ECE 342
Solid-State Devices & Circuits
5. PN Junctions and Diodes
Jose E. Schutt-Aine
Electrical & Computer Engineering
University of Illinois
jschutt@emlab.uiuc.edu
ECE 342 – Jose Schutt-Aine
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Definitions
B: material dependent parameter = 5.4  1031 for Si
EG: Bandgap energy = 1.12 eV
k: Boltzmann constant=8.6210-5 ev/K
ni: intrinsic carrier concentration
At T = 300 K, ni = 1.5  1010 carriers/cm3
Jp: current density A/m2
q: electron charge
Dp: Diffusion constant (diffusivity) of holes
mp: mobility for holes = 480 cm2 /V sec
mn: mobility for electrons = 1350 cm2 /V sec
ND: concentration of donor atoms
nno: concentration of free electrons at thermal equilibrium
NA: concentration of acceptor atoms
ppo: concentration of holes at thermal equilibrium
D
D
kT
Einstein Relation : n  p 
 VT : thermal voltage
mn m p
q
ECE 342 – Jose Schutt-Aine
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PN Junction
•
When a p material is connected to an n-type material, a junction is formed
– Holes from p-type diffuse to n-type region
– Electrons from n-type diffuse to p-type region
– Through these diffusion processes, recombination takes place
– Some holes disappear from p-type
– Some electrons disappear from n-type
A depletion region consisting of bound charges is thus formed
Charges on both sides cause electric field  potential = Vo
ECE 342 – Jose Schutt-Aine
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PN Junction
•
•
Potential acts as barrier that must be overcome for holes to diffuse into the
n-region and electrons to diffuse into the p-region
Open circuit: No external current
Junction built-in voltage
From principle of detailed balance and
equilibrium we get:
N N 
Vo  VT ln  A 2 D 
 ni 
For Si, Vo is typically 0.6V to 0.8V
 s : silicon permittivity
 s  11.7 o  1.04 108 F/m
xn N A

xp ND
Charge equality in depletion region gives:
qx p AN A  qxn AN D
A: cross-section of junction
xp: width in p side
xn : width in n side
Wdep  xn  x p 
ECE 342 – Jose Schutt-Aine
2 s  1
1 


Vo
q  N A ND 
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Example
Find the barrier voltage across the depletion region of a silicon
diode at T = 300 K with ND=1015/cm3 and NA=1018/cm3.
Use
N N 
Vo  VT ln  A 2 D 
 ni 
@ 300K,
ni  1.5 1010 /cm3
VT  0.026 V
 1018 1015 
 1013 
Vo   o  0.026ln 
  0.026ln 

 1.52  1020 
2.25




Vo   o  0.026  29.12  0.7571 volts
Vo   o  0.7571 volts
ECE 342 – Jose Schutt-Aine
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PN Junction under Reverse Bias
•
When a reverse bias is applied
– Transient occurs during which depletion capacitance is charged to new
bias voltage
– Increase of space charge region
– Diffusion current decreases
– Drift current remains constant
– Barrier potential is increased
– A steady state is reached
– After transient: steady-state reverse current = IS-ID (ID is very small) 
reverse current ~ IS ~10-15 A
Under reverse bias the current in the diode is negligible
ECE 342 – Jose Schutt-Aine
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Depletion Layer Stored Charge
q j  qN  qN D xn A
A: cross section area
qj: stored charge
Let Wdep= depletion-layer width
qj  q
N AND
AWdep
N A  ND
The total voltage across the depletion layer is Vo + VR
Wdep 
2 s  1
1 


 Vo  VR 
q  N A ND 
ECE 342 – Jose Schutt-Aine
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Depletion Capacitance
Cj 
dq j
dVR
Q is bias point
VR is reverse voltage
VR VQ
Cj 
s A
Wdep
 sq  N AND   1 
C jo  A

 
2  N A  N D  Vo 

C jo
1
VR
Vo
Cj 
C jo
 VR 
1  
 Vo 
m
m is the grading coefficient and depends on how
the concentration varies from the p side to the n
side
1/3 <m <1/2
For an abrupt junction, m=0.5
ECE 342 – Jose Schutt-Aine
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Forward-Biased Junction Carrier Distribution
pn, np
P
Region
pn(xn)
Depletion
Region
np(-xp)
N
Region
Excess concentration
pn(x)
pno
np(x)
Thermal Equilibrium Value
npo
-xp
0
xn
x
NA >> ND
Barrier voltage is now lower than Vo
In steady state, concentration profile of excess minority carriers
remains constant
ECE 342 – Jose Schutt-Aine
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Forward-Biased PN Junction
Diode equation:


