1-Semiconductor Diodes
Majority and Minority Carriers
In an n-type material the electron is called the majority carrier and hole the minority carrier.
In a p-type material the hole is the majority carrier and the electron is the minority carrier
Fig1-1 (a) n-type material (b) p-type material
The region of uncovered positive and negative ions is called the depletion region due to the
depletion of carriers in this region.
Fig 1-2 p-n junction with no external bias
No Applied Bias
In the absence of an applied bias voltage, the net flow of charge in any one direction for a
semiconductor diode is zero
Fig 1-3 No-bias conditions for a semiconductor diode
Reverse-Bias condition (VD < 0V)
Fig1-4 Reverse-biased p-n junction
1
The current that exist under reverse-bias conditions is called the reverse saturation current and
is represented by IS
Fig1-5 Reverse-bias conditions for semiconductor diode
Forward-Bias condition (VD > 0V)
A semiconductor diode is forward-biased when the association p-type is positive and n-type is
negative has been established
Fig1-6 forward-biased p-n junction
Fig 1-7 Ideal diode (a) symbol; (b) characteristics
The characteristics of an ideal are those of a switch that can conduct current in only one
direction
Therefore the ideal diode is a short circuit for the region of conduction, and an open circuit in
the region of non conduction.
2
Fig 1-8 (a) Conduction (b) non conduction state of ideal diode
[1-1]
Fig1-9 Si semiconductor diode characteristics
Fig1-10 Forward-bias conditions for a semiconductor diode
3
Resistance Levels
DC or Static Resistance
[1-2]
Fig1-11 determining the dc resistance of a diode at particular
operating point, the lower the current through a diode the higher the dc resistance level
AC or Dynamic Resistance
Fig1-12 defining the dynamic or ac resistance
[1-3]
Fig1-13 determining the ac resistance at a Q-point
The lower the Q-point of operation (smaller current or lower voltage) the higher the ac
resistance
The derivative of a function of a point is equal to the slope of the tangent line drawn at
that point
4
[1-4]
[1-5]
Average AC Resistance
The resistance determined by a straight line drawn between the two intersections established
by maximum and minimum values of input voltage
[1-6]
Fig1-14 the average ac resistance
5
As with the dc and ac resistance levels, the lower the level of currents used to determine the
average resistance the higher the resistance level.
Summary Table
Table 1-1 Resistance Levels
Diode Equivalent Circuit
Table 1-2 Diode Equivalent Circuit (Models)
6
2-Diode Application
Load line analysis
Fig2-1 Series diode configuration (a) circuit (b) characteristics
Fig2-2 load line& finding the point of operating
Example1:
For the Fig2-3; Determine:
Fig2-3 (a) circuit (b) characteristics
7
Solution:
The intersection between the load line and the characteristic curve defines the Q-point as
shown in fig2-4:
Fig2-4 solution of Ex1:
Example2:
Repeat the analysis of Ex1: with R=2KΩ
Solution:
Note: the reduced slope and levels of diode current for the increasing loads
The Q-point is defined by:
Fig2-5 Solution of Ex2:
Example3:
Repeat Ex2: using approximate equivalent for the silicon semiconductor diode.
8
Solution:
Fig2-6 Solution of Ex3:
Diode Approximations
Table 2-1 Approximate and ideal semiconductor diode model
Series diode configurations with DC inputs
In general, a diode is in the on state if the current established by the applied sources is such
that its direction matches that of the arrow in the diode symbol, and VD≥ 0.7V for silicon and
VD ≥ 0.3V for germanium
Fig 2-7 Series diode configuration
9
Fig2-8 determining the state of the diode of fig2-7
Fig2-9 the equivalent model for the on diode of fig2-7
Due to the open circuit, the diode current is 0A and the voltage across the resistance is:
(a)
(b)
(c)
Fig2-10 (a) Reversing the diode of fig2-7 (b) the state of diode (c) the equivalent model of the
off diode of (a)
Example4:
Solution:
Example5:
Repeat Ex4: with the diode reversed
Solution:
10
Example6:
Determine ID, VD, and VO for the circuit of fig
Solution:
Example 7:
Determine I, V1, V2 and VO for the circuit of fig
Solution:
11
Parallel and series-parallel configurations
Example8: Determine I 1, ID1, ID2 and VO for the circuit of fig
Solution:
Example9: Determine the current I for the network of fig
Solution:
Diode switching circuit
Diode switching circuits typically contain two or more diodes, each of which is connected to an
independent voltage source. Understanding the operation of a diode switching circuit depends
on determining which diodes, if any, are forward biased and which, if any, are reverse biased.
