PN junction and diode

1. 1
Topic 3
PN Junction and Diode

2. 2
Semiconductor
• Semiconductor - materials with characteristics which fall
between insulator and conductors.
– It will not pass current as readily as a conductor nor will it block
current as effectively as an insulator.
– E.g. Germanium (Ge), Silicon (Si) and Carbon (C).
• These materials have atomic structures which may be easily
altered to obtain specific electrical characteristics.

3. 3
• The bandgap energy largely determines the
electrical properties of solids:
• EG ≥ 3eV Insulator
• EG ≥ 0.1 - 3eV Semiconductor
• EG = 0 Conductor

4. 4
Semiconductor atom
• All atoms have a nucleus where protons and neutrons are located. It carries
positive charges, which are surrounded by electrons orbiting in certain
shells around the nucleus.
• Electrons occupy energy bands (shells). The 1st shell can only hold 2
electrons, the 2nd 8 electrons, the 3rd 18 electrons, the 4th 32 electrons and
so on.
• The outermost shell of an atom is called the valence shell and the electrons
orbiting within this shell are called valence electrons.

5. 5
• When an atom contains exactly 4 electrons in its outer
shell, it does not readily give up or accept electrons.
– Elements which contain atoms of this type do not make good
conductors or insulators.
• They are called semiconductor.

6. 6
• Each atom shares its 4 valence
electrons with 4 neighbouring atoms.
– covalent bond.
• Each of the atoms in the structure tries
to take an additional electron to fill its
valence shell with 8 electrons to
achieve stability.
– Therefore each atom tends to be stable
and will not easily give up or accept
electrons.
• Due to the nature of the covalent bond
in semiconductor, semiconductor
materials have crystalline structure.

7. 7
Conduction in intrinsic (pure)
semiconductor
(1) Effect of Temperature
• The electrical characteristic of pure semiconductor material is highly
dependent upon temperature.
• At extremely low temperature, the valence electrons are held tightly
to their parent atoms.
– Therefore, the material cannot support current flow at this temperature.
– Pure Ge and Si crystal function like an insulator.
• As the temperature of a Ge and Si crystal is increased, the valence
electrons became agitated.
– Some with more energy will occasionally break away from the covalent bonds.
– They will be free to drift from one atom to the next in a random manner.
– These free moving electrons are able to support a small amount of electrical
current if a voltage is applied to the semiconductor.
– If they are exposed to extremely high temperature, they conduct as well as an
ordinary conductor.

8. 8
(2) Electron-Hole Pair
• When an electron receives enough energy, it breaks away from
a covalent bond, a vacancy called hole exist in the bond.
– A hole represents the absence of an electron.
– Since electron has a negative charge, the hole has the characteristics of
a positively charged particle.
– Each corresponding free electron and hole generated is referred to as an
electron-hole pair.
• A hole will exerts an attraction force on an electron and
therefore, will be filled by a nearby passing electron.
– This process is known as recombination and it causes a continued loss
of holes and free electrons.
• At any given temperature, the rate of recombination of
electron-hole pairs is always equal to the generation of new
electron-hole pairs so that the total number of free electrons
and holes is constant.

9. 9
(3) Current flow in pure semiconductor
• If a voltage is applied across a pure semiconductor, the free
electrons are attracted to the positive terminal of the voltage source
and the holes drift towards the negative terminal.
– The movement of particles is shown in the following diagram.

10. 10
Extrinsic (impure) semiconductor
• Pure semiconductor materials contain only a small
number of electrons and holes at room temperature.
• Impurities (pentavalent or trivalent materials) are added
to the semiconductor material through doping to
increase the conductivity of semiconductor.
• Doping creates two types of extrinsic semiconductor: n-
type and p-type.

11. 11
(1) n–type Semiconductor
• A pure semiconductor material doped with a pentavalent
materials.
– E.g. arsenic (As)
– the arsenic atom will replace one of the Ge or Si atoms and share 4 of
its valence electrons with adjacent atoms in a covalent bond.
– However, its 5th valence electron is loosely attached to the nucleus of
the arsenic atom and can be easily set free.
Note: no holes are generated when the 5th electron leaves the
valence band.
• Since the As atom is called a donor atom.
– There are many donor atoms within the semiconductor material, there
are also many additional free electrons within the material.
– Due to the presence of these extra electrons in the doped Ge and Si
crystal, the material is called an n-type semiconductor.

12. 12
• If a voltage is applied to an n–type semiconductor, the free
electrons will flow towards the positive terminal of the battery.
• The free electrons in the n–type come from two sources:
– Donor atoms (only electrons are generated, not electron-hole pairs) due
to doping process.
– Thermally generated electron-hole pairs by breaking of covalent bond.
• The holes move towards the negative terminal.

13. 13
• At normal room temperature the number of free electrons
provided by the donor atom will greatly exceed the number
of holes and electrons that are produced by the breaking of
covalent bonds.
• Therefore the total number of electrons flowing in n–type
semiconductor greatly exceeds the number of holes.
– Hence, electrons are called the majority carriers while holes are
called minority carriers.
• The material is still electrically neutral.

14. 14
(2) p–type Semiconductor
• A pure semiconductor material added with a trivalent material.
– E.g. Aluminium (Al).
– The Al atom shared its electron with 3 adjacent atoms in the crystal
structure but cannot share with the 4th adjacent atom.
– This creates a hole in the covalent bond.
– The trivalent atom can take/accept an electron, it is also called acceptor
atom.
– A hole created by this doping process is not accompanied by a free
electron.

15. 15
• Since many Al atoms are added, a large number of holes
will be present in the material.
– Holes can be thought of as positive charges because the absence of
an electron leaves a net positive charge on the atom.
– Holes are the majority carriers in p–type material.
– There are a few free electrons that are created when electron-pairs
are thermally generated.
– Electrons in p–type material are called minority carriers.

16. 16
The pn – junction
• A pn–junction is formed at
the boundary between the
two regions (n–type and p–
type semiconductors)
– A diode is created.
– It is the electrical
characteristics of the junction
rather than the semiconductor
sample material that
determines the electrical
behavior of the entire
structure.