I D  I S eV / nVT  1
 Dp
D 
I S  Aqni2 
 n 
 Lp N D Ln N A 


2
since ni is a strong function of temperature; thus Is is a
strong function of temperature
n has a value between 1 and 2. Diodes made using
standard IC process have n=1; discrete diodes have n=1
In general, assume n=1
If V
VT , we can use I D  I S eV /VT
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Diode Characteristics
• Three distinct regions
– The forward-bias region, determined by v > 0
– The reverse-bias region, determined by v < 0
– The breakdown region, determined by v < -VZK
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Diode I-V Relationship
Breakdown
– Electric field strong enough in depletion layer to break covalent bonds and
generate electron-hole pairs. Electrons are then swept by E-field into the nside. Large number of carriers for a small increase in junction voltage
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The Diode
+
I
V
•
Diode Properties
– Two-terminal device that conducts current freely in one direction but blocks
current flow in the opposite direction.
– The two electrodes are the anode which must be connected to a positive voltage
with respect to the other terminal, the cathode in order for current to flow.
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Ideal Diode Characteristics
+
I
+
V
V
-
-
V<0
OFF
ECE 342 – Jose Schutt-Aine
I
I>0
ON
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Ideal Diode Characteristics
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Diode Models
Exponential
Piecewise Linear
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Constant-Voltage-Drop
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Diode Models
Small-signal
Ideal-diode
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Piecewise-Linear Model
for vD  VD 0 : iD  0
1
for vD  VD 0 : iD   vD  VD 0 
rD
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Piecewise-Linear Model
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Constant-Voltage-Drop Model
for iD  0 : vD  0.7 V
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Constant-Voltage-Drop Model
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Diodes Logic Gates
OR Function
Y  A B C
AND Function
Y  A B C
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Diode Circuit Example 1
IDEAL Diodes
Assume both diodes are on; then
VB  0 and V  0
I D2 
10  0
 1 mA
10
At node B
0  (10)
I 1 
 I  1 mA, V  0 V
5
D1 is conducting as originally
assumed
ECE 342 – Jose Schutt-Aine
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Diode Circuit Example 2
IDEAL Diodes
Assume both diodes are on; then
VB  0 and V  0
10  0
I D2 
 2 mA
5
At node B
I 2
0  (10)
 I  1 mA  wrong
10
original assumption is not correct …
assume D1 is off and D2 is on
I D2 
10  (10)
 1.33 mA
15
VB  10  10 1.33  3.3 V
D1 is reverse biased as assumed
ECE 342 – Jose Schutt-Aine
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Example
The diode has a value of IS = 10-12 mA at room temperature (300o K)
(a) Approximate the current I assuming the voltage drop across the
diode is 0.7V
(b) Calculate the accurate value of I
(c) If IS doubles for every 6o C increase in temperature, repeat part
(b) if the temperature increases by 40o C
(a) The resistor will have an approximate
voltage of 6-0.7 = 5.3 V. Ohm’s law then
gives a current of
5.3
I
 2.65 mA
2
(b) The current through the resistor must
equal the diode current; so we have
6 V
I
(resistor current )
2
I  I S eV /VT (diodecurrent )
ECE 342 – Jose Schutt-Aine
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Example (cont’d)
6 V
 1012 eV /0.026
2
Nonlinear equation  must be solved iteratively
Solution: V = 0.744 V
Using this value of the voltage, we can calculate the current
6  V 6  0.744
I

 2.63 mA
2
2
When the temperature changes, both Is and VT will change. Since
VT=kT/q varies directly with T, the new value is:
340
VT (340)  VT (300) 
 0.0295
300
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Example (cont’d)
The value of Is doubles for each 6o C increase, thus the new value
of Is is
I S (340)  I S (300)  240/6  1.016 1010 mA
The equation for I is then
6 V
I
 1.016 1010  eV /0.0295
2
Solving iteratively, we get
V  0.640 V
and
I  2.68 mA
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Example
Two diodes are connected in series as shown in the figure with Is1
=10-16 A and Is2 =10-14 A. If the applied voltage is 1 V, calculate the
currents ID1 and ID2 and the voltage across each diode VD1 and VD2.
The diode equations can be written as:
VD 1 VD 2
I
I
S1
I D1  I S1eVD1 / VT I D 2  I S 2eVD 2 / VT
e VT  D1  1
IS 2
I D2
I 
from which VD1  VD 2  VT ln  S 1   0.12
 IS 2 
Using KVL, we get VD1  VD 2  1 from which VD 2  0.44 V and VD1  0.56 V
I D1  1016 e0.56 / 0.026  0.22 m A=I D 2
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Small Signal Model
Approximation - valid for
small fluctuations about
bias point
iD  I Devd / nVT
1
rd 
 iD 
 v 
 D i
nVT

