




Continuity of E-Field Across Junction
 Recall that E-Field diverges on charge. For a sheet
charge at the interface, the E-field could be
discontinuous
 In our case, the depletion region is only populated
by a background density of fixed charges so the E-
Field is continuous
 What does this imply?
 Total fixed charge in n-region equals fixed charge
in p-region! Somewhat obvious result.
)
0
(
)
0
( 0
0






 x
E
x
qN
x
qN
x
E p
no
s
d
po
s
a
n


no
d
po
a x
qN
x
qN 

Potential Across Junction
 From our earlier calculation we know that the
potential in the n-region is higher than p-region
 The potential has to smoothly transition form high
to low in crossing the junction
 Physically, the potential difference is due to the
charge transfer that occurs due to the concentration
gradient
 Let’s integrate the field to get the potential:





x
x
po
s
a
po
p
dx
x
x
qN
x
x
0
'
)
'
(
)
(
)
(



x
x
po
s
a
p
p
x
x
x
qN
x
0
'
2
'
)
(
2
















Potential Across Junction
 We arrive at potential on p-side (parabolic)
 Do integral on n-side
 Potential must be continuous at interface (field
finite at interface)
2
0 )
(
2
)
( p
s
a
p
p
o x
x
qN
x 





2
0 )
(
2
)
( n
s
d
n
n x
x
qN
x 





)
0
(
2
2
)
0
( 2
0
2
0 p
p
s
a
p
n
s
d
n
n x
qN
x
qN





 





Solve for Depletion Lengths
 We have two equations and two unknowns. We are
finally in a position to solve for the depletion
depths
2
0
2
0
2
2
p
s
a
p
n
s
d
n x
qN
x
qN



 


no
d
po
a x
qN
x
qN 
(1)
(2)










d
a
a
d
bi
s
no
N
N
N
qN
x


2










a
d
d
a
bi
s
po
N
N
N
qN
x


2
0


 p
n
bi 



Sanity Check
 Does the above equation make sense?
 Let’s say we dope one side very highly. Then
physically we expect the depletion region width for
the heavily doped side to approach zero:
 Entire depletion width dropped across p-region
0
2
lim
0 




a
d
d
d
bi
s
N
n
N
N
N
qN
x
d



a
bi
s
a
d
d
a
bi
s
N
p
qN
N
N
N
qN
x
d



 2
2
lim
0 













Total Depletion Width
 The sum of the depletion widths is the “space
charge region”
 This region is essentially depleted of all mobile
charge
 Due to high electric field, carriers move across
region at velocity saturated speed












d
a
bi
s
n
p
d
N
N
q
x
x
X
1
1
2
0
0
0


μ
1
10
1
2
15
0 







q
X bi
s
d


cm
V
10
μ
1
V
1 4


pn
E

Have we invented a battery?
 Can we harness the PN junction and turn it into a
battery?
 Numerical example:
2
ln
ln
ln
i
A
D
th
i
A
i
D
th
p
n
bi
n
N
N
V
n
N
n
N
V 











 


mV
600
10
10
10
log
mV
60
ln
mV
26 20
15
15
2




i
A
D
bi
n
N
N

?

Contact Potential
 The contact between a PN junction creates a
potential difference
 Likewise, the contact between two dissimilar
metals creates a potential difference (proportional
to the difference between the work functions)
 When a metal semiconductor junction is formed, a
contact potential forms as well
 If we short a PN junction, the sum of the voltages
around the loop must be zero:
mn
pm
bi 

 


0
p
n
mn

pm

+
−
bi
 )
( mn
pm
bi 

 



PN Junction Capacitor
 Under thermal equilibrium, the PN junction does
not draw any (much) current
 But notice that a PN junction stores charge in the
space charge region (transition region)
 Since the device is storing charge, it’s acting like a
capacitor
 Positive charge is stored in the n-region, and
negative charge is in the p-region:
no
d
po
a x
qN
x
qN 

Reverse Biased PN Junction
 What happens if we “reverse-bias” the PN
junction?
 Since no current is flowing, the entire reverse
biased potential is dropped across the transition
region
 To accommodate the extra potential, the charge in
these regions must increase
 If no current is flowing, the only way for the charge
to increase is to grow (shrink) the depletion regions
+
−
D
bi V