17. 17
(1) Formation of the Space-Charge
Region (Depletion Region)
• At the instant the pn–junction is formed, the electrons near
the junction in the n region diffuse (because of high
concentration) across the junction into the p region and
combine with the holes near the junction.

18. 18
• Electrons leaving the n region leaves a layer of donor positive ions.
• When these electrons combine with the holes in the p region, a layer of
acceptor negative ions is formed.
• This region is known as space-charge region or depletion region. It is
depleted of charge carriers.
• The main purpose of a depletion region is to control current flow.

19. 19
(2) Barrier Potential
• In the depletion region, the forces between the negative ions
and positive ions form an electric field.
• This electric field prevents the majority carriers from one side
to cross over to the other side.
• Therefore, there is no net current flow in pn device.
– The potential difference across the electric field is known as barrier
potential.
– It is the amount of voltage required by the majority carriers to get
through the electric field. Hence, we need to apply a voltage source
across the pn junction that is greater than the barrier potential so that
electrons from n region can flow through the device.

20. 20
• The barrier potential depends on
– the type of semiconductor material,
– amount of doping and
– temperature.
• The minority carriers are aided by the electric field to cross the
junction.
– For example, minority carriers in the p region, that is, the electrons
which are near to the depletion region will be swept across by the
electric field to the n region and become the majority carrier there.
• By controlling the width of the depletion layer, we are able to
control the resistance of the pn junction and thus the amount of
current that can pass through the device.
– Bias is a potential applied to a pn junction to control the width of the
depletion layer.

21. 21
(1) Forward Bias
• A pn–junction is forward biased when the n–type is more
negative than the p–type.
• There are two ways in which a pn–junction can be forward
biased:

22. 22
• In forward bias mode and when VF >
barrier potential, the pn–junction
begins to conduct.
• The depletion region becomes
narrower.
• The majority carriers, for example,
electrons, from n side cross the pn–
junction and move into the p side
with little opposition and conduction
occurs.
• There is always a large majority of
electrons on the n side and holes on
the p side because the battery
replenishes them.
IF

23. 23
• Once a pn–junction begins to conduct, it provides a slight
opposition to current.
• This is the bulk resistance: it is the combined resistance of the
n–type and p–type in forward–biased mode.
• Since RB is extremely low (typically, 25 or less), very little
voltage drop across this resistance.
• Therefore, IR drop across RB is usually ignored in circuit
calculations.
npB RRR 

24. 24
• When a forward–biased pn–junction begins to conduct, the
forward voltage, VF across the junction is slightly greater
than the barrier potential.
• When the diode conducts, VF are approximated as
VF = 0.7V for Si
VF = 0.3V for Ge
• IF is the forward current.
• VF is the voltage drop across the diode.

25. 25
(2) Reverse Bias
• A pn–junction is reverse biased when the n–type is more
positive than the p–type.
• There are two ways in which a pn–junction can be reverse
biased:

26. 26
• Reverse bias and its effects are
illustrated in the following
figures.
• Electrons in the n–type will
head toward the positive
terminal of the source.
– This will further deplete the
electrons in the n–type near the
junction.
• The same principle applies to
the holes in p–type.
• The depletion layer is then
widened.
• This increases the resistance of
the junction and conduction
drops to near zero.

27. 27
• The increase in the depletion width is extremely conductive to
the minority carriers.
• The electrons in the p–type which are near to the depletion
region will be swept across the junction by the electric field
into the n region.
• Therefore, there is a current flow - the reverse current, IR.
• It is very small compare to IF.
• IR depends on temperature because the electron-hole pairs
(minority carriers) are generated from the breaking of covalent
bond which is temperature dependant.
– It is not affected by the magnitude of the reverse bias VR.

28. 28
PNJUNCTION DIODE
• A diode is forward biased even though a resistor, R, is
connected in series with it.
– Its p–type is connected to the positive terminal of the source and its n–
type to the negative terminal.
– The p–type is guaranteed to be more positive than the n–type.
• But whether the diode conducts or not, will have to depend on
the voltage drop across it.

29. 29
THE DIODE
• There are various types of diode but the ones we will be
studying are the pn–junction diode (normally, we just call
it “diode”) and zener diode.
• Diode conducts current in one direction (when it is in
forward bias mode) and blocks current in the other
direction (when it is in reverse bias mode).
• The symbol is shown here.

30. 30
• When reverse biased, the diode conduction drops to nearly
zero. What is the voltage drop across the diode?
– A diode does not conduct when the symbol points to the more positive
of the diode potentials.

31. 31
DIODE MODELS
• A model is a representation of a component.
• It is used to show the characteristics of the component
when the component is used in a particular application.
• The diode has 3 models. Each has its particular use in a
particular application.
– The ideal diode model.
– The practical diode model.
– The complete diode model.

32. 32
(1) The Ideal Diode Model
• This model represents the diode as a switch:
– close (forward bias) or open (reverse bias).
• When the diode is forward bias (closed switch), we can say
that:
– The diode has no resistance: it acts as a short circuit.
– There is no voltage drop across the diode.
• When the diode is reverse bias (opened switch), we can
say that:
– The diode has infinite resistance; therefore, it acts as an open
circuit.
– The applied voltage drops across the diode terminals.

33. 33
• Normally, the ideal
model of the diode is
used in the initial
stages of circuit
troubleshooting.
– We are only interested
in whether the diode is
acting as a one-way
conductor.
– If it is not, then the
diode is faulty and
needs to be replaced.

34. 34
Example
Solution
Since D1 is reverse biased, it acts as an open circuit.
Therefore,
IT = 0A
VD1 = VS
VR1 = 0A

35. 35
Example
Solution
Since D1 is forward biased, it acts as a short circuit.
Therefore,
VD1 = 0V
VR1 = 5V
IT = VR1 / R1 = 5 / 1000 = 5mA

36. 36
(2) The Practical Diode Model
• It is a bit more complex than the ideal diode model.
• The model includes characteristics
– forward voltage VF,
– peak reverse voltage VRRM,
– average forward current IO,
– forward power dissipation PD(max))
which are useful for mathematical analysis to help
determine which diode will be used for a given circuit.

