ECE 271
Electronic Circuits I
Topic 3
Diodes and Diodes Circuits
NJIT ECE-271 Dr. S. Levkov
Chap 3 -1
Chapter Goals
• Develop electrostatics of the pn junction
• Define regions of operation of the diode (forward bias, reverse bias,
and reverse breakdown)
• Explore various diode models including the mathematical model, the
ideal diode model, and the constant voltage drop model
• Apply the various types of models in circuit analysis
• Explore diode applications
• Explore different special types of diodes
• Understand the SPICE representation and model parameters for the
diode. Practice simulating diode circuits using SPICE
NJIT ECE-271 Dr. S. Levkov
Chap 3 -2
Diode Introduction
• A diode is formed by joining
an n-type semiconductor
with a p-type semiconductor.
Diode symbol
NJIT ECE-271 Dr. S. Levkov
• A pn junction is the
interface between n and p
regions.
• Study of diodes will consist
of three sections:
1. PN-junction physics
2. Diode circuits analysis
3. Diode applications
Chap 3 -3
1. The pn junction
A simple (or doped) semiconductor does not in itself possess properties that
make it useful for electronic circuits.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -4
1. The pn junction
A simple (or doped) semiconductor does not in itself possess properties that
make it useful for electronic circuits.
However, when a section of p-type material and a section of n-type material
are brought in contact to form a pn junction, a number of interesting
properties arise.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -5
1. The pn junction
A simple (or doped) semiconductor does not in itself possess properties that
make it useful for electronic circuits.
However, when a section of p-type material and a section of n-type material
are brought in contact to form a pn junction, a number of interesting
properties arise.
The major effect happens in the small section around the contact area –
depletion region, where the holes and electrons recombine leaving no
charge carriers.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -6
1. The pn junction
A simple (or doped) semiconductor does not in itself possess properties that
make it useful for electronic circuits.
However, when a section of p-type material and a section of n-type material
are brought in contact to form a pn junction, a number of interesting
properties arise.
The major effect happens in the small section around the contact area –
depletion region, where the holes and electrons recombine leaving no
charge carriers.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -7
pn Junction Electrostatics
Donor and acceptor concentration on
either side of the junction.
Concentration gradients give
rise to diffusion current - ID
NJIT ECE-271 Dr. S. Levkov
Chap 3 -8
Space-Charge (Depletion) Region
at the pn Junction
•
•
•
•
If the diffusion process was to continue
unabated, this would result eventually in the
uniform concentration of e. & h. through the
entire semiconductor and pn junction would
disappear.
This does not happen because the competing
process appears that oppose the diffusion
current.
As holes diffuse from p-region they leave (-)
ionized atoms of acceptors. As electrons
diffuse from n-region they leave (+)ionized
atoms of donors
This creates the space-charge or depletion
region with localized (-) and (+) charge 
electric field E(x) drift current that oppose
the diffusion current.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -9
Drift Currents
• Gauss’ Law: an electric field due to the charge distribution is
E 
c
s
• Assuming constant permittivity*  s and one dimension
1
E(x) 
s

  (x)dx
• Resulting electric field gives rise to a drift current IS of minority
carriers. With no external circuit connections, drift and diffusion

currents cancel:
IS = ID . (There is no actual current, since this would
imply power dissipation, rather the electric field cancels the diffusion
current ‘tendency.’)
• The equilibrium condition is maintained by the barrier voltage V0, or ,
in another words, junction potential  j .
*In electromagnetism, permittivity is the measure of how much resistance is encountered when forming
an electric field in a medium
NJIT ECE-271 Dr. S. Levkov
Chap 3 -10
Potential Across the Junction
 j  V0
qN A x p  qN D xn
Charge Density
Using
qN A x p  qN D xn
Electric Field
(charge neutrality), it can be shown that
 N A ND
V0   j   E ( x)dx  VT ln 
2
n
 i
NJIT ECE-271 Dr. S. Levkov
Potential

kT
 , VT 
q

Chap 3 -11
Width of Depletion Region
From the previous expressions, the expression for the width of
the space-charge region, or depletion region can be obtained. It
is called the depletion region since the excess holes and electrons
are depleted from the dopant atoms on either side of the junction.
2 s
wd 0  ( xn  x p ) 
q
NJIT ECE-271 Dr. S. Levkov
 1
1 


 j
 N A ND 
Chap 3 -12
Example 1. Width of Depletion Region
Problem: Find built-in potential and depletion-region width for given diode
Given data:On p-type side: NA = 1017/cm3 on n-type side: ND = 1020/cm3
Assumptions: Room-temperature operation with VT = 0.025 V
Analysis:
 1017 /cm3 1020 /cm3  
 N A ND 
  0.979V
 j  VT ln  2   (0.025 V)ln 
20
6

10 /cm  
 ni 
with  s  11.7 0  11.7  8.85 1014
q  1.6 1019
NJIT ECE-271 Dr. S. Levkov


2 s  1
1
w



  0.113m
 d0
q N
N  j
D
A

Chap 3 -13
Example 2. Diode Electric Field and p,n
depletion width
• Problem: Find the size of the individual depletion layers on either side
of a pn junction and the max. value of the electric field for a given diode.
• Given data: On the p-type side: NA = 1017/cm3 on the n-type side: ND =
wd0 0.113 m
 j  0.979 V
1020/cm3 from earlier example,
• Assumptions: Room-temperature operation

From charge neutrality
and
x p  xn
ND
NA
xn  x p

NA
ND
we have
xp
xn

ND
NA
,
.


n 


N D 
N A 

wd 0  xn  x p  x 1
 x 1
N A  p  N D 
Thus
Hence
, and
qN A x p  qN D xn
xn 
NJIT ECE-271 Dr. S. Levkov
wd 0
1.1310-4  m


ND 

1





N
A 

and
xp 
wd 0
0.113  m


NA 

1





N
D 

Chap 3 - 14
Example 2. Diode Electric Field and p,n
depletion width (cont.)
For the electric field, we have that from
diagram on slide 3.8 it follows that
j
 j    E ( x)dx and the
is equal to the area under triangle.
Since wd 0 is the base of the triangle, it follows
2 j 2(0.979V )
EMAX  w 
173 kV / cm
0.113

m
d0
NJIT ECE-271 Dr. S. Levkov
Chap 3 - 15
The equilibrium in pn junction - summary
Result of diffusion created by the difference of concentration of
holes and electrons in different parts of pn junction: holes
will diffuse from left to the right and electrons from right to
the left. This process results in diffusion current I D that
creates the difference of potential at the depletion region.
This is the majority carrier diffusion current.
 
NJIT ECE-271 Dr. S. Levkov
Chap 3 -16
The equilibrium in pn junction - summary
Result of diffusion created by the difference of concentration of
holes and electrons in different parts of pn junction: holes
will diffuse from left to the right and electrons from right to
the left. This process results in diffusion current I D that
creates the difference of potential at the depletion region.
This is the majority carrier diffusion current.
 
NJIT ECE-271 Dr. S. Levkov
However, this difference of potential acts like a battery making
an electric field that creates drift current. This is the
minority carrier current also known as reverse saturation
current I s
Chap 3 -17
The equilibrium in pn junction - summary
Result of diffusion created by the difference of concentration of
holes and electrons in different parts of pn junction: holes
will diffuse from left to the right and electrons from right to
the left. This process results in diffusion current I D that
creates the difference of potential at the depletion region.
This is the majority carrier diffusion current.
 
However, this difference of potential acts like a battery making
an electric field that creates drift current. This is the
minority carrier current also known as reverse saturation
current I s
Under equilibrium conditions, the diffusion current is
exactly balanced by the drift current so that the net
current across the p–n junction is zero, I S  I D
NJIT ECE-271 Dr. S. Levkov
Chap 3 -18
The equilibrium in pn junction - summary
Result of diffusion created by the difference of concentration of
holes and electrons in different parts of pn junction: holes
will diffuse from left to the right and electrons from right to
the left. This process results in diffusion current I D that
creates the difference of potential at the depletion region.
This is the majority carrier diffusion current.
 
