Download Student Biographies

Electronics - Diodes
Introduction





The simplest and most fundamental nonlinear circuit element is the diode. It is a two
terminal device like a resistor but the two terminals are not interchangeable.
We will start by describing an “ideal” diode and then look at how closely a real diode
approximates the ideal situation. We will be considering silicon diodes throughout this
book.
As electrical engineers we can analyze diode circuits if we have equations which
describe the terminal characteristics of the device, but we need to look further and
understand physically how the diode works since the diode is also the basis of the BJT
and MOSFET devices we will be studying later in this course.
One of the most common uses of diode is in rectifier circuits (conversion of ac signals
to dc) so we will spend some time on examples and then look at some other diode
applications
We also need to look at how diode model parameters can be extracted for use in
simulation programs such as SPICE. The parameters can then be used to simulate
some of the application circuit examples.
© REP 5/23/2017 EGRE224
Page 3.1-1
Electronics - Diodes
The Ideal Diode


The diode symbol and terminal voltage and
current definitions are shown to the right. The
quantity VA is referred to as the Applied voltage.
YOU MUST MEMORIZE THIS FIGURE!
The i-v characteristic for the ideal diode passes
no current when the applied voltage (with the
polarity given in the definition) is negative, and
when the applied voltage is positive the diode
is a perfect short circuit (zero resistance).
anode
“Cut off”
 vA 
i
Reverse Bias Circuit Model
i  vA 
vA  0  i  0
© REP 5/23/2017 EGRE224
Forward Bias
“ON”
n
p
i
Reverse Bias
cathode
The external circuit must limit the current
under Forward Bias conditions since the
diode will have no resistance
vA
Forward Bias Circuit Model
i  vA 
i  0  vA  0
Page 3.1-2
Electronics - Diodes
The diode is polarity dependent!
Forward Bias Current Limit Example
(resistor limits the current)
 10V
 10V
1mA  10V

10V
1k
P
N
Reverse Bias Current Limit Example
(diode, in cut off, limits the current)

Short circuit
1k

0V

© REP 5/23/2017 EGRE224
 10V

0V
1k
N
P

Open circuit
0mA
1k

10V

Page 3.1-3
Electronics - Diodes
A Simple Application: The Rectifier
 vD 
vI

vO
R
iD

vI
vP
for vI  0
t
 vD  vI 
 vD  0 
vI
iD

vO  vI
R
The positive half-cycle
is transmitted

vO
vP
vI
iD  0
for vI  0
R

vO  0

The negative half-cycle
is blocked
t
© REP 5/23/2017 EGRE224
Page 3.1-4
Electronics - Diodes
Exercises Involving Rectification

Exercise 3.1 Sketch the transfer characteristic of simple rectifier. The transfer
characteristic is vO vs. vI . We see that when vI is negative vO zero and when vI is
positive vO is equal to vI
vO

vO  vI
45 degrees
vI
Exercise 3.2 Find the waveform of vD. Well we know that the input voltage has to divide
across the diode and the resistor, so when there is no voltage across the resistor (vO =0)
then it must be across the diode and vice-versa. The diode voltage will be the exact
complement of the output voltage in this case.
t
 vP
 vD

If vI has a peak value of 10V and R=1k, find the peak value of iD and the dc component

of vO.


10 sin  
5

vOdc
 cos  0
vP
10
iP 

 10mA
2

R 1,000
5
10
vOdc  1  1 
 3.18V
0

© REP 5/23/2017 EGRE224
vP
t

Page 3.1-5
Electronics - Diodes
Battery Charger Rectification Example

Example 3.1 The circuit below is used to charge a 12V battery, where vS is a sinusoid
with a 24V peak amplitude. Find the fraction of each cycle during which the diode
conducts, also find the peak value of the diode current and the maximum reverse-bias
voltage that appears across the diode.
R  100
iD
vS

12V
24V
12V

  cos1 0.5  60
24 cos  12
vS
iD
2
iD
Vd max
t
 24V
2  120 or one third of the cycle
Id 
24VP  12V
 0.12 A
100
Vd max  12V  24VP  36V
© REP 5/23/2017 EGRE224
Page 3.1-6
Electronics - Diodes
Another Application: Diode Logic Gates




Diodes and resistors can be used to implement digital logic functions
0V is a Low and +5V is a high
In the circuit on the left below if any one of the three inputs is at +5V the output vQ will
also be at +5V and there will be a current flowing through the resistor. If all three input
are zero the diodes will be cut off and the output will be grounded through the resistor.
The results are summarized in the OR gate truth table next to the circuit
In the circuit on the right below, if any of the inputs are zero that diode will be on and the
output will be at zero volts. If all three inputs are at +5V the diodes will be cut off and the
output will be at +5V. The results are summarized in the AND gate table.
OR Gate
v A v B vC v Q
vA
vB
vQ
vC
R
© REP 5/23/2017 EGRE224
0
0
0
0
5
5
5
5
0
0
5
5
0
0
5
5
0
5
0
5
0
5
0
5
0
5
5
5
5
5
5
5
Output
Inputs
Output
Inputs
v A v B vC v Q
AND Gate
 5V
R
vA
vB
vC
vQ
0
0
0
0
5
5
5
5
0
0
5
5
0
0
5
5
0
5
0
5
0
5
0
5
0
0
0
0
0
0
0
5
Page 3.1-7
Electronics - Diodes
Simple DC Analysis of Ideal Diode Circuits