37. 37
• VF is the voltage drop across the diode.
• The following figures show the diode characteristics curve.

38. 38
• IF suddenly increases when the voltage across the diode VF 
VK (knee voltage).
– Knee voltage is a general term to describe the situation where current
suddenly increase/decrease.
VK  0.7 V (Si)
– In an actual circuit, VK = 0.7V to 1.1V, depending on the current
through the device.
– As long as the diode is conducting, even if the applied voltage VS is
increased further, the voltage drop across the diode will remain VF =
VK.

39. 39
Example
Solution
D1 is forward biased. It acts as a short circuit.
VD1 = 0.7V
VR1 = 5 – 0.7 = 4.3V
IT = VR1 / R1
= 4.3 / 1000
= 4.3mA

40. 40
Peak Reverse Voltage VRRM
• When the diode is reverse-biased, the
maximum reverse voltage that will
not force the diode to conduct is
called VRRM.
• When VRRM is exceeded, the depletion
layer breaks down and the diode
conducts in the reverse direction.
– Any insulator will conduct if an applied
voltage is high enough to cause the
insulator to break down.
• The diode conducts in the reverse
direction when the VR > VRRM.
Normally, when a diode is forced to
conduct in the reverse direction, the
device is destroyed.

41. 41
• The reverse current in this case is known as the avalanche
current.
• Avalanche breakdown is the result of collisions of high –
energy minority carriers with atoms to break the covalent
bond.
– When the reverse bias is large enough, some of the minority carriers
will gain sufficient energy and when these carriers collide with the
atoms, covalent bonds are broken.
– So, more electron – hole pairs are generated and will gain high –
energy (due to the large reverse bias) to cause further broken covalent
bonds.
– As the process is repeated, the saturation current quickly increases and
breakdown takes place.
• VRRM is a very important limiting parameter.
• In practice, the VRRM should at least be 20% greater than the
maximum voltage that the diode is expected to block.
– The reason for 20% is because in practical applications, there will be
minor variations in the source amplitude.

42. 42
Example
During the negative alternation of the input, the peak value Vpk (50
V) is dropped across the reverse-biased diode.
Solution
VRRM = 1.2 VR(pk)
= (1.2) (50)
= 60V (minimum)
The VRRM rating for D1 must be 60 V or greater to ensure that
the component isn’t damaged by the voltage source.
As long as VRRM rating is 20% greater than the maximum
reverse voltage in a circuit, we are not concerned about the
diode’s exact value.

43. 43
Average Forward Current IO
• This rating indicates the maximum allowable value of dc
forward current.
• Again, IO must be taken as 20% greater than the forward
current IF.

44. 44
Example
Determine the minimum average forward
current rating that would be required for
the diode shown.
Solution
IF = (VS – 0.7) / RL
= (50 – 0.7) / 200 = 246.5mA
Therefore, D1 must have an IO > 246.5mA.
IO = 1.2 IF = (1.2) (246.5) = 295.8mA (minimum)
Note: when dealing with ac circuits, you will need to
determine the average (dc equivalent) current for the circuit.

45. 45
Forward Power Dissipation PD(max)
• This rating indicates the maximum possible power dissipation
of the device when it is forward biased.
Example
Calculate the minimum forward power dissipation rating for
any diode that would be used in the circuit shown.
Solution
IF = (VS – 0.7) / RL
= (10 – 0.7) / 100 = 93mA
PF = IFVF
= (93mA) (0.7) = 65.1mW
PD(max) = 1.2 PF
= (1.2) (65.1) = 78.12 mW (minimum)

46. 46
(3) The Complete Diode Model
• The complete diode model most accurately represents the
true operating characteristics of the diode.
• It includes characteristics e.g.
– bulk resistance RB,
– reverse current IR
that are meant for circuit development, high-frequency
analysis, etc.
• The model is used to explain many of the differences
between predicted and measured circuit values.

47. 47
Bulk Resistance RB
• The effect due to RB is that VF become non-constant.
– VF is constant in the practical diode model.

48. 48
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Diode Behaviour

49. 49
• The diode equivalent circuit looks as follows.
Note: VB and RB only exist when the diode is in
forward bias, not reverse bias.
• As IF increases, IFRB increases too.
– The total voltage across a diode varies with IF.
– The measured values of VF generally vary between 0.7 V and 1.1 V.
– In a low-current circuit, the VF will tend to be closer to 0.7 V because
lesser voltage drop across RB.
– In a high-current circuit, the VF will tend to be closer to 1.1 V.
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50. 50
Reverse Current IR
• In reverse bias, the diode is supposed to stop current flow.
• However, there is a small current IR that flows because of the
minority carriers.
• IR is made of two components
IR = IS + ISL
where IS = the reverse saturation current (It varies with
temperature)
ISL = the surface-leakage current (It varies with the
reverse voltage).

51. 51
• Since ISL << IS, it is safe to assume that IR  IS.
– This makes IR dependent on temperature as well.
– The voltage developed across any series resistance by IR is usually
insignificant. However, knowing it helps us to explain:
• Any voltage across the series resistance may increase dramatically
with increases in operating temperature.
• We often do not measure the full applied voltage across a reverse-
biased diode.

52. 52
Diode Capacitance
• When a diode is reversebiased,
it forms a depletion layer
(insulator) between two
semiconductor materials.
– It has some measurable amount of
junction capacitance.
• Junction capacitance is very
important in high-frequency
operation of a diode.
– There is a type of diode designed to
make use of its junction
capacitance: it is called a varactor.

53. 53
Temperature Effects on Diode
Operation
• An increased in temperature
means increased thermal activity
and decreased diode resistance.
– This is true for both forward bias
and reverse bias.
• As temperature increases, at a
fixed value of VF, IF increases.
• As temperature increases, at a
fixed value of IF, VF decreases.

54. 54
• As temperature increases, IR increases for all values of VR (<
VRRM).