However, this difference of potential acts like a battery making
an electric field that creates drift current. This is the
minority carrier current also known as reverse saturation
current I s
Under equilibrium conditions, the diffusion current is
exactly balanced by the drift current so that the net
current across the p–n junction is zero, I S  I D
When no external current or voltage is applied to the p–n
junction, the potential gradient (barrier voltage V0 )
forms an energy barrier that prevents further diffusion
of charge carriers across the junction.
NJIT ECE-271 Dr. S. Levkov
V0
Chap 3 -19
The pn junction under the applied voltage
condition
When an external battery is connected across a pn junction, the amount of
current flow is determined by the polarity of the applied voltage and its
effect on the space–charge region.
When the battery polarity is applied in the same direction as barrier voltage
we say that a pn junction is in reverse bias.
When the battery polarity is applied in the opposite direction to barrier
voltage we say that a pn junction is in forward bias.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -20
The pn junction in reverse bias
+
-
+
+
+
-
The free electrons in the n–type material are attracted toward the positive terminal of the
battery and away from the junction
NJIT ECE-271 Dr. S. Levkov
Chap 3 -21
The pn junction in reverse bias
+
-
+
+
+
-
The free electrons in the n–type material are attracted toward the positive terminal of the
battery and away from the junction, creating new holes
NJIT ECE-271 Dr. S. Levkov
Chap 3 -22
The pn junction in reverse bias
+
-
+
+
+
-
The free electrons in the n–type material are attracted toward the positive terminal of the
battery and away from the junction, creating new holes and the positive charge on the right of
depletion region becomes even more positive.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -23
The pn junction in reverse bias
+
-
+
+
+
-
The free electrons in the n–type material are attracted toward the positive terminal of the
battery and away from the junction, creating new holes and the positive charge on the right of
depletion region becomes even more positive.
The holes from the p–type material are attracted toward the negative terminal of the battery
and away from the junction,
NJIT ECE-271 Dr. S. Levkov
Chap 3 -24
The pn junction in reverse bias
+
-
+
+
+
+
The free electrons in the n–type material are attracted toward the positive terminal of the
battery and away from the junction, creating new holes and the positive charge on the right of
depletion region becomes even more positive.
The holes from the p–type material are attracted toward the negative terminal of the battery
and away from the junction, creative negatively charged ions
NJIT ECE-271 Dr. S. Levkov
Chap 3 -25
The pn junction in reverse bias
+
-
+
+
+
-
The free electrons in the n–type material are attracted toward the positive terminal of the
battery and away from the junction, creating new holes and the positive charge on the right of
depletion region becomes even more positive.
The holes from the p–type material are attracted toward the negative terminal of the battery
and away from the junction, creative negatively charged ions, and the negative charge on the
left of depletion region becomes more negative.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -26
The pn junction in reverse bias
+
-
+
+
+
-
The free electrons in the n–type material are attracted toward the positive terminal of the
battery and away from the junction, creating new holes and the positive charge on the right of
depletion region becomes even more positive.
The holes from the p–type material are attracted toward the negative terminal of the battery
and away from the junction, creative negatively charged ions, and the negative charge on the
left of depletion region becomes more negative.
As a result, the space–charge region at the junction becomes effectively wider, and the
potential gradient increases until it approaches the potential of the external battery.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -27
The pn junction in reverse bias
+
-
+
+
+
-
The free electrons in the n–type material are attracted toward the positive terminal of the
battery and away from the junction, creating new holes and the positive charge on the right of
depletion region becomes even more positive.
The holes from the p–type material are attracted toward the negative terminal of the battery
and away from the junction, creative negatively charged ions, and the negative charge on the
left of depletion region becomes more negative.
As a result, the space–charge region at the junction becomes effectively wider, and the
potential gradient increases until it approaches the potential of the external battery.
Current flow is then extremely small because no voltage difference ( electric field ) exists
across either the p–type or the n–type region.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -28
The pn junction in forward bias
-
The free electrons in the p–type material near the positive terminal of the battery break their
electron–pair bonds and enter the battery,
NJIT ECE-271 Dr. S. Levkov
Chap 3 -29
The pn junction in forward bias
-
The free electrons in the p–type material near the positive terminal of the battery break their
electron–pair bonds and enter the battery, creating new holes,
NJIT ECE-271 Dr. S. Levkov
Chap 3 -30
The pn junction in forward bias
-
The free electrons in the p–type material near the positive terminal of the battery break their
electron–pair bonds and enter the battery, creating new holes, which reduce the negative
space charge on the left .
NJIT ECE-271 Dr. S. Levkov
Chap 3 -31
The pn junction in forward bias
-
The free electrons in the p–type material near the positive terminal of the battery break their
electron–pair bonds and enter the battery, creating new holes, which reduce the negative
space charge on the left .
The electrons from the negative terminal of the battery enter the n–type material
NJIT ECE-271 Dr. S. Levkov
Chap 3 -32
The pn junction in forward bias
-
-
The free electrons in the p–type material near the positive terminal of the battery break their
electron–pair bonds and enter the battery, creating new holes, which reduce the negative
space charge on the left .
The electrons from the negative terminal of the battery enter the n–type material, diffuse
toward the junction
NJIT ECE-271 Dr. S. Levkov
Chap 3 -33
The pn junction in forward bias
-
The free electrons in the p–type material near the positive terminal of the battery break their
electron–pair bonds and enter the battery, creating new holes, which reduce the negative
space charge on the left .
The electrons from the negative terminal of the battery enter the n–type material, diffuse
toward the junction, and reduce the positive space charge on the right.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -34
The pn junction in forward bias
The free electrons in the p–type material near the positive terminal of the battery break their
electron–pair bonds and enter the battery, creating new holes, which reduce the negative
space charge on the left .
The electrons from the negative terminal of the battery enter the n–type material, diffuse
toward the junction, and reduce the positive space charge on the right.
As a result, the space charge region becomes effectively narrower, and the energy barrier
decreases to an insignificant value.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -35
The pn junction in forward bias
-
The free electrons in the p–type material near the positive terminal of the battery break their
electron–pair bonds and enter the battery, creating new holes, which reduce the negative
space charge on the left .
The electrons from the negative terminal of the battery enter the n–type material, diffuse
toward the junction, and reduce the positive space charge on the right.
As a result, the space charge region becomes effectively narrower, and the energy barrier
decreases to an insignificant value.
Excess electrons from the n–type material can then penetrate the space charge region, flow
across the junction, and move by way of the holes in the p–type material toward the positive
terminal of the battery, creating large diffusion current
This electron flow continues as long as the external voltage is applied, with resulting diode
current.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -36
The pn junction in diode summary
 
 
Is
 
 
 
- reverse saturation current of minority carriers
NJIT ECE-271 Dr. S. Levkov
Chap 3 -37
Diode Junction Potential for Different
Applied Voltages
V0
NJIT ECE-271 Dr. S. Levkov
Chap 3 -38
Diode i-v Characteristics
NJIT ECE-271 Dr. S. Levkov
Chap 3 -39
Diode i-v Characteristics
NJIT ECE-271 Dr. S. Levkov
Chap 3 -40
Diode i-v Characteristics
NJIT ECE-271 Dr. S. Levkov
Chap 3 -41
Diode i-v Characteristics
Figure 3.9
NJIT ECE-271 Dr. S. Levkov
Pspice model here
Chap 3 -42
Diode Equation

 qvD
iD  I S exp 
 nkT

where
IS
vD
q
k
T
n
VT
=
=
=
=
=
=
=

 vD  
 
  1  I S exp  nV   1
 
 T  

reverse saturation current (A)
voltage applied to diode (V)
electronic charge (1.60 x 10-19 C)
Boltzmann’s constant (1.38 x 10-23 J/K)
absolute temperature
nonideality factor (dimensionless)
kT/q = thermal voltage (V) (25 mV at room temp.)
IS is typically between 10-18 and 10-9 A, and is strongly temperature dependent due
to its dependence on ni2. The nonideality factor is typically close to 1, but
approaches 2 for devices with high current densities. It is assumed to be 1 in this
text.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -43
Diode Voltage and Current Calculations
(Example)
Problem: Find diode voltage for diode at room-temperature dc operation for
different specifications with VT = 0.025 V.
If IS = 0.1 fA, ID = 300 A
-4
I D 
VD VT ln 1  1(0.0025V )ln(1 310 A)  0.718 V
I S 
10-16 A




NJIT ECE-271 Dr. S. Levkov
Chap 3 -44
Diode Voltage and Current Calculations
(Example)
Problem: Find diode voltage for diode at room-temperature dc operation for
different specifications with VT = 0.025 V.
If IS = 0.1 fA, ID = 300 A
-4
I D 
VD VT ln 1  1(0.0025V )ln(1 310 A)  0.718 V
I S 
10-16 A




If IS = 10 fA, ID = 300 A
VD 0.603V

NJIT ECE-271 Dr. S. Levkov
Chap 3 -45
Diode Voltage and Current Calculations
(Example)
Problem: Find diode voltage for diode at room-temperature dc operation for
different specifications with VT = 0.025 V.
If IS = 0.1 fA, ID = 300 A
-4
I D 
VD VT ln 1  1(0.0025V )ln(1 310 A)  0.718 V
I S 
10-16 A