Example 3.2(a) Given the following circuit, Find the indicated values of I and V.
How do we know which diodes are conducting and which are not? It might be hard to
tell, so we make an assumption (always write down your assumption), then proceed
with your analysis and then check to see if everything is consistent with your initial
assumption. If things are not consistent then our assumption was invalid. NOTE, this
does not mean that all your work was in vain, sometimes it is just as important to prove
what is incorrect as what is correct.
For now, lets assume that both diodes are conducting
 10V
I D2
I
D1
B
5k
 10V
© REP 5/23/2017 EGRE224
If D1 is on VB=0 and the output V=0 also.
We can now find the current through D2
10k
D2

V

I D2 
10V  0V
 1mA
10,000
We can write a node equation at node B,
looking at the sum of the currents
I  I D 2  I 5k
0V   10V 
5,000
I  1mA Therefore D1 is on as assumed
I  1mA 
Page 3.1-8
Electronics - Diodes
Another Circuit



Example 3.2(b) This is the same circuit as the previous one except that the values of the
two resistors have been exchanged.
Again I will assume that both diodes are on, do the analysis and check the results.
Again VB=0 and V=0.
I D2 
We can write a node equation at node B,
looking at the sum of the currents
I  I D 2  I10k
 10V
I D2
I
D1
B
10k
 10V
© REP 5/23/2017 EGRE224
5k
D2
10V  0V
 2mA
5,000

V

I  1mA 
I  1mA
0V   10V 
10,000
Not possible, therefore
assumption was wrong
Now assume D1 is off and D2 is on
I D2 
10V   10V 
 1.33mA
10,000  5,000
Now solving for VB we get 3.33V and I=0 since D1 is off
Page 3.1-9
Electronics - Diodes
Diode Terminal Characteristics
An Analog sweep has been converted to Digital (discrete values)
Rd =
DV
1
= slope
DI
I
slope =
5mA
Rd is the dynamic (changing) resistance
4mA
DI
3mA
Breakdown
Voltage
from -6 to -hundreds of volts
Reverse
Bias Va < 0
The higher the
doping levels of the
n and p sides of the
diode, the lower
the breakdown
voltage.
Calculations
DV
2mA
1mA
slightly negative
- 1mA
- 2mA
(+0.2 volt increments)
DI
DV
Forward
Bias Va > 0
1.0
Va
- 3mA
- 4mA
Turn-on
Voltage
The turn-on voltge is a
function of the
semiconductor used.
~ 0.7V for Si and ~ 1.7V
for GaAs
Rd
breakdown
“Off”
R is
high
The resistance of the
“closed”
diode is not
switch
constant, it depends
on the polarity
“on” R is low
and magnitude of the
applied voltage
Va
© REP 5/23/2017 EGRE224
Page 3.1-10
Electronics - Diodes
Diode Analogy

A Diode can be thought of as a one-way valve (one-way street!)
 When no force (voltage) is applied to the valve, no current
flows

When a force (voltage) greater than a particular threshold is
applied in one direction, a current can flow
IForward

When a force is applied in the opposite direction no (very little)
current can flow unless the diode undergoes breakdown.
Ireverse
Breakdown
© REP 5/23/2017 EGRE224
Page 3.1-11
Electronics - Diodes
Determining the Polarity of a Diode
Curve tracer
The connections are correct
Va is being applied to the
p-side of the diode.
I
red
P
N
Va
1.0
black
Va
I
Va
-1.0
Reverse the connections
to the diode, Va is being
applied to the n-side of the
diode
© REP 5/23/2017 EGRE224
N
P
Va
Page 3.1-12
Electronics - Diodes
The Forward Bias Region


Forward-bias is entered when va>0
The i-v characteristic is closely approximated by
i  I S  e nV  1


 Is, saturation current or scale current, is a constant for a given diode at a given
temperature, and is directly proportional to the cross-sectional area of the diode
 VT, thermal voltage, is a constant given by
v
T
V
T

kT
q
 K = Boltzman’s constant = 1.38 x 10-23 joules/kelvin
 T = the absolute temperature in kelvins = 273 + temp in C
 q = the magnitude of electronic charge = 1.60 x 10-19 coulomb

For appreciable current i, i >>IS, current can be approximated by
v
i  I S e nV T
or alternatively
n
p
v  nV T ln
 vA 
i
I
i
I
S
1.0
© REP 5/23/2017 EGRE224
Page 3.1-13
Va
Electronics - Diodes
The Reverse Bias Region

Reverse-bias is entered when va < 0 and the diode current becomes
i  IS

Real diodes exhibit reverse currents that are much larger than IS. For instance, IS for a
small signal diode is on the order of 10-14 to 10-15 A, while the reverse current could be
on the order of 1 nA (10-9 A).
A large part of the reverse current is due to leakage effects, which are proportional to
the junction area.

breakdown voltage
VZK
I
Va
reverse-bias region
© REP 5/23/2017 EGRE224
Page 3.1-14
Electronics - Diodes
The Breakdown Region


