part1 electronics

7/22/2019 part1 electronics

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1-Semiconductor DiodesMajority and Minority CarriersIn an n-typematerial the electronis called the majoritycarrier and holethe minoritycarrier.In ap-typematerial the holeis the majoritycarrier and the electronis the minoritycarrier

Fig1-1 (a) n-type material (b) p-type material

The region of uncovered positive and negative ions is called the depletion regiondue to the

depletion of carriers in this region.

Fig 1-2 p-n junction with no external biasNo Applied BiasIn the absence of an applied bias voltage, the net flow of charge in any one direction for asemiconductor 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


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The current that exist under reverse-bias conditions is called the reverse saturation current andis 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 isnegative 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 onedirectionTherefore the ideal diode is ashort circuitfor the region of conduction, and anopen circuitinthe region of non conduction.


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Fig 1-8 (a) Conduction (b) non conduction state of ideal diode

[1-1]

Fig1-9 Sisemiconductor diode characteristics

Fig1-10 Forward-bias conditions for a semiconductor diode


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Resistance LevelsDC or Static Resistance

[1-2]

Fig1-11 determining the dc resistance of a diode at particularoperating 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-pointThe 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 atthat point


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[1-4]

[1-5]

Average AC ResistanceThe resistance determined by a straight line drawn between the two intersections establishedby maximum and minimum values of input voltage

[1-6]

Fig1-14 the average ac resistance


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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 TableTable 1-1 Resistance Levels

Diode Equivalent CircuitTable 1-2 Diode Equivalent Circuit (Models)


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2-Diode ApplicationLoad line analysis

Fig2-1 Series diode configuration (a) circuit (b) characteristics

Fig2-2 load line& finding the point of operatingExample1:

For the Fig2-3; Determine:

Fig2-3 (a) circuit (b) characteristics


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Solution:

The intersection between the load line and the characteristic curve defines the Q-point asshown in fig2-4:

Fig2-4 solution ofEx1:Example2:Repeat the analysis of Ex1: with R=2KSolution:

Note: the reduced slope and levels of diode current for the increasing loadsThe Q-point is defined by:

Fig2-5 Solution of Ex2:

Example3:Repeat Ex2: using approximate equivalent for the silicon semiconductor diode.


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Solution:

Fig2-6 Solution of Ex3:

Diode ApproximationsTable 2-1 Approximate and ideal semiconductor diode model

Series diode configurations with DC inputsIn general, a diode is in the onstate if the current established by the applied sources is suchthat its direction matches that of the arrow in the diode symbol, and VD0.7Vfor silicon andVD 0.3Vfor germanium

Fig 2-7 Series diode configuration


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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 theoff diode of (a)Example4:

Solution:

Example5:Repeat Ex4: with the diode reversed

Solution:


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Example6:

Determine ID, VD, and VOfor the circuit of figSolution:

Example 7:

Determine I, V1, V2and VOfor the circuit of figSolution:


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Parallel and series-parallel configurationsExample8: Determine I1, ID1, ID2and VOfor the circuit of fig

Solution:

Example9: Determine the current I for the network of fig

Solution:

Diode switching circuitDiode switching circuits typically contain two or more diodes, each of which is connected to anindependent 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 itsanode 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/ORGates


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OR gate: is such that the output voltage level will be a 1 if either or both input is a 1. The10V level is assigned a 1for Boolean algebra while the 0V input is assigned a 0Example10:Determine Vofor the network in fig (2-12)

Fig (2-12) Fig (2-13)

D1is in the onstate due to the applied voltage (10V) while D2is in the off state. Vo = E - V D = 10v - 0.7 = 9.3v = (E - VD) / R = (10-0.7) / 1K = 9.3mA

The output voltage level is not 10V as defined for an input of 1,but the 9.3V is sufficiently at a1level 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 logicAND gate of fig (2-14).

Fig (2-14) Fig (2-15)

With 10v at the cathode D1, is assumed that D1is in the off state.D2 is assumed to be in the on state due to the low voltage at the cathode side and theavailability of the 10v source through 1K resistor.The voltage at Vo is 0.7v due to forwardbiased diode D2 i.e. I = (E-Vo) / R = (10-0.7) / 1K = 9.3mA


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Half-Wave RectifierHalf-wave rectification: is the process of removing one half of the input signal to establisha dc level.The cct of the fig(2-16) called a half wave rectifiers will generates a waveform Vothat will havean average value of particular use in the ac todc conversion process.

