Unit 3

EDC

UNIT IIIHalf wave rectifier, ripple factor, full wave rectifier, Harmonic components in a

rectifier circuit, Inductor filter, Capacitor filter, L- section filter, ∏- section filter,

Multiple L- section and Multiple ∏-section filter, and comparison of various filters

circuits in terms of ripple factors, Simple circuit of a regulator using zener diode,

Series and Shunt voltage regulators

1

Rectifier:

A rectifier is an electrical device that converts alternating current (AC) to direct

current (DC), a process known as rectification. Rectifiers have many uses including as

components of power supplies and as detectors of radio signals. Rectifiers may be made

of solid state diodes, vacuum tube diodes, mercury arc valves, and other components.

A device which performs the opposite function (converting DC to AC) is known

as an inverter. When only one diode is used to rectify AC (by blocking the negative or

positive portion of the waveform), the difference between the term diode and the term

rectifier is merely one of usage, i.e., the term rectifier describes a diode that is being used

to convert AC to DC. Almost all rectifiers comprise a number of diodes in a specific

arrangement for more efficiently converting AC to DC than is possible with only one

diode. Before the development of silicon semiconductor rectifiers, vacuum tube diodes

and copper (I) oxide or selenium rectifier stacks were used.

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Half wave Rectifier:

In half wave rectification, either the positive or negative half of the AC wave is

passed, while the other half is blocked.

Because only one half of the input waveform reaches the output, it is very

inefficient if used for power transfer.

Half-wave rectification can be achieved with a single diode in a one-phase supply,

or with three diodes in a three-phase supply.

In positive half cycle, D is forward biased (ideal) and output voltage is same the

input voltage. In the negative half cycle, D is reverse biased, and output voltage is

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zero. When the diode is reverse biased, entire transformer voltage appears across

the diode. The maximum voltage across the diode is Vm.

The diode must be capable to withstand this voltage. Therefore PIV half

waverating of diode should be equal to Vm in case of single phase rectifiers. The

avg-current rating must be greater than Iavg

In reverse bias the voltage v is less than cut in voltage vγ that is v< vγ.

In forward bias the voltage v is greater than cut in voltage vγ that is v>vγ.

DC AMMETER READING: By definition average value of periodic function is

given by the area of one cycle of the curve divided by the base then

Idc = 1/2π ∫02π I dθ

= 1/2π ∫0π Im sinθ dθ

=Im/2π ∫0π sinθ dθ

=Im/2π (2)

Idc =Im/π

AC AMMETER READING:

By definition RMS value is squared of periodic function of time is given by the

area of one cycle of the curve which represents the square of the function divided by the

base then

I rms = (1/2π ∫02π I 2dθ)1/2

= 1/2π (∫0π Im2sin2θ dθ)1/2

=Im/2π (∫0π(1-cos2θ) dθ

I rms =Im/2

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Similarly for dc volt meter reading is given by: Vdc=vm/π

Similarly for dc volt meter reading is given by: Vrms=vm/2

RIPPLE FACTOR (γ):

Ripple factor is nothing but the measurement of fluctuating components

This is defined as,

γ = rms value of alternating component of wave

Average value of wave

γ = I | rms/Idc or v |

rms/vdc

Where I | rms and v |

rms denotes the rms value of the ac components.

In order to measure the ripple voltage or ripple current in the out put should be

made with instruments that respond to higher than power frequency.

A capacitor must be used in series with the input to the meter to block the dc

component then I | = I-Idc

But I |rms= (1/2π ∫0

2π (I-Idc) 2dθ)1/2

= (1/2π ∫02π (I2-2IdcI+Idc

2) dθ)1/2

From our definition Idc = 1/2π ∫02π I dθ

Then I | rms=(I rms

2-Idc2)1/2

Then γ = I | rms/Id c= (I rms

2-Idc2)1/2/Idc

= ((I rms/Idc)2-1)1/2

We have I rms =Im/2 Idc =Im/π

Then ripple factor γ =1.21

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EFFICIENCY (OR) RATIO OF RECTIFICATION (η):

It is defined as the ratio of dc output power to ac input power.

η=dc output power/ac input power.

= pdc/pac

where pdc=Idc Vdc and pac=I rmsVrms

then η=IdcVdc / I rmsVrms

we know that Idc =Im/π I rms =Im/2 Vdc=vm/π Vrms=vm/2

Then η=4/π2 =0.406

Then efficiency of half wave rectifier is given by 40.6%.