The key to this determination is remembering that is a diode is forward biased only if its
anode is positive with respect to it's cathode.
Fig (2-11)
One of the very import applications of diode switching circuits is diode logic circuits AND/OR
Gates
12
OR gate: is such that the output voltage level will be a 1 if either or both input is a 1. The
10V level is assigned a 1 for Boolean algebra while the 0V input is assigned a 0
Example10: Determine Vo for the network in fig (2-12)
Fig (2-12)
Fig (2-13)
D1 is in the on state due to the applied voltage (10V) while D2 is in the off state
.
Vo = E - VD = 10v - 0.7 = 9.3v
= (E - VD ) / R = (10-0.7) / 1KΩ = 9.3mA
The output voltage level is not 10V as defined for an input of 1, but the 9.3V is sufficiently at a
1 level with only one input.
AND gate: is such that the output voltage level is will be 1 if both inputs are a 1.
Example11: Determine the output level for the positive logic AND gate of fig (2-14).
Fig (2-14)
Fig (2-15)
With 10v at the cathode D1 , is assumed that D1 is in the off state.
D2 is assumed to be in the on state due to the low voltage at the cathode side and the
availability of the 10v source through 1KΩ resistor.The voltage at Vo is 0.7v due to forward
i.e. I = (E-Vo) / R = (10-0.7) / 1KΩ = 9.3mA
biased diode D2
13
Half-Wave Rectifier
Half-wave rectification: is the process of removing one half of the input signal to establish
a dc level.
The cct of the fig(2-16) called a half wave rectifiers will generates a waveform Vo that will have
an average value of particular use in the ac– to–dc conversion process.
Fig (2-16) half wave Rectifier.
During the interval (t=0 to T/2) the polarity of the input voltage Vi is shown in fig (2-17).
Fig (2-17) conducting region (0 toT/2)
The result that for period 0 to T/2,
Vo=Vi.
For period T/2 to T, the polarity of the input voltage Vi is shown in fig(2-18 ) and the ideal diode
produces in off state, Vo=0V.
Fig (2-18) Non conducting region (T/2 to T)
An average value determined by average value determined by average dc value =
0.318Vm
Fig (2-19) half wave rectified signal.
14
The effect of using a silicon diode with VT=0.7V is shown by fig (2-20) for the forward bias. The
input must now be at least 0.7V before the diode conducts.
When conducting Vo =Vi - VT
If Vm > VT …
if Vm is close to VT …
i.e. Vdc= 0.318Vm
i.e. Vdc ≈ 0.318(Vm – VT)
Fig (2-20) Effect of VT on Half-wave rectified signal.
Eexample 1: for Half wave Rectifier:
(a) Sketch the output vo and determine the dc level of the output for the network of Fig (2-21),
(b) Repeat part (a) if the ideal diode is replaced by a silicon diode.
Fig (2-21)
Solution:
(a) In this situation the diode will conduct during the negative part of the input and vo will
appear as shown in fig (2-22)
Fig (2-22)
For the full period, the dc level is
Vdc = -0.318Vm = -0.318(20) = -6.36 V
(b) Using a silicon diode, the output has the appearance of Fig (2-23)
Fig (2-23)
Vdc = -0.318(Vm - 0.7) = -0.318(19.3) = -6.14 V
15
Full-Wave Rectification
The dc level obtained from a sinusoidal input by half wave rectifier can be improved using a
process called Full-Wave Rectification. Four diodes in a bridge configuration can be used as
Full Wave Rectifier as shown in fig (2-24).
Fig (2-24) Full-wave bridge rectifier
Fig (2-25) Network for period (0 to T/2)
For the positive region of the input the conducting diodes are D2 & D3 while D1& D4 are in the
off state as shown in fig (2-25).
Fig (2-25)
For the negative region of the input the conducting diodes are D1 & D4 while D2 & D3 are in
the off state as shown in fig (2-26).
Fig (2-26)
The dc level for Full wave rectifier is twice that obtained for a half wave system
i.e. average(d.c) level= 0.636Vm
Over one full cycle the input and output voltage is shown in fig (2-27)
Fig (2-27)
The effect of Vo has also doubled, as shown in fig (2-28) for silicon diodes during the
conduction state (for positive region).
16
i.e. Vd.c=0.636Vm (Vm >> 2VT)
And if Vm is close to 2VT
i.e. Vd.c=0.636(Vm- 2VT)
A second popular full wave rectifier used only two diodes but requiring a centre tapped (CT)
transformer to establish the input signal across each section of the secondary of the
transformer as shown in fig (2-28).