The breakdown region is entered when the magnitude of the reverse voltage exceeds
the breakdown voltage, a threshold value specific to the particular diode. The value
corresponds to the “knee” of the i-v curve and is denoted VZK. Z stands for Zener, which
will be discussed later, and K stands for knee.
In the breakdown region, the reverse current increases rapidly, with the associated
increase in voltage drop being very small.
breakdown voltage
VZK
I
Va
reverse-bias region
breakdown region
© REP 5/23/2017 EGRE224
Page 3.1-15
Electronics - Diodes
Conductors and Insulators
Ohm’s Law: V = I * R
E
e-
i
+
Conductor: small V  large i \ R is small
Insulator: large V  small i \ R is large ( )
© REP 5/23/2017 EGRE224
Page 3.1-16
Electronics - Diodes
Semiconductors
Tetrahedron Covalent Bonds in a Semiconductor
Figure taken from Semiconductor Devices, Physics and Technology, S. M. Sze,1985, John Wiley & Sons
© REP 5/23/2017 EGRE224
Page 3.1-17
Electronics - Diodes
Semiconductors (cont.)
Bonds, Holes, and Electrons in Intrinsic Silicon
Figure taken from Semiconductor Devices, Physics and Technology, S. M. Sze,1985, John Wiley & Sons
© REP 5/23/2017 EGRE224
Page 3.1-18
Electronics - Diodes
Doped Semiconductors
Bonds, Holes, and Electrons in Doped Silicon
Figure taken from Semiconductor Devices, Physics and Technology, S. M. Sze,1985, John Wiley & Sons
© REP 5/23/2017 EGRE224
Page 3.1-19
Electronics - Diodes
The Diode
B
A
Al
SiO2
p
n
Cross-section of pn -junction in an IC process
A
Al
A
p
n
B
One-dimensional
representation
B
diode symbol
Figure taken from supplemental material for Digital Integrated Circuits, A Design Perspective, Jan M. Rabaey,1996, Prentice Hall
© REP 5/23/2017 EGRE224
Page 3.1-20
Electronics - Diodes
Carrier Motion


Carriers move due two two different mechanisms
 Carrier drift in response to an electric field
 Carriers diffuse from areas of high concentration to areas of lower concentration
Since both carrier types (electrons and holes) can be present and there are two
mechanisms for each carrier there are four components to the overall current, as shown
below

J p  J p|drift  J p|diffusion  qp pE  qDpp
J n  J n|drift  J n|diffusion
© REP 5/23/2017 EGRE224
 drift  diffusion

 qn nE  qDnn
Page 3.1-21
Electronics - Diodes
Diffusion Current

Carriers move from areas of high concentration to low concentration
hole conc.
J p   qD p
dp
dx
current flow
++ +
+
+ + +
+
+
++ +
+
+
hole motion
dp
is negative
dx
electron conc.
J n  qDn
dn
dx
current flow
electron motion
dn
is negative
dx
© REP 5/23/2017 EGRE224
Page 3.1-22
Electronics - Diodes
Carrier Drift

Definition - Drift is the motion of a charged particle in response to an applied electric
field.
 Holes are accelerated in the direction of the applied field
 Electrons move in a direction opposite to the applied field
 Carriers move a velocity known as the thermal velocity, uth
typical value of ~ 5x106 cm/s


The carrier acceleration is frequently interrupted by scattering events 
E
 Between carriers
 Ionized impurity atoms
e Thermally agitated lattice atoms
vth
 Other scattering centers
The result is net carrier motion, but in a disjoint fashion
vdrift
 Microscopic motion of one particle is hard to analyze
 We are interested in the macroscopic movement of many, many particles
 Average over all the holes or all the electrons in the sample
 The resultant motion can be described in terms of a drift velocity, vd
© REP 5/23/2017 EGRE224
Page 3.1-23
Electronics - Diodes
Carrier Drift, continued


Definition - Current, is the charge per unit time crossing an arbitrarily chosen plane of
observation oriented normal to the direction of current flow.
Consider a p-type bar of semiconductor material, with cross-sectional area A.

E
A
+

vd t
v d tA
pvd tA
qpvd tA
qpvd A
The current can be written as:
I P drift  qpvd A
or in vector form










All holes this distance from plane A will cross the plane in time t
All holes in this volume will cross the plane in a time t
Holes crossing the plane in time t
Charge crossing the plane in a time t
Charge crossing the plane per unit time
J P drift  qpvd
where J is the current density, current per unit area

We seek to directly relate J to the field.
 For small to moderate values of the electric field the measured drift velocity is
directly proportional to the applied field, we can write
vdrift


 pE
cm2
where  p is the hole mobility in
V  sec
The mobility is the constant of proportionality between the drift velocity and the electric
field
© REP 5/23/2017 EGRE224
Page 3.1-24
Electronics - Diodes
Drift Velocity vs Electric Field
proportionality constant is the Mobility

Some typical values for carrier mobilites in silicon at 300K and doping levels of 1015 cm3
cm 2
n  1350
V sec
cm 2
 p  480
V sec
velocity saturation
typical value of ~ 107 cm/s
7
1 10
Carrier Drift Velocity
6
1 10
cm/sec
v dn ( Efield )

v dp ( Efield )
vdrift
 
E
5
1 10
4
1 10
100
3
1 10
4
1 10
5
1 10
6
1 10
Efield
Electric Field in Volts/cm
© REP 5/23/2017 EGRE224
Page 3.1-25
Electronics - Diodes
Abrupt Junction Formation