PEAK INVERSE VOLTAGE:

It is defined as the maximum reverse voltage that a diode can withstand with out

destroying the junction. The peak inverse voltage across a diode is peak of the negative

half cycle.

Fro half wave rectifier PIV is vm .

TRANSFORMAR UTILIZATION FACTOR (TUF):

That means how much amount of power used by the transformer. It is the ratio of

dc power delivered to the load to the rating of transformer secondary. Then,

TUF= dc power delivered to the load / ac rating of the transformer secondary

Then TUF= pdc/pac(rated)

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= Idc Vdc/ I rmsVrms(rated)

=2(2) 1/2/π2

TUF=0.286

Form factor:

The form factor is given by the ratio of I rms to Idc

Then Form factor= I rms/Idc

=1.57.

Peak factor:

It is the ratio of maximum value of current to the rms value of current.

Peak factor=Im/I rms

=2

Regulation:

The variation of dc out put voltage varies as a function of dc load current. In

general the percentage of regulation for ideal power supply is zero.

The percentage regulation is defined as,

% regulation = (VNL-VFL/VFL)*100

We know that VDC=Vm/π-Idc *Rf

VNL =Vm/π and VFL =Vm/π-Idc *Rf

% regulation= [((Vm/π)-Vm/π+Idc *Rf)/ (Vm/π-Idc *Rf )]*100

= (Idc *Rf)/(Vm/π-Idc *Rf )*100

=(Rf/RL)*100

Disadvantages of half wave rectifier :

High ripple factor (1.21).

Low ratio of rectification (0.406).

Low TUF (0.286).

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FULL WAVE RECTIFIER:

In this Circuit that the conduction takes place through one diode during one half

of the power cycle and through the other diode during the second half of the power cycle.

During the positive half of the input signal, anode of diode D1 becomes positive

and at the same time the anode of D2 become negative. Hence D1 conducts and D2 does

not conduct. The load Current flows through D1 and the voltage drop across RL will be

equal to the input voltage.

During the negative half of the input signal the anode of D1 becomes negative and

the anode of D2 becomes positive. Hence D1 does not conduct and D2 conducts. The load

current flows through D2 and the voltage drop across RL will be equal to input voltage.

The sum of these two currents, has the forms as shown

Then we will get Vo =+Vm sinωt 0≤ωt≤π

=-Vm sinωt π ≤ωt≤2π

8

Figure: out put wave forms of full wave rectifier

Here minus sign appears in second equation becomes during the second half-

cycle. The wave is still sinusoidal, but inverted.

Hence IL=(Vm/RL)sinωt =Im sinωt 0≤sinωt≤π

=-Im sinωt π≤sinωt≤2π

AVERAGE OF DC VOLTAGE:-

For half wave rectifier Idc = 1/2π ∫02π Im sinωt dt

For half wave rectifier Idc is twice of that half wave rectifier.

I.e.: Idc=2Im/π

Now Idc =1/2π ∫02π Io dθ

=1/2π [ ∫0π Imsinθ dθ +∫

π2π - Imsinθ dθ]

Idc =2Im/π

Similarly Vdc=2Vm /π

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RMS value:-

Irms= (1/2π ∫02π I 2dθ)1/2

= (1/2π ∫0π Im2sin2θ dθ+∫

π2πIm2sin2θ dθ)1/2

Irms =Im/√2

Similarly Vrms=Vm/√2

RIPPLE FACTOR (γ):

γ = ((I rms/Idc)2-1)1/2

But Irms =Im/√2

Idc =2Im/π

γ=0.486

EFFICIENCY (OR) RATIO OF RECTIFICATION:-

η=dc output power/ac input power

= pdc/pac

Where pdc=Idc Vdc and pac=I rmsVrms

Then η=IdcVdc / I rmsVrms

We know that Idc =2Im/π I rms =Im/√2

Vdc=2Vm/π Vrms=Vm/√2

Then η=8/π2 =81.2%

Then efficiency of half wave rectifier is given by 81.2%.

10

PEAK INVERSE VOLTASGE:-

The maximum voltage that can be occurred across the Full Wave rectifier circuit

is given by “2Vm”

TRANSFORMER UTILIZATION FACTOR:-

In Full wave rectifier we are using a center tapped transformer. The secondary

current flows through each half separately in every half cycle. While the primary of

transformer carries current continuously. Hence TUF is calculated separately for primary

and secondary windings.

