The simplest and most fundamental nonlinear circuit element is the diode' It is a two terminal devic

1
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.

2
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
cathode
n
p
The external circuit must limit the current under
Forward Bias conditions since the diode will have
no resistance
Forward Bias
Reverse Bias
Cut off
ON
Reverse Bias Circuit Model
Forward Bias Circuit Model
3
The diode is polarity dependent!
Forward Bias Current Limit Example (resistor
limits the current)
Reverse Bias Current Limit Example (diode,
in cut off, limits the current)
P N
N P
Short circuit
Open circuit
4
A Simple Application The Rectifier
t
The negative half-cycle is blocked
The positive half-cycle is transmitted
t
5
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
• 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.
• If vI has a peak value of 10V and R1kW, find the
  peak value of iD and the dc component of vO.

45 degrees
t
t
6
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.

t
7
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.

Output
Inputs
Output
Inputs
OR Gate
AND Gate
0 0 0 0 0 0 5 0 0 5 0
0 0 5 5 0 5 0 0 0 5 0 5
0 5 5 0 0 5 5 5 5
0 0 0 0 0 0 5 5 0 5 0
5 0 5 5 5 5 0 0 5 5 0 5
5 5 5 0 5 5 5 5 5
8
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

If D1 is on VB0 and the output V0 also. We can
now find the current through D2
We can write a node equation at node B, looking
at the sum of the currents
B
Therefore D1 is on as assumed
9
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 VB0 and V0.

We can write a node equation at node B, looking
at the sum of the currents
Not possible, therefore assumption was wrong
Now assume D1 is off and D2 is on
B
Now solving for VB we get 3.33V and I0 since D1
is off
10
Diode Terminal Characteristics
An Analog sweep has been converted to Digital
(discrete values)
I
DV
1
Rd
DI

slope
DI
slope
DV
5mA
Rd is the dynamic (changing) resistance
4mA
DI
3mA
Forward Bias Va gt 0
Breakdown Voltage
from -6 to -hundreds of volts
2mA
DV
1mA
Va
The turn-on voltge is a function of the
semiconductor used. 0.7V for Si and 1.7V for
GaAs
- 1mA
slightly negative
1.0
Reverse Bias Va lt 0
- 2mA
(0.2 volt increments)
- 3mA
Turn-on Voltage
- 4mA
The higher the doping levels of the n and p sides
of the diode, the lower the breakdown voltage.
Rd
breakdown
The resistance of the diode is not constant, it
depends on the polarity and magnitude of the
applied voltage
Off R is high
closed switch
on R is low
Calculations
Va
11
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
• When a force is applied in the opposite direction
  no (very little) current can flow unless the
  diode undergoes breakdown.

IForward
Ireverse Breakdown
12
Determining the Polarity of a Diode
Curve tracer
I
The connections are correct Va is being applied
to the p-side of the diode.
red
black
P N
Va
Va
1.0
I
N P
Va
-1.0
Va
13
The Forward Bias Region

• Forward-bias is entered when vagt0
• The i-v characteristic is closely approximated by
  
• 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
• K Boltzmans 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 gtgtIS, current can be
  approximated by
• or alternatively

14
The Reverse Bias Region

• Reverse-bias is entered when va lt 0 and the diode
  current becomes
• 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
15
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
16
Conductors and Insulators
Ohms Law V I R
Conductor small V large i \ R is small
Insulator large V small i \ R is large ( )
17
SemiconductorsTetrahedron Covalent Bonds in a
Semiconductor
Figure taken from Semiconductor Devices, Physics
and Technology, S. M. Sze,1985, John Wiley Sons
18
Semiconductors (cont.)Bonds, Holes, and
Electrons in Intrinsic Silicon
Figure taken from Semiconductor Devices, Physics
and Technology, S. M. Sze,1985, John Wiley Sons
19
Doped SemiconductorsBonds, Holes, and Electrons
in Doped Silicon
Figure taken from Semiconductor Devices, Physics
and Technology, S. M. Sze,1985, John Wiley Sons
20
The Diode
B
A
Al
SiO
2
p
n
Cross-section of
pn
-junction in an IC process
A
Al
A
p
n
B
B
One-dimensional
representation
diode symbol
Figure taken from supplemental material for
Digital Integrated Circuits, A Design
Perspective, Jan M. Rabaey,1996, Prentice Hall
21
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