Junction Formation
N
P
0


X
The Depletion Region
 + represents an immobile donor impurity (i.e. P+ )
 - represents an immobile acceptor impurity (i.e. B- )
 - represents a mobile electron
 + represents a mobile hole
p type
- + - - + - ++
-+ - + - +- + + + + ++
- - - - + + +
+
- - + - + -+ -
Carrier Concentrations
 pp ~ Na
 np0 ~ (ni )2 / Na
 nn ~ Nd
 pn0 ~ (ni )2 / Nd
- - - - -
n type
+ + + -+ - +
+ +
+ + ++ + + + - +
+-
+
++ +
-+
-
x
pp
np0
© REP 5/23/2017 EGRE224
nn
pn0
Depletion
Region
Page 3.1-26
Electronics - Diodes
The Depletion or Space Charge Region
hole diffusion
electron diffusion
-
+
+
+
+
+
hole drift
electron drift
charge density
(Coulombs/cm-3)
xp
Q p = -qNaxp
-qNa
Electric field (x)
-
xp
+
Abrupt depletion approximation
+qNd
  q p  n  N D
Q n = qNdxn
x
xn
xd = xn + xp
C
(Volts/cm)
xn
 r Si O  0 Area
 N A
   Electric _ field  
2
xd

K s 0
d ( Electric _ field )


dx
K s 0
Maximum Field (Emax )
Electrostatic
potential V(x) (Volts)
xp
© REP 5/23/2017 EGRE224
Potential    Electric _ field
Vbi
Electric _ field   
xn
Page 3.1-27
Electronics - Diodes
Reverse Bias
pn0
np0
p-region
-W1 0
W2
x
n-region
diffusion
© REP 5/23/2017 EGRE224
Page 3.1-28
Electronics - Diodes
pn (W2)
Forward Bias
pn0
Lp
np0
p-region
-W1 0
W2
n-region
x
diffusion
© REP 5/23/2017 EGRE224
Page 3.1-29
Electronics - Diodes
Analysis of Forward Biased Diode Circuits



We have already looked at the ideal diode model for forward bias (short circuit). In this
section we will work with a detailed model and then explore simplifying assumptions
that allows us to work back towards our ideal case.
We will use a simple circuit consisting of a dc source VDD and a resistor and a diode in
series. We want to determine the exact current through the circuit, ID and the exact
voltage dropped across the diode VD.
If we assume that the voltage source VDD is greater than ~0.5 volts the diode will
obviously be in the forward mode of operation and the current through the diode will be
given by the following equation
ID  ISe

VD
nVT
Note we do not know the exact value of VD but we can relate it to other values in our
circuit, for example we can write a Kirchhoff’s loop equation
ID
VDD  VD
ID 
R

If we assume that IS and n are known, we
have two equations and two unknowns (IS
and VD) and we can solve for them by
 Graphical means
 Iterative (mathematical) means
© REP 5/23/2017 EGRE224
R
VDD +
+
VD
-
Page 3.1-30
Electronics - Diodes
Graphical (Load Line) Analysis





Our circuit has two components (not
counting the voltage source), a
resistor and a diode, which are
connected to each other.
D
Each device constrains (or puts
limits on) the other
Consider a toy slot car race track
with a battery powered car. The car
could go any where if put on a wide
open surface but when placed on
the track it is constrained to follow
the course. The car will only be
found on the course (the track
constraint) and the motor will
determine where on the course (car
constraint). The exact position
depends on both constraints
We will plot the characteristics of
0
each device separately in the circuit
0
as if the other device was not there
and then combine our constraints
for the final solution
We have already looked at the diode
and its characteristic is repeated
here
i (mA)
© REP 5/23/2017 EGRE224
vD is linked to VDD
as VDD , iD and vD 
Since the n side of the diode is
grounded the characteristic looks
like our typical characteristic
(already presented)
vD (V)
Page 3.1-31
Electronics - Diodes
Graphical (Load Line) Analysis continued



Lets look at the resistor
characteristic now
In this case one terminal of the
resistor is at the voltage VDD and the
other is at some unknown voltage
VD at the diode
We can determine this unknown
voltage (operating point) by
superimposing the graphs of the
expressed for diode current.
iR (mA)
VDD
R
i (mA)
vR (V)
VDD
VDD
R
ID 
ID  ISe
VD
nVT
v (V)
0
© REP 5/23/2017 EGRE224
VDD  VD
R
0
VDD
Page 3.1-32
Electronics - Diodes


The straight line is known
as the load line.
The load line intersects the
diode curve at point Q, the
operating point. The
coordinates of Q are ID and
VD.
i (mA)
VDD
R
I
Q
D
slope  
1
R
v (V)
0
© REP 5/23/2017 EGRE224
0
V
D
VDD
Page 3.1-33
Electronics - Diodes
Iterative Analysis




Example 3.4 Assume that the resistor in our graphical analysis circuit is 1k and VDD is
5V
The diode has a current of 1mA if it is at a voltage of 0.7 volts and the voltage drops by
0.1 volt for every decade decrease in current.
Find the current through the circuit and the exact voltage across the diode.
We can start by assuming we have set up the conditions so that the voltage across the
diode is 0.7 volts, we do this so that we can do some calculations about our diode that
we can use later to zero in on our actual conditions
ID 

VDD  VD 5  0.7

 4.3mA
R
1000
This current is larger than the 1mA current at 0.7 volts so we conclude that the actual
diode voltage will be larger than 0.7 volts. Since the relationship between the current
and the voltage is exponential we can adjust our voltage estimate slightly using an
equation we derived earlier relating the voltage change to the current ratio, namely
I 
 0.0043 
V2  V1  2.3nVT log  2  so V2  V1  0.1 log 

I
0.001


 1
V2  0.763V
note : If 2.3nVT  0.1 then n  1.74

Now using this value in our original equation we get
ID 
VDD  VD 5  0.763

 4.237mA
R
1000
© REP 5/23/2017 EGRE224
Converged to
ID=4.237mA
VD=0.762V
 0.004237 
V2  0.763  0.1 log 
  0.762V
0.0043