If IS = 10 fA, ID = 300 A
VD 0.603V

With IS = 0.1 fA, ID = 1 mA,
VD  0.748V

NJIT ECE-271 Dr. S. Levkov
Chap 3 -46
Diode Current for Reverse, Zero, and
Forward Bias
• Reverse bias: iD  I S exp  vD / VT   1  I S  0  1   I S
for example:
vD  4VT  0.1V  iD  I S exp  vD / VT   1  I S exp  4   1  I S 0.018  1  I S
NJIT ECE-271 Dr. S. Levkov
Chap 3 -47
Diode Current for Reverse, Zero, and
Forward Bias
• Reverse bias: iD  I S exp  vD / VT   1  I S  0  1   I S
for example:
vD  4VT  0.1V  iD  I S exp  vD / VT   1  I S exp  4   1  I S 0.018  1  I S
• Zero bias:
NJIT ECE-271 Dr. S. Levkov
iD  I S exp  vD / VT   1  I S 1  1  0
Chap 3 -48
Diode Current for Reverse, Zero, and
Forward Bias
• Reverse bias: iD  I S exp  vD / VT   1  I S  0  1   I S
for example:
vD  4VT  0.1V  iD  I S exp  vD / VT   1  I S exp  4   1  I S 0.018  1  I S
• Zero bias:
iD  I S exp  vD / VT   1  I S 1  1  0
• Forward bias: iD  I S exp  vD / VT   1  I S exp  vD / VT

vD  4VT  0.1V  iD  I S exp  vD / VT   1  I S exp  4   1  I S 55  1  55 I S
NJIT ECE-271 Dr. S. Levkov
Chap 3 -49
Semi-log Plot of Forward Diode Current and
Current for Three Different Values of IS
Only small 60mV increase in the forward voltage is
needed to raise the diode current by 10.
NJIT ECE-271 Dr. S. Levkov
I S [ A]  10I S [B]  100 I S [C]
Chap 3 -50
Reverse Bias -details
External reverse bias adds to the built-in potential of the pn
junction. The shaded regions below illustrate the increase in the
characteristics of the space charge region due to an externally
applied reverse bias, vD.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -51
Reverse Bias (cont.)
External reverse bias also increases the width of the depletion region
since the larger electric field must be supported by additional charge.
2 s  1
1 
w d  (x n  x p ) 
 
 j  v R
q N A N D 


wd  wd 0
v
1 R
j

2 s  1
1 
where wd 0  (x n  x p ) 
 
 j
q N A N D 
Since the reverse saturation results from thermal generation of electronhole pairs in the depletion region, it is dependent on the volume of the
space charge region. It can be shown that the reverse saturation current
 gradually increaseswith increased reverse bias:
I S  I S0 1
NJIT ECE-271 Dr. S. Levkov
vR
j
However, IS is approximately constant
at IS0 under forward bias.
Chap 3 -52
Reverse Breakdown
Increased reverse bias
eventually results in the diode
entering the breakdown
region, resulting in a sharp
increase in the diode current.
The voltage at which this
occurs is the breakdown
voltage, VZ.
2 V < VZ < 2000 V
This is not destructive
phenomenon, diode can be
repeatedly operated in this
condition.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -53
Reverse Breakdown Mechanisms
• Avalanche Breakdown
Si diodes with VZ greater than about 5.6 volts breakdown according to
an avalanche mechanism. As the electric field increases, accelerated
carriers begin to collide with fixed atoms. As the reverse bias
increases, the energy of the accelerated carriers increases, eventually
leading to ionization of the impacted atoms. The new carriers also
accelerate and ionize other atoms. This process feeds on itself and
leads to avalanche breakdown.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -54
Reverse Breakdown Mechanisms (cont.)
•
Zener Breakdown
Zener breakdown occurs in heavily doped diodes. The heavy doping results in
a very narrow depletion region at the diode junction. Reverse bias leads to
carriers with sufficient energy to tunnel directly between conduction and
valence bands  moving across the junction. Once the tunneling threshold is
reached, additional reverse bias leads to a rapidly increasing reverse current.
•
Quantum tunneling refers to the quantum mechanical phenomenon where a particle tunnels through
a potential barrier that it classically could not surmount because its total energy kinetic is lower than
the potential energy of the barrier. This tunneling plays an essential role in several physical
phenomena, including radioactive decay, and has important application to modern devices such
as flash memory, the tunneling diode and scanning tunneling microscope.
Quantum tunneling is a consequence of the wave particle duality of matter and is often explained
using the Heisenberg uncertainty principle: an elementary particle such as an electron has a nonzero
probability of moving from one side of any physical barrier to the other, regardless of the height or
width of the barrier, or in other words, regardless of whether the potential energy of the barrier is
greater than the kinetic energy of the particle.
Other names for this effect are quantum-mechanical tunneling, wave-mechanical
tunneling, electron tunneling and tunnel effect.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -55
2. Diode Circuits Analysis
•
The diode is the most simple and fundamental example of nonlinear circuit elements.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -56
2. Diode Circuits Analysis
•
•
The diode is the most simple and fundamental example of nonlinear circuit elements.
Like a resistor, a diode has two terminals.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -57
2. Diode Circuits Analysis
•
•
•
The diode is the most simple and fundamental example of nonlinear circuit elements.
Like a resistor, a diode has two terminals.
Unlike the resistor, which has a linear (straight line) relationship between current i
and voltage v, the diode has a nonlinear i-v characteristics.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -58
2. Diode Circuits Analysis
•
•
•
•
The diode is the most simple and fundamental example of nonlinear circuit elements.
Like a resistor, a diode has two terminals.
Unlike the resistor, which has a linear (straight line) relationship between current i
and voltage v, the diode has a nonlinear i-v characteristics.
The main idea: diode conducts when it is forward biased and does not conduct when it
is reverse biased.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -59
2. Diode Circuits Analysis
•
•
•
•
The diode is the most simple and fundamental example of nonlinear circuit elements.
Like a resistor, a diode has two terminals.
Unlike the resistor, which has a linear (straight line) relationship between current i
and voltage v, the diode has a nonlinear i-v characteristics.
The main idea: diode conducts when it is forward biased and does not conduct when it
is reverse biased.
To forward bias a diode, the
anode (p) must be more
positive than the cathode (n) or
less negative.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -60
2. Diode Circuits Analysis
•
•
•
•
The diode is the most simple and fundamental example of nonlinear circuit elements.
Like a resistor, a diode has two terminals.
Unlike the resistor, which has a linear (straight line) relationship between current i
and voltage v, the diode has a nonlinear i-v characteristics.
The main idea: diode conducts when it is forward biased and does not conduct when it
is reverse biased.
To forward bias a diode, the
anode (p) must be more
positive than the cathode (n) or
less negative.
To reverse bias a diode, the
anode (p) must be less
positive than the cathode (n)
or more negative.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -61
A Diode Puzzle 1
Which lamps are alight? Some may not be full brightness.
+
-
+
-
A Diode Puzzle 1
Which lamps are alight? Some may not be full brightness.
+
-
+
-
A Diode Puzzle 2
Which lamps are alight? Some may not be full brightness.
+
-
+
-
A Diode Puzzle 2
Which lamps are alight? Some may not be full brightness.
+
-
+
-
Diode Models
Since the diode has a nonlinear i-v characteristics, the circuit analysis is more
complicated.
Depending on the needs and type of the circuit (simple/complex) models of different degree
of accuracy can be used:
•The full nonlinear model - graphical solution (load line analysis) or numerical solution
•The piecewise linear approximation at the Q-point (represent diode as a resistor and source)
• The ideal diode model.
• The constant voltage drop diode model (offset model) .
The first two are “small signal models”, used for higher accuracy and simpler circuits.
The last two are “large signal models” , used for more complicated circuits and less accuracy.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -66
Diode Circuit Analysis: Basics
The loop equation for the diode circuit is:
V  I D R  VD
This is also called the load line for the diode.
We need also to add the diode equation:
I D  I S eVD /VT  1
V and R may represent the Thévenin
equivalent of a more complex 2terminal network. The objective of
diode circuit analysis is to find the
quiescent operating point for the
diode.
The solution to these equations can be found
by different methods using models of
different complexity mentioned earlier.
Q-Point = (ID, VD)
Explanation here
NJIT ECE-271 Dr. S. Levkov
Chap 3 -67
Nonlinear Model
1. Graphical Load-Line Analysis
• This method uses the full nonlinear model of a diode.
• The loop equations for the Thévenin equivalent of a 2-terminal
network with a diode are
VDD  I D R  VD
and
or
I D  VDD / R  VD /R
I D  I S eVD /VT  1
(1)
(2)
where ID and VD are to be found.
• Second of these two equations is nonlinear, so some
specific methods have to be used to solve it.
• Load line analysis uses graphical method.
• Equation (1) is a line with a slope 1/R,
equation (2) is the iv-curve of a diode.
• Their intersection represent the solution –
a diode’s Q-point.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -68
Nonlinear Model
1. Graphical Load-Line Analysis -Example
Problem: Find diode Q-point
Given data: V = 10 V, R = 10kW.
Analysis:
10  I D104  VD
To define the load line we use two points:
For VD  0, ID  10V 10kW  1 mA
For VD  5V, ID  5V 10kW  0.5 mA