55. 55
ZENER DIODE
• A pn–junction diode operated in
this region is usually destroyed
by the excessive reverse current
and the heat it produces.
– But the zener diode is designed to
work in the reverse breakdown
region.
• The reverse voltage across the
diode VR remains near-constant
voltage (called zener voltage,
VZ) despite a drastic change in
current flow through the diode.

56. 56
• Zener diode is useful as a voltage regulator.
– A voltage regulator is a circuit designed to maintain a relatively
constant voltage despite anticipated variations in load current or
input voltage.
• Zener diode is useful in the reverse bias mode, not in the
forward bias mode.
• There are two types of reverse breakdown: avalanche
breakdown and zener breakdown.
• The materials of the zener diode are designed in such a
way that the zener breakdown occurs at relatively low VR.

57. 57
• Zener breakdown is a type of reverse breakdown that
occurs at relatively low reverse voltages.
• The n–type and p–type materials of zener diode are heavily
doped, resulting in a narrow depletion layer.
• Therefore, the depletion layer can break down at a lower
VR than the depletion layer in a pn–junction diode.
• Zener diodes with low VZ ratings experience zener
breakdown, while those with high VZ ratings will
experience avalanche breakdown.

58. 58
Zener Operating Characteristics
• A Zener diode maintains a
near-constant VR for a
range of IR values.
• Zener knee current IZK is
the minimum current
required to maintain
voltage regulation
(constant voltage).
• To use the zener as a
voltage regulator, the
current through the diode
must never be lesser than
IZK.
VZT

59. 59
• The maximum Zener knee current IZM is the maximum
amount of current the diode can tolerate without being
damaged.
• Zener test current IZT is the current level at which the VZ
rating of the diode is measured.
– It is an important parameter in circuit design.
– If you need to have a zener voltage that is as close to the rated
value of VZ as possible, you need to design the circuit to have a
current value equal to IZT.

60. 60
• Zener impedance ZZ is the zener diode’s opposition to any
change in current.

61. 61
• There are two equivalent circuits for the zener diode.
• The ideal model considers the zener to be a voltage source
equal to VZ.
– When placed in a circuit, this voltage source opposes the applied
circuit voltage.

62. 62
• The practical model includes a series resistor, ZZ.
• This model is mainly used for predicting the response of the
diode to a change in circuit current.

63. 63
Example
A 1N4736 zener diode has a ZZ of 3.5. The data sheet gives
VZT = 6.8V at IZT = 37mA and IZK = 1mA. What is the voltage
across the zener terminals when the current is 50mA? When
the current is 25mA?
Solution
For IZ = 50mA,
ΔIZ = IZ − IZT = 13mA
ΔVZ = ΔIZZZ = (13mA) (3.5) = 45.5mV
The change in voltage due to the increase in current
above the IZT value causes the zener terminal voltage to
increase.
The zener voltage for IZ = 50mA is
VZ = 6.8 + ΔVZ = 6.8 + 0.0455 = 6.85V

64. 64
Example
A 1N4736 zener diode has a ZZ of 3.5. The data sheet gives
VZT = 6.8V at IZT = 37mA and IZK = 1mA. What is the voltage
across the zener terminals when the current is 50mA? When
the current is 25mA?
Solution
For IZ = 25mA,
ΔIZ = IZ − IZT = −12mA
ΔVZ = ΔIZZZ = (−12mA) (3.5) = −42mV
The change in voltage due to the decrease in current
below IZT value causes the zener terminal voltage to
decrease. The zener voltage for IZ = 25mA is
VZ = 6.8 − ΔVZ = 6.8 − 0.042 = 6.76V

65. 65
Zener Regulation with a Variable Load
• The zener diode maintains a nearly constant voltage across RL as long as the zener
current is greater than IZK and less than IZM.
– When the output terminals of the zener regulator are open (RL = ), the load current
IL is zero and all current is through the zener.
– When a load is connected, part of the total current is through the zener and part
through RL.
– When RL decreases, IL increases and IZ decreases.
• The zener diode continues to regulate the voltage until IZ reaches its minimum value IZK.
At this point IL is at maximum and the load is called a full load.

66. 66
Negative Series Clipper
• It eliminates the negative alternation of the ac input.
– When D1 is forward bias, VL = Vin – 0.7.
– When D1 is reverse bias, VD1 = Vin, VL = 0V.

67. 67
COMMON DIODE APPLICATION
CLIPPERS (LIMITERS)
• A clipper (or limiter) is a diode circuit that is used
to eliminate some portion of a waveform.
– For example, the half-wave rectifier is a clipper that
eliminates one of the alternations of an ac signal.
• There are two basic clippers:
– Series clipper: series clippers contain a diode in series
with the load.
– Shunt clipper: shunt clippers contain a diode in parallel
with the load.

68. 68
(1) Series Clippers
• The half-wave rectifier is actually a series clipper.
• The series clippers provide an output when the diode is
forward biased and no output when the diode is reverse biased.
– Two types: the negative and the positive series clippers.

69. 69
Positive Series Clipper
• Works the same way as negative series clippers with only
the following differences:
– The output voltage polarities are reversed.
– The current directions through the circuit are reversed.

70. 70
(2) Shunt Clippers
• The shunt clipper provides an output when the diode is
reverse biased and shorts the output signal to ground when
the diode is forward biased.

71. 71
VR(S) = Vin + 0.7 V

72. 72
Example
Determine the value for VL for each for the input
alternations.

73. 73
Solution
• When the input is positive, the diode is forward biased.
Thus, VL = 0.7 V.
VR(S) = Vin – 0.7 = 10V – 0.7 = 9.3 Vpk
• When the input is negative, the diode is reverse biased. Thus,
VL = (RL / RL + RS) Vin
= (1.2k / 1.4k)(–10V)
= − 8.45Vpk

74. 74
• What is the use of RS?
• It is used as a current-limiting resistor.
• During forward biased, the diode shorts the signal source
to the ground during positive alternation of the input
signal.
– The diode and the components in the signal source may be
damaged by the excessive forward current.
• Note: Make the RS much lower than the RL.

