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EECS 105 Spring 2005, Lecture 27
Lecture 27:
PN Junctions
Prof. Niknejad
Department of EECS
University of California, Berkeley
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?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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:
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Diffusion Equations


Assume that the mean free path is λ
Find flux of carriers crossing x=0 plane
n ( 0)
n ( )
n(  )
1 
dn  
dn  
F  vth  n(0)     n(0)    
2 
dx  
dx  
1
dn
n( )vth
2
F  vth
dx
1
n(  )vth
2

Department of EECS
1
F  vth n( )  n( ) 
2
0

J  qF  qvth
dn
dx
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Einstein Relation

The thermal velocity is given by kT
1
2
mn*vth2  12 kT
  vth c
Mean Free Time
kT q c
vth  v   kT * 
mn
q mn*
2
th c
c
 kT  dn
dn
J  qvth
 q
 n 
dx
 q
 dx
 kT 
Dn     n
 q 
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Total Current and Boundary Conditions

When both drift and diffusion are present, the total
current is given by the sum:
J  J drift  J diff


dn
 q n nE  qDn
dx
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
J1 (1 )
J 2 ( 2 )
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J1  J 2
1E1   2 E2
E1  2

E2  1
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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:
dno
J n  0  qn0  n E0  qDn
dx
 n 
dno
 q  d
  no E0   no 0
dx
 kT  dx
 Dn 
 kT  dno
dn0
d0   
 Vth
n0
 q  n0
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Carrier Concentration and Potential (2)

We have an equation relating the potential to the
carrier concentration
 kT  dno
dn
d0   
 Vth 0
n0
 q  n0

If we integrate the above equation we have
n0 ( x)
0 ( x)  0 ( x0 )  Vth ln
n0 ( x0 )

We define the potential reference to be intrinsic Si:
0 ( x0 )  0
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n0 ( x0 )  ni
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Carrier Concentration Versus Potential

The carrier concentration is thus a function of
potential
n0 ( x)  ni e0 ( x ) / Vth


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
p0 ( x)  ni e 0 ( x ) / Vth

Note that the law of mass action is upheld
n0 ( x) p0 ( x)  ni2 e 0 ( x ) / Vth e0 ( x ) / Vth  ni2
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
The Doping Changes Potential

Due to the log nature of the potential, the potential changes
linearly for exponential increase in doping:
n0 ( x)
n0 ( x)
n0 ( x)
0 ( x)  Vth ln
 26mV ln
 26mV ln 10 log 10
ni ( x0 )
ni ( x0 )
10
n0 ( x)
0 ( x)  60mV log 10
10
p0 ( x)
0 ( x)  60mV log 10
10


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
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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
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p-type
NA
−+−+−+−+−+− −
−−−−−− V
++++
−+
+−+−+−+−+− +
ND
n-type
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
PN Junction Currents


Consider the PN junction in thermal equilibrium
Again, the currents have to be zero, so we have
dno
J n  0  qn0  n E0  qDn
dx
dno
qn0  n E0  qDn
dx
E0 
dno
dx   kT 1 dn0
n0  n
q n0 dx
 Dn
dpo
Dp
kT 1 dp0
dx
E0 

n0  p
q p0 dx
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
PN Junction Fields
p-type
n-type
NA
ND
p0  N a
p0 ( x)
J diff
E0
 x p0
ni2
n0 
Na
xn 0
ni2
p0 
Nd
n0  N d
J diff
E0
– –++
Transition Region
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Total Charge in Transition Region

To solve for the electric fields, we need to write
down the charge density in the transition region:
0 ( x)  q( p0  n0  N d  N a )

In the p-side of the junction, there are very few
electrons and only acceptors:
0 ( x)  q( p0  N a )

 x p0  x  0
Since the hole concentration is decreasing on the pside, the net charge is negative:
N a  p0
Department of EECS
 0 ( x)  0
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Charge on N-Side

Analogous to the p-side, the charge on the n-side is
given by:
0 ( x)  q(n0  N d )

0  x  xn 0
The net charge here is positive since:
 0 ( x)  0
N d  n0
n0  N d
ni2
n0 
Na
J diff
E0
– –++
Department of EECS
Transition Region
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
“Exact” Solution for Fields

Given the above approximations, we now have an
expression for the charge density
q(ni e 0 ( x ) /Vth  N a )
 0 ( x)  
0 ( x ) / Vth
q
(
N

n
e
)
d
i


 x po  x  0
0  x  xn 0
We also have the following result from
electrostatics
dE0
d 2  0 ( x)
 2 
dx
dx
s


Notice that the potential appears on both sides of
the equation… difficult problem to solve
A much simpler way to solve the problem…
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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
 qN a  x po  x  0
 0 ( x)  
  qN d 0  x  xn 0

The solution for electric field is now easy
dE0  0 ( x)