Fig (2-16) half wave Rectifier.

During the interval (t=0 to T/2) the polarity of the input voltage Viis 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 Viis shown in fig(2-18 ) and the ideal diodeproduces 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.


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The effect of using a silicon diode with VT=0.7V is shown by fig (2-20) for the forward bias. Theinput must now be at least 0.7V before the diode conducts.

When conducting Vo =Vi- VT

If Vm> VT i.e. Vdc= 0.318Vm

if Vmis close to VT i.e. Vdc0.318(Vm VT)

Fig (2-20) Effect of VTon Half-wave rectified signal.

Eexample 1: for Half wave Rectifier:

(a) Sketch the output voand 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 silicondiode.

Fig (2-21)

Solution:(a) In this situation the diode will conduct during the negative part of the input and vo willappear as shown in fig (2-22)

Fig (2-22)

For the full period, the dc level isVdc= -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


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Full-Wave RectificationThe dc level obtained from a sinusoidal input by half wave rectifier can be improved using aprocess called Full-Wave Rectification. Four diodes in a bridge configuration can be used asFull 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 conductingdiodes are D2& D3while D1& D4are in the

off stateas shown in fig (2-25).

Fig (2-25)

For the negative region of the input the conductingdiodes are D1 & D4while D2 & D3are inthe off stateas shown in fig (2-26).

Fig (2-26)

The dc level for Full wave rectifier is twice that obtained for a half wave systemi.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 theconduction state (for positive region).


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i.e. Vd.c=0.636Vm (Vm >> 2VT)

And if Vm is close to 2VT i.e. Vd.c=0.636(Vm- 2VT)

A secondpopular 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 Viapplied to the transformer, the diode D1is short circuit andthe diode D2 is open circuit.

Fig (2-29)

During the negative portion of Viapplied to the transformer, the diode D1is open circuit andthe diode D2is short circuitas 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)


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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 theavailable dc level to the following:

Vdc= 0.636(5) = 3.18 volt

ClippersClippers: is the network that has the ability to clip off a portion of the input signal withoutdistorting the remaining part.

1-Series clipper: The diode is in the series with loadas 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)


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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 thediode. Applying the condition:

id= 0 at vd= 0

The level of vithat will cause a transition in state is:

vi = Vdc

For an input voltage greater than Vdcvoltsthe diode is in the short-circuit state, while forinput voltages less than Vdcvolts it is in the open-circuitor (off state).

Fig (2-36)

3-Defined terminals and polarity of vowhen the diode is in the short circuit state, the outputvoltage vocan be determined by using KVL.

Fig (2-37)

vi- V - vo = 0 (CW direction)

And vo= vi-V

4-Sketch the input signals above the output and determine the output at instantaneous valuesof the input by using the formula for each case.

Keep in mind that at an instantaneous value of vithe input can be treated as a dc supply ofthat value and the corresponding dc (the instantaneous) value of the output determined.


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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 statefor the positive region of vi , aiding effect of V = 5 volts.

Vo= vi+ 5.

Fig (2-40)


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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 forvoltages morepositive than -5 Vthe 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= 20volt (0 to T/2) the network of Fig (2-44) will result.

Fig (2-44)

The diode is in the short circuitstate (on state)

And Vo= 20 + 5 = 25 volt.


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For vi= -10volt the network of Fig (2-45) will result,

Fig (2-45)Placing the diode in the open circuitstate (off state),

And Vo= iRR = (0) R = 0 volt.The resulting output voltage appears in Fig(2-46)

Fig (2-46)

Note: the clipper not only clipped off5volt from the total swing but raised the dclevel of thesignal by 5volts.

2-parallel clipper: The diode is in theparallel to the loadas shown in fig (2-47).

Fig (2-47)

Example 1: Determine vofor 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)


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Where the defined terminals for vorequire that

vo= V = 4 volts.

The condition id= 0A at vd= 0V 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 vo = vi.

Completing the sketch of voresults in the waveform of Fig (2-52).