75. 75
(3) Biased Clippers
• The level of ac voltage in a series or a shunt clipper is
limited to approximately zero.
• In a biased clipper, the level to which an ac voltage is
limited can be adjusted by adding a bias voltage VB in
series with the diode.

76. 76
• In practice, a potentiometer is used to provide an adjustable
value of VB.

77. 77
(4) Clipper Applications
• They are generally used to perform one of the two
functions:
– Alter the shape of a waveform
– Provide circuit transient protection
• E.g. the half-wave rectifier is an example where the circuit
alters the shape of an ac signal, changing a sine-wave to a
pulsating dc.

78. 78
Transient Protection
• A transient is an abrupt current or voltage spike that has an
extremely short duration.
– A current surge is an example of a transient.
– Transients can damage circuits whose input must stay within certain
voltage or current limits.
• For example, digital circuits have inputs that can handle only
voltages that fall within a specified range.
– A voltage transient that goes outside the specified voltage range cannot
be allowed to reach the input.
– A clipper can be used in this case to prevent such a transient from
reaching a digital circuit.

79. 79
Digital
circuits
The transient occurs on the input line
It will forward bias D2, causing the diode to conduct
With D2 conducting, the transient will be shorted to ground,
protecting the input to circuit A.
Input square wave goes above 5V, D1 will be FB.
It shorts any input greater than 5V back to the +5V supply.

80. 80
• The figure shows a clipper designed to protect the output of
circuit B from transients produced when a square wave is used
to drive a speaker.
• A speaker is essentially a big coil (inductor).
– When a square wave is used to drive the coil in a speaker, that coil
produces a counter emf.
– This counter emf is produced each time that the output voltage makes
the transition from +5V to 0V. Its magnitude is greater than the original
voltage and opposite in polarity.
– Each time circuit B tries to drive the output line to 0V, the speaker tries
to force the line to continue to some negative value.
– This counter emf could destroy the driving circuit B.
– The clipping diode shorts the counter emf produced by the speaker coil
to ground before it can harm the driving circuit.

81. 81
AM Detector
• Another common application for the diode clipper is in a
typical AM receiver.
• Purpose: produce dc output voltage that varies with the peak of
the input signal.
• The input signal has an average value of 0V.
– This is because each positive peak has an equal and opposite negative
peak.
• The clipping diode is used to eliminate the negative portion of
the input waveform.
• With the negative portion eliminated, the capacitor charges
and discharges at the rate the peak input varies.

82. 82
CLAMPERS (DC RESTORER)
• A clamper is a diode circuit that is used to shift a
waveform either above or below a given reference voltage
without distorting the waveform.
• There are two types:
– positive clamper and
– negative clamper.
• A positive clamper shifts its input waveform so that the
negative peak of the waveform is approximately equal to
the clamper dc reference voltage.

83. 83

84. 84
• In the figure above, the clamper dc reference voltage is at 0V.
• The peak-to-peak value of the output waveform remains at
20Vpp but the 10Vpk becomes 20Vpk and the −10Vpk becomes
the clamper dc reference, 0V.
Note: The dc value for the output waveform is no longer the
same as the input waveform: Vin(dc) = 0 V and Vout(dc) = 10 V.
• In summary, the Vdc normally changes but the Vpp does not
when the waveform goes through the clamper circuit.
• For sine-wave, the dc (average) value falls halfway between
the peak to peak.
• A negative clamper does the opposite to that of the positive
clamper.

85. 85
(1) Clamper Operation
• The clamper works on the basis of switching time constants.
When the diode is forward biased, it provides a charging path
for the capacitor as shown.
• C1 is completely charged when TC = 5RD1C1
where RD1 is the bulk resistance of the diode.

86. 86
• When the diode is reverse biased, C1 discharge through RL. C1
is completely discharged when TD = 5RLC1.
• For the clamper circuit to work properly, TD >> TC.

87. 87
Consider a clamper has values RD1 = 10 , C1 = 1 mF and RL =
10 k. Therefore,
TC = 5RD1C1 = 5(10)(1 X 10-6) = 50 µs
TD = 5RLC1 = 5(10000)(1 X 10-6) = 50 ms
A square-wave is applied to the input of the clamper circuit.

88. 88
• When the input is at 5V, D1 is
forward biased.
• C1 charges quickly to 5 V. The
voltage drop across RL, VL = 0V.
• When the input is at −5V, D1 is
reversed biased.
• C1 will discharge through RL.
• The voltage drop across RL is the
sum of Vin and VC1 which is −10
V.

89. 89
Note: choose C1 so that it will discharge very slowly, to maintain a
constant −10 V.
• A clamper provides an output square wave that varies between 0 V
and −10 V.
• When designed properly, clampers work for a variety of input
waveforms, not just square-wave.
Input
Output

90. 90
(2) Biased Clampers
• Biased clampers shift a
waveform so that it falls
above or below a dc
reference other than 0 V.
• Figure shows a negative-
biased clamper.
– If VB is −20 V and R1 is set
so that the potential applied
to D1 is −10 V.
– The input waveform would
be shifted so that its positive
peak voltage is
approximately −10 V.

91. 91
• The positive-biased clamper
shifts its input waveform so
that the negative peak of the
output waveform is
approximately equal to the
potential applied to D1.
• The Zener diodes are used in
Zener clampers to set the dc
reference voltages of the
circuits.
– If let’s say a 10 V Zener is
used, the dc reference voltage is
approximately 10V.

92. 92
• Let’s say a 10 V Zener is used in
a positive zener clamper, the dc
reference voltage is
approximately −10 V.
• The operation of any clamper
depends on the one–way
conduction of the pn–junction
diode, not the Zener diode.
– Zener diode conducts in the forward
and reverse bias mode. Therefore, it
cannot be the diode that performs
the one-way conduction.
– During the positive alternation of
the input, the diode branch
conducts, clamping the positive
peak to (VZ + 0.7 V).
– During the negative alternation of
the input, D1 is reverse biased,
blocking conduction through the
diode branch.