22
Diffusion Current

• Carriers move from areas of high concentration to
  low concentration

hole conc.
current flow








hole motion






electron conc.
current flow
electron motion
23
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
• The carrier acceleration is frequently
  interrupted by scattering events
• Between carriers
• Ionized impurity atoms
• Thermally agitated lattice atoms
• Other scattering centers
• The result is net carrier motion, but in a
  disjoint fashion
• 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

typical value of 5x106 cm/s
e-
vth
vdrift
24
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.
• The current can be written as
• 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
• The mobility is the constant of proportionality
  between the drift velocity and the electric field

A

25
Drift Velocity vs Electric Fieldproportionality
constant is the Mobility

• Some typical values for carrier mobilites in
  silicon at 300K and doping levels of 1015 cm-3

velocity saturation
typical value of 107 cm/s
Carrier Drift Velocity
cm/sec
m
Electric Field in Volts/cm
26
Abrupt Junction Formation

• Junction Formation
• Carrier Concentrations
• pp Na
• np0 (ni )2 / Na
• nn Nd
• pn0 (ni )2 / Nd

• 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

N
P
X
0
pp
nn
Depletion Region
pn0
np0
27
The Depletion or Space Charge Region
hole diffusion
electron diffusion
hole drift
electron drift
charge density
Abrupt depletion approximation
(Coulombs/cm-3)
qNd
xp

Q n qNdxn
x
xn
Q p -qNaxp
-
xd xn xp
-qNa
Electric field (x)
(Volts/cm)
xp
xn
Maximum Field (Emax )
Electrostatic potential V(x) (Volts)
Vbi
xn
xp
28
Reverse Bias
29
Forward Bias
30
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
• 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 Kirchhoffs loop equation

• 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

31
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.
• 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 each device
  separately in the circuit 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

iD (mA)
0
0
vD (V)
Since the n side of the diode is grounded the
characteristic looks like our typical
characteristic (already presented)
32
Graphical (Load Line) Analysis continued
iR (mA)

• 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.

i (mA)
vR (V)
v (V)
0
0
33

• 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)
v (V)
0
0
34
Iterative Analysis

• Example 3.4 Assume that the resistor in our
  graphical analysis circuit is 1kW 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
• 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
• Now using this value in our original equation we
  get

Converged to ID4.237mA VD0.762V
35
A graphical view of the iterative analysis
i (mA)
4.3 mA
4.237 mA
END
1.0 mA
v (V)
0
0
0.7V
0.762V
START
0.763V
36
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

iD (mA)
A, slope 0
0
vD (V)
37
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)
B, vertical
A, horizontal
0
vD (V)
38
Example

• Exercise 3.16 For the circuit shown below, find
  ID and VD for VDD5V and R10kW. Assume that the
  diode has a voltage of 0.7V at 1mA current and
  the voltage changes by 0.1V / decade of current
  change.
• Use (a) iteration, (b) the piecewise linear model
  with VD00.65V and rD20W, and (c) the constant
  voltage-drop model with VD0.7V.

ID
VD -
10kW
5 V
39
Example, continued

• Iteration

40
Example, continued

• The piecewise-linear model

41
Example, continued

• The constant-voltage-drop model

10kW
VD -
0.7 V
5 V
42
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 characterisitcs

Bias Point - Q
ID
1.0
id (t)
t
0
0.75
0.55
0.65
0
vD (V)
iD(t)
(DCac)
VD0
ac
vD(t) -
vd (t)
vd(t)
VD
VD0.7
(DCac)
DC
43
Small Signal Analysis

• If we set the ac signal to zero, the current
  through the diode due to the DC bias is given by
• When we add in the ac small signal to the DC
  voltage bias the total signal is
• The total (DC ac) instantaneous current is
• Which we can re-arrange to get
• Substituting in the DC equation from above, we get

• If we keep the amplitude of the ac signal small,
  such that
• 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.
• This IS the small signal approximation, valid for
  amplitudes less than about 10mV
• 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

Where
44
Small Signal Resistance (incremental resistance)

• On the previous page we found
• And since,
• 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

iD (mA)
Bias Point - Q
ID
tangent at Q
vD (V)
0
0
VD0
The equation of the tangent line is given by
45
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)
• The total model has the components shown below
• The incremental voltage across the diode is

iD
vD -
ideal
VD0
tangent at Q
rd
Bias Point - Q
ID
vD (V)
0
0
VD0
VD
46
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

iDIDid
vDVDvd -
R
vs
VDD
iDIDid
vDVDvd -
R
vs
ideal
VD0
VDD
rd
47
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