Fig (2-52)

Example2:Repeat Example 1using a silicon diode with VT= 0.7 volts.

Solution:Thetransition voltagecan first be determined by applying the condition

id= 0A at Vd= VD= 0.7 volts


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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 VTwas to dropthe (on state) level to 3.3 volt from 4 volt.

Fig (2-55)


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A variety of series and parallel clippers with the resulting output for the sinusoidal input areprovided in Fig (2-56)

. Fig (2-56)


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ClampersThe clamping network will clamp a signal to a different dc level. The network must have acapacitor, a diode, and a resistive element, the magnitude of R & C must be chosen such thatthe time constant (t=RC) is large enough to ensure that the voltage across the capacitor doesnot dischargeduring off state of the diode. We will assume that the capacitor will fully chargeor 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 theresistor 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 theinput 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 biasthe diode2-During this period assume that the capacitor will charge up to a level determined by thevoltage across the capacitor in its equivalent open-circuit state.


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3-Assume that during the period the diode is an open circuitoff statethe capacitor will hold onits charge and therefore voltage.4-Applying KVL to determine Vofor both state on & off.5-The general rule that theswing of the output must match the swing of the input signal

Eexample 1:Determine vofor 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 t1to t2of the input signal since the diode is in it's short-circuit 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= 5volts 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 onvo, 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: = RC= (100 k) (0.1F) = 0.01 s = 10 ms

The total discharge time is therefore 5 = 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


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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 vocan be determined by KVL in the output section.

+5 - 0.7 - v o= 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 t2to t3the 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)


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A number of clamping circuits shown in fig (2-68)


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Zener RegionWhen the applied reverse potential becomes more and more negative, a few free minoritycarriers have developed sufficient velocity to liberate additional carriers through ionization.When VZdecreases to very low levels,this mechanism, called Zener breakdown, will contributeto the sharp change in the characteristic. It occurs because there is a strong electric field in theregion of the junction that can destroy the bonding forces within the atom and "generate"

carriers. Diodes employing this unique portion of the characteristic of ap-njunction are calledZener diodes.

Fig (2-69) Comparison of Si and Ge semiconductor diodesZener DiodeThe Zener diode is a device that is designed to make full use of this Zener region. zener regionoccurs at a reverse biaspotential of VZ.

Fig (2-70) Zener diodes (a) Zener potential (b) characteristic and notation

Any voltage from 0 to VZwill result in an open-circuit as occurred below VT for the silicondiode. Silicon diode maintains its open-circuitin the reverse-biasregion but the Zener diodeassumes a short-circuit state once the reverse voltage is reached(i.e. Vd=VZ for shortcircuit). Zener device switchfrom the open-circuit state to short circuit

Fig (2-71)Zener equivalent cct(a)complete(b) approximate.


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Zener Diode Applications1- AC Voltage Regulators (limiters or Clippers)Two back-to-back zeners can be used as an ac regulator or a simple square-wavegeneratoras 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 ReferenceTwo 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 Vz1and Vz2, both diodes will be in breakdownstate(on state) and the three reference voltages will be available

Fig (2-75)


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3- DC Voltage Regulatorsa- 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 RLand ILthat will result in VRLbeingmaintained 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 RLthat will turn the Zener diode


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A plot of VLversus RLappears in Fig (2-78a) and VLversus ILin Fig (2-78b)

Fig (2-78)

b- Fixed RL, Variable ViFor fixed value of RLin Fig (2-79) the voltage Vimust be sufficiently to turn the Zener diodeon. The turn-on voltage is determined by:

Fig (2-79)

[2-4]

[2-5]

[2-6]

Example 2:Determine the range of values of Vithat will maintain the zener diode in the (on state).

Fig (2-80)Solution:


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A plot of VLversus Viis provided in fig (2-81)

Fig (2-81)


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Regulators (Filters and Power Supplies)Voltage regulator providea fixed output voltageThe operation is convertingan acvoltage into a dcvoltage using transformer, rectifier, andfilter

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 voltagewhich 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 anonzero average. The resulting dc signal is not pure dc or even a good representation of it

3-A simple capacitor filteris often sufficient to provide this smoothing action.