Fig (2-28)
During the positive portion of Vi applied to the transformer, the diode D1 is short circuit and
the diode D2 is open circuit.
Fig (2-29)
During the negative portion of Vi applied to the transformer, the diode D1 is open circuit and
the diode D2 is short circuit as shown in fig (2-30).
Fig (2-30)
Example 2: for Full-wave rectifier.
Determine the output wave-form for the network of Fig (2-31) and calculate the output dc level
Fig (2-31)
17
Solution:
The network will appear as shown in Fig (2-32) for the positive region of the input voltage,
Fig (2-32)
Where
vo=1/2vi or
Vo(max) = 1/2Vi(max) = 1/2(10) = 5 volt
For the negative region of the input voltage the network will be appear as shown in Fig (2-33).
Fig (2-33)
The effect of removing two diodes from the bridge configuration was therefore to reduce the
available dc level to the following:
Vdc = 0.636(5) = 3.18 volt
Clippers
Clippers: is the network that has the ability to clip off a portion of the input signal without
distorting the remaining part.
1-Series clipper: The diode is in the series with load as shown in fig (2-34).
Fig (2-34) Series clipper
The addition of a dc supply can have a clear effect on the output of a clipper
.
Fig (2-35)
18
1- Determine the diode is an open or short circuit (off or on state).
2-Determine the applied voltage (transition voltage) that will cause a change in state for the
diode. Applying the condition:
id = 0 at vd = 0
The level of vi that will cause a transition in state is:
vi = Vdc
For an input voltage greater than Vdc volts the diode is in the short-circuit state, while for
input voltages less than Vdc volts it is in the open-circuit or (off state).
Fig (2-36)
3-Defined terminals and polarity of vo when the diode is in the short circuit state, the output
voltage vo can be determined by using KVL.
Fig (2-37)
vi - V - vo = 0
And
(CW direction)
vo = v i -V
4-Sketch the input signals above the output and determine the output at instantaneous values
of the input by using the formula for each case.
Keep in mind that at an instantaneous value of vi the input can be treated as a dc supply of
that value and the corresponding dc (the instantaneous) value of the output determined.
19
Fig (2-38)
Eexample 1: Determine the output waveform for the network of Fig (2-39)
Fig (2-39)
Solution:
The diode will be in the on state for the positive region of vi , aiding effect of V = 5 volts.
Vo = vi + 5.
Fig (2-40)
20
Substituting id = 0 at Vd = 0 for the transition voltage, we obtain the network of Fig(2-41)
Fig (2-41)
Vi = -5 V.
For voltages more negative than -5 V the diode will enter its open-circuit state, while for
voltages more positive than -5 V the diode is in the short-circuit state as in Fig (2-42).
Fig (2-42)
Example2:
Repeat Example 1: for the square-wave input of fig (2-43).
Fig (2-43)
Solution:
For vi = 20 volt (0 to T/2) the network of Fig (2-44) will result.
Fig (2-44)
The diode is in the short circuit state (on state)
And
Vo = 20 + 5 = 25 volt.
21
For vi = -10 volt the network of Fig (2-45) will result,
Fig (2-45)
Placing the diode in the open circuit state (off state),
And
Vo = iR R = (0) R = 0 volt.
The resulting output voltage appears in Fig(2-46)
Fig (2-46)
Note: the clipper not only clipped off 5volt from the total swing but raised the dc level of the
signal by 5volts.
2-parallel clipper: The diode is in the parallel to the load as shown in fig (2-47).
Fig (2-47)
Example 1: Determine vo for the network of Fig (2-48)
Fig (2-48)
Solution:
The polarity of the dc supply and the direction of the diode suggest that the diode will be in the
(on state) for the negative region of the input signal. Fig (2-49)
Fig (2-49)
22
Where the defined terminals for vo require that
vo = V = 4 volts.
The condition id = 0A at vd = 0 V has been imposed. The result is:
vi (transition) = V = 4 volts.
Fig (2-50)
The input voltage must be greater than 4 volt for the diode to be in the (off state).
Any input voltage less than 4 volt will result in a short-circuited diode (on state).
For the open-circuit state the network will appear as shown in Fig (2-51),
Fig (2-51)
Where
v o = vi .
Completing the sketch of vo results in the waveform of Fig (2-52).
Fig (2-52)
Example2:
Repeat Example 1 using a silicon diode with VT = 0.7 volts.