TUF (primary) =Pdc/Pac(rated)

=(Idc)2RL/ I rmsVrms(rated)

We know that Idc =2Im/π

Irms =Im/√2 and Vrms (rated) = (Im*RL)/ √2

Then TUF (primary) = 0.812

For secondary winding TUF is given by

TUF (secondary) =2* TUF (secondary) of HWR

=2*0.287

=0.574

Then TUF of FWR is given by = [TUF (primary) + TUF (secondary)]/2

= 0.692

Form factor:

The form factor is given by the ratio of I rms to Idc

Then Form factor= I rms/Idc

=1.11

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Peak factor:

It is the ratio of maximum value of current to the rms value of current.

Peak factor=Im/I rms

=√2

REGULATION for FWR:-

The regulation indicates that how DC Voltage Varies as a function if load current.

The Percentage of regulation is defined by

% regulation = (VNL-VFL/VFL)*100

The Variation of DC voltage as a function of load current as follows.

We know that VDC=2Vm/π-Idc *Rf

VNL =2Vm/π and VFL =2Vm/π-Idc *Rf

% regulation= [((2Vm/π)-2Vm/π+Idc *Rf)/ (2Vm/π-Idc *Rf)]*100

= (Idc *Rf)/ (2Vm/π-Idc *Rf)*100

= (Rf/RL)*100

ADVANTAGES:-

1) Low ripple factor (0.483).

2) High efficiency (81%).

3) High TUF (0.693).

4) Current flows for both the half cycles.

DISADVANTAGES:-

1) The voltage at the secondary is 2Vm for each half cycle while the output is only Vm.

Hence the output voltage is only half of the secondary voltage.

2) Diodes with high PIV rating are needed.

3) The Center tapped transformer used in the circuit has following disadvantages:

i) It is bulky.

ii) It is costly.

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iii) It is difficult to adjust the Center Point accurately.

BRIDGE RECTIFIER:-

A more widely used full wave rectifier circuit is the bridge rectifier. There is no

need for center tapped transformer.

Circuit diagram for bridge rectifier

During the positive half cycle

During the negative half cycle

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It requires four diodes instead of two, but avoids the need for a center tapped

transformer.

During the positive half cycle of the secondary voltage, diodes D1 and D3 are

conducting and D2 and D4 are no conducting. There fore, current flows through the

secondary winding, diode D1 and D3, resistor RL.

During the negative half cycle of the secondary voltage, the diodes D2 &D4

conduct and diodes D1 and D3 do not conduct. Then current flows through the

secondary winding, diode D2 D4 and Resistor RL.

In both cases current passes through the load resistor in the same direction.

Then bridge rectifier is similar in all aspects of a full wave rectifier.

AVERAGE OF DC VOLTAGE:-

For half wave rectifier Idc = 1/2π ∫02π Im sinωt dt

For half wave rectifier Idc is twice of that half wave rectifier.

I.e.: Idc=2Im/π

Now Idc =1/2π ∫02π Io dθ

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=1/2π [∫0π Imsinθ dθ +∫

π2π - Imsinθ dθ]

Idc =2Im/π

Similarly Vdc=2Vm /π

RMS value:-

Irms= (1/2π ∫02π I 2dθ)1/2

= (1/2π ∫0π Im2sin2θ dθ+∫

π2πIm2sin2θ dθ)1/2

Irms =Im/√2

Similarly Vrms=Vm/√2

RIPPLE FACTOR (γ):

γ = ((I rms/Idc)2-1)1/2

But Irms =Im/√2

Idc =2Im/π

γ=0.486

EFFICIENCY (OR) RATIO OF RECTIFICATION:-

η=dc output power/ac input power

= pdc/pac

Where pdc=Idc Vdc and pac=I rmsVrms

Then η=IdcVdc / I rmsVrms

We know that Idc =2Im/π I rms =Im/√2

Vdc=2Vm/π Vrms=Vm/√2

Then η=8/π2 =81.2%

Then efficiency of half wave rectifier is given by 81.2%.

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PEAK INVERSE VOLTASGE:-

The maximum voltage that can be occurred across the Full Wave rectifier circuit

is given by “Vm”. Here no Center tapped transformer then PIV of bridge rectifier is Vm

TRANSFORMER UTILIZATION FACTOR:-

TUF=Pdc/Pac (rated)

= (Idc)2RL/ I rmsVrms (rated)

We know that Idc =2Im/π

Irms =Im/√2 and Vrms (rated) = (Im*RL)/ √2

Then TUF= 0.812

Form factor:

The form factor is given by the ratio of Irms to Idc

Then Form factor= I rms/Idc m =1.11

Peak factor:

It is the ratio of maximum value of current to the rms value of current.