4-The resulting dc voltagewith some rippleor ac voltage variationis then provided as inputto an IC regulator

5-IC regulatorthat provides as output a well-defined dc voltage level with extremely low ripplevoltage 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 of25V and an ac ripple voltage of 1.5V rms are obtained. Calculate the ripple of the filter output.


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Solution:

Voltage Regulation

Another factor of importance in a voltage supply is the amount of change in the output dcvoltage over the range of the circuit operation.This voltage change with respect to either the loadedor unloaded is described by a factorcalled voltage regulation.

Example 2:A dc voltage supply provides 60V when the output is unloaded. When full-load current is drawnfrom the supply, the output voltage drops to 56V. Calculate the value of voltage regulationSolution:

The output voltage from most supplies decreasesas the amount of current drawn from thevoltage supply is increased.The smaller the voltage decreases, the smaller the percent of V.R. and the better theoperation of the voltage supply circuit.

Ripple Factor of Rectified SignalThe rectified voltage is not a filtered voltage; it contains a dc component and a ripplecomponent. By calculate these components we obtain the ripple factor.

1-Half wave Rectified SignalSince 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,


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Fig2-86(a) capacitor filter circuit (b) output voltage waveform

If no loadwere connected to the filter, the output waveform a constant dc level equalin valueto thepeak voltage Vmfrom the rectifier circuit.

For the full-wave-rectified signal indicated in Fig (2-86b)T1is the time during which a diode of the full-wave rectifier conducts and charges capacitorup to the peak rectifier output voltage Vm.

T2 is the time during which the rectifier voltage drops below the peak voltage, and thecapacitor dischargesthrough the load.

The capacitor filter circuit provides a large dcvoltage with little ripplefor light loadsand asmaller dcvoltage with larger ripplefor heavyloads.

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


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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:


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f = frequency of the sinusoidal ac power supply voltage usually 60 HzIdc= average current drawn from the filter by the load,C = filter capacitor value.

For light loads Vdc Vm

f= 60 Hz

Idc in mA , Cinf, RL ink.

Example 3:Calculate the ripple voltage of a full-wave rectifier with a 100 F filter capacitor connected to aload of 50 mA.Solution:

DC Voltage, Vdc:

f = 60 Hz

Vm = the peak rectified voltage, V

Idc = the load current, mAC = the filter capacitor, f

Example 4:The peak rectified voltage for the filter circuit of Ex 3: is 30 volts, calculate the filter dc voltage.Solution:

Note:

The largervalue of average currentdrawn from the filter the less value of output dc voltageThe larger the value of the filter capacitorthe closerthe output dc voltageapproaches thepeak value of Vm.


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Filter-Capacitor RippleUsing the definition of ripple, we obtain the expression for the ripple factor of a full wavecapacitor filter

Since Vdcand Idcrelate to the filter load RL, we can also express the ripple as

Idc in mAC inf,Vdc in volts

RL in k.larger load current, larger ripple factorand inversely with the capacitor size.

Example 5:a load current of 50mA is drawn from a capacitor filter circuit C=100 F.if the peak rectifiedvoltage is 30V, calculate r.Solution:Using results of Ex 3 and Ex4, we get

RC FilterIt is possible to further reduce the amount of ripple across a filter capacitorwhile reducingthe 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 is1-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 reducethe amount of ripplein relation to the dc level, providing better filteroperationthan for the simple-capacitor filter.

Since the rectifier feeds directly into a capacitor, the peak currents through the diodes aremany timesthe average current drawn from the supply.


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The voltage developed across capacitor Cl is then further filtered by the resistor-capacitorsection (R,C2)providing an output voltage having less percent of ripple than that across Cl.

The load RL draws dc current through resistor Rwith an output dc voltage across the loadbeing less than that across Cldue to the voltage drop across R.

Fig (2-89) Full wave rectifier and RC filter circuit.

DC Operation of RC Filter SectionThe equivalent circuit used for considering the dc voltage and current in the filter and loadshown in fig (2-90). The two filter capacitors are open circuitfor dcand

Fig (2-90) (a) dc equivalent circuit (b) ac equivalent circuit

Fig (6-90a) show that the voltage Vdcacross capacitor Cl is attenuated by a resistor-dividernetwork of R andRL , the resulting dc voltage across the load being V'dc

Example 6:The addition of an RCfilter section with R=120, reduces the dc voltage across the initial filter

capacitor from 60 V (Vdc). If the load resistance is 1k, calculate the value of the output dcvoltage (Vdc) from the filter circuit.