Page 3.1-34
Electronics - Diodes
A graphical view of the iterative analysis
i (mA)
2
VDD
R
4.3 mA
4.237 mA
3
END
1 1.0 mA
0
v (V)
0
VDD
0.7V
START 1
0.762V
3
0.763V
2
© REP 5/23/2017 EGRE224
Page 3.1-35
Electronics - Diodes
Approximating the diode forward characteristic
with two straight lines

The analysis of a diode circuit can be greatly simplified by approximating the
exponential i-v curve with two straight lines. One line, A, has a zero slope and the
second line, B, has a slope of 1/rD

The piecewise-linear model is described as follows:
i
i
iD (mA)
D
D
 0,
v V
 v V  r
D
D
D0
D0
D
,
v V
D
D0
i
D

B, slope =
1
r
D
A, slope = 0
0
V
© REP 5/23/2017 EGRE224
D0
v

ideal
D
V
r
D0
D
vD (V)
Page 3.1-36
Electronics - Diodes
Constant-Voltage Drop Model



This model is even simpler than the piecewise-linear or battery-plus-resistance model
shown on the previous slide. Here, we use a vertical straight line, B, to approximate the
fast-rising part of the exponential i-v curve of the diode.
We assume that a forward-conducting diode exhibits a constant voltage drop, VD, which
is approximately 0.7 V.
This model is used in the initial phases of analysis and design to give a rough estimate
of circuit behavior.
iD (mA)
i
D
B, vertical

v

A, horizontal
0
© REP 5/23/2017 EGRE224
V
D
ideal
D
V
D
 0.7 V
vD (V)
Page 3.1-37
Electronics - Diodes
Example

Exercise 3.16 For the circuit shown below, find ID and VD for VDD=5V and R=10k.
Assume that the diode has a voltage of 0.7V at 1mA current and the voltage changes by
0.1V / decade of current change.
ID
10k
+
VD
-
5V

Use (a) iteration, (b) the piecewise linear model with VD0=0.65V and rD=20, and (c) the
constant voltage-drop model with VD=0.7V.
© REP 5/23/2017 EGRE224
Page 3.1-38
Electronics - Diodes
Example, continued

Iteration
V D  0.7 V
10k
5V
5  0.7
 0.43 mA
10
0.43

0
.
7

n
ln
VD
V T 1 , where nV T ln 10  0.1
Thus, V D  0.7  0.043 ln 0.43  0.663 V
+
VD
-
ID 
nV T  0.0434 V
5  0.663
 0.434 mA
10
0.434

0
.
7

0
.
0434
ln
 0.663 V
VD
1
ID 
© REP 5/23/2017 EGRE224
Page 3.1-39
Electronics - Diodes
Example, continued

The piecewise-linear model
10k
5V
+
VD
-
0.65 V
20
5  0.65
 0.434 mA
0.02
V D  0.65  0.434 * 0.02  0.659 V
ID 
© REP 5/23/2017 EGRE224
Page 3.1-40
Electronics - Diodes
Example, continued

The constant-voltage-drop model
10k
5V
+
VD
-
0.7 V
5  0.7
 0.43 mA
10
VD  0.7 V
ID 
© REP 5/23/2017 EGRE224
Page 3.1-41
Electronics - Diodes
DC Forward Bias with an ac small signal
tangent at Q
iD (mA)
The DC bias level determines the
ac parameters
By restricting the input signal
swing to small values we can
“linearize” the characteristic like
we did for amplifier transfer
1.0
characterisitcs
slope 
1
rd
Bias Point - Q
ID
id (t)
0
iD(t)
ac
vd(t)
VD +
DC
0
0.55
0.65
0.75
vD (V)
(DC+ac)
+
vD(t)
(DC+ac)
© REP 5/23/2017 EGRE224
VD0
t
vd (t)
VD=0.7
Page 3.1-42
Electronics - Diodes
Small Signal Analysis

If we set the ac signal to zero, the current
through the diode due to the DC bias is
given by
ID  ISe


vd
 1
nVT
VD
nVT

When we add in the ac small signal to the
DC voltage bias the total signal is
vD (t )  VD  vd (t )

iD (t )  I S e


VD
nVT
e
vd
nVT
Substituting in the DC equation from above,
we get
vd
iD (t )  I D e
© REP 5/23/2017 EGRE224
nVT




This IS the small signal approximation,
valid for amplitudes less than about 10mV
iD (t )  I D 
Which we can re-arrange to get
iD (t )  I S e

iD (t )  I S e
VD  vd
nVT
We can expand the exponential in an infinite
series, but we find that a sufficiently
accurate expression can be found using
only the first two terms.

v
iD (t )  I D 1  d
 nVT
The total (DC +ac) instantaneous current is
vD
nVT
If we keep the amplitude of the ac signal
small, such that
I D vd
nVT
We find that the total current is made up of
a DC component and an ac component that
is directly proportional to the small signal
voltage AND the DC bias level
I v
iD  I D  id Where
id 
D d
nVT
Page 3.1-43
Electronics - Diodes
Small Signal Resistance (incremental resistance)
On the previous page we found
id 
I D vd
nVT
iD (mA)
Bias Point - Q
And since,
id
I
 gd  D
vd
nVT
\ rd 
1 nVT

gd
ID
The ac small-signal resistance is inversely
proportional to the DC bias current ID
In the graphical representation we find
that about the Q point
rd 
1
 iD 
 v 
 D i
© REP 5/23/2017 EGRE224
ID
0
slope 
1
rd
tangent at Q
0
VD0
vD (V)
The equation of the tangent line is given
by:
1
iD 
rd
vD  VD 0 
D ID
Page 3.1-44
Electronics - Diodes
The Equivalent Circuit Model for the Diode