Peak factor=Im/I rms

=√2

Frequency is 2F.

ADVANTAGES:-

1 Low ripples (0.482).

2 High efficiency (81%).

3 High TUF (0.693).

4 No transformer core saturation.

5 Does not require Center tapped transformer, which reduces bulkiness & cost.

6 PIV rating of the diode used is only Vm.

7 As its TUF is high, it is extensively used in high power applications.

DISADVANTAGES:-

1. Uses 4 diodes.

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COMPARISION OF RECTIFIER CIRCUITS:-

Sl.no. Particulars HWR FWR B-FWR

1 No. of diodes required 1 2 4

2 Avg Value of Current, Idc Im/π 2Im/π 2Im/π

3 RMS Value of Current, Irms Im/2 Im/√2 Im/√2

4 Peak inverse voltage (PIV) Vm 2Vm 2Vm

5 Peak load current, Im Vm/RL+Rf Vm/RL+Rf Vm/RL+2Rf

6 DC output voltage, Vdc Vm/ π 2Vm/ π 2Vm/ π

7 Rectification efficiency η 40.6% 81.2% 81.2%

8 Ripple factor 1.21 0.482 0.482

9 Fundamental frequency of ripple F 2f 2f

10 Voltage regulation Good Better Good

11 TUF 0.286 0.692 0.812

12 Form factor 1.57 1.11 1.11

13 Peak factor 2 √2 √2

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HARMONIC COMPONENTS IN RECTIFIERS CIRCUIT:

The representation of the output current wave in a rectifier is obtained by means

of a Fourier series.

For half wave rectifier circuit expression for current wave form is

I = Im {1/П+ ½sinωt- 2/П∑K=2, 4, 6 [coskωt/ (k+1) (k-1)]}

The lowest angular frequency present in the expression is that of the primary

source of ac power.

Except for the angular frequency ω all other terms have even harmonics of the

power frequency.

The corresponding equation for full wave rectifier is given by that the full wave

rectifier having two half wave circuits which are so arranged that one circuit

conducts during one half of the cycle and second operates during the second half

cycle.

It is clear that the currents are functionally related by the expression

I1(α) =I2 (α+П)

Then total current I=I1+I2 then

I = Im {2/П- 4/П ∑K=2, 4, 6, [cosk\ωt/ (k+1) (k-1)]}

We observe that the fundamental angular frequency ω has been eliminated

from the equation, the lowest frequency in the output being 2ω, a second

harmonic term.

The second desirable feature of the full wave circuit is the fact that the current

pulses in the two halves of the transformer winding are in such directions that the

magnetic cycle through which the iron of the core is taken is essentially that of the

alternating current.

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This eliminates any dc saturation of the transformer Core which would give rise

to additional harmonics in the output.

The power supply must provide an essentially ripple free source of power from an

ac lines, then it is demonstrated above that the output of rectifier contains ripple

components in addition to a dc term.

Hence we need filter placed in between the rectifier and load in order to attenuate

the ripple components.

The rectifier is non linear device, no simple exact method of solution of the power

supply problem exist. However, for each type of filter used, a reasonable linear

approximation is made which allows the circuit to be analyzed by the usual

methods of ac circuit theory.

FILTERS:-

A power supply must provide ripple free source of power from an ac line but the

output of a rectifier circuit contains ripple components in addition to a dc term.

Hence it is necessary to include a filter between the rectifier and the load in order

to attenuate these ripples components. Capacitor and inductors are used as filters.

INDUCTOR FILTER:-

The operation of inductor filter depends on the fundamental property of an

inductor to oppose any change of current. As a result, any sudden change that might

occur in a circuit with out an inductor is smoothed out by the presence of an inductor in

the circuit.

HALF WAVE RECTIFIER:-

Suppose that an inductor or choke filter is connected in series with the load in a

half-wave circuit.

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Assume that the diode and choke resistance are negligible. then the controlling the

current in the circuit during the time that current flows is

Vi = Vm sinωt= (L*di/dt) +RL* i

An exact solution is obtained when the differential equation is subjected to initial

conditions that I =0 and t=0.

This solution is valid only as long as it provides a positive value of current since

the diode can conduct only in one direction.

The time at which the current falls to zero is called “cutout point”.

This shows the effect of the inductance on the wave form of the output circuit in a

half wave rectifier with an inductor filter. The load resistance RL is assumed

constant.