These points and the resulting load line
are plotted. Q-point is given by
intersection of load line and diode
characteristic:
Q-point = (0.95 mA, 0.6 V)
NJIT ECE-271 Dr. S. Levkov
Chap 3 -69
Nonlinear Model
2. Numerical Solution
• This method uses the full nonlinear model of a diode and
numerical solution of nonlinear equation.
• Substituting diode equation (2) into the circuit equation (1) we
get one nonlinear equation for variable VD
VDD  I S eVD /VT  1 R  VD
• Solving this equation gives diode’s Q-point.
• The solution is usually numerical using some iterative method.
Can be obtained using software with a numerical solver (e.g.
Matlab), Newton method using Excel, manual simple trial and
error, manual iterations using Newton method.
• After obtaining solution, the linear two element diode model
can be obtained.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -70
Nonlinear Model
2. Numerical Solution - Example
2) Make initial guess VD0 .
3) Evaluate f and its derivative f’ for this
value of VD.
4) Calculate new guess for VD using
Given data: IS =10-13 A, VT =0.0025 V.


I D  I S eVD /VT  1  1013 exp 40VD   1 
10  104 1013 exp 40VD   1  VD
 
0
VD  VD
1
0
f VD

0
f ' (VD )
5) Repeat steps 3 and 4 till convergence.
Using a spreadsheet we get :
Q-point = ( 0.9426 mA, 0.5742 V)
The solution is given by a transcendental
equation. A numerical answer can be found
by using Newton’s iterative method.
1) Create a function
f  10  1041013exp 40VD   1  VD
Since, usually we don’t have accurate
saturation current values and significant
tolerances exist for sources and passive
components, we need answers precise to
only 2or 3 significant digits.
Spreadsheet example here
NJIT ECE-271 Dr. S. Levkov
Chap 3 -71
Linearisation
Two Element Model of Diode
• After obtaining the exact solution, the two element linear diode model can be obtained.
• Differentiating expression for ID:
GD 
RD 
  ID 
 (VD )

 V  I  IS
I
1
I S exp  D   D
 D
VT
VT
VT
 VT 
VT
ID
• Example (previous slide)
RD 
NJIT ECE-271 Dr. S. Levkov
VT
0.025

 26.523W
I D 0.0009426
Chap 3 -72
Small Signal Diode Model
The last model is often called the “small signal model” – when the diode is biased to
operate at a point on the forward i-v characteristic and a small ac signal is
superimposed on the DC quantities.
If the source signal is vD (t )  VD  vd (t ) , then
1. Find the diode’s Q-point (VD , ID ) for VD
and find VD0 and rD
2. Make two element small signal model
VD0
3. Use it to find id (t ) 
NJIT ECE-271 Dr. S. Levkov
vd (t )
rD
and iD (t )  I D  id (t )
Chap 3 -73
Ideal Diode Model
If an ideal diode is forward-biased, the voltage across
the diode is zero.
If an ideal diode is reverse-biased, the current through
the diode is zero.
vD =0 for iD >0
iD =0 for vD < 0
Thus, the diode is assumed to be either
on or off.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -74
Analysis using Ideal Model for Diode
Need to determine only the conduction state of an ideal diode.
1. Assume a diode conduction state (on or off).
NJIT ECE-271 Dr. S. Levkov
Chap 3 -75
Analysis using Ideal Model for Diode
Need to determine only the conduction state of an ideal diode.
1. Assume a diode conduction state (on or off).
2. Substitute ideal diode circuit model into circuit (short circuit if “on,”
open circuit if “off”).
NJIT ECE-271 Dr. S. Levkov
Chap 3 -76
Analysis using Ideal Model for Diode
Need to determine only the conduction state of an ideal diode.
1. Assume a diode conduction state (on or off).
2. Substitute ideal diode circuit model into circuit (short circuit if “on,”
open circuit if “off”).
3. Solve for diode current and voltage, using linear circuit analysis techniques.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -77
Analysis using Ideal Model for Diode
Need to determine only the conduction state of an ideal diode.
1. Assume a diode conduction state (on or off).
2. Substitute ideal diode circuit model into circuit (short circuit if “on,”
open circuit if “off”).
3. Solve for diode current and voltage, using linear circuit analysis techniques.
4. If the solution is consistent with the assumption, then the initial assumption
was correct; if not, the diode conduction state is opposite to that initially
assumed.
For example, if the diode has been assumed to be “off” but the diode voltage computed after
replacing the diode with an open circuit is a forward bias, then it must be true that the actual state
of the diode is “on.”
NJIT ECE-271 Dr. S. Levkov
Chap 3 -78
Ideal Model for Diode
Example 1 (simple) – Find Q-point
NJIT ECE-271 Dr. S. Levkov
Chap 3 -79
Ideal Model for Diode
Example 1 (simple) – Find Q-point
Since source appears to force positive
current through diode, assume diode is
on.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -80
Ideal Model for Diode
Example 1 (simple) – Find Q-point
Since source appears to force positive
current through diode, assume diode is
on.
ID 
(10  0)V
 1 mA
10kW
| ID  0
Our assumption is correct
Q-point = (1 mA, 0 V)
NJIT ECE-271 Dr. S. Levkov
Chap 3 -81
Ideal Model for Diode
Example 1 (simple) – Find Q-point
Since source appears to force positive
current through diode, assume diode is
on.
ID 
(10  0)V
 1 mA
10kW
| ID  0
Our assumption is correct
Q-point = (1 mA, 0 V)
NJIT ECE-271 Dr. S. Levkov
Chap 3 -82
Ideal Model for Diode
Example 1 (simple) – Find Q-point
Since source appears to force positive
current through diode, assume diode is
on.
ID 
(10  0)V
 1 mA
10kW
Since source is forcing current backward
through diode assume diode is off.
| ID  0
Our assumption is correct
Q-point = (1 mA, 0 V)
NJIT ECE-271 Dr. S. Levkov
Chap 3 -83
Ideal Model for Diode
Example 1 (simple) – Find Q-point
Since source appears to force positive
current through diode, assume diode is
on.
(10  0)V
ID 
 1 mA
10kW
Since source is forcing current backward
through diode assume diode is off.
Hence ID = 0 . Loop equation is:
10  VD  10 4 ID  0
| ID  0
Our assumption is correct
VD  10V | VD  0
Our assumption is correct.
Q-point = (1 mA, 0 V)
Q-point = (0 mA, -10 V)
NJIT ECE-271 Dr. S. Levkov

Chap 3 -84
Ideal Model for Diode
Example 2 - average
KCL better:
NJIT ECE-271 Dr. S. Levkov
v1  12 v1  11

0
5
4
Chap 3 -85
Ideal Model for Diode
Example 2 - average
KCL better:
NJIT ECE-271 Dr. S. Levkov
v1  12 v1  11

0
5
4
Chap 3 -86
Ideal Model for Diode
Example 3 - two-diode circuit
Find diodes Q-points using the ideal diode model.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -87
Ideal Model for Diode
Example 3 - two-diode circuit
Analysis: The ideal diode model is chosen.
Assumption 1. Since the 15-V source appears to force
positive current through D1 and D2, and the -10-V
source is forcing positive current through D2, assume
both diodes are on.
Since the voltage at node D is zero due to the short
circuit of ideal diode D1,
(15  0)V
 1.50 mA
10kW
0  10V 
ID 2 
 2.00 mA
5kW
I1  ID1  ID 2 | ID1  1.50  2.00  0.500 mA
I1 