The equation of the tangent line is a model of the diode operation for small signal
changes about the bias DC point (Q point)
iD 

1
vD  VD 0 
rd
The total model has the components shown below
iD
+
vD
-
vD  VD 0  iD rd
ideal
VD0
rd
vD  VD 0  I D  id rd
vD  VD 0  I D rd   id rd
tangent at Q
vD  VD  id rd
slope 
Bias Point - Q

The incremental voltage across the diode is
ID
id
vd  id rd
0
© REP 5/23/2017 EGRE224
1
rd
vd
0
VD0 VD
vD (V)
Page 3.1-45
Electronics - Diodes
Application of the Diode Small-Signal Model
Consider the circuit shown at the right,
with combined DC and ac voltage input
causing a DC and ac current. We can
analyze the response of the circuit by
using the diode model developed on the
previous page and performing the circuit
analysis
iD=ID+id
R
vs
VDD +
VDD  vs  iD R  VD 0  iD rd
+
vD=VD+vd
-
VDD  vs  I D  id R  VD 0  I D  id rd
VDD  vs  I D R  VD 0  I D rd   id R  rd 
VDD  vs  I D R  VD  id R  rd 
vs
VDD
iD=ID+id
R
+
ideal
vD=VD+vd
VD0
rd
-
© REP 5/23/2017 EGRE224
Page 3.1-46
Electronics - Diodes
Application of the Diode Small-Signal Model continued

Separate the result from the previous page into a DC response and model and an ac
response and model

The small-signal analysis is done by eliminating all DC sources and replacing the diode
with the small-signal resistance. Using ac voltage division of the ac signal voltage we
get the small-signal voltage across the diode to be
VDD  vs  I D R  VD  id R  rd 
rd
R  rd
vd  v s
Circuit for DC Analysis
Circuit for small-signal Analysis
VDD  I D R  VD
vs  id R  rd 
ID
id
R
VDD
+
ideal
VD0 VD
rd
-
© REP 5/23/2017 EGRE224
vs
+
R
rd
vd
Page 3.1-47
Electronics - Diodes
Power Supply Ripple Example





Example 3.6 The power supply has a 10V DC value and a 1V peak-topeak sinusoidal ripple at a frequency of 60 Hz.
The ripple is an imperfection of the DC power supply design (we will
talk about this in more detail in a later section)
Calculate the dc voltage across the diode and the magnitude of the
sine-wave signal appearing across it
Assume the diode has a 0.7V drop at a current of 1mA and that the
ideality factor n=2
Calculate the dc diode current by assuming VD=0.7V
ID 

Since this value is close to 1mA the diode voltage will be close to the
assumed value of 0.7V. At this DC operating point we can calculate
the incremental (dynamic) resistance rd as follows
rd 

10  0.7
 0.93mA
10,000
nVT
20.025

 53.8
3
ID
0.93x10
The peak-to-peak small signal voltage across the diode can be found
using the ac model and the voltage divider rule
vd  peak  to  peak   2

rd
53.8
2
 10.7mV
R  rd
10,000  53.8
+V=10V+ripple
R=10k
+
vd
-
+V=10V+ripple
R=10k
+
rd=53.8 v
d
-
This value is quite small and our use of the “small signal” model is
justified
© REP 5/23/2017 EGRE224
Page 3.1-48
Electronics - Diodes
Voltage Regulation Using Diode Forward Voltage Drops




Example 3.7 The string of three diodes shown in the
figure provide a voltage of about 2.1V
We want to see
a) how much of a fluctuation (percentage change in
regulation) there is in the output for a 1V (10%)
change in the power supply voltage
b) percentage change in regulation when there is a 1k
load resistance. Assume n=2
With no load the nominal dc current is given by
10  2.1
ID 
 7.9mA
1,000
R=1k
+
vo
RL=1k
-
Thus the dynamic resistance of each diode is
rd 


10V + 1V
nVT 20.025

 6.3
3
ID
7.9 x10
The total resistance of the diodes will be 3rd or 18.9
Using voltage division on the 1V p-p change (10%) we get
Dvo  2
3rd   2 18.9  37.1mV
R  3rd 
1,000  18.9
© REP 5/23/2017 EGRE224
Page 3.1-49
Electronics - Diodes
Voltage Regulation continued

When the load resistor is connected it draws current a current from the node that the
diodes are connected to which reduces the dc current in the diode string.
IL 

2.1
 2.1mA
1,000
If it is assumed that the dynamic resistance stays the same, then the output small signal
change is given by
Dvo  0.00213rd   0.002118.9  39.7mV

But when the dc current in the diode string is decreased the dynamic resistance
changes
I Diodes  I R  I L  7.9mA  2.1mA  5.8mA

nVT 20.025

 8.6
ID
5.8 x10 3
This leads us to
Dvo

3r R 
25.8
2
2
 49.1mV
R  3r R 
1,000  25.8
d
L
d

rd 
L
It appears that the small signal model is not entirely justified
© REP 5/23/2017 EGRE224
Page 3.1-50
Electronics - Diodes
Diode Model for High Frequencies