93. 93
• Zener clampers come only in two varieties:
– Negative clampers with positive dc reference voltages
– Positive clampers with negative dc reference voltages
• We cannot have a positive zener clamper with a positive dc
reference voltage.
• If D1 is reversed, there is nothing to prevent the diode
branch from conducting during the negative alternation of
the input signal.
• The clamping action is therefore eliminated. This is not
desirable.

94. 94
VOLTAGE MULTIPLIERS
• A voltage multiplier produces a dc output voltage that is a
multiple of an ac peak input voltage.
– For example, a voltage doubler provides a dc output voltage that is
twice its peak input voltage.
• They are not power generators.
– This is because the input peak current is decreased.
• The device has a low current capability.
– The output current capability is reduced every time the voltage is
increased.
• Therefore, it is usually used in high-voltage, low-current
applications.
– For example, the cathode-ray tube in a television.

95. 95
(1) Half-Wave Voltage Doublers
• The half-wave voltage doubler consists of two diodes and two
capacitors.
• Electrolytic capacitors are normally used because of their high
capacity ratings.
• In practice, multiplier circuits are used in power supply
applications.

96. 96
• During the negative alternation of the input cycle, D1 is
forward biased, D2 is reverse biased.
• C1 charges to the peak value of input signal.
• C2 discharges through RL.

97. 97
• During the positive alternation of the input cycle, D1 is reverse
biased, D2 is forward biased.
• VS and C1 are effectively connected in series with C2.
• C2 charges to approximately 2VS(pk).

98. 98
• The time constant of the above circuit is such that C2
discharges very little across RL during the negative alternation
of the input.
• During the positive alternation of the input, C2 charges to
approximately 2VS(pk).
• In this way, the output waveform closely resembles that of a
filtered half-wave rectifier

99. 99
• The circuit has a dc output voltage and ripple voltage.
• The dc output voltage is approximated as Vdc  2VS(pk).
• The output ripple from the multiplier is calculated in the same
manner as for a filtered half-wave rectifier.
• The amount of ripple depends on the values of the capacitors
and the current demand of the load.
– High capacitor values and low current demand reduce the amount of
ripple.
• To get a negative voltage doubler, reverse the direction of the
diodes.
• The circuit output is a negative voltage, -2VS(pk).

100. 100
(2) Full-Wave Voltage Doublers
• During the positive half-cycle
of the input, D1 is forward
biased, D2 is reverse biased.
– C1 charges to VS(pk).
• When the input polarity is
reversed, D1 is reverse biased,
D2 is forward biased.
– C2 charges to VS(pk).
• Since C1 and C2 are in series,
the total voltage across the
two components is Vdc 
2VS(pk).

101. 101
BASIC POWER SUPPLY
CIRCUITS
• The power supply is a group of circuits that convert the ac
energy provided by the wall outlet to dc energy for our
daily use.

102. 102
• C3 is a filter capacitor. It is used to reduce the ripple at the
output.
– A full-wave voltage doublers is used as a dual-polarity power supply.
• A basic linear power supply can be broken into three circuit
groups:
– Rectification
– Filtering
– Voltage regulation

103. 103
TRANSFORMER
• Transformer is not a solid-state device, but they are closely
related to the operation of power supply.
The above figure shows the schematic for a transformer.
• Transformers are made up of inductors that are in close
proximity to each other, yet are not electrically connected.
• An alternating voltage applied to the primary induces an
alternating voltage in the secondary.
• Thus, a transformer provides ac coupling from primary to
secondary while providing physical isolation between the two
circuits.

104. 104
• Three types of transformer:
– Step-up.
– Step-down.
– Isolation: the output voltage is equal to the input voltage.
• The turns ratio is the ratio of the number of turns in the primary
to the number of turns in the secondary.
– The turns ratio is equal to the voltage ratio of the component.
where NS = the number of turns in the secondary,
NP = the number of turns in the primary,
VS = the secondary voltage, and
VP = the primary voltage
p
S
p
S
V
V
N
N


105. 105
• Ideally, transformers are 100% efficient. This means 100%
of the input power is transferred to the output.
PS = PP
• Therefore,
VSIS = VPIP
• From the formula,
– For a step-down transformer, IP < IS
– For a step-up transformer, IP > IS
• Current varies in the opposite way that voltage varies.
P
S
S
P
V
V
I
I

S
P
p
S
I
I
N
N


106. 107
RECTIFIER
• A rectifier is a diode circuit that converts the ac to
pulsating dc.
• There are 3 basic types of rectifier circuits:
– Half-wave rectifier
– Full-wave center-tapped rectifier
– Full-wave bridge rectifier (the most commonly used)

107. 108
(1) Half-Wave Rectifier
• Half-wave rectifier is a simple diode.
• It is connected in series between a transformer and its load.

108. 109
• The diode is used to eliminate either the negative or positive
alternation of the input.
– A sine-wave at the input, we should get a half-wave rectified output
waveform.
• The output will drop across RL.
– Therefore we can observe (using a scope) or measure (using a
multimeter) the output across RL, not the diode. The diode direction
determines which half-cycle is eliminated.

109. 110
Basic Circuit Operation
• During positive cycle of the input, the diode is forward
biased and provides a path for the current.
• Therefore, VD1  0V. VS will drop across RL (VL = VS).

110. 111
• During negative cycle of the input, the diode is reverse biased
and acts as an open.
• Therefore, VD1 = VS.
• No current flow in the circuit, so, no voltage developed across
RL, VL = 0V.

111. 112
Load Voltage and Load Current
• What we have discussed so far is based on an ideal diode.
Take VF into account, the peak load voltage, VL(pk) is given as
VL(pk) = VS(pk) – VF
• We can get VS(pk) from the transformer turns ratio formula.
NS / NP = VS(pk) / VP(pk)
• Usually, source voltages are given as rms values. Convert to
the peak value using
Vpk = Vrms / 0.707
• The peak load current is found as
IL(pk) = VL(pk) / RL

112. 113
Example
What is the peak load current for the circuit shown in the
figure with a 120 Vac input?