Field zero outside
dx
s
transition region
x
 0 ( x' )
E0 ( x)  
dx'  E0 ( x p 0 )
xp0
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s
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Depletion Approximation (2)

Since charge density is a constant
E0 ( x)  
x
xp0

 0 ( x' )
qN a
dx'  
( x  x po )
s
s
If we start from the n-side we get the following
result
E0 ( xn 0 )  
xn 0
x
Field zero outside
transition region
Department of EECS
 0 ( x' )
qN d
dx'  E0 ( x) 
( xn 0  x)  E0 ( x)
s
s
E0 ( x)  
qN d
s
( xn 0  x )
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Plot of Fields In Depletion Region
p-type
NA
–––––
–––––
–––––
–––––
+++++
n-type
+++++
+++++
ND
+++++
Depletion
Region
E0 ( x)  





qN a
s
( x  x po )
E0 ( x)  
qN d
s
( xn 0  x )
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?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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 EField is continuous
What does this imply?
E 0 ( x  0)  
n
qN a
s
x po  
qN d
s
xno  E 0p ( x  0)
qN a x po  qN d xno

Total fixed charge in n-region equals fixed charge
in p-region! Somewhat obvious result.
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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 )  
x
qN a
s
xp0
( x' x po )dx'
x

qN a  x'

 ( x)   p 
 x' x po 
s  2
 xp0
2
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Potential Across Junction

We arrive at potential on p-side (parabolic)
qN a
 ( x)   p 
( x  x p0 )2
2 s
p
o

Do integral on n-side
n ( x )  n 

qN d
( x  xn 0 ) 2
2 s
Potential must be continuous at interface (field
finite at interface)
qN d 2
qN a 2
n (0)  n 
xn 0   p 
x p 0   p (0)
2 s
2 s
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Solve for Depletion Lengths

We have two equations and two unknowns. We are
finally in a position to solve for the depletion
depths
qN d 2
qN a 2
n 
xn 0   p 
x p0
2 s
2 s
(1)
qN a x po  qN d xno
(2)
2 sbi  N a

xno 
qN d  N a  N d



2 sbi  N d 


x po 
qN a  N d  N a 
bi  n   p  0
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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:
xn 0  lim
N d 
x p 0  lim
N d 

2 sbi N d
0
qN d N d  N a

2 sbi  N d 
2 sbi

 
qN a  N d  N a 
qN a
Entire depletion width dropped across p-region
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Total Depletion Width

The sum of the depletion widths is the “space
charge region”
Xd0


2 sbi  1
1 


 x p 0  xn 0 

q  Na Nd 
This region is essentially depleted of all mobile
charge
Due to high electric field, carriers move across
region at velocity saturated speed
Xd0
2 sbi  1 

 15   1μ
q  10 
Department of EECS
E pn 
1V
V
 104
1μ
cm
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Have we invented a battery?

Can we harness the PN junction and turn it into a
battery?
 ND
NA 
ND N A


bi  n   p  Vth  ln
 ln
 Vth ln
2

n
n
n
i
i 
i

?

Numerical example:
ND N A
10151015
bi  26mV ln
 60mV  log
 600mV
2
20
ni
10
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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:
+
bi
−
Department of EECS
mn
 pm
0  bi   pm  mn
n
p
bi  ( pm  mn )
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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:
qN a x po  qN d xno
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Reverse Biased PN Junction

What happens if we “reverse-bias” the PN
junction?
+
 bi  VD
−



VD
VD  0
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
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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 builtin potential with the effective reverse bias:
2 s (bi  VD )  N a

xn (VD ) 
qN d
 Na  Nd

V
  xn 0 1  D
bi

2 s (bi  VD )  N d

x p (VD ) 
qN a
 Na  Nd

VD
  x p 0 1 
bi

2 s (bi  VD )  1
1 


X d (VD )  x p (VD )  xn (VD ) 


q
 Na Nd 
VD
X d (VD )  X d 0 1 
bi
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Charge Versus Bias

As we increase the reverse bias, the depletion
region grows to accommodate more charge
QJ (VD )  qN a x p (VD )  qN a 1 



VD
bi
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
QJ (VD  vD )  QJ (VD )  q(vD )
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Derivation of Small Signal Capacitance

From last lecture we found
dQD
QJ (VD  vD )  QJ (VD ) 
dV
C j  C j (VD ) 
dQ j
dV
Cj 
V VD
d

dV
qN a x p 0
2bi 1 

Notice that
C j0 
qN a x p 0
2bi
Department of EECS
qN a

2bi
VD
bi
vD  
VD


V
  qN a x p 0 1 




bi

V VR

C j0
1
 2 sbi  N d


 qN a  N a  N d
VD
bi

q s N a N d
 
2bi N a  N d

University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Physical Interpretation of Depletion Cap
q s N a N d
C j0 
2bi N a  N d