Solution:
The transition voltage can first be determined by applying the condition
id = 0 A at Vd = VD = 0.7 volts
23
Fig (2-53)
Applying KVL around the output loop in the clockwise direction, we find
Vi + VT - V = 0
And
Vi = V - VT = 4 - 0.7 = 3.3 volts
For input voltages greater than 3.3 volts, the diode will be an open circuit and
Vo = Vi.
For input voltages less than 3.3 volts, the diode will be in the (on state) Fig (2-54) where
Vo = 4 - 0.7 = 3.3 volts
Fig (2-54)
The resulting output waveform appears in Fig (2-55) Note that the only effect of VT was to drop
the (on state) level to 3.3 volt from 4 volt.
Fig (2-55)
24
A variety of series and parallel clippers with the resulting output for the sinusoidal input are
provided in Fig (2-56)
.
Fig (2-56)
25
Clampers
The clamping network will clamp a signal to a different dc level. The network must have a
capacitor, a diode, and a resistive element, the magnitude of R & C must be chosen such that
the time constant (t=RC) is large enough to ensure that the voltage across the capacitor does
not discharge during off state of the diode. We will assume that the capacitor will fully charge
or discharge in five time constant (t).
Fig (2-57)
During the interval (0 to T/2) the diode is in short circuit (on state) shorting out the effect of the
resistor R, thus the capacitor will charge to V very quickly and Vo = 0, as shown in fig (2-58),
Fig (2-58)
During the interval (T/2 toT) the diode is in open circuit (off state) now R is back in the network,
Fig (2-59)
The capacitor holds the charge and therefore voltage (V=Q/C). Thus by using KVL:
- V – V – Vo = 0
Vo= -2V
The output signal is as shown in fig (2-60)
Fig (2-60)
For clamping networks the total swing of the output is equal to the total swing of the
input signal.
The following steps may be used for analyzing clamping networks:1- Start the analysis from the period of the input signal that will forward bias the diode
2-During this period assume that the capacitor will charge up to a level determined by the
voltage across the capacitor in its equivalent open-circuit state.
26
3-Assume that during the period the diode is an open circuit off state the capacitor will hold on
its charge and therefore voltage.
4-Applying KVL to determine Vo for both state on & off.
5-The general rule that the swing of the output must match the swing of the input signal
Eexample 1: Determine vo for the network of Fig (2-61) for the indicated input.
Fig (2-61)
Solution:
T=1/f = 1/1000Hz=1ms
and 1/2 T = 0.5 ms
The analysis will begin with the period t1 to t2 of the input signal since the diode is in it's shortcircuit state .For this interval the network will appear as shown in Fig (2-62)
Fig (2-62)
The output is across R, but it is also directly across the 5volts battery.
The result is vo = 5 volts for this interval.
Applying KVL around the input loop will result in
-20 + Vc - 5 = 0
Vc = 25 volts
The capacitor will therefore charge up to 25 volt, as stated in comment 2.
For the period t2 to t3 the network will appear as shown in Fig (2-63).
Fig (2-63)
The open-circuit equivalent for the diode will remove the 5volt battery from having any effect on
vo, and applying KVL around the outside loop of the network will result in
+ 10 + 25 - Vo = 0
Vo = 35 volts
The time constant of the discharging network is determined by the product RC:
τ = RC = (100 kΩ) (0.1 µF) = 0.01 s = 10 ms
The total discharge time is therefore …5τ = 5(10 ms) = 50 ms.
Since the interval t2 to t3 = 0.5 ms, i.e. the capacitor will hold its voltage during the discharge
27
period between pulses of the input signal.
The resulting output appears in Fig (2-64) with the input signal.
Fig (2-64)
Example 2: Repeat Example 1 using a silicon diode with VT = 0.7 volts.
Solution:
For the short-circuit state vo can be determined by KVL in the output section.
+5 - 0.7 - vo= 0
vo= 5 - 0.7 = 4.3 volts
Fig (2-65)
For the input section KVL will result in
-20 + Vc + 0.7 - 5 = 0
Vc = 25 - 0.7 = 24.3 volts
For the period t2 to t3 the network as in Fig (2-66), Applying KVL
Fig (2-66)
+ 10 + 24.3 - Vo= 0
Vo= 34.3 volts
The resulting output appears in Fig (2-67).
Fig (2-67)
28
A number of clamping circuits shown in fig (2-68)
29
Zener Region
When the applied reverse potential becomes more and more negative, a few free minority
carriers have developed sufficient velocity to liberate additional carriers through ionization.