FULL WAVE RECTIFIER :-

The inductive filter is short circuit for DC and offers some impedance for AC. So

it can be used as a rectifier. AC voltage dropped across inductor where as DC pass

through it.

Full wave chock input filter

20

Load voltage forms for L=0 and L≠0

An inductor has the fundamental property of opposing only change in current

flowing through it. This property is used in the series inductor filter as shown in

figure.

Whenever the current through an inductor tends to change, a “back emf” is

induced in the conductor. This induced back emf prevents the current from

changing its value.

Since the reactance of the inductor increases with frequency i.e. (XL=2ПfL) then it

gives lowest ripple frequency and large dc component and small ac component.

The operation of the series inductor filter depends upon the current through it that

means it requires a continuous flow of current, therefore, increases in load current

results in reduced ripple. Then it is used FWR since it gives constant output.

The output of the full wave rectifier is given by

I = Im {2/П- 4/П ∑K=2, 4, 6, [cosk\ωt/ (k+1) (k-1)]}

I = Im2/П- 4Im /П ∑K=2, 4, 6, [cosk\ωt/(k+1)(k-1)]}

To analyze this filter for a FWR the Fourier series can be written as,

I = Im2/П- 4Im /П [1/3cos2ωt+1/15 3cos4ωt+1/353cos6ωt+………]

Assuming that the third and higher terms contribute little output voltage then,

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I = 2Im/П- 4Im /П [1/3cos2ωt]

Neglecting the diode, chock, and transformer resistances since they are very small

compared with RL. There fore the dc component of current is,

Im = Vm/RL.

Now the impedance of L and RL in series with the ac component at a frequency of

2ω is,

Z=√ [RL2+ (2ωL)2]

For the ac component,

Im= Vm/√ [RL2+ (2ωL)2]

Then total load current is given by,

I = 2Im/П- 4Im /3П [cos2ωt] Then,

I = 2Vm/RLП- 4Vm /3П [cos (2ωt-ψ)]/ √ [RL2+ (2ωL)2]

Where ψ is the phase angle by which inductor current IL lags inductor voltage VL.

Then ψ=tan-1(2ωL/ RL). γ

The ripple factor: - the ripple factor is defined as the ratio of the rms value of the ripple

to the dc voltage of the wave is given by,

γ = {4Vm /3П√2√ [RL2+ (2ωL)2]}/2Vm/RLП

γ=2/3√2*1/ [1+4ω2L2/RL2]

The expression shows that filtering improves with decreased circuit resistance

with increased currents. If the ratio of 4ω2L2/RL2 is large compared with unity then ripple

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factor is reduced to RL/3√2ωL. In case of load resistance is infinite then ripple factor is

given by 0.47.

REGLULATION:-

The dc out put voltage is given by Vdc =Idc*RL

=2Vm/П

=0.637Vm

=0.90 Vrms

Where Vrms is the transformer secondary voltage measured to the center tap.

Then Vdc =Idc*RL

=2ImRL /П =2Vm/П

Then Vdc=2Vm/П- IdcRf

THE CAPACITOR FILTERS:-

Half wave Rectifier:

The filtering effect is done by shunting a capacitor across the

load resistor. The capacitor offers low resistance for ac components

and high resistance for dc components that means the dc current

appears at the output.

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Half wave Rectifier with capacitor filter

Out put wave forms

We know that in case of half wave rectifier rectifies one half of the ac

input and hence current passes only for half cycle.

During the positive half cycle diode D conducts and charges the

capacitor to the peak value of the input signal. When the voltage

across the capacitor C is Vc=Vm

During the negative half cycle diode D is in non conducting stage in this duration

the capacitor discharges through RL and loss charge. That means the output

voltage and charge both are being equal and decreases exponentially with time

constant RLC.

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Because of large discharge time constant CRL the capacitor does not have

sufficient time to discharge appreciably, Due to this fact the capacitor maintains a

sufficiently large voltage across RL even during the negative half cycle.

When the rectified voltage exceeds the capacitor voltage Vc, the capacitor C

again gets charged quickly to Vm.

For fixed value of capacitance larger the load resistance RL larger will be the

discharge time constant CRL and lower ripples will occur and vice versa.

We know that charge acquired by the capacitor is equal to the charge lost by the

capacitor and it is necessary to express that Vγ as a function of load current and

capacitance.