Circuit for DC Analysis
Circuit for small-signal Analysis
ID
VD -
R
ideal
VDD
VD0
rd
48
Power Supply Ripple Example

• Example 3.6 The power supply has a 10V DC value
  and a 1V peak-to-peak 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 n2
• Calculate the dc diode current by assuming
  VD0.7V
• 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
• The peak-to-peak small signal voltage across the
  diode can be found using the ac model and the
  voltage divider rule
• This value is quite small and our use of the
  small signal model is justified

V10Vripple
R10kW
vd -
V10Vripple
R10kW
vd -
rd53.8W
49
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 1kW load resistance. Assume n2
• With no load the nominal dc current is given by
• Thus the dynamic resistance of each diode is
• The total resistance of the diodes will be 3rd or
  18.9W
• Using voltage division on the 1V p-p change (10)
  we get

10V 1V
R1kW
vo -
RL1kW
50
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.
• If it is assumed that the dynamic resistance
  stays the same, then the output small signal
  change is given by
• But when the dc current in the diode string is
  decreased the dynamic resistance changes
• This leads us to
• It appears that the small signal model is not
  entirely justified

51
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 diodes 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
52
Small-Signal Resistance Calculation and Model

• Exercise 3.20 Find the value of the diode
  small-signal resistance rd at bias currents of
  0.1, 1, 10mA (assume n1)
• Exercise 3.21 For a diode that conducts 1 mA at a
  forward voltage drop of 0.7V (with n1), find
  the equation of the straight line tangent at
  ID1mA.
• From the question above we find that rd is 25W
  and we can substitute in the bias point values
  and solve for VD0

53
Exercise 3.22, How small is small in the
Small-signal model

• Consider a diode with n2 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)

54
Diode Characteristic in the Reverse Breakdown
Region - Zener Diodes
i
VZ0
VZ

• 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
• The incremental (dynamic) resistance in reverse
  breakdown is given by rZ

VZKnee
v
IZKnee
Bias Point - Q
IZT
Test current
DI
DV
Circuit symbol for a Zener diode
IZ
VZ -
DVDI(rZ)
55
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

IZ
VZ -
VZ0
rz
intercept
56
A Shunt Regulator
i
Zener regulator
Load
IL
I
v
R
VO -
IZ
i
Zener regulator
Load
IL
I
v
R
VO -
IZ
57
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 20W, 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 kW
• (d) Find the value of VO when RL 0.5 kW
• (e) What is the minimum value of RL for which
  the diode still operates in the breakdown region.
• 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

V (10V 1V)
R 0.5 kW
vo -
6.8V zener
RL 1 kW
58
Zener example continued

• With no load connected, the current through the
  zener diode is given by
• We can now find V0, the voltage at the operating
  point current
• Now, for a /- 1V change in V can be found from
• When a load resistance of 2kW is connected, the
  load current will be approximately 6.8V/2000W 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,
• A more accurate result comes from analysis of
  this circuit

V
R 0.5 kW
vo -
RL 2 kW
59
Zener Example continued

• If we change the load resistance to 500W the load
  current would increased to 6.8V/500W 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
  cant 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.
• For the zener to be at the edge of the breakdown
  region, the current has to be IZIZK0.2mA and
  VZVZK6.7V. 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

60
Shunt Regulator
Zener regulator
IL
I
R
VO -
VS
IZ
VSmax
VO
VS
Load
reduced ripple
VSmin
t
t
IL
I
R
IZ
VO -
VZ0
VS
rz
61
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
  W.
• 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.

designing for Izmin(1/3)ILmax
62
Temperature Effects

• Temperature Coefficient (TC) is expressed in
  mV/degree C
• depends on Zener voltage and operating current
• For VZlt5V 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 -
VD -
63
Rectifier Circuits
Power transformer
IL



ac line 120V (rms) 60 Hz
vO
VO
-
-
-
t
t
t
t
t
64
Half-Wave Rectifier
Ideal
VD0
rD
D




vs
vs
R
vo
R
vo
-
-
-
-
vS
vo
VD0
v
VS
vo
t
VD0
vS
VD0
0
65
Full-Wave Rectifier with Center Tapped Transformer
D1


vs
R
vo
-
-

vs
D2
-
vS
-vS
vo
VD0
v
VS
vo
t
vS
VD0
-VD0
0
66
Full-Wave Bridge Rectifier

D4
D1
vo

-
vs
R
D2
D3
-
vS
-vS
2VD0
v
VS
vo
t