The small signal model that we have developed is a
resistive one and it applies for low frequencies
where the charge storage is negligible.
Charge storage effects were modeled by two
capacitances
 The diode depletion layer capacitance (Cj)
 The forward biased diffusion capacitance (Cd)
When we include these two capacitances in parallel
with the diode’s dynamic resistance (rd) we get the
high frequency diode model shown at the right
The formulas for the model parameters are also
shown at right
For high frequency digital switching applications
large signal equations for Cj and Cd are used
rd
Cj
Cd
DC Bias Point : I D , VD
rd 
nVT
ID
Cd 
Cj 
T
VT
ID
C j0
m
for VD  0
 VD 
1  
 V0 
C j  2C j 0 for VD  0
© REP 5/23/2017 EGRE224
Page 3.1-51
Electronics - Diodes
Small-Signal Resistance Calculation and Model

rd 1 
Exercise 3.20 Find the value of the diode small-signal resistance rd at bias currents of
0.1, 1, 10mA (assume n=1)
nVT 10.025

 250
ID
0.0001

rd 2 
nVT 10.025

 25
ID
0.001
rd 3 
nVT 10.025

 2.5
ID
0.01
Exercise 3.21 For a diode that conducts 1 mA at a forward voltage drop of 0.7V (with
n=1), find the equation of the straight line tangent at ID=1mA.
 From the question above we find that rd is 25 and we can substitute in the bias
point values and solve for VD0
iD 
1
vD  VD 0   0.001  1 0.7  VD 0 
rd
25
VD 0  0.025  0.7
VD 0  0.675V
© REP 5/23/2017 EGRE224
Page 3.1-52
Electronics - Diodes
Exercise 3.22, How small is small in the Small-signal model


Consider a diode with n=2 biased at a dc current of 1mA. Find the change in current as
a result of changing the voltage by (a) -20mV (b) -10mV (c) -5mV (d) +5mV (e) +10mV (f)
+20mV. Do the calculations using both the small-signal model and the exponential
model
(a)
I D1  I D 2
1
1
 vD  VD 0   vD  VD 0 
rd
rd
I D1  I D 2  I S e
I D1  I D 2
I D1  I D 2
VD 1
nVT
 ISe
VD 1  0.02
nVT
 ISe
VD 1
nVT
0.02


1  e nVT 




 0.02


2  0.025 

 I D1 1  e




 0.001 
© REP 5/23/2017 EGRE224
Page 3.1-53
Electronics - Diodes
Diode Characteristic in the Reverse Breakdown
Region - Zener Diodes
If the Zener diode is biased
in the reverse breakdown
region of operation the
current can fluctuate wildly
about the Q point and the
voltage across the diode will
remain relatively unchanged
The knee current and knee
voltage is usually specified
on a zener diode data sheet
VZ0 V
ZKnee
VZ
IZKneev
Bias Point - Q
IZ
+
VZ
© REP 5/23/2017 EGRE224
slope 
1
rZ
IZT
The incremental (dynamic)
resistance in reverse
breakdown is given by rZ
Circuit symbol
for a Zener diode
i
DV
DI
Test
current
DVDIrZ
Page 3.1-54
Electronics - Diodes
The Reverse Bias Zener Model



We can see from the previous
page that we can model the
zener diode in the breakdown
region as straight line having
an x (voltage) intercept at VZ0
and a slope of 1/rZ. The model
is shown at the right
the reverse breakdown
characteristic of a Zener diode
is very steep (low resistance).
For a very small change in
voltage biased in the
breakdown region the current
changes significantly.
The zener diode can be used to
absorb or buffer a load from
large current changes, I.e. keep
the voltage across the load
approximately constant
© REP 5/23/2017 EGRE224
IZ
+
VZ
-
VZ0
rz
VZ  VZ 0  rz I Z
intercept
slope 
1
rZ
Page 3.1-55
Electronics - Diodes
A Shunt Regulator
i
v
I
Zener
regulator I
L
R
i
v
I
+
VO
-
Zener
regulator I
L
R
© REP 5/23/2017 EGRE224
IZ
IZ
Load
Load
+
VO
-
Page 3.1-56
Electronics - Diodes
Zener Voltage Regulation



Example 3.8 A 6.8 V Zener diode in the circuit
shown below is specified to have VZ = 6.8V at IZ =
5mA and rZ =20, and IZK = 0.2mA.
The supply voltage is nominally +10V but can vary
by plus or minus 1 V.
(a) Find the output VO with no load and V+ at 10V
(b) Find the value of VO resulting from the +/- 1 V
change in V+
(c) Find the change in VO resulting from connecting
a load resistance RL= 2 k
(d) Find the value of VO when RL =0.5 k
(e) What is the minimum value of RL for which the
diode still operates in the breakdown region.
V+ (10V + 1V)
R = 0.5 k
+
6.8V
zener
vo
RL= 1 k
-
We can start by determining the value of VZ0. VZ0 is
the x-axis intercept of the line tangent to the
characteristic at the reverse bias operating point
VZ  VZ 0  rz I Z
VZ 0  6.8  200.005
© REP 5/23/2017 EGRE224
VZ 0  VZ  rz I Z
VZ 0  6.7V
Page 3.1-57
Electronics - Diodes
Zener example continued

With no load connected, the current through the zener diode is given by
V   VZ 0 10  6.7
IZ  I 

 6.35mA
R  rZ
500  20

We can now find V0, the voltage at the operating point current

Now, for a +/- 1V change in V+ can be found from
V0  VZ 0  I Z rZ  6.7  6.350.02  6.83V
DV0  DV 

rZ
20
  1
 38.5mV
R  rZ
500  20
When a load resistance of 2k is connected, the load current will be approximately
6.8V/2000 or 3.4mA. This current will not be flowing through the zener diode if it is
flowing through the load so the change in the zener current is -3.4mA. The
corresponding change in the zener voltage (which is also the output voltage) is,
DV0  rZ DI Z  20 3.4mA  68mV