Here T2 is the total discharging time of the capacitor ,when discharging at

constant rare of current Idc, and the capacitor will loss an amount of charge

Idc*T2. The charge it has acquired=VγC

The charge it has lost =Idc*T2

Hence charge in the capacitor voltage is Vγ= Idc*T2/C

For better discharging action the smaller will be the conduction time T1 and

closer to T2. That is T discharging>>T charging

Then T2≈T=1/f

Then Vγ= Idc/fC

If we observe the output wave form then it appears like a “triangular wave form”

then ripple factor for triangular wave form is Vrms= Vγ/2√3

Then ripple factor is γ=Vrms/Vdc

= Vγ/2√3*Vdc

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= Vγ/2√3VγfCRL

=1/2√3fCRL .

FULL WAVE RECTIFIER:

A capacitor filter connected directly across the load is shown a. The property of a

capacitor is that it allows ac component and blocks dc component.

The operation of the capacitor filter is to short the ripple to ground but leave the

dc to appear at output when it is connected across the pulsating dc voltage.

During the positive half cycle, the capacitor charges unto the peak vale of the

transformer secondary voltage, Vm and will try to maintain this value as the full

wave input drops to zero. Capacitor will discharge through RL slowly until the

transformer secondary voltage again increases to a value greater than the

capacitor voltage.

The diode conducts for a period, which depends on the capacitor voltage. The

diode will conduct when the transformer secondary voltage becomes more than

the diode voltage. This is called the cut in voltage.

The diode stops conducting when the transformer voltage becomes less than the

diode voltage. This is called cut out voltage.

Referring to the figure below, with slight approximation the ripple voltage can be

assumed as triangular.

From the cut-in point to the cut-out point, whatever charge the capacitor acquires

is equal to the charge the capacitor has lost during the period of non-conduction,

i.e., from cut-out point to the next cut-in point.

The charge it has acquired=VγC

The charge it has lost =Idc*T2

Hence charge in the capacitor voltage is Vγ= Idc*T2/C

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Full wave rectifier with capacitor filter

Out put wave forms

Here the total charging time T= (T1+T2) + (T1+T2)

T=2T2

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Then T=1/2f

Now Vγ= Idc/2fC

If we observe the output wave form then it appears like a “triangular wave form” then

ripple factor for triangular wave form is Vrms= Vγ/2√3

Then ripple factor is γ=Vrms/Vdc

= Vγ/2√3*Vdc

= Vγ/2√3Vγ2fCRL

=1/4√3fCRL .

L-SECTION FILTER (OR) LC FILTER:-

The two types of filtering actions considered above may be combined into a

single L-section filter.

The ripple factor is directly proportional to the load resistance RL in the inductor

filter and inversely proportional to RL in the capacitor filter. Therefore if these two

filters are combined as LC filter or L section filter as shown in figure the ripple

factor will be independent of RL.

If the value of inductance is increased it will increase the time of conduction. At

some critical value of inductance, one diode, either D1 or D2 will always

conducting.

From Fourier series, the output voltage can be expressed as

V = 2Vm/П- 4Vm /3П [cos2ωt]

And the out put voltage equals to Vdc= 2Vm/П

Now the overall resistance in the circuit is R then Vdc= 2Vm/П-Idc*R

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Ripple factor:-

Here the reactance of the chock is must be large compared with impedance of the

capacitor and resistor .that is, XL>>XC.And also XC<<RL.

That is a little error occurred then by assuming that the entire alternating current

passes through the capacitor and none through the resister.

Under these conditions the net impedance is given by XL=2ωL.

Then alternating current through the circuit is,

Irms=4Vm /3√2ПXL

Then Irms=2Vdc/3√2XL

Irms=√2Vdc/3XL

Where the resistance R is neglected then the voltage across the load is the voltage across

the capacitor then,

Vrms=Irms*XC

Vrms=√2VdcXC /3XL

Then γ=Vrms/Vdc

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= [√2VdcXC /3XL]/Vdc

=√2XC/3XL

=√2/3*4ω2LC

From the above formula it is clear that the ripple factor is independent of load resistor RL.

CRITICAL INDUCTANCE:-

Consider that condition when no inductor is used, as already found current will

flow in the diode circuit for a small portion & the cycle, and the capacitor will

become charged to the peak transformer voltage in each cycle.

Suppose that a small inductance is inserted then there will be some what cutout

may still occur.

As the value of inductance is increased and supplies current to the load

continuously and no cutout occurs. This value of inductance is referred as critical

inductance LC.

Under these circumstances each diode conducts for one – half of the cycle and the

input voltage to filter circuit is given by,

V = 2Vm/П- 4Vm /3П [cos2ωt]

The peak value of current of the peak √2Irms of the ac component of the current must hot

exceed the direct current.