113. 114
Solution
• VP(pk) = VP(rms) / 0.707 =
• VS(pk) = (NS/NP) VP(pk) =
• VL(pk) = VS(pk) – VF =
• IL(pk) = VL(pk)/RL = 5.59mA

114. 115
Average Voltage and Average
Current
• The average voltage Vave is the dc equivalent value of the
ac waveform.
– In most cases, Vave and Vdc are used to describe the same value.
• For half-wave rectifier:
Vave = Vpk / 
Iave = Ipk / 

115. 116
Example
Determine the dc load voltage and current for the rectifier shown.
What role does the value of Iave play in component substitution?

116. 117
Solution
VS(pk) = 24 / 0.707 =
VL(pk) = VS(pk) – VF =
Vave = VL(pk) /  =
IL(pk) = VL(pk) / RL =
Iave = IL(pk) /  = 528.39 μA
The maximum dc forward current that can flow through a diode is equal
to the average forward current IO rating of the device. A faulty diode
from a rectifier may need to be substituted. Therefore, Iave can be
calculated as estimation or measured, so as to make sure that Iave < IO.

117. 118
Peak Inverse Voltage, PIV
• It is the maximum amount of reverse bias that will be
applied to a diode in a rectifier circuit.
• For a half-wave rectifier, the PIV is given as
PIV = VS(pk)
– When the diode is reverse biased, no voltage drop across RL.
– All of VS is dropped across the diode.
– Thus, PIV determines the VRRM of the diode.

118. 119
(2) Full-Wave Center-Tapped Rectifier
• This type of rectifier circuit requires the transformer to be
a center-tapped transformer.
• The transformer has a lead connected to the center of the
secondary winding.
– For a 24V center-tapped transformer, voltage from the center tap to
each of the outer winding terminals is 12V.
Center-tapped

119. 120

120. 121
Basic Circuit Operation
• The load voltage is approximately equal to half the secondary
voltage because the transformer is center tapped.
VL(pk)  VS(pk)/2

121. 122
Load Voltage and Load Current
• The peak load voltage is,
VL(pk) = [VS(pk)/2] – 0.7
• The full-wave rectifier produces twice as many output
pulses (per input cycle) as the half-wave rectifier.
• The average load voltage for a full-wave rectifier,
Vave = 2(Vpk/)

122. 123
Example
Determine the dc load voltage, the peak current and also
the dc load current for the circuit shown.

123. 124
Solution
• VS(pk) = 30 / 0.707 =
• VL(pk) = ½VS(pk) – VF =
• Vave = 2(VL(pk)/) =
• IL(pk) = VL(pk)/RL =
• Iave = Vave/RL = 2.56mA

124. 125
Peak Inverse Voltage PIV
• When one of the diodes is reverse biased, the voltage across
that diode is approximately equal to VS = 34 Vpk.
• The secondary voltages are +17 Vpk and 17 Vpk.
• Assuming D1 is ideal, then VD1 = 0V and the cathode of D1 is
at +17 Vpk.
• Since the cathode of D1 is connected directly to the cathode of
D2, the cathode of D2 is also at +17 Vpk.
• Since D2 is off, it will cause an open in the circuit. Therefore
VD2 = +34 Vpk.
• Therefore, the reverse voltage across either diode is
PIV = 2 VL(pk) = VS(pk)

125. 126
Full-Wave versus Half-Wave
• For a 24Vac (rated) transformer, the dc output voltages for the
two circuits are calculated as
Vave = 10.58 Vdc (half-wave rectifier)
Vave = 10.36 Vdc (full-wave rectifier)
• The two circuits produce nearly identical dc output voltages
the same values of transformer secondary voltage.
– Question: why bother with the full-wave rectifier when it has the same
dc output values as the half-wave rectifier?
• The reasons are
– If VL(pk) for the two circuits are equal, the full-wave rectifier will have
twice the dc load voltage and power efficiency of the half-wave
rectifier. We need to use a step-up transformer with turns ratio 1:2.
– The full-wave rectifier has twice the output frequency of the half-wave
rectifier, which has an impact on the filtering of the rectifier output.

126. 127
(3) Full-Wave Bridge Rectifier
• It has the following advantage:
– It does not require a center-tapped transformer and therefore, can be
directly coupled to the ac power line.
– It produces nearly double VL(pk) of the full-wave center-tapped rectifier.
So, we get a higher dc output voltage from the supply.

127. 128
• The current direction through the load does not change, nor
has the resulting polarity of the load voltage.

128. 129
Load Voltage and Load Current
• The center-tapped transformer is essential for the center-
tapped rectifier to work.
• But the output voltage in a center-tapped rectifier is reduced to
half the secondary voltage by the center tap on the transformer
secondary.
• The bridge rectifier does not require a center tap transformer to
work. It can be coupled directly to the ac line input, just like
the half-wave rectifier.
• The output voltage, across the load, is
VL(pk)  VS(pk) (ideal diode)
• For practical diode model,
VL(pk) = VS(pk) – 2(0.7)

129. 130
Example
Determine the dc load voltage and current values for the
circuit shown.
?
?
?
?
?

130. 131
Solution
• The peak secondary voltage is VS(pk) = 12 / 0.707 =
• The peak load voltage is VL(pk) = VS(pk) – 2VF =
• The dc load voltage is Vave = 2(VL(pk) / ) =
• The dc load current is Iave = Vave/RL = 825.8µA

131. 132
Example
• A center-tapped rectifier has a 12Vac transformer and a 12k
load. Determine the dc load voltage and current values for the
circuit.
• Use the values for comparison with the bridge rectifier.
• The table below summarizes the results of the two examples
above:
Value Bridge Rectifier Center-tapped Rectifier
VL(pk) 15.57V 7.79V
Vave 9.91V 4.96V
Iave 825.8µA 413.3μA

132. 133
Peak Inverse Voltage PIV
• Ideal diode :
PIV = VS(pk)
• Practical diode :
PIV = VS(pk) – 0.7
Why not VS(pk) – 2(0.7)?

