3.
SEMICONDUCTOR DEVICES I
Summary so Far
1. Examined some of the basic physics behind
semiconducting materials. In particular, have been able
to calculate:
carrier concentrations, nc and pv
conductivity 
in terms of g, me*, mh*, T etc.
for both:
intrinsic semiconductors
extrinsic semiconductors
2. For extrinsic semiconductors:
ncand can be controlled over a wide range by
doping (addition of small amounts of impurities).
Ec
EF
Ec
EF
Ev
Ev
n-type
Group V impurities
e.g. P
At room temp’: nc is large
p-type
Group III impurities
e.g. B
At room temp’: pv is large
Can build semiconductor devices by combining n-type
and p-type semiconductors.
Initially, will look at 2 basic device types.
1.
METAL-SEMICONDUCTOR JUNCTION
All devices require connection to metal wires to form
electronic circuits
2.
PN JUNCTION
Device formed by joining together p-type and n-type
semiconductors.
Very Important!
Many devices based on this.
3.1 METAL-SEMICONDUCTOR JUNCTION –
SCHOTTKY BARRIER
Consider what happens when metal is brought into
contact with an n-type semiconductor.
Band Diagram before contact
m
Fermi
level
s
conduction
band
Fermi
level
valence
band
Metal
m s -
n-type Semiconductor
work function of metal
work function of semiconductor
Fermi levels of metal and semiconductor at different
energies unstable energetically

upon contact, electron flow between materials
until Fermi levels equilibriate
If m > s (as drawn), electrons flow from semiconductor
to metal.
Band diagram after contact
depletion region
m - s
+
+
Fermi
level
x = x0
x=0
x
Metal
Semiconductor
Electron density in semiconductor is much less than in
metal
 electron flow from semiconductor to metal
depletes a substantial volume of the semiconductor of
charge carriers leaving positively ionised donors

Energy barrier
depletion layer.
-
m –s
This arises due to an electrostatic dipole layer between
electrons transferred to metal and the positively ionised
donors in depletion layer
How wide is barrier? Can get this using Poisson’s
equation in depletion region (between x = 0 and x = x0).
WIDTH OF DEPLETION REGION
Poisson’s equation in 1-D
dV


dx

2
2


0
[relates potential V to charge density ]


 = dielectric constant
 = Nde
In depletion region:
[Nd – density of ionised donors]
Electron energy,  = - eV

d Ne

dx

2
2
d
2
0
Solving subject to the boundary conditions
 = (d/dx) = 0 at x =0
 = m – s at x = x0
find that
2
2
0
0
N xe
  
2
d
m
s
0
So:
Width of depletion layer (x0)  N
Typically
100 Å < x0 < 1000 Å
1
2
d
CURRENT-VOLTAGE (I-V) BEHAVIOUR
So previous result tells us:
For heavy doping (large Nd)
 barrier becomes thin compared to electron
mean free path (x0 very small)

can effectively consider junction as 2 thermionic
emitters [Unit 2]facing one another across barrier.
In zero applied bias (voltage) – currents equal and
opposite
Ism
Ims
Metal
Semiconductor
Ism = current from semiconductor to metal
Ims = current from metal to semiconductor
Recall from Unit 2 that equilibrium current I0 for
thermionic emitter (at temp’ T) can be written as
    
I0 = A exp
kT 

(c.f. Dushman-Richardson eq’n)
m
s
Consider applying forward bias (voltage) V to junction.
Fermi level of semiconductor is raised relative to that in
metal  current from semiconductor to metal Ism is
increased
       eV 
Ism = A exp

kT

m
s
 eV 
= I exp 
 kT 
0
Work function effectively lowered.
Current from metal to semiconductor Ims remains
constant and is therefore I0.
So net current I in forward bias is given by
I = Ism - Ims
 eV 
I = I [exp   1]
 kT 
0
Similarly, in reverse bias –V, can show
  eV 
I  I [1  exp
]
 kT 
0
Fermi level in semiconductor is lowered relative to Fermi
level in metal.
I
forward
bias
V
reverse bias
Rectifying action  diode behaviour
3.2 P-N JUNCTION
Consider what happens when an n-type semiconductor
and a p-type semiconductor are brought into contact.
Band diagram just before contact
c
Fn
Fp
v
n
p
Energetically unstable (once n and p are in contact).
On n-side - have large no. of electrons in cond’n band
On p-side - have large no. of holes in valence band
Need to consider flow of both electrons and holes.
Electrons close to interface flow from n to p
Holes close to interface flow from p to n
- until Fermi levels equilibriate.
These recombine  get layer on both sides of barrier
depleted of free carriers
Band diagram after contact
n
p
+
+
+
c
F
Lp
V
-Ln
x=0
x
Depletion layer  electrostatic dipole layer
positively charged donor ions on n-side
negatively charged acceptor ions on p-side
Width of depletion layer - Ln + Lp
Can determine width of depletion layer by solving
Poisson’s equation in interface region.
WIDTH OF DEPLETION REGION OF PN JUNCTION
dV