Idc=Vdc/RL then

Vdc/RL≥4Vm /3ПXL

Vdc/RL≥2*2Vm /3ПXL

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Vdc/RL≥2Vdc /3XL

XL≥2RL/3

Then the value for critical inductance is given by,

LC=RL/3ω since XL=2ωLC

Design considerations (or) Bleeder Resistor:-

When the load current IL=0, the output voltage is Vm and when the load current IL

is increased slightly the output falls to 2Vm/П .

In order to avoid this abrupt change at no load and load conditions, some current

is always made to flow in the load circuit by using a resistor called “bleeder

resistor” in parallel with the capacitor filter.

The bleeder resistor is also used to discharge the capacitor filter when the supply

is off.

The reactors whose inductance decreases when the dc current through it increases

are called “swinging chokes”.

Problem:-

An L-section filter is used in the output of a full-wave rectifier that is fed from 40-0-40V

transformer. The load current is 0.2A. Two 40uF capacitors and two 2H chokes are

available. The diodes used are assumed to be ideal. Assume that the input power-line

frequency applied to the transformer primary winding is 50Hz.

(a) Show that a single L-section filter can be designed using the available

components which can supply a continuous current to he load at all times

(b) Calculate the 100Hz ripple voltage when the filter consists of one coke and one

capacitor.

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(c) Compare the 100Hz ripple voltage when the two chokes are in series and the two

capacitors are in parallel to provide the resultant inductance and capacitance in the

filter respectively.

(d) Compare the ripple factors of parts (a) and (b)

Solution:-

Since the rms value of the transformer secondary is 40,

Then Vm=40√2=56.57.

Vdc = 2Vm/П=36v.

Since the desired load current is 0.2A then RL is

Given by RL=Vdc/IL=180ohms

Then critical inductance Lc= LC=RL/3ω =0.19H

Clearly each choke has inductance of 2H greater than the critical inductance Lc=0.19H

then L. section filter can be designed.

(b) Then XC=1/2ωC=39.79ohms.

XL=1/2ωL=1256.64ohms.

Then Vrms=√2VdcXC /3XL=0.537v.

(c) When the two chokes are in series and two capacitors are in parallel the resultant

inductance and capacitance is 4H & 80uF.

XC=1/2ωC=19.89ohms.

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XL=1/2ωL=2513.28ohms.

Vrms=√2VdcXC /3XL=0.134v

(d)Let rb rb be the ripple factor of Part (a) & part (b) respectively, then

rb/ rb=4.

MULTIPLE L-SECTIONS FILTER:-

This type of filtering is much more complete through the use of two L-section

filters in cascade a shown in figure.

It is assumed that the reactance of the all the chokes are much larger than the

reactance of the capacitors.

Pulse assumes that the reactance of the last capacitor is small compared with the

resistance of the load.

Multiple L-section filters

Then the impedance between A3 and B3 is effectively XC2 and between A2 & B2 is XC1 and

between A1 & B1 is XL1 then

We know that Vrms=√2VdcXC /3XL

The ac voltage across C1 is approximately,

VA2B2=I1*X1

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The alternating current I2 is approximately, through L2 is,

I2=VA2B2/XL

The ac voltage across C2 and hence across the load is

I2*C2=I1 XC1XC2/XL2

=√2VdcXC1XC2 /3XL1XL2

Then the ripple factor is given by dividing this expression by Vdc hence,

γ=√2XC1XC2 /3XL1XL2

For multiple similar sections, ripple factor becomes.

γ=√2/3[XC/XL] n

From the above the product Lc for a specified ripple factor may be evaluated with

frequency 60Hz then,

LC=1.76(0.471/γ) 1/n

П – SECTION OR CLC FILTER:-

A very smooth output may be obtained by using a filter that consists of two

capacitors separated by an inductor.

П -section filter

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These П filters shows by highly peaked diode currents and poor regulation as for

thsimple capacitor input filter.

The action of a П – section filter can best be understood by considering the

inductor and the second capacitor as an L-section filter that acts upon the

triangular output voltage wave form from the first capacitor.

The ripple voltage can be calculated by analyzing the triangular wave in the

capacitor filter in to Fourier series and multiplying each component by XC1/XL1

for this harmonic.

