Transcript Lecture 9 - University of California, Berkeley

EECS 105 Fall 2003, Lecture 9
Lecture 9:
PN Junctions
Prof. Niknejad
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 9
Prof. A. Niknejad
Lecture Outline
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
Department of EECS
PN Junctions Thermal Equilibrium
PN Junctions with Reverse Bias
University of California, Berkeley
EECS 105 Fall 2003, Lecture 9
Prof. A. Niknejad
PN Junctions: Overview


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
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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
Department of EECS
p-type
NA
−+−+−+−+−+− −
−−−−−− V
++++
−+
+−+−+−+−+− +
ND
n-type
University of California, Berkeley
EECS 105 Fall 2003, Lecture 9
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 9
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 9
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 9
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 9
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 9
Prof. A. Niknejad
Depletion Approximation
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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
Department of EECS
s
University of California, Berkeley
EECS 105 Fall 2003, Lecture 9
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 9
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 9
Prof. A. Niknejad
Continuity of E-Field Across Junction
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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 9
Prof. A. Niknejad
Potential Across Junction
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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 9
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 9
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 9
Prof. A. Niknejad
Sanity Check
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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 9
Prof. A. Niknejad
Total Depletion Width
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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 9
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 9
Prof. A. Niknejad
Contact Potential
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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 9
Prof. A. Niknejad
PN Junction Capacitor
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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 9
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 9
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 9
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 9
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 9
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 9
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 9
Prof. A. Niknejad
“Diffusion” Resistor
N-type Diffusion Region
Oxide
P-type Si Substrate

Resistor is capacitively isolation from substrate
–
Must Reverse Bias PN Junction!
Department of EECS
University of California, Berkeley