NJIT ECE-271 Dr. S. Levkov
But, ID1 < 0 is not allowed by the diode, so try again.
Chap 3 -88
Ideal Model for Diode
Example 3 - two-diode circuit (cont.)
Since the current in D1 is zero, ID2 = I1,
15 10,000I1  5,000I D2 10  0
25V
I1 
1.67 mA
15,000W
Assumption 2.
Since current in D2 is valid,
but that in D1 is not, the
second guess is D1 off and
D2 on.
NJIT ECE-271 Dr. S. Levkov
VD1 15 10,000I1 15 16.7  1.67 V
Q-Points are D1 : (0 mA, -1.67 V):off
D2 : (1.67 mA, 0 V) :on
Now, the results are consistent with the
assumptions.
Chap 3 -89
Constant Voltage Drop (CVD) Model for Diode
(Offset Model)
vD = Von for iD >0
iD = 0 for vD < Von.
Typically, if Von = 0.6V:
Ideal diode = CVD with Von =0
NJIT ECE-271 Dr. S. Levkov
Chap 3 -90
Constant Voltage Drop Model for Diode
Example 1 - simple
Since the 10-V source appears
to force positive current
through the diode, assume
diode is on.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -91
Constant Voltage Drop Model for Diode
Example 1 - simple
Since the 10-V source appears
to force positive current
through the diode, assume
diode is on.
KVL :  10  I D *10k  Von  0
(10  Von )
10k
(10  0.6)

 0.940 mA
10k
ID 
NJIT ECE-271 Dr. S. Levkov
Chap 3 -92
Constant Voltage Drop Model for Diode
Example 1 - simple
Since the 10-V source appears
to force positive current
through the diode, assume
diode is on.
KVL :  10  I D *10k  Von  0
(10  Von )
10k
(10  0.6)

 0.940 mA
10k
ID 
Solution shows the current is positive, so
assumption is correct, diode is on.
Q-point = (0.94 mA, 0.6 V)
NJIT ECE-271 Dr. S. Levkov
Chap 3 -93
Constant Voltage Drop Model for Diode
Example 2 - average
Find the lowest value of v1 such that D1 conducts
if VB = 2V:
a) use the CVD diode model;
b) compare with ideal diode model.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -94
Constant Voltage Drop Model for Diode
Example 2 - average
Find the lowest value of v1 such that D1 conducts
if VB = 2V:
a) use the CVD diode model;
b) compare with ideal diode model.
i
a) Assume the diode D1 is off. Then there is no current.
i = 0  KVL: -v1 + vD1 + 0.6 + 2 = 0  vD1 = v1 – 2.6
Diode conducts if vD1 > 0, so v1 > 2.6 is the value of v1
required.
b) The diagram will look similarly but without the
0.6V battery.
Assume D1 is off. Similarly to (a), we have vD1 = v1 –
2  v1 > 2.0 is the required v1 .
NJIT ECE-271 Dr. S. Levkov
Chap 3 -95
Constant Voltage Drop Model for Diode
Example 3 – multiple diodes circuit
Find the Q-point for D1 , D2, D3 .
NJIT ECE-271 Dr. S. Levkov
Chap 3 -96
Constant Voltage Drop Model for Diode
Example 3 – multiple diodes circuit (cont. 1)
a) Assume all diodes are off. Calculating the
voltages in the circuit on the left (zero
currents) we see that all diodes have large
forward biases. Assumption does not hold.
b) Assume all diodes are on. Going from right
to left, we can calculate the node voltages:
VC = - 0.6
VB = - 0.6 + 0.6 = 0
VA = 0 - 0.6 = - 0.6
10  0
 0.6  (20)
 1mA , I 2 
 1.94mA
3
10 10
10 103
 0.6  (10)
I3 
 0.94mA
10  103
Then I1 
And I D1  I 2 , I1  I D1  I D 2 , I 3  I D 2  I D 3 
I D1  1.94 mA , I D 2  0.94 mA , I D 3  1.86 mA
OK
No
OK
Assumption does not hold for D2, since its current is
negative.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -97
Constant Voltage Drop Model for Diode
Example 3 – multiple diodes circuit (cont 2)
c) Assume D1 , D3 - on, D2 - off.
1) KVL for the loop from +10V source to 20V source:
10  10 4 I1  0.6  10 4 I 2  20  0

 I1  I 2  I D1
29.4
 I D1 
 1.47 mA - OK
20 103
2) From KVL for the loop from -10V source I D 3  I 3 
to the ground:
3) KVL for the loop from the ground to
+10V source:
Assumption holds and D1 : (1.47,0.6V),
on
NJIT ECE-271 Dr. S. Levkov
 0.6  (10)
 0.94 mA  OK
10 103
10  10 4 I1  VD 2  0.6  0
VD 2  4.10V  0  OK
D2 : (0  4.10V), D3 : (0.94mA,0.6V)
off
on
Chap 3 -98
Models comparison
We can compare the four models on the simple example that was used to
illustrate all of them
All four results are quite similar. Even the simplest ideal model
overestimates the current only by 10%.
We see that the current is quite insensitive to the to the actual choice of the
actual choice of the diode voltage. This is a result of the exponential
dependence for the current and quite large source voltage.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -99
3. Diode Applications
NJIT ECE-271 Dr. S. Levkov
Chap 3 -100
Rectifier Circuits
• A basic rectifier converts an ac voltage to a pulsating dc voltage.
• A filter then eliminates ac components of the waveform to
produce a nearly constant dc voltage output.
• Rectifier circuits are used in virtually all electronic devices to
convert the 120-V 60-Hz ac power line source to the dc voltages
required for operation of electronic devices.
• In rectifier circuits, the diode state changes with time and a given
piecewise linear model is valid only for a certain time interval.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -101
Half-Wave Rectifier Circuit with
Resistive Load
NJIT ECE-271 Dr. S. Levkov
Chap 3 -102
Half-Wave Rectifier Circuit with
Resistive Load
For the positive half-cycle of the input, the source forces
positive current through the diode, the diode is on, and
vO = vS.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -103
Half-Wave Rectifier Circuit with
Resistive Load
For the positive half-cycle of the input, the source forces
positive current through the diode, the diode is on, and
vO = vS.
During the negative half cycle, negative current can’t exist
in the diode. The diode is off, current in resistor is zero, and
vO =0 .
NJIT ECE-271 Dr. S. Levkov
Chap 3 -104
Half-Wave Rectifier Circuit with
Resistive Load
For the positive half-cycle of the input, the source forces
positive current through the diode, the diode is on, and
vO = vS.
During the negative half cycle, negative current can’t exist
in the diode. The diode is off, current in resistor is zero, and
vO =0 .
This is for the ideal diode model. What will be for CVD?
NJIT ECE-271 Dr. S. Levkov
Chap 3 -105
Half-Wave Rectifier Circuit with
Resistive Load (cont.)
Using the CVD model, during the on-state of the
diode vO = (VP sinwt)- Von.
The output voltage is zero when the diode is off.
VD = Von.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -106
Half-Wave Rectifier Circuit with
Resistive Load (cont.)
Using the CVD model, during the on-state of the
diode vO = (VP sinwt)- Von.
The output voltage is zero when the diode is off.
Often a step-up or step-down transformer is used to
convert the 120-V, 60-Hz voltage available from the
power line to the desired ac voltage level as shown.
VD = Von.
http://www.falstad.com/circuit/e-rectify.html
NJIT ECE-271 Dr. S. Levkov
Chap 3 -107
Half-Wave Rectifier Circuit with
Resistive Load (cont.)
Using the CVD model, during the on-state of the
diode vO = (VP sinwt)- Von.
The output voltage is zero when the diode is off.
Often a step-up or step-down transformer is used to
convert the 120-V, 60-Hz voltage available from the
power line to the desired ac voltage level as shown.
VD = Von.
NJIT ECE-271 Dr. S. Levkov
The rectifier output signal is very far from being DC.
How to improve?
Chap 3 -108
Half-Wave Rectifier Circuit with
Resistive Load (cont.)
Using the CVD model, during the on-state of the
diode vO = (VP sinwt)- Von.
The output voltage is zero when the diode is off.
Often a step-up or step-down transformer is used to
convert the 120-V, 60-Hz voltage available from the
power line to the desired ac voltage level as shown.
VD = Von.
The rectifier output signal is very far from being DC.
How to improve?
Time-varying components in the rectifier output
are removed using a filter capacitor.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -109
Rectifier with Capacitor Load
Peak Detector Circuit
To understand better the operation of a
filter, first consider the peak detector
circuit.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -110
Rectifier with Capacitor Load
Peak Detector Circuit
To understand better the operation of a
filter, first consider the peak detector
circuit.
Von  0
NJIT ECE-271 Dr. S. Levkov
As the input voltage rises, the diode is on,
and the capacitor (initially discharged)
charges up to the input voltage minus the
diode voltage drop.
Chap 3 -111
Rectifier with Capacitor Load
Peak Detector Circuit
To understand better the operation of a
filter, first consider the peak detector
circuit.
Von  0
As the input voltage rises, the diode is on,
and the capacitor (initially discharged)
charges up to the input voltage minus the
diode voltage drop.
At the peak of the input voltage, diode
current tries to reverse, and the diode cuts
off. The capacitor has no discharge path
and retains a constant voltage providing a
constant output voltage:
Vdc = VP - Von
NJIT ECE-271 Dr. S. Levkov
Chap 3 -112
Half-Wave Rectifier Circuit with RC Load
NJIT ECE-271 Dr. S. Levkov
Chap 3 -113
Half-Wave Rectifier Circuit with RC Load
As the input voltage rises during the first
quarter cycle, the diode is on and the
capacitor (initially discharged) charges up
to the peak value of the input voltage.
Vdc =
Von  0
NJIT ECE-271 Dr. S. Levkov
Chap 3 -114
Half-Wave Rectifier Circuit with RC Load
As the input voltage rises during the first
quarter cycle, the diode is on and the
capacitor (initially discharged) charges up
to the peak value of the input voltage.
Vdc =
Von  0
NJIT ECE-271 Dr. S. Levkov
At the peak of the input, the diode current
tries to reverse, the diode cuts off, and the
capacitor discharges exponentially
through R. Discharge continues till the
input voltage exceeds the output voltage
which occurs near the peak of next cycle.
Chap 3 -115
Half-Wave Rectifier Circuit with RC Load
As the input voltage rises during the first
quarter cycle, the diode is on and the
capacitor (initially discharged) charges up
to the peak value of the input voltage.
Vdc =
Von  0
At the peak of the input, the diode current
tries to reverse, the diode cuts off, and the
capacitor discharges exponentially
through R. Discharge continues till the
input voltage exceeds the output voltage
which occurs near the peak of next cycle.
This process then repeats once every
cycle.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -116
Half-Wave Rectifier Circuit with RC Load
As the input voltage rises during the first
quarter cycle, the diode is on and the
capacitor (initially discharged) charges up
to the peak value of the input voltage.
Vdc =
Von  0
At the peak of the input, the diode current
tries to reverse, the diode cuts off, and the
capacitor discharges exponentially
through R. Discharge continues till the
input voltage exceeds the output voltage
which occurs near the peak of next cycle.
This process then repeats once every
cycle.
This circuit can be used to generate
negative output voltage if the top plate of
capacitor is grounded instead of bottom
plate. In this case, Vdc = -(VP - Von)
NJIT ECE-271 Dr. S. Levkov
Chap 3 -117
Half-Wave Rectifier Circuit with RC Load
The following diagram shows the input and output voltage for the real diode (or
CVD model), when Von  0
NJIT ECE-271 Dr. S. Levkov
Ripple Voltage and Conduction Interval
The output voltage is not constant as in an ideal peak detector, but has a ripple voltage Vr.
Von  0
NJIT ECE-271 Dr. S. Levkov
Chap 3 -119
Ripple Voltage and Conduction Interval
The output voltage is not constant as in an ideal peak detector, but has a ripple voltage Vr.
The diode conducts for a short time DT called the conduction interval during each cycle,
and its angular equivalent is called the conduction angle c.
Von  0
NJIT ECE-271 Dr. S. Levkov
Chap 3 -120
Ripple Voltage and Conduction Interval
The output voltage is not constant as in an ideal peak detector, but has a ripple voltage Vr.
The diode conducts for a short time DT called the conduction interval during each cycle,
and its angular equivalent is called the conduction angle c.
It’s usually assumed that Vr << VDC and DT <<T, where VDC = VP - Von is the ideal DC
voltage in the absence of ripple.
How to find DT?
Von  0
NJIT ECE-271 Dr. S. Levkov
Chap 3 -121
Ripple Voltage and Conduction Interval
The output voltage is not constant as in an ideal peak detector, but has a ripple voltage Vr.
The diode conducts for a short time DT called the conduction interval during each cycle,
and its angular equivalent is called the conduction angle c.
It’s usually assumed that Vr << VDC and DT <<T, where VDC = VP - Von is the ideal DC
voltage in the absence of ripple.
How to find DT
The discharge of C:
vo (t ')  (V Von )e
P