 D
V 0

D
V

Voltage Dependence of Depletion Width
 Can redo the math but in the end we realize that the
equations are the same except we replace the built-
in potential with the effective reverse bias:













d
a
D
bi
s
D
n
D
p
D
d
N
N
q
V
V
x
V
x
V
X
1
1
)
(
2
)
(
)
(
)
(


bi
D
n
d
a
a
d
D
bi
s
D
n
V
x
N
N
N
qN
V
V
x















 1
)
(
2
)
( 0
bi
D
p
d
a
d
a
D
bi
s
D
p
V
x
N
N
N
qN
V
V
x















 1
)
(
2
)
( 0
bi
D
d
D
d
V
X
V
X


 1
)
( 0

Charge Versus Bias
 As we increase the reverse bias, the depletion
region grows to accommodate more charge
 Charge is not a linear function of voltage
 This is a non-linear capacitor
 We can define a small signal capacitance for small
signals by breaking up the charge into two terms
bi
D
a
D
p
a
D
J
V
qN
V
x
qN
V
Q





 1
)
(
)
(
)
(
)
(
)
( D
D
J
D
D
J v
q
V
Q
v
V
Q 



Derivation of Small Signal Capacitance
 From last lecture we found
 Notice that




 D
V
D
D
J
D
D
J v
dV
dQ
V
Q
v
V
Q
D
)
(
)
(
R
D V
V
bi
p
a
V
V
j
D
j
j
V
x
qN
dV
d
dV
dQ
V
C
C

 

















1
)
( 0
bi
D
j
bi
D
bi
p
a
j
V
C
V
x
qN
C


 



1
1
2
0
0
d
a
d
a
bi
s
d
a
d
a
bi
s
bi
a
bi
p
a
j
N
N
N
N
q
N
N
N
qN
qN
x
qN
C


























 2
2
2
2
0
0

Physical Interpretation of Depletion Cap
 Notice that the expression on the right-hand-side is
just the depletion width in thermal equilibrium
 This looks like a parallel plate capacitor!
d
a
d
a
bi
s
j
N
N
N
N
q
C




2
0
0
1
0
1
1
2 d
s
d
a
bi
s
s
j
X
N
N
q
C



 











)
(
)
(
D
d
s
D
j
V
X
V
C



A Variable Capacitor (Varactor)
 Capacitance varies versus bias:
 Application: Radio Tuner
0
j
j
C
C

EECS 105 Fall 2003, Lecture 27
Part II: Currents in PN Junctions

Diode under Thermal Equilibrium
 Diffusion small since few carriers have enough energy to penetrate barrier
 Drift current is small since minority carriers are few and far between: Only
minority carriers generated within a diffusion length can contribute current
 Important Point: Minority drift current independent of barrier!
 Diffusion current strong (exponential) function of barrier
p-type n-type
D
N A
N
-
-
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
0
E
bi
q
,
p diff
J
,
p drift
J
,
n diff
J
,
n drift
J
−
−
+
+
−
−
Thermal
Generation
Recombination
Carrier with energy
below barrier height
Minority Carrier Close to Junction

Reverse Bias
 Reverse Bias causes an increases barrier to
diffusion
 Diffusion current is reduced exponentially
 Drift current does not change
 Net result: Small reverse current
p-type n-type
D
N A
N
-
-
-
-
-
-
-
+
+
+
+
+
+
+
( )
bi R
q V
 
+
−

Forward Bias
 Forward bias causes an exponential increase in
the number of carriers with sufficient energy to
penetrate barrier
 Diffusion current increases exponentially
 Drift current does not change
 Net result: Large forward current
p-type n-type
D
N A
N
-
-
-
-
-
-
-
+
+
+
+
+
+
+
( )
bi R
q V
 
+
−

Diode I-V Curve
 Diode IV relation is an exponential function
 This exponential is due to the Boltzmann distribution of carriers versus
energy
 For reverse bias the current saturations to the drift current due to minority
carriers
1
d
qV
kT
d S
I I e
 
 
 
 
d
qV
kT
d
s
I
I
1

( )
d d S
I V I
   

Minority Carriers at Junction Edges
Minority carrier concentration at boundaries of
depletion region increase as barrier lowers …
the function is