Then output voltage can be represented by,

V=Vdc-Vγ/П (sin2ωt-sin4ωt/2+sin6ω/3-…………)

We know that Vγ=Idc/2fc

Then rms of second – harmonic voltage is

Vrms= Vγ/ П√2

=√2IdcXC

Where XC=1/2ωc and XC is reactance of ‘C’ at second harmonic frequency

The output of ‘C’ filter is the input for the L – section filter.

Then current through inductor is Irms

Then the output for the П – section filter is that the product of the output of capacitor and

L-section filter.

Then ripple factor γ=Vrms/Vdc

=√2IdcXC/Vdc (XC1/XL1)

=√2XCXC1/RLXL1

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From the filter DC output voltage = (The DC voltage for a capacitor filter) – (the drop

across L1)

If we consider all the harmonics then for a frequency of 60Hz then

γ =3300/CC1L1RL

Where ‘C’ is in uF & ‘L’ is in H and ‘R’ is in ohms

If the II – section filter is followed by another L-section whose parameters are like L2 and

C2 then the above expression leads to

γ=√2 (XC/RL) (XC1/XL1)n

VOLTAGE REGULATION USING ZENER DIODE:-

Voltage regulator is an electrical regulator designed to automatically maintain a

constant voltage level.

The basic characteristics of a zener diode are observed. The two diodes are

avalanche; zener diodes (break down diodes).They is specially designed pn-

junction diodes to operate in the break down region.

Zener diode maintains a constant output voltage in its break down region even

though the current flowing through it is varied.

It is the important property of the zener diodes to maintain a constant output

voltage. So the zener diode some times called as voltage regulator and zener

diodes used as a regulator of voltage is called as zener voltage regulator. Or zener

regulator.

Our electronic circuit maintains a nearly constant output voltage, but in practice the

output will be varied due to the following reasons.

i) Change in input supply voltage

ii) Change in load resistance

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iii) Change in temperature.

ZENER DIODE SHUNT VOLTAGE REGULATOR:-

The circuit diagram for zener shunt voltage regulator is shown below.

Shunt voltage regulator

Since the zener diode is connected in parallel with the load, the circuit is known

as shunt regulator. The resistance R is series resistance it is current limiting

resistor. Vz is the break down voltage.

Now the output voltage Vi of the section is varies in between Vsmax and Vsmin

If the break down voltage Vz is chosen to be equal to the output voltage VL i.e. Vz

= VL

That means the filter output voltage Vi reverse biases the zener diode and that will

be always operates on the breakdown region.

Let Ii, IL, Iz are the current passing through Ri, RL& diode respectively then

current through RL is given by

IL=VL/RL

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How ever in practical zener diode has a finite value of resistance called zener

resistance (rz). Then voltage drop across is VZ=IZ* rz, then

By applying KVL loop we get Vi=RIi+ VZ

Then Ii= (Vi-VZ)/R

Then we know that Ii=IL+IZ

IZ=Ii -IL

Then = (Vi-VZ)/R-VL/RL

SERIRS VOLTAGE REGULATOR:-

The physical reason for the improvement in voltage regulation with the circuit lies

in the fact that a large fraction of increase in input voltage appears across the

control transistor, so that the output voltage tries to remain constant.

If the input increases, the output must also increase (but to a much smaller extent),

because it is this increase in output that acts to bias the control transistor toward

less current.

This additional bias causes an increase in collector-to-emitter voltage which tends

to compensate for the increase input.

From the foregoing explanation it follows that if the change in output were

amplified before being applied to the control transistor, better stabilization would

result. The improvement is demonstrated with reference to Figure.

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ro ΔI=i Q1 ID I1 IL + + R3 ΔIC2=iC2 IB1 RD

R1 RL

IB2 + Q2 R2 ΔVO=vo

ΔVi = vi V2 + - D RZ VR

- - -

Here a fraction of the output voltage b Vo needs increase only slightly, and yet Q2

may cause a large current change in R3. Thus it is possible for almost all of ΔVi

to appear across R3 (and since the base-to-emitter voltage is small, also across Q1)

and for Vo to remain essentially constant. These considerations are now mad more

quantitative.

From the figure out put voltage is given by

Hence a convenient method for changing the output is adjusting the ratio R1/R2 by

means of a resistance divider as indicated in Figure.

An approximate expression for Sv (sufficiently accurate for most applications) is

obtained as follows: The input-voltage change Vi is very much larger than the

output change Vo. Also, by the definition ΔIL =0, and to a first approximation we

can neglect the ac voltage drop across ro.

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Hence ΔVi=vi appears as shown in Figure. Neglecting the small change in base-

to-emitter voltage of Q1, the current change ΔI=i in.

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