dx

2
Potential V
(Poisson’s Eq’n)
2
0
Charge density 

Nde
Lp
x
-Ln
-Nae
n-side
 = d = Nde

p-side
 = a = -Nae
Integrate Poisson’s eq’n (twice) on BOTH SIDES
(i.e. for both n and p)
Solutions for n and p sides must match at x = 0!
i.e. V and (dV/dx) both continuous at x = 0).
Boundary Conditions
At x = -Ln:
At x = Lp:
(dV/dx) = 0, V = Vn
(dV/dx) = 0, V = Vp
At x = 0:
(dV/dx) and V are continuous.
[V = V0]
Solve as outlined above (and Tanner pages 145-146).
Potential difference V across junction
V = Vn – Vp
eN L  N / N   1
V
2
2
d
n
d
a
0
This applies whether junction is under external bias V or
under zero bias.
V  g/e
For zero applied bias:
If external bias V is applied, use: c = dQ/dV
Charge Q per unit area is:
Q = NdLne
So for capacitance (per unit area) c, can show

c
1
V
1
2
Width of depletion layer depends on applied bias V
 capacitance of layer varies on V
 pn junction can act as variable capacitor
[See Seminar Question 4!]
I-V BEHAVIOUR OF PN JUNCTION
Zero Applied Voltage (Bias)
Recall earlier band diagram
np0
Ec
+
+
nn0
eVB
+
F
Ev
pn0
pp0
n-type
p-type
Away from depletion region:
n-side provides supply of free electrons (nn0 large)
p-side provides supply of free holes (pp0 large)
However, due to thermal excitation of charge carriers
across the junction, we get:
small conc’n np0 of free electrons on p-side
small conc’n pn0 of free holes on n-side
Potential drop VB across junction (in zero applied bias):

np0/nn0 = pn0/pp0 = exp(-eVB/kT)
[1]
External Forward Bias
What happens if external bias voltage Vext is applied?
Band diagram
np
nn
Ec
EF
eVj
eVext
Ev
pn
pp
Forward Bias - potential drop across junction is reduced.
So np (and pn) increase.
If Vj is the potential drop across the junction
np/nn = pn/pp = exp(-eVj/kT)
We have that
Vj = VB - Vext
Hence
np/nn = pn/pp = exp(-eVB/kT)exp(eVext/kT)
[2]
For zero applied bias we had that
np0/nn0 = pn0/pp0 = exp(-eVB/kT)
[1]
Can substitute for exp(-eVB/kT) using eq’ns [1] and [2].
np/nn = np0/nn0 exp(eVext/kT)
pn/pp = pn0/pp0 exp(eVext/kT)
nn is large - plenty of free electrons on n-side
pp is large - plenty of free holes on p-side
They change only slightly when we apply a bias voltage.
nn  nn0
i.e.
pp  pp0
So to good approximation we can write
np/np0 = pn/pn0 = exp(eVext/kT)
Can now recall argument used for metal-semiconductor
junction - i.e. treat both halves of pn junction as 2
thermionic emitters facing each other.

I  I expeV / kT   1
0
ext
External Reverse Bias
Band diagram
np
Ec
EF
nn
eVext
Ev
pn
pp
Reverse Bias - potential drop across junction is
increased.
So np (and pn) decrease.
Can use similar argument to show that in reverse bias:

I  I 1  exp eV / kT 
0
ext
Band Diagrams - Summary
p
n
zero applied
external bias
eVext
applied forward
bias
np increases
n
p
eVext
applied reverse
bias
np decreases
I-V Characteristic

I  I expeV / kT   1 - forward bias

I  I 1  exp eV / kT  - reverse bias
0
0
ext
ext
N.B. - 2 carrier types.
I-V characteristics for holes and electrons identical
current flow is additive –
negative electrons flow from n to p
positive holes flow in opposite direction.
I
V
So pn junction behaves as a rectifier (diode).
Assumed that junction was narrow i.e. negligible
probability of electrons and holes recombining in
junction region.
More rigorous treatment – considers diffusion of carriers
across junction, taking account of recombination
probability
 leads to same form of I-V
characteristic though
In practice often get
I
V
threshold
I in forward bias doesn’t “turn on” until V > threshold.
Due to presence of traps in junction region – I > 0 once
traps are saturated.
Zener Diode
pn diode with very heavy doping levels.
Recall that depletion layer width  as doping conc’n 
So voltage drop across junction is over small distance
 E-field very high in depletion region.
Hence, electrons injected into depletion region subject to
v. high acceleration.
If E-field high enough, electrons gain enough energy to
ionise atoms in semiconductor lattice
 rapid increase in I in for small changes in V for
reverse bias
I
Vz
Breakdown occurs at specific value of V
V
-
Vz
Can taylor Vz by altering doping level
Useful in protection circuits for delicate components.