)
(
)
(
p
p
n
n
x
x
p
x
x
p (minority) hole conc. on n-side of barrier
(majority) hole conc. on p-side of barrier
kT
Energy
Barrier
e /
)
(


(Boltzmann’s Law)
kT
V
q D
B
e /
)
( 

 
A
n
n
N
x
x
p )
( 

“Law of the Junction”
Minority carrier concentrations at the edges of the
depletion region are given by:
kT
V
q
A
n
n
D
B
e
N
x
x
p /
)
(
)
( 


 
kT
V
q
D
p
p
D
B
e
N
x
x
n /
)
(
)
( 



 
Note 1: NA and ND are the majority carrier concentrations on
the other side of the junction
Note 2: we can reduce these equations further by substituting
VD = 0 V (thermal equilibrium)
Note 3: assumption that pn << ND and np << NA

Minority Carrier Concentration
The minority carrier concentration in the bulk region for
forward bias is a decaying exponential due to recombination
p side n side
-Wp Wn
xn
-xp
0
A
qV
kT
n
p e
0 0
( ) 1
A
p
x
qV
L
kT
n n n
p x p p e e

 
  
 
 
0
n
p
0
p
n
0
A
qV
kT
p
n e
Minority Carrier
Diffusion Length

Steady-State Concentrations
Assume that none of the diffusing holes and
electrons recombine  get straight lines …
p side n side
-Wp Wn
xn
-xp
0
A
qV
kT
n
p e
0
n
p
0
p
n
0
A
qV
kT
p
n e
This also happens if the minority carrier
diffusion lengths are much larger than Wn,p
, ,
n p n p
L W


Diode Current Densities
0 1
A
p
qV
p
diff n kT
n n p
p
x x
dn D
J qD q n e
dx W

 
  
 
 
0 0
( )
( )
A
qV
kT
p p p
p p
dn n e n
x
dx x W


  
p side n side
-Wp Wn
xn
-xp
0
A
qV
kT
n
p e
0
n
p
0
p
n
0
A
qV
kT
p
n e
0 1
A
n
qV
p
diff n kT
p p n
x x n
D
dp
J qD q p e
dx W

 
    
 
 
2
1
A
qV
p
diff n kT
i
d n a p
D D
J qn e
N W N W
  
  
  
 
 
 
2
0
i
p
a
n
n
N


Diode Small Signal Model
 The I-V relation of a diode can be linearized
( )
1
d d d d
q V v qV qv
kT kT kT
D D S S
I i I e I e e

 
   
 
 
( )
1 d d
D D D
q V v
I i I
kT

 
   
 
 
2 3
1
2! 3!
x x x
e x
    
d
D d d
qv
i g v
kT
 

Diode Capacitance
 We have already seen that a reverse biased diode
acts like a capacitor since the depletion region
grows and shrinks in response to the applied field.
the capacitance in forward bias is given by
 But another charge storage mechanism comes into
play in forward bias
 Minority carriers injected into p and n regions
“stay” in each region for a while
 On average additional charge is stored in diode
0
1.4
S
j j
dep
C A C
X

 

Charge Storage
 Increasing forward bias increases minority charge density
 By charge neutrality, the source voltage must supply equal
and opposite charge
 A detailed analysis yields:
p side n side
-Wp Wn
xn
-xp
( )
0
d d
q V v
kT
n
p e

0
n
p
0
p
n
( )
0
d d
q V v
kT
p
n e

1
2
d
d
qI
C
kT


Time to cross junction
(or minority carrier lifetime)

Forward Bias Equivalent Circuit
 Equivalent circuit has three non-linear elements: forward
conductance, junction cap, and diffusion cap.
 Diff cap and conductance proportional to DC current
flowing through diode.
 Junction cap proportional to junction voltage.

Fabrication of IC Diodes
 Start with p-type substrate
 Create n-well to house diode
 p and n+ diffusion regions are the cathode and annode
 N-well must be reverse biased from substrate
 Parasitic resistance due to well resistance
p-type
p+
n-well
p-type
n+
annode
cathode
p

Diode Circuits
 Rectifier (AC to DC conversion)
 Average value circuit
 Peak detector (AM demodulator)
 DC restorer
 Voltage doubler / quadrupler /…

Lecture27_ali.ppt

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