A more accurate result comes from analysis of this circuit
DV0  70mV
© REP 5/23/2017 EGRE224
V+
R = 0.5 k
VZ0
rz
+
vo
-
RL= 2 k
Page 3.1-58
Electronics - Diodes
Zener Example continued

If we change the load resistance to 500 the load current would increased to 6.8V/500
= 13.6mA, but the most current we could get through the pull up resistor and still have
the zener in breakdown would be (10-6.8)/500 or 6.4 mA, so we can’t approach 13.6mA
unless the zener diode is off (reverse biased but not in breakdown). With the diode off
we have a simple voltage divider between the pull up resistor and the load resistor.
VO  V 
RL
500
 10
 5V
R  RL
500  500
which is lower than the zener breakdown voltage
10
IL 
 10mA
500  500

For the zener to be at the edge of the breakdown region, the current has to be
IZ=IZK=0.2mA and VZ=VZK=6.7V. At this point the current supplied through the resistor R
is (9-6.7)/500 or 4.6 mA. The load current would be this current minus the current
flowing through the zener to just keep it at the breakdown knee (0.2mA), or 4.6mA 0.2mA = 4.4mA. We can now find the value of RL for to cause this
RL 
© REP 5/23/2017 EGRE224
6.7
 1,500
0.0044
Page 3.1-59
Electronics - Diodes
Shunt Regulator
Line Regulation 
DVO
DVS
Load Regulation 
I
VS
Zener
regulator I
L
R
VSmax
VS
VSmin
t
IZ
+
VO Load
-
VO
DVO
DI L
reduced ripple
t
R
rz
 VS
 I L rz R 
R  rz
R  rz
r
Line Regulation  z
R  rz
VO  VZ 0
Load Regulation  rZ R
R
VS min  VZ 0  rZ I Z min
I Z min  I L max
© REP 5/23/2017 EGRE224
IL
I
R
VS
IZ
+
VZ0 V
rz - O
Page 3.1-60
Electronics - Diodes
Example 3.9 Design of a Zener Shunt Regulator


It is required to design a zener shunt regulator to provide a voltage of approximately 7.5
volts. The original supply varies between 15 and 25 volts and the load current varies
between 0 and 15 mA. The zener diode we have available has a VZ of 7.5 V at a current of
20 mA and its rZ is 10 .
Find the required value of R and determine the line and load regulation. Also determine
the percentage change in VO corresponding to the full change in VS and IL.
VZ  VZ 0  rZ I Z  7.5V  VZ 0  10(20mA)
R
VS min  VZ 0  rZ I Z min
I Z min  I L max
Line Regulation 
R
VZ 0  7.3V
15V  7.3V  10(5mA)
 383
5mA  15mA
designing for Izmin=(1/3)ILmax
rZ
10

 25.4mV / V
rZ  R 10  383
Line Regulation  (rZ // R)  (10 // 383)  9.7mV / mA
Full Change in VS - DVo  2.54mV / V 10  0.254V or 3.4%
Full Change in VS - DVo  9.7mV / mA15mA  0.15V or  2%
© REP 5/23/2017 EGRE224
Page 3.1-61
Electronics - Diodes
Temperature Effects


Temperature Coefficient (TC) is expressed in mV/degree C
 depends on Zener voltage and operating current
 For VZ<~5V the TC is typically negative and those greater than 5V the TC is positive
 for certain current levels and VZ around 5V the TC and be made zero which makes a
temperature insensitive supply
Another technique for making a temperature insensitive supply is to use one zener with
a positive TC (say 2 mV/degree C) and a regular diode with a negative TC (say 2mV/degree C) and design a circuit in which the effects cancel
+
VZ
+
VD
-
© REP 5/23/2017 EGRE224
+
VD
-
Page 3.1-62
Electronics - Diodes
Rectifier Circuits
Power
transformer
+
IL
+
+
ac line
120V (rms)
Diode
rectifier
vO
60 Hz
-
Filter
Voltage
Regulator VO
-
-
t
Load
t
© REP 5/23/2017 EGRE224
t
t
t
Page 3.1-63
Electronics - Diodes
Half-Wave Rectifier
Ideal VD0 rD
D
+
+
vs
R
-
+
vs
vo
+
-
R
-
vo
vo
-
v
VS
VD0
vS
vo
R
Slope 
R  rD
t
VD0
vS
0 VD0
vo 
R
vS  VD 0
R  rD
vo  0, vs  VD 0
R
, vs  VD 0
R  rD
© REP 5/23/2017 EGRE224
rD  R
vO  vS  VD 0
PIV  VS
Page 3.1-64
Electronics - Diodes
Full-Wave Rectifier with Center Tapped Transformer
D1
+
+
vs
R
+
vs
vo
-
D2
-
vo
v
R
Slope 
R  rD
-VD0 0 VD0
VS
VD0
vS
-vS
vo
t
vS
PIV  2VS  VD 0
© REP 5/23/2017 EGRE224
Page 3.1-65
Electronics - Diodes
Full-Wave Bridge Rectifier
+
D4
+
vs
D2
-
vo
D1
-
R
D3
v
2VD0
vS
-vS
VS
vo
t
PIV  VS  2VD0  VD0  VS  VD0
© REP 5/23/2017 EGRE224
Page 3.1-66