t'
RC ,


t '   t  T   0
4






Vr  (V Von )  v o(t ')  (V Von ) 1 e
P
P
Using exp(  x)  1  x :
T  DT
V  (V  V )
1
r
P on RC 
T
DT 
1
w

T DT 
RC 



Von  0
 (VP  Von ) T

R
C

2V
2V
r ,   wDT 
r
c
V
V
P
P
NJIT ECE-271 Dr. S. Levkov
Chap 3 -122
Half-Wave Rectifier Analysis
Example
Problem: Find the dc output voltage, output current, ripple voltage,
conduction interval, and conduction angle for a half-wave rectifier.
Given data: secondary voltage Vrms = 12.6 (60 Hz), R = 15 W,
C = 25,000 F, Von = 1 V.
Analysis: Using discharge interval T=1/60 s,
Vdc VP Von  (???1)V
NJIT ECE-271 Dr. S. Levkov
Chap 3 -123
Half-Wave Rectifier Analysis
Example
Problem: Find the dc output voltage, output current, ripple voltage,
conduction interval, and conduction angle for a half-wave rectifier.
Given data: secondary voltage Vrms = 12.6 (60 Hz), R = 15 W,
C = 25,000 F, Von = 1 V.
Analysis: Using discharge interval T=1/60 s,
Vdc  VP Von  (12.6 2 1)V 16.8 V
Idc 
Vr 
VP Von 16.8V

1.12 A
R
15W
(VP Von ) T
T
 Idc  0.747 V
R
C
C
NJIT ECE-271 Dr. S. Levkov
Chap 3 -124
Half-Wave Rectifier Analysis
Example
Problem: Find the dc output voltage, output current, ripple voltage,
conduction interval, and conduction angle for a half-wave rectifier.
Given data: secondary voltage Vrms = 12.6 (60 Hz), R = 15 W,
C = 25,000 F, Von = 1 V.
Analysis: Using discharge interval T=1/60 s,
Vdc  VP Von  (12.6 2 1)V 16.8 V
Idc 
Vr 
VP Von 16.8V

1.12 A
R
15W
c  wDT 
(VP Von ) T
T
 Idc  0.747 V
R
C
C

DT  wc 
2Vr
 0.290 rad 16.6o
VP
c 0.29

 0.769 ms
2f 120

NJIT ECE-271 Dr. S. Levkov
Chap 3 -125
Peak Diode Current
Von  0
In rectifiers, nonzero current exists in the
diode for only a very small fraction of period
T, yet an almost constant dc current flows
out of the filter capacitor to load.
The total charge lost from the filter capacitor
in each cycle is replenished by the diode
during a short conduction interval causing
high peak diode currents. If the repetitive
current pulse is modeled as a triangle of
height IP and width DT, then, equating the
charge supplied during conduction interval to
the charge lost during the whole period
Q  IP
DT
 I dcT  I P  I dc D2TT
2
Using the values from the previous example:
I P  Idc 2T  48.6 A
DT
NJIT ECE-271 Dr. S. Levkov
Chap 3 -126
Surge Current
In addition to the peak diode currents, there is an even larger current
through the diode called the surge current that occurs when power is
first turned on.
During first quarter cycle, current through diode is approximately
 d


id (t )  ic (t )  C  VP sin w t   w CVP cosw t
 dt



Peak values of this initial surge current occurs at t = 0+:
ISC  w CVP 168 A
using values from previous example.
Actual values of surge current won’t be as large as predicted above
because of the neglected series resistances associated with the rectifier
diode and transformer.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -127
Peak Inverse Voltage Rating
The peak inverse voltage (PIV) rating of
the rectifier diode is the diode
breakdown voltage.
Von  0
When the diode is off, the reverse-bias
across the diode is Vdc - vS. When vS
reaches negative peak,
PIVVdc vsmin VP Von (VP )2VP

NJIT ECE-271 Dr. S. Levkov
The PIV value corresponds to the
minimum value of Zener breakdown
voltage required for the rectifier diode.
Chap 3 -128
Diode Power Dissipation
Average power dissipation in a diode is given by
I
1T
1T
PD   v D (t)iD (t)dt   Von iD (t)dt Von P DT Von Idc
2 T
T0
T0
The simplification is done by assuming a triangular approximation for the

diode current and that the voltage across diode is constant at Von.
Average power dissipation in the diode series resistance is given by
1T 2
1
4
2 R
PD   iD (t)RS dt  ID2 RS DT  T Idc
S
T 3 DT
T0
3
This power dissipation can be reduced by minimizing peak current

through the use of a proper size of filter capacitor or by using full-wave
rectifiers.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -129
Full-Wave Rectifiers
Full-wave rectifiers utilize both have periods
of the input sin signal and reduce power
dissipation.
All specifications are the same as for halfwave rectifiers.

