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 p n
The external circuit must limit the current
under Forward Bias conditions since the
diode will have no resistance
Reverse Bias
Forward 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
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 R=1kW, 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

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 VB=0 and the output V=0 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 We can write a node equation at node B,
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.
We can write a node equation at node B,
looking at the sum of the currents
Not possible, therefore
assumption was wrong B
Now assume D1 is off and D2 is on
Now solving for VB we get 3.33V and I=0 since D1 is off

10. Diode Terminal Characteristics
An Analog sweep has been converted to Digital (discrete values)
Rd = DV 1 I = DI DI
slope
slope =
5mA DV
Rd is the dynamic (changing) resistance
4mA DI
3mA
Breakdown
Voltage
Forward
Bias Va > 0
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
slightly negative
- 1mA
1.0
Reverse
Bias Va < 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
“Off”
R is
high
The resistance of the diode is not
constant, it depends on the polarity
and magnitude of the applied voltage
“closed”
switch
“on” R is low
Calculations Va

11. Diode Analogy IForward Ireverse Breakdown
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
Reverse the connections
to the diode, Va is being
applied to the n-side of the
diode Va

13. The Forward Bias Region
Forward-bias is entered when va>0
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 = Boltzman’s constant = 1.38 x joules/kelvin
T = the absolute temperature in kelvins = temp in C
q = the magnitude of electronic charge = 1.60 x coulomb
For appreciable current i, i >>IS, current can be approximated by
or alternatively p n Va I
1.0

14. The Reverse Bias Region
Reverse-bias is entered when va < 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 to 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 VZK I Va
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
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 (» ¥)

17. Semiconductors Tetrahedron 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 Semiconductors Bonds, 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 Al A SiO p n Cross-section of pn2 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 hole conc. current flow + + + + + + + + hole motion
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 e- vth vdrift
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 Field proportionality 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
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 + P N - X + -
n type
p type x 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) xp
+qNd +
Q n = qNdxn x
Q p = -qNaxp xn -
xd = xn + xp
-qNa
Electric field (x)
(Volts/cm) xp xn
Maximum Field (Emax )
Electrostatic
potential V(x) (Volts)
Vbi xp xn

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 Kirchhoff’s loop equation + VD - ID
VDD + 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

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)
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
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)
i (mA)
vR (V)
v (V)

33. i (mA) v (V) 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)

34. Iterative Analysis Converged to ID=4.237mA VD=0.762V
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
ID=4.237mA
VD=0.762V

35. A graphical view of the iterative analysis
i (mA) 2
4.3 mA
4.237 mA 3
END 1
1.0 mA
v (V)
0.7V
0.762V 3
START 1
0.763V 2

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)
B, slope =
A, slope = 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
vD (V)

38. Example
Exercise 3.16 For the circuit shown below, find ID and VD for VDD=5V and R=10kW. 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 VD0=0.65V and rD=20W, and (c) the constant voltage-drop model with VD=0.7V. ID + VD -
10kW
5 V

39. Example, continued
Iteration + VD -
5 V
10kW

40. Example, continued 10kW 0.65 V 5 V 20W + VD -
The piecewise-linear model
10kW
5 V + VD -
20W
0.65 V

41. Example, continued 10kW 0.7 V 5 V + VD -
The constant-voltage-drop model
10kW + VD -
0.7 V
5 V

42. DC Forward Bias with an ac small signal
iD (mA)
tangent at Q
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
0.55
0.65
0.75
vD (V)
iD(t)
(DC+ac) ac
VD0 +
vD(t) -
vd (t)
vd(t)
VD +
VD=0.7
(DC+ac) DC

43. Small Signal Analysis Where
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)
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)
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
iD=ID+id R +
vD=VD+vd - vs
VDD +
iD=ID+id +
vD=VD+vd - 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 + vd - vs id R rd 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 n=2
Calculate the dc diode current by assuming VD=0.7V
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
+V=10V+ripple
R=10kW + vd -
+V=10V+ripple
R=10kW + vd -
rd=53.8W

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 n=2
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
R=1kW + vo -
RL=1kW

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 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

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 n=1)
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 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 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)

54. Diode Characteristic in the Reverse Breakdown Region - Zener Diodes
VZ0 i
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 VZ
VZKnee v
IZKnee
Bias Point - Q
IZT
Test
current DI
Circuit symbol
for a Zener diode DV IZ + VZ -
DV=DI(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 v i v + VO - + VO - Zener regulator Load I IL R 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 - rz
VZ0
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 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.
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

60. Shunt Regulator VZ0 rz + VO - + VO - Zener regulator I IL R VS IZ VS
VSmax IZ VS VO
Load
reduced ripple
VSmin t t I IL 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 + VZ + - VD - + VD -
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 - + VD -

63. Rectifier Circuits IL Diode rectifier Filter Voltage Regulator Load +
Power
transformer IL
Diode
rectifier
Filter
Voltage
Regulator
Load + + +
ac line
120V (rms)
60 Hz vO VO - - - t t t t t

64. Half-Wave Rectifier vs vs vo vo D + + + + R R - - - - Ideal VD0 rD vo
VD0 vS

65. Full-Wave Rectifier with Center Tapped Transformer+ + vs R vo - - + vs D2 - vo vS
-vS
VD0 v VS vo t
-VD0
VD0 vS

66. Full-Wave Bridge Rectifier+ D4 D1 vo + - vs R D2 - D3 vS
-vS
2VD0 v VS vo t