When VZ decreases to very low levels,this mechanism, called Zener breakdown, will contribute
to the sharp change in the characteristic. It occurs because there is a strong electric field in the
region of the junction that can destroy the bonding forces within the atom and "generate"
carriers. Diodes employing this unique portion of the characteristic of a p-n junction are called
Zener diodes.
Fig (2-69) Comparison of Si and Ge semiconductor diodes
Zener Diode
The Zener diode is a device that is designed to make full use of this Zener region. zener region
occurs at a reverse bias potential of VZ.
Fig (2-70) Zener diodes (a) Zener potential (b) characteristic and notation
Any voltage from 0 to VZ will result in an open-circuit as occurred below VT for the silicon
diode. Silicon diode maintains its open-circuit in the reverse-bias region but the Zener diode
assumes a short-circuit state once the reverse voltage is reached (i.e. Vd=VZ for short
circuit ). Zener device switch from the open-circuit state to short circuit
Fig (2-71)Zener equivalent cct(a)complete(b) approximate.
30
Zener Diode Applications
1- AC Voltage Regulators (limiters or Clippers)
Two back-to-back zeners can be used as an ac regulator or a simple square-wave
generator as shown in examples below:
a- Sinusoidal ac regulator
Fig (2-72)
Fig (2-73)
b- Simple square-wave generator
Fig (2-74)
2- DC Voltage Reference
Two or more levels can be established by placing zener diodes in series as shown in fig (2-75)
as long as vi (E) is greater than the sum of Vz1 and Vz2, both diodes will be in breakdown
state(on state) and the three reference voltages will be available
Fig (2-75)
31
3- DC Voltage Regulators
a- Fixed Vi, Variable RL
Fig (2-76)
[2-1]
[2-2]
[2-3]
Example 1:
(a) For the network of Fig (2-77), determine the range of RL and IL that will result in VRL being
maintained at 10 V.
(b) Determine the maximum wattage & rating of the diode as a regulator.
Fig (2-77)
Solution:
(a)To determine the value of RL that will turn the Zener diode
32
A plot of VL versus RL appears in Fig (2-78a) and VL versus IL in Fig (2-78b)
Fig (2-78)
b- Fixed RL, Variable Vi
For fixed value of RL in Fig (2-79) the voltage Vi must be sufficiently to turn the Zener diode
on. The turn-on voltage is determined by:
Fig (2-79)
[2-4]
[2-5]
[2-6]
Example 2:
Determine the range of values of Vi that will maintain the zener diode in the (on state).
Fig (2-80)
Solution:
33
A plot of VL versus Vi is provided in fig (2-81)
Fig (2-81)
34
Regulators (Filters and Power Supplies)
Voltage regulator provide a fixed output voltage
The operation is converting an ac voltage into a dc voltage using transformer, rectifier, and
filter
Fig (2-82) Block diagram showing parts of a power supply.
1-The ac voltage, typically 120 volts rms, is connected to a transformer which steps that
voltage up, or down to the level for the desired dc output.
2-A diode rectifier then provides a half wave or, more typically, full-wave-rectified voltage
which is applied to a filter to smooth the varying signal.
A rectifier circuit is necessary to convert a signal having zero average value to one that has a
nonzero average. The resulting dc signal is not pure dc or even a good representation of it
3-A simple capacitor filter is often sufficient to provide this smoothing action.
4-The resulting dc voltage with some ripple or ac voltage variation is then provided as input
to an IC regulator
5-IC regulator that provides as output a well-defined dc voltage level with extremely low ripple
voltage over a range of load
Ripple Voltage.
The filter output voltage of fig (2-83), has a dc value and some ac variation (ripple).
Fig (2-83) Filter voltage wave form showing dc & ripple voltages
Consider measuring the output voltage of the filter circuit using a dc voltmeter and an ac (rms)
voltmeter.
Example 1 :
Using a dc and ac voltmeter to measure the output signal from a filter circuit, a dc voltage of
25V and an ac ripple voltage of 1.5V rms are obtained. Calculate the ripple of the filter output.
35
Solution:
Voltage Regulation
Another factor of importance in a voltage supply is the amount of change in the output dc
voltage over the range of the circuit operation.
This voltage change with respect to either the loaded or unloaded is described by a factor
called voltage regulation.
Example 2:
A dc voltage supply provides 60V when the output is unloaded. When full-load current is drawn
from the supply, the output voltage drops to 56V. Calculate the value of voltage regulation
Solution:
The output voltage from most supplies decreases as the amount of current drawn from the
voltage supply is increased.