Reversing polarity of the diodes gives a fullwave rectifier with negative output voltage.



NJIT ECE-271 Dr. S. Levkov
Chap 3 -130
Full-Wave Rectifiers
Full-wave rectifiers cut capacitor discharge
time in half and require half the filter
capacitance to achieve a given ripple voltage.
All specifications are the same as for halfwave rectifiers.


Von  0


NJIT ECE-271 Dr. S. Levkov
Chap 3 -131
Full-Wave Rectifier Equations
Von  0
Vdc  VP Von , Vr  ???
DT  ???
NJIT ECE-271 Dr. S. Levkov
Chap 3 -132
Full-Wave Rectifier Equations
Von  0
Vdc  VP Von , Vr 
1
DT  w
(VP Von ) T
R
2C
T (VP Von ) 1 2Vr
w
VP
RC VP
c  wDT 
Vr and discharge time is one half of those in halfwave rectifier.
NJIT ECE-271 Dr. S. Levkov
PIV  ???
2Vr
VP
I  ???
P
Chap 3 -133
Full-Wave Rectifier Equations
Von  0
Vdc  VP Von , Vr 
1
DT  w
(VP Von ) T
R
2C
T (VP Von ) 1 2Vr
w
VP
RC VP
c  wDT 
Vr and discharge time is one half of those in halfwave rectifier.
Conduction interval and peak current are also
reduced.

PIV rating is the same for each diode.
NJIT ECE-271 Dr. S. Levkov
PIV  2V
2Vr
VP
T
I I
P DC DT
P
For negative output voltage – reverse the polarity
of both diodes.
Chap 3 -134
Full-Wave Bridge Rectifier
The requirement for a centertapped transformer in the fullwave rectifier is eliminated
through use of 2 extra diodes.



All other specifications are the
same as for a half-wave rectifier
except
PIV  VP  Von  VP
For negative output voltage –
reverse the polarity of both
diodes.

http://www.falstad.com/circuit/e-fullrect.html
NJIT ECE-271 Dr. S. Levkov
Chap 3 -135
Rectifier Topology Comparison
• Filter capacitors are a major factor in determining cost, size and weight
in design of rectifiers.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -136
Rectifier Topology Comparison
• Filter capacitors are a major factor in determining cost, size and weight
in design of rectifiers.
• For a given ripple voltage, a full-wave rectifier requires half the filter
capacitance as that in a half-wave rectifier. Reduced peak current can
reduce heat dissipation in diodes. Benefits of full-wave rectification
outweigh increased expenses and circuit complexity (an extra diode and
center-tapped transformer).
NJIT ECE-271 Dr. S. Levkov
Chap 3 -137
Rectifier Topology Comparison
• Filter capacitors are a major factor in determining cost, size and weight
in design of rectifiers.
• For a given ripple voltage, a full-wave rectifier requires half the filter
capacitance as that in a half-wave rectifier. Reduced peak current can
reduce heat dissipation in diodes. Benefits of full-wave rectification
outweigh increased expenses and circuit complexity (an extra diode and
center-tapped transformer).
• The bridge rectifier eliminates the center-tapped transformer, and the
PIV rating of the diodes is reduced. Cost of extra diodes is negligible.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -138
Rectifier Topology Comparison and
Design Tradeoffs
NJIT ECE-271 Dr. S. Levkov
Chap 3 -139
Rectifier Design Analysis
Problem: Design a rectifier with specifications: Vdc = 15 V, Vr < 0.15 V, Idc = 2 A
Assumptions: Von = 1 V, Vr << VDC , DT<<T , 60Hz
Analysis: Use a full-wave bridge rectifier that needs a smaller value of filter
capacitance, smaller diode PIV rating, and no center-tapped transformer.
VP Vdc  2Von 15 2


V 12 V
 required transformer voltage
rms
2
2
2
 T /2 
 1

1 





C  Idc 
s
 0.111 F
  2 A
 120   0.15V 
 V 



r 

V
2(0.15V )
1 2Vr
1

 0.352 ms  conduction interval
17V
w VP 120
1/ 60  s



I P  Idc  2   T   2 A 
 94.7 A  peak diode current
D
T
2
0.352
ms




I surge  wCVP 120 (0.111)(17)  711 A | PIV =VP 17 V
DT 
NJIT ECE-271 Dr. S. Levkov
Chap 3 -140
Diode Logic Gates
• Diodes together with resistors can be used to implement the digital logic
functions.
• Consider a positive logic system with low (logical 0) as 0V and high (logical 1)
as 5V.
• Circuit (a) represent logical OR function:
Y  A B  C
• Circuit (b) represent logical AND function
Y  A B C
NJIT ECE-271 Dr. S. Levkov
Chap 3 -141
Limiter Circuits
• For the input in the range L / K  vi  L / K the limiter acts as a linear
circuit: vo = Kvi (if K<1 – passive limiter).
• If vi is beyond that range, the output is limited by (L-, L+ )
Transfer characteristic for a limiter circuit
NJIT ECE-271 Dr. S. Levkov
Chap 3 -142
Limiter Circuits
• For the input in the range L / K  vi  L / K the limiter acts as a linear
circuit: vo = Kvi (if K<1 – passive limiter).
• If vi is beyond that range, the output is limited by (L-, L+ )
Y  A B C
Transfer characteristic for a limiter circuit
NJIT ECE-271 Dr. S. Levkov
Chap 3 -143
Diode Limiter Circuits Examples
NJIT ECE-271 Dr. S. Levkov
Chap 3 -144
Diode Limiter Circuits Examples
1
NJIT ECE-271 Dr. S. Levkov
2
3
4
Chap 3 -145
Diode Limiter Circuits Examples
1
NJIT ECE-271 Dr. S. Levkov
2
3
4
Chap 3 -146
Diode Limiter Circuits Examples
1
NJIT ECE-271 Dr. S. Levkov
2
3
4
Chap 3 -147
Diode Limiter Circuits Examples
1
NJIT ECE-271 Dr. S. Levkov
2
3
4
Chap 3 -148
Diode Limiter Circuits Examples
NJIT ECE-271 Dr. S. Levkov
Chap 3 -149
Diode Demodulator and Envelop Detector
NJIT ECE-271 Dr. S. Levkov
Chap 3 -150
Diode Demodulator and Envelop Detector
NJIT ECE-271 Dr. S. Levkov
Chap 3 -151
4. Special Diodes and Applications
NJIT ECE-271 Dr. S. Levkov
Chap 3 -152
Zener Diode
• As could be seen on diode’s ivcharacteristics, in the reverse breakdown
region the curve is very steep (almost
vertical). This suggests using it in the
design of voltage regulators.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -153
Zener Diode
• As could be seen on diode’s ivcharacteristics, in the reverse breakdown
region the curve is very steep (almost
vertical). This suggests using it in the
design of voltage regulators.
• Diodes designed to operate in reverse
breakdown are called Zener diodes and
use the indicated symbol.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -154
Zener Diode
• As could be seen on diode’s ivcharacteristics, in the reverse breakdown
region the curve is very steep (almost
vertical). This suggests using it in the
design of voltage regulators.
• Diodes designed to operate in reverse
breakdown are called Zener diodes and
use the indicated symbol.
• In breakdown, the diode is modeled
with a voltage source, VZ, and a series
resistance, RZ. RZ models the slope of
the i-v characteristic.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -155
Breakdown Region Diode Model
IZK – knee current, below it ivcurve is almost straight line.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -156
Breakdown Region Diode Model
IZK – knee current, below it ivcurve is almost straight line.
VZ is usually specified at the
test current IZT . 6V Zener diode
will exhibit 6V at specified test
current of, say, 10ma.
VZ can be within the range of a
few volts to few hundreds.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -157
Breakdown Region Diode Model
IZK – knee current, below it ivcurve is almost straight line.
VZ is usually specified at the
test current IZT . 6V Zener diode
will exhibit 6V at specified test
current of, say, 10ma.
VZ can be within the range of a
few volts to few hundreds.
rZ = DV/DI - dynamic
resistance, is in the range of
few ohms to few tens.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -158
Analysis of Zener Diode
Using load-line analysis:
0  20 VD  5000I D
NJIT ECE-271 Dr. S. Levkov
Chap 3 -159
Analysis of Zener Diode
Using load-line analysis:
0  20 VD  5000I D
Choose 2 points:
(0V, -4mA)
NJIT ECE-271 Dr. S. Levkov
and
Chap 3 -160
Analysis of Zener Diode
Using load-line analysis:
0  20 VD  5000I D
Choose 2 points:
(0V, -4mA)
NJIT ECE-271 Dr. S. Levkov
and (-5V, -3mA)
Chap 3 -161
Analysis of Zener Diode
Using load-line analysis:
0  20 VD  5000I D
Choose 2 points:
(0V, -4mA)
and (-5V, -3mA)
to draw the load line.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -162
Analysis of Zener Diode
Using load-line analysis:
0  20 VD  5000I D
Choose 2 points:
(0V, -4mA)
and (-5V, -3mA)
to draw the load line. It intersects the i-v
characteristic at the Q-point: (-2.9 mA, -5.2 V).
NJIT ECE-271 Dr. S. Levkov
Chap 3 -163
Analysis of Zener Diode
Using load-line analysis:
0  20 VD  5000I D
Choose 2 points:
(0V, -4mA)
and (-5V, -3mA)
to draw the load line. It intersects the i-v
characteristic at the Q-point: (-2.9 mA, -5.2 V).
Using the piecewise linear model:
VZ = 5V, rZ =100 ohm, IZ ID 0
20 5100IZ  5  0