133. 134
FILTER
• Aim of power supply –
produce a constant dc
output voltage.
• Filter: reduce variations
in the rectifier output
signal.
– A little bit of voltage variation
still exist after filtering 
ripple voltage, Vr.
• It is not desirable
• In audio amplifier, ripple can
produce an annoying “hum” at
60 Hz or 120 Hz.
• Ripple voltage depends
on
1.Type of rectifier
2.Filter component values
3.Load resistance
Vr
Half-wave rectifier
Filter

134. 135
Basic Capacitive Filter
• Most basic type and most commonly used.
• It is a capacitor connected in parallel with the
load resistance.
Diode is forward-biased
Capacitor charged to VS(pk)
Voltage decrease to zero (diode is
FB) & increase in negative direction
(Diode is RB).
Capacitor discharges through RL.

135. 136
• The capacitor charges through the diode.
– Let’s say the bulk resistance of diode is 5 , it takes 5RDC = 2.5
ms to fully charge the capacitor.
– The discharge path for the capacitor is through the load resistor.
– For the capacitor to fully discharge, it takes 5RLC = 500 ms.
• In other words, the capacitor charges instantaneously and discharges
very slowly.
– For a 60 Hz signal, the capacitor discharges for 16.7 ms before it
is provided with another charging voltage.

136. 137
• Vr can be minimized by increasing the
discharging time.
– Use a high capacitance filter together with a high
resistance load.
• However, the load of a power supply may
consist of any kind of circuit, each with its own
input resistance characteristics.
– It is impractical to vary RL in to control Vr.
• Choosing the suitable capacitance depends on
1. The amount of surge current Isurge that the rectifier
diodes can stand.
2. The cost of using a larger-than-required capacitance.

137. 138
Surge Current
• When the power supply is first turn on, the filter
capacitor acts as a short circuit.
• The current through the diode is only limited by
the resistance of the transformer secondary
winding, RW, and the bulk resistance of the
diode, RW, which are very low.
• The surge current,
Isurge = VS(pk) / [RW + RB]

138. 139
Example
• If the input of the circuit shown in the the previous slide is
170 Vpk, a turns ratio of 2:1, RW = 0.8 , RB = 5 , what it
the initial value of surge current for the circuit?
Solution
VS(pk) = (NS / Np) VP(pk) =
Isurge = VS(pk) / [RB + RW] =
= 14.66 A
This value may damage the diode. If the surge current is
too large for the diode, a resistor can be added to limit the
current as shown in the next slide.

139. 140
• For the circuit above, its surge current is given as
Isurge = VS(pk) / [RW + 2RB + Rsurge ]
• Adding the surge resistor limits the current to a
safe level for the diode to operate. However, this
reduces the voltage drop across the output.
– The value of the surge resistor, Rsurge, must be small
compare to the load resistor, RL.
• Another option is to use a lower capacitance
filter. It charges faster thus reducing the duration
of the surge.
– The trade off is greater ripple voltage.

140. 141
Filtered Output Voltages
• Output from a filter has peak, average (dc) and
ripple voltages.

141. 142
• The dc output voltage can be calculated as
Vdc = Vpk  (Vr / 2)
• The ripple voltage is
Vr = (ILt) / C
where
IL = Vdc / RL
IL = the dc load current
t = the time between charging peaks.
• Note the equations are in a loop. To solve this problem,
assume Vdc  Vpk.
• The ripple factor, r, is an indication of the effectiveness of the
filter.
r = Vr / Vdc

142. 143

143. 144
Example
• Determine the value of Vdc for the circuit shown.
The input voltage has a frequency of 60 Hz.
Vdc = 16.16 V
r = 0.015

144. 145
Filter Effects on Diode PIV
• The filter does not have significant effect on the
PIV across each diode of the full-wave rectifier.
• However, the PIV across the diode for a half-
wave rectifier is twice the secondary voltage.
PIV = 2VS(pk)

145. 146
Example
The zener diode shown has values of IZK = 3 mA
and IZM = 100 mA. What is the minimum
allowable value of RL?
RL(min) = 241 

146. 147
Voltage Regulation
• Voltage regulator is used to maintain a constant power supply
output voltage.
– There are many types, we will look at Zener diode.
• The zener current must be kept within the range
IZK < IZ < IZM

147. 148
• The current through RS
IT = (Vin  VZ) / RS
• The current through RL
IL = VZ / RL
• The current through D1
IZ = IT  IL
• To maintain zener regulation, IZ > IZK
IL(max) = IT  IZK
• The minimum load resistance RL(min)
RL(min) = VZ / IL(max)
• Likewise, the maximum load resistance RL(min)
RL(min) = VZ / (IT  IZM)

148. 149
• Zener diode reduces the amount of ripple voltage
present at the filtered output.
• To the ripple waveform, there is a voltage divider
present in the regulator. The voltage divider is
made up of RS series with ZZ||RL.
• The ripple output of the regulator is
  r
SLz
Lz
outr V
RRZ
RZ
V


||
||
)(

149. 150
Example
The filtered output from a full-wave rectifier has
a peak-to-peak ripple voltage of 1.5 V. What will
the ripple at the load equal?
0.129 Vpp

150. 151
• Zener regulator is rarely used. It wastes a large amount of
power.
• Ideally, a voltage regulator is able to maintain a constant dc
output voltage even though there are changes in the input
voltage of the load current demand.
– In practice, a change in the input voltage does cause a change in the
regulator’s output voltage.
• Line regulation indicates how much change occurs in the
output voltage for a given change in the input voltage.
Line regulation = Vout / Vin
– A 2 mV/V means that the output voltage will change by 2 mV for each 1
V change in the regulator’s input voltage.
• The smaller the line regulation value, the closer the regulator is
to the ideal case (0V / Vin).

151. 152
• A practical voltage regulator experiences slight change in
output voltage where there is a change in load current
demand.
• Load regulation specifies the change that occurs in the
output voltage over a certain range of load current values,
from minimum current (no load) to maximum current (full
load).
Load regulation = (VNL  VFL) / IL
= Vout / IL
– A 10 mV/mA means that the output voltage will change by 10 mV for
each 1 mA change in load current.