The smaller the voltage decreases, the smaller the percent of V.R. and the better the
operation of the voltage supply circuit.
Ripple Factor of Rectified Signal
The rectified voltage is not a filtered voltage; it contains a dc component and a ripple
component. By calculate these components we obtain the ripple factor.
1-Half wave Rectified Signal
Since the ac voltage component of a signal containing a dc level is
Vr(rms) is the rms value of the total voltage. For the half wave-rectified signal,
36
The ripple factor of a voltage is defined by:
This can be expressed as
For a half-wave-rectified signal the output dc voltage is:
Vdc = 0.318 Vm
2-Full wave Rectified Signal: For the full-wave-rectified signal
For the full-wave rectifier the value of Vdc is:
Vdc = 0.636 Vm
Simple-Capacitor Filter
A popular filter circuit is the simple-capacitor filter circuit, the capacitor is connected across the
rectifier output this filtered voltage has a dc level with some ripple voltage riding on it .
Fig (2-84) Simple capacitor filters
.
Fig (2-85) capacitor filter operation (a) full-wave rectifier voltage (b) filtered output voltage.
37
Fig2-86(a) capacitor filter circuit (b) output voltage waveform
If no load were connected to the filter, the output waveform a constant dc level equal in value
to the peak voltage Vm from the rectifier circuit.
For the full-wave-rectified signal indicated in Fig (2-86b)
T1 is the time during which a diode of the full-wave rectifier conducts and charges capacitor
up to the peak rectifier output voltage Vm .
T2 is the time during which the rectifier voltage drops below the peak voltage, and the
capacitor discharges through the load.
The capacitor filter circuit provides a large dc voltage with little ripple for light loads and a
smaller dc voltage with larger ripple for heavy loads.
Fig (2-87a) Approximate output voltage of capacitor filter circuit.
Figure (2-87a) shows the output waveform approximated by straight line charge and discharge.
From an analysis of this voltage waveform the following relation can be:
Since the form of the ripple waveform for half-wave is the same as for full-wave
38
Ripple Voltage Vr(rms)
Assuming a triangular ripple waveform approximation as shown in fig (2-87b)
Fig (2-87b) Approximate triangular ripple voltage for capacitor filter
Fig (2-87c) Ripple voltage
During capacitor-discharge the voltage change across C is:
39
f = frequency of the sinusoidal ac power supply voltage usually 60 Hz
Idc = average current drawn from the filter by the load,
C = filter capacitor value.
For light loads Vdc ≈ Vm
f = 60 Hz
Idc in mA , C in μf, RL in kΩ.
Example 3:
Calculate the ripple voltage of a full-wave rectifier with a 100 µF filter capacitor connected to a
load of 50 mA.
Solution:
DC Voltage, Vdc:
f = 60 Hz
Vm = the peak rectified voltage, V
Idc = the load current, mA
C = the filter capacitor, μf
Example 4:
The peak rectified voltage for the filter circuit of Ex 3: is 30 volts, calculate the filter dc voltage.
Solution:
Note:
The larger value of average current drawn from the filter the less value of output dc voltage
The larger the value of the filter capacitor the closer the output dc voltage approaches the
peak value of Vm.
40
Filter-Capacitor Ripple
Using the definition of ripple, we obtain the expression for the ripple factor of a full wave
capacitor filter
Since Vdc and Idc relate to the filter load RL , we can also express the ripple as
Idc in mA
C in μf,
Vdc in volts
RL in kΩ.
larger load current, larger ripple factor and inversely with the capacitor size.
Example 5:
a load current of 50mA is drawn from a capacitor filter circuit C=100 µF.if the peak rectified
voltage is 30V, calculate r.
Solution:
Using results of Ex 3 and Ex4, we get
RC Filter
It is possible to further reduce the amount of ripple across a filter capacitor while reducing
the dc voltage by using an additional RC filter section as shown in Fig (2-88)
Fig (2-88) RC filter stage
The purpose of the added network is
1-to pass the dc component of the voltage developed across the first filter capacitor Cl .
2-and to attenuate the ac component of the ripple voltage developed across Cl .
This action would reduce the amount of ripple in relation to the dc level, providing better filter
operation than for the simple-capacitor filter.
Since the rectifier feeds directly into a capacitor, the peak currents through the diodes are
many times the average current drawn from the supply.
41
The voltage developed across capacitor Cl is then further filtered by the resistor-capacitor
section (R,C2) providing an output voltage having less percent of ripple than that across Cl.