Notice that the expression on the right-hand-side is
just the depletion width in thermal equilibrium
C j0   s

 1
1 



2 sbi  N a N d 
q
1

s
Xd0
This looks like a parallel plate capacitor!
C j (VD ) 
Department of EECS
s
X d (VD )
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
A Variable Capacitor (Varactor)

Capacitance varies versus bias:
Cj
C j0

Application: Radio Tuner
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Part II: Currents in PN Junctions
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Diode under Thermal Equilibrium
Minority Carrier Close to Junction
Thermal
+
p-type
Generation
−
- +
+
- +
+
−
- +
+
- +
J
p
,
drift
+
+
J n ,drift - - ++
ND
- +
+
n-type
J n ,diff
J p ,diff
−
NA
E0
Recombination
+
qbi
−




Carrier with energy
below barrier height
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
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Reverse Bias


Reverse Bias causes an increases barrier to
diffusion
Diffusion current is reduced exponentially
p-type
-
ND
-
-
+
+
+
+
+
+
+
n-type
NA
+
−
q(bi  VR )


Drift current does not change
Net result: Small reverse current
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Forward Bias


Forward bias causes an exponential increase in
the number of carriers with sufficient energy to
penetrate barrier
Diffusion current increases exponentially
p-type
-
ND
-
-
+
+
+
+
+
+
+
n-type
NA
+
−
q(bi  VR )


Drift current does not change
Net result: Large forward current
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Diode I-V Curve
Id
Is
d
 qV

kT
I d  I S  e  1


I d (Vd  )   I S
1
qVd
kT

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
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Minority Carriers at Junction Edges
Minority carrier concentration at boundaries of
depletion region increase as barrier lowers …
the function is
(minority) hole conc. on n-side of barrier
p n ( x  xn )

p p ( x   x p ) (majority) hole conc. on p-side of barrier
 e ( Barrier Energy ) / kT
p n ( x  xn )
 e q(B VD ) / kT
NA
Department of EECS
(Boltzmann’s Law)
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
“Law of the Junction”
Minority carrier concentrations at the edges of the
depletion region are given by:
pn ( x  xn )  N Ae q ( B VD ) / kT
n p ( x   x p )  N De
q (B VD ) / kT
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
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Minority Carrier Concentration
pn 0 e
p side
qVA
kT
n side
n p 0e
qVA
kT
x
 qVkTA
  Lp
pn ( x)  pn 0  pn 0  e  1 e


pn 0
np0
-Wp
-xp
xn
Minority Carrier
Diffusion Length
Wn
The minority carrier concentration in the bulk region for
forward bias is a decaying exponential due to recombination
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Steady-State Concentrations
Assume that none of the diffusing holes and
electrons recombine  get straight lines …
pn 0 e
p side
qVA
kT
n side
n p 0e
qVA
kT
pn 0
np0
-Wp
-xp
xn
Wn
This also happens if the minority carrier
Ln , p  Wn , p
diffusion lengths are much larger than Wn,p
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Diode Current Densities
pn 0 e
p side
n p 0e
qVA
kT
dn p
n side
qVA
kT
dx
( x) 
n p 0e
qVA
kT
 np0
 x p  (Wp )
pn 0
np0
np0
-Wp
-xp
xn
Wn
ni2

Na
qVA


D
diff
n
kT
J n  qDn
q
n p 0  e  1
dx x  x
Wp


p
qVA
D


dp
p
J pdiff  qDp n
 q
pn 0 1  e kT 
dx x  xn
Wn


dn p
J
Department of EECS
diff
 Dp
Dn
 qn 

 N dWn N aW p

2
i
  qVA

kT
  e  1


University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Diode Small Signal Model

The I-V relation of a diode can be linearized
qVd qvd
d  vd )
 q (VkT

I D  iD  I S  e
 1  I S e kT e kT


2
3
x
x
ex  1  x   
2! 3!
 q (Vd  vd )
I D  iD  I D 1 

kT




qvd
iD 
 g d vd
kT
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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
Cj  A



S
X dep
 1.4C j 0
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
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Charge Storage
pn 0 e
p side
n p 0e
q (Vd  vd )
kT
n side
q (Vd  vd )
kT
pn 0
np0
-Wp



-xp
xn
Wn
Increasing forward bias increases minority charge density
By charge neutrality, the source voltage must supply equal
and opposite charge
1 qI d

A detailed analysis yields: Cd 
2 kT
Time to cross junction
(or minority carrier lifetime)
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
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.
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Fabrication of IC Diodes
p+
cathode
annode
p
n+
n-well
p-type
p-type





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
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 27
Prof. A. Niknejad
Diode Circuits





Rectifier (AC to DC conversion)
Average value circuit
Peak detector (AM demodulator)
DC restorer
Voltage doubler / quadrupler /…
Department of EECS
University of California, Berkeley