(20 5)V
IZ 
 2.94 mA
5100W
NJIT ECE-271 Dr. S. Levkov

Chap 3 -164
Voltage Regulator Using the Zener Diode
NJIT ECE-271 Dr. S. Levkov
Chap 3 -165
Voltage Regulator Using the Zener Diode
The Zener diode keeps the
voltage across load resistor
RL constant. For Zener
breakdown operation, IZ > 0.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -166
Voltage Regulator Using the Zener Diode
VS VZ (205)V

 3 mA
R
5kW
V
5V
IL  Z 
1 mA | IZ  IS  IL  2 mA
RL 5kW
IS 

The Zener diode keeps the
voltage across load resistor
RL constant. For Zener
breakdown operation, IZ > 0.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -167
Voltage Regulator Using the Zener Diode
VS VZ (205)V

 3 mA
R
5kW
V
5V
IL  Z 
1 mA | IZ  IS  IL  2 mA
RL 5kW
IS 

The Zener diode keeps the
voltage across load resistor
RL constant. For Zener
breakdown operation, IZ > 0.
NJIT ECE-271 Dr. S. Levkov
For proper regulation, Zener current must be
positive. If the Zener current < 0, the Zener
diode no longer controls the voltage across
the load resistor and the voltage regulator is
said to have “dropped out of regulation”.


VS
R
1
1

IZ  VZ   0 | RL  
R

R
R
min
R

L 
V

 S 1


V

 Z


Chap 3 -168
Voltage Regulator Using a Zener Diode:
Example Including Zener Resistance
Problem: Find the output voltage and
Zener diode current for a Zener diode
regulator.. What is the variation in the
load voltage if the source voltage
varies by 25%?
Given data: VS  20  5 V
R = 5 kW, RZ = 0.1 kW,VZ = 5 V
NJIT ECE-271 Dr. S. Levkov
Chap 3 -169
Voltage Regulator Using a Zener Diode:
Example Including Zener Resistance
Nominal value:
VL  20V VL  5V
VL


0
5000W 100W 5000W
VL  5.19 V
IZ 
Problem: Find the output voltage and
Zener diode current for a Zener diode
regulator.. What is the variation in the
load voltage if the source voltage
varies by 25%?
Given data: VS  20  5 V
R = 5 kW, RZ = 0.1 kW,VZ = 5 V
VL  5V 5.19V  5V

1.9 mA  0
100W
100W
Worst case analysis.
VS=15V:
VL 15V VL  5V
VL


0
5000W 100W 5000W
VL  5.10 V I Z 1 mA  0
VS=25V:
VL  25V VL  5V
VL


0
5000W 100W 5000W
VL  5.28 V I Z  2.8 mA  0
25% change in VS  2% change in VL
NJIT ECE-271 Dr. S. Levkov
Chap 3 -170
Photo Diodes and Photodetectors
If the reverse biased depletion region of a pn junction diode is
illuminated with light with sufficiently high frequency, photons
can provide enough energy to cause electrons to jump the
semiconductor bandgap to generate electron-hole pairs and
create the electric field and photocurrent.
EP  h 
hc

 EG
h =Planck’s constant = 6.626 x 10-34 J-s

 = frequency
of optical illumination
= wavelength of optical illumination
c = velocity of light = 3 x 108 m/s
Photon-generated current can be used in photodetector circuits
to generate an output voltage
vO iPHR

NJIT ECE-271 Dr. S. Levkov
Chap 3 -171
Solar Cells and Light-Emitting Diodes
Without the reverse bias, the illuminated
photodiode functions as a solar cell.
In solar cell applications, optical illumination
is constant, and dc current IPH is generated.
The goal is to extract power from the cell, and
the i-v characteristics are plotted in terms of
cell current and cell voltage. For a solar cell to
supply power to an external circuit, the ICVC
product must be positive, and the cell should
be operated near the point of maximum output
power Pmax.
Light-Emitting Diodes (LEDs) use
recombination processes in the forward-biased
pn junction diode to produce light. When a
hole and electron recombine, an energy equal
to the bandgap of the semiconductor is
released as a photon.
NJIT ECE-271 Dr. S. Levkov
Chap 3 -172
Diode Spice Model
Rs represents the inevitable series
resistance of a real device structure. The
current controlled current source
models the ideal exponential behavior
of the diode. Capacitor C includes
depletion-layer capacitance for the
reverse-bias region as well as diffusion
capacitance associated with the junction
under forward bias.
Typical default values: Saturation
current IS = 10 fA, Rs = 0 W, transit
time TT = 0 seconds, N = 1
NJIT ECE-271 Dr. S. Levkov
Chap 3 -173
Schottky Barrier Diode
One semiconductor region of the pn
junction diode can be replaced by a
non-ohmic rectifying metal contact. A
Schottky contact is easily formed on
n-type silicon. The metal region
becomes the anode. An n+ region is
added to ensure that the cathode
contact is ohmic. Current is conducted
by major carriers only – electrons
NJIT ECE-271 Dr. S. Levkov
Schottky diodes turn on at a lower
voltage than pn junction diodes and
have significantly reduced internal
charge storage under forward bias.
Uses: bipolar logic circuits, fast
switching applications.
Chap 3 -174
Diode Layout
NJIT ECE-271 Dr. S. Levkov
Chap 3 -175
PN-junction Diode Capacitance
Typical capacitance in diodes is quite small (tens of pF- few nF ).
However it exists and influences the dynamic switching properties
if a diode.
Reverse Bias Capacitance (cont.)
Changes in voltage lead to changes in depletion width and charge.
This leads to a capacitance that we can calculate from the chargevoltage dependence.
 N A ND 
Qn  qN D xn A  q 
 wd A Coulombs.
 N A  ND 
C j0 A
dQn
Cj 

dvR
v
1 R
Where Cj0 is the zero bias junction capacitance per unit area
NJIT ECE-271 Dr. S. Levkov
F/cm2
j
C j0 
s
wd 0
Chap 3 -176
Forward Bias Capacitance
In forward bias operation, additional charge is stored in neutral
region near edges of space charge region.
QD  iD T Coulombs
T is called diode transit time and depends on size and type of
diode.
Additional diffusion capacitance, associated with forward region
of operation is proportional to current and becomes quite large at
high currents.
dQD iD  IS  T iD  T
Cj 


F
dv D
VT
VT
NJIT ECE-271 Dr. S. Levkov
Chap 3 -177
Dynamic Switching Behavior of Diodes
The non-linear depletion-layer
capacitance of the diode prevents the
diode voltage from changing
instantaneously and determines turn-on
and recovery times. Both forward and
reverse current overshoot the final
values when the diode switches on and
off as shown
NJIT ECE-271 Dr. S. Levkov
Chap 3 -178
Diode Temperature Coefficient
Diode voltage under forward bias:
iD  kT iD  kT iD 
v D  VT ln   1
ln   1
ln  
IS  q IS  q IS 
Taking the derivative with respect to temperature yields
 dvD  k lni D  kT 1 dI S  v D  V 1 dI S  v D  VGO  3VT
 
T
dT

q
I S 
q I S dT
T
I S dT
T
V/K
Assuming iD >> IS, IS  ni2, and VGO is the silicon bandgap energy at 0K.
For a typical silicon diode
dvD 0.65 1.12  0.075V

= -1.82 mV/K  - 1.8 mV/ C
dT
300K
NJIT ECE-271 Dr. S. Levkov
Chap 3 -179