The load RL draws dc current through resistor R with an output dc voltage across the load
being less than that across Cl due to the voltage drop across R.
Fig (2-89) Full wave rectifier and RC filter circuit.
DC Operation of RC Filter Section
The equivalent circuit used for considering the dc voltage and current in the filter and load
shown in fig (2-90). The two filter capacitors are open circuit for dc and
Fig (2-90) (a) dc equivalent circuit (b) ac equivalent circuit
Fig (6-90a) show that the voltage Vdc across capacitor Cl is attenuated by a resistor-divider
network of R and RL , the resulting dc voltage across the load being V'dc
Example 6:
The addition of an RC filter section with R=120Ω, reduces the dc voltage across the initial filter
capacitor from 60 V (Vdc). If the load resistance is 1kΩ, calculate the value of the output dc
voltage (Vdc) from the filter circuit.
42
Solution:
The drop across the filter resistor and the load current drawn:
AC Operation of RC Filter Section
Fig (2-90b) is the equivalent circuit for analyzing the ac operation of the filter cct.
Vr (rms) is the input to the filter stage, it is a ripple or ac signal part of the voltage across C1.
Both the RC filter stage R, C2 and the load resistance RL affect the ac signal at the output.
For C2 = 10µF at a ripple voltage frequency f = 60Hz, the ac impedance of the capacitor is:
This capacitive impedance is in parallel with the load resistance RL
If RL= 2 kΩ, the parallel combination of the two components would an impedance Z
This is close to the value of the capacitive impedance (Xc) alone.
Note:
In parallel combination we can neglecting the loading RL if RL > 5Xc
The ripple frequency = 60 Hz for the ripple voltage from a half-wave rectifier.
The ripple frequency = 120 Hz for the ripple voltage from a full-wave rectifier
Xc = 1/ wC
We have value of w = 377 for 60 Hz and of w = 754 for 120 Hz.
Example7:
Calculate the impedance of a 15µF capacitor used in the filter section of a circuit using fullwave rectification.
Solution:
Using the simplified relation that the parallel combination of the load resistor and the capacitive
impedance equals, approximately, the capacitive impedance, the ac attenuation in the filter is
If RL > 5Xc then :
43
Example 8:
The output of a full-wave rectifier and capacitor filter is further filtered by an RC filter section
(Fig2-91). The component values of the RC section are R = 500 Ω & C = 10 µF.
If the initial capacitor filter develops 150V dc with a 15V ac ripple voltage, calculate the
resulting dc and ripple voltage across a 5kΩ load.
Fig (2-91) RC filter circuit for Example 8:
Solution:
DC Calculations:
AC Calculations:
Calculating the value of the capacitive impedance first (for full-wave operation):
Since this impedance is not quite 5 times smaller than that of the filter resistor (R = 500 Ω),
Voltage-Multiplier Circuits
1-Voltage Doubler
A modification of the capacitor filter circuit allows building up a larger voltage than Vm
The use of this type of circuit allows keeping the transformer peak voltage rating low while
stepping up the peak output voltage to two, three, four, or more times Vm
Fig (2-92) half wave voltage doubler
During the positive-voltage half-cycle across the transformer,
Dl conducts D2 is cut off, charging Cl up to the Vm
During the negative half-cycle of the secondary voltage,
44
Dl is cut off and D2 conducts charging C2
Fig (2-93) Double operation, (a) positive half-cycle (b) negative half-cycle
We can sum the voltages around the outside loop (Fig.2-93b):
From which
On the next positive half-cycle:
D2 is none conducting and C2 will discharge through the load.
If no load is connected across C2 both capacitors stay charged Cl to Vm and C2 to 2Vm.
If there is a load connected to the output , the voltage across C2 drops during the positive
half-cycle and the capacitor is recharged up to 2Vm during the negative half-cycle.
Another doubler circuit is the full-wave doubler of Fig (2-94).
Fig (2-94) full wave voltage doubler
Fig (2-95) Alternate half-cycle of operation for full wave voltage doubler
45
During the positive half-cycle of transformer secondary voltage
D1 conducts charging Cl to a peak voltage Vm , D2 is non conducting at this time.
During the negative half-cycle
D2 conducts charging C2 while Dl is non conducting.
One difference is that the effective capacitance is that of Cl and C2 in series, which is less
than the capacitance of either Cl or C2 alone.