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Solution:

The drop across the filter resistor and the load current drawn:

AC Operation of RC Filter SectionFig (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 rippleor acsignal part of the voltage across C1.Both the RCfilter stage R, C2and theloadresistance RLaffect the acsignal at the output.

For C2= 10F at a ripple voltage frequency f= 60Hz, the acimpedance of the capacitor is:

This capacitive impedance is inparallelwith the load resistance RLIfRL= 2 k, theparallel combinationof 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 neglectingthe loadingRL 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 15F capacitor used in the filter section of a circuit using full-wave rectification.Solution:

Using the simplified relation that the parallel combination of the load resistor and the capacitiveimpedance equals, approximately, the capacitive impedance, the ac attenuationin the filter is

If RL> 5Xc then :


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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 RCsection are R= 500 &C= 10 F.If the initial capacitor filter develops 150V dc with a 15V ac ripple voltage, calculate theresulting dc and ripple voltage across a 5k 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 smallerthan that of the filter resistor (R = 500),

Voltage-Multiplier Circuits1-Voltage DoublerA 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 lowwhilestepping upthepeak output voltageto two, three, four, or more times Vm

Fig (2-92) half wave voltage doubler

During thepositive-voltage half-cycleacross the transformer,

Dl conducts D2is cut off, charging Clup to the Vm

During the negative half-cycleof the secondary voltage,


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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 C2will dischargethrough the load.

If no loadis connected across C2both capacitors stay charged Cl to Vm and C2to 2Vm.If there is a loadconnected to the output , the voltage across C2 drops during the positivehalf-cycle and the capacitor is recharged up to 2Vmduring 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


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During the positive half-cycleof transformer secondary voltageD1 conducts charging Cl to a peak voltage Vm, D2 is non conductingat this time.

During the negative half-cycleD2 conducts charging C2 while Dlis non conducting.

One difference is that the effective capacitance is that of Cl and C2 in series, which is lessthan the capacitance of either Clor C2alone.The lowercapacitor value will providepoorerfiltering action than thesingle-capacitor filter

In summary: The half-wave or full-wave voltage doubler provide twice the peak voltage 2Vm

2-Voltage Tripler and QuadruplerFig (2-96) shows an extension of the half-wave voltage doubler, which develops three and fourtimes VmIt should be obvious from the pattern of the circuit connection how additional diodes andcapacitors 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 thepositive half-cycleCl charges through D1to a peak voltage Vm

During the negative half-cycleC2charges to twice the peak voltage 2Vm

During thepositive half-cycle:D3conducts and the voltage across C2charges C3to the same 2Vm

On the negative half-cycleD2 and D4conduct with C3 charging C4to 2Vm.

The voltage across C2 is 2VmThe voltage across Cl and C3 is 3VmThe voltage across C2 and C4 is 4Vm.

If additional sections of diode and capacitor are used, each capacitor will be charged to 2VmMeasuring from the topof the transformer winding, will provide oddmultiples of VmMeasuring from the bottomof the transformer will provide evenmultiples Vm.


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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 junction3- 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 value6- 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/ 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 calledclippers.

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 direction25-The ideal diode is a short in the region of conduction and an open circuit in the region of

non conduction26-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 thediode.

30-The threshold voltage is about 0.7 V for silicon diodes and 0.3 V for germanium diode31-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 network33-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 same34-Zener diodes are diodes that make effective use of the Zener breakdown potential of an

ordinaryp-njunction characteristic to provide a device of wide importance and application.For Zener conduction, the direction of conventional flow is opposite to the arrow in thesymbol. 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 thenetwork, and determine the open-circuit voltage between the two points where the Zener


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diode was originally connected. If it is more than the Zener potential and has the correctpolarity, the Zener diode is in the "on" state.

36-A half-wave or full-wave voltage doubler employs 2 capacitors a tripler, 3 capacitors aquadrupler, 4 capacitors. for each, the number of diodes equals the number of capacitors.

Equations

Ideal: VT= 0VHalf-wave rectifier: Vdc= 0.318 VmFull-wave rectifier: Vdc= 0.636 Vm

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