ID
D ID
iD  I D  id
Total
DC
applied (small)
vD  VD  vd
ECE 342 – Jose Schutt-Aine
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Diodes as Voltage Regulators
• Objective
–
–
–
–
Provide constant dc voltage between output terminals
Load current changes
Dc power supply changes
Take advantage of diode I-V exponential behavior
Big change in current
correlates to small
change in voltage
ECE 342 – Jose Schutt-Aine
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Voltage Regulator - Example
Assume n=2 and calculate % change caused by a ±10%
change in power-supply voltage with no load.
Nominal value of current is:
10  2.1
I
 7.9 mA
1
Incremental resistance for each diode:
nVT 2  25
rd 

 6.3
I
7.9
Resistance for all 3 diodes:
r  3rd  18.9 
Voltage change
r
0.0189
vo  2
2
 37.1 mV  18.5 mV  0.9%
rR
0.0189  1
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Diode as Rectifier
While applied source alternates in
polarity and has zero average
value, output voltage is
unidirectional and has a finite
average value or a dc component
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Diode as Rectifier
vs is a sinusoid with 24-V peak amplitude. The diode conducts when vs
exceeds 12 V. The conduction angle is 2 where  is given by
24cos  12    60
The conduction angle is 120o, or one-third of a cycle. The peak value of
the diode current is given by
Id 
24  12
 0.12 A
100
The maximum reverse voltage across the diode occurs when vs is at its
negative peak: 24+12=36 V
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Half-Wave Rectifier
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Full-Wave Rectifier
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Bridge Rectifier
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Bridge Rectifier
• Properties
– Uses four diodes.
– vo is lower than vs by two diode drops.
– Current flows through R in the same direction during both
half cycles.
The peak inverse voltage (PIV) of each diode:
PIV  vs  2vD  vD  vs  vD
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Peak Rectifier
Filter capacitor is
used to reduce the
variations in the
rectifier output
ECE 342 – Jose Schutt-Aine
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Rectifier with Filter Capacitor
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Rectifier with Filter Capacitor
•
Operation
– Diode conducts for brief interval t
– Conduction stops shortly after peak
– Capacitor discharges through R
– CR>>T
– Vr is peak-to-peak ripple
iL  vo / R
iD  iC  iL  C
I L  Vp / R
dvI
 iL
dt

iDav  I L 1   2Vp / Vr
vo  V p e  t / CR
Vr
Vp
T
IL
Vp


CR fCR fC


iD max  I L 1  2 2Vp / Vr
ECE 342 – Jose Schutt-Aine

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Diode Circuits - Rectification
Vin  Asin t
Rectification with ripple reduction.
C must be large enough so that RC time
constant is much larger than period
ECE 342 – Jose Schutt-Aine
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Diode Circuits
Vout  VD


I D  I S eVD / VT  1
VS  RI D  VD  RI D (VD )  VD
Nonlinear transcendental system  Use graphical method
ID
Diode characteristics
Vs/R
Load line
(external characteristics)
Vout
VS
VD
Solution is found at itersection of load line characteristics
and diode characteristics
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Diode Circuits – Iterative Methods
Newton-Raphson Method
Vout  VD
Wish to solve f(x)=0 for x
Use: xk 1  xk   f '( xk ) f ( xk )
1
1
x( k 1)  x( k )   f '( x( k ) )  f ( x( k ) )
f (VD ) 
VD  VS
 I S eVD / VT  1  0
R
1 I
f '(VD )   S eVD / VT
R VT


VD( k )  VS
( k 1)
D
V
V
(k )
D

R

V ( k ) / VT
 IS e
D

1
1 I S VD( k ) / VT
 e
R VT
Where VD( k ) is the value of VD at the kth iteration
Procedure is repeated until convergence to final (true) value of VD which is the
solution. Rate of convergence is quadratic.
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