The lower capacitor value will provide poorer filtering action than the single-capacitor filter
In summary: The half-wave or full-wave voltage doubler provide twice the peak voltage 2Vm
2-Voltage Tripler and Quadrupler
Fig (2-96) shows an extension of the half-wave voltage doubler, which develops three and four
times Vm
It should be obvious from the pattern of the circuit connection how additional diodes and
capacitors may be connected so that the output voltage may also be five, six, seven, etc.,
times Vm
Fig (2-96) voltage tripler and quadrupler.
During the positive half-cycle
Cl charges through D1 to a peak voltage Vm
During the negative half-cycle
C2 charges to twice the peak voltage 2Vm
During the positive half-cycle:
D3 conducts and the voltage across C2 charges C3 to the same 2Vm
On the negative half-cycle
D2 and D4 conduct with C3 charging C4 to 2Vm.
The voltage across C2 is 2Vm
The voltage across Cl and C3 is 3Vm
The voltage across C2 and C4 is 4Vm.
If additional sections of diode and capacitor are used, each capacitor will be charged to 2Vm
Measuring from the top of the transformer winding, will provide odd multiples of Vm
Measuring from the bottom of the transformer will provide even multiples Vm.
46
SUMMARY (Diode)
1- The depletion region is a region adjacent to the p-n junction containing no majority carriers.
2- Forward bias permits majority carrier current through the p-n junction
3- Reverse bias prevents majority carrier current.
4- A p-n structure is called a diode.
5- Reverse breakdown occurs when the reverse-biased voltage exceeds a specified value
6- The single diode in a half-wave rectifier conducts for half of the input cycle.
7- The output frequency of a half-wave rectifier equals the input frequency.
8- The average (dc) value of a half-wave rectified signal is 0.318 or 1/π times its peak value.
9- Each diode in a full-wave rectifier conducts for half of the input cycle.
10-The output frequency of a full-wave rectifier is twice the input frequency.
11- The basic types of full-wave rectifier are center-tapped and bridge.
12-The output voltage of a center-tapped full-wave rectifier is1/2 of the total secondary voltage.
13-The output voltage of a bridge rectifier equals the total secondary voltage.
14-A capacitor filter provides a dc output approximately equal to the peak of the input.
15- Ripple voltage is caused by the charging and discharging of the filter capacitor.
16- The smaller the ripple, the better the filter.
17-Diode limiters cut off voltage above and below specified levels. Limiters are also called
clippers.
18-Diode clampers add a dc level to an ac signal.
19-The zener diode operates in reverse breakdown.
20-A zener diode maintains an essentially constant voltage across its terminals over a
specified range of zener currents.
21- Zener diodes are used as shunt voltage regulators.
22-Regulation of output voltage over a range of load currents is called load regulation.
23-The smaller the percent regulation, the better.
24-The characteristics of an ideal diode are a close match with those of a simple switch except
for the important fact that an ideal diode can conduct in only one direction
25-The ideal diode is a short in the region of conduction and an open circuit in the region of
non conduction
26-The region near the junction of a diode that has very few carriers is called the depletion
region.
27-In the absence of any externally applied bias. The diode current is zero.
28-In the forward-bias region the diode current will increase exponentially with in crease in
voltage across the diode.
29-In the reverse-bias region the diode current is the very small reverse saturation current until
Zener breakdown is reached and current will flow in the opposite direction through the
diode.
30-The threshold voltage is about 0.7 V for silicon diodes and 0.3 V for germanium diode
31-Rectification is a process whereby an applied waveform of zero average value changed to
one that has a dc level.
32-Clippers are networks that "clip" away part of the applied signal either to create a specific
type of signal or to limit the voltage that can be applied to a network
33-Clampers are networks that "clamp" the input signal to a different dc level, in any event, the
peak-to-peak swing of the applied signal will remain the same
34- Zener diodes are diodes that make effective use of the Zener breakdown potential of an
ordinary p-n junction characteristic to provide a device of wide importance and application.
For Zener conduction, the direction of conventional flow is opposite to the arrow in the
symbol. The polarity under conduction is also opposite to that of the conventional diode.
35-To determine the state of a Zener diode in a dc network, simply remove the Zener from the
network, and determine the open-circuit voltage between the two points where the Zener
47
diode was originally connected. If it is more than the Zener potential and has the correct
polarity, the Zener diode is in the "on" state.
36-A half-wave or full-wave voltage doubler employs 2 capacitors a tripler, 3 capacitors a
quadrupler, 4 capacitors. for each, the number of diodes equals the number of capacitors.
Equations
Ideal:
VT = 0V
Half-wave rectifier: Vdc = 0.318 Vm
Full-wave rectifier: Vdc = 0.636 Vm
48