Analysis of L, C, LC & π filters. · 2015-08-19 · Differentiator and Integrator circuits...

Analog Circuit-1 Notes

Muhammed Riyas A.M,Asst. Professor,Dept. of ECE,MCET Pathanamthitta

Module-1

Contents

Analysis of half wave rectifiers full wave rectifiers.

Analysis of L, C, LC & π filters.

Zener voltage regulator.

Transistor series and shunt voltage regulators.

Short circuit and fold back protection.

Response of high pass RC circuits to sine wave, step inputs pulse and square wave inputs.

Response of Low pass RC circuits to sine wave, step inputs pulse and square wave inputs.

Differentiator and Integrator circuits Clipping and Clamping Circuits.

DC Power Supply:

The electric energy available in our country is in the form of AC voltage 230V, 50 Hz. But

most of electronic devices operate in DC power supplies. Almost all basic household

electronic circuits need an unregulated AC to be converted to constant DC. All devices will

have a certain power supply limit and the electronic circuits inside these devices must be

able to supply a constant DC voltage within this limit. That is, all the active and passive

electronic devices will have a certain DC operating point (Q-point or Quiescent point), and

this point must be achieved by the source of DC power. The DC power supply is practically

converted to each and every stage in an electronic system. Thus a common requirement

for all this phases will be the DC power supply. Regulated DC power supply is an electronic

circuit that is designed to provide a constant dc voltage of predetermined value across load

terminals irrespective of ac mains fluctuations or load variations. The block diagram of a

regulated DC power supply is shown in figure.



Fig:Block Diagram of a Power Supply

The first block in the DC power supply is a transformer. A transformer is a device which

transforms high voltage AC into low voltage AC or vice versa. Our goal is to convert high voltage

AC into low voltage DC. So there is absolutely no reason to use step-up transformer. The

transformer that is used in power supply is step-down transformer, which steps down the input

AC voltage. The main reason why we use transformer in the system are as follows.

We want to reduce the voltage level which we get from the AC mains. Transformer can do

the job of reducing the voltage level in a simple and efficient manner.

The diodes used in the rectifier block cannot handle such a high level of voltage from the AC

mains. So the voltage is first stepped down by the transformer and the reduced voltage is

applied to the rectifier section.

The output of the transformer is given to the input of the rectifier.

The rectifier converts the input sinusoidal voltage to a unipolar output or pulsating DC. Full

wave rectifiers are the most commonly used rectifiers in power supply. The output of the

rectifier is given to input of the filter circuit. The output after being processed by full wave

rectifier is not a pure DC. The output is a pulsating DC. The output contains large fluctuations in

voltages. The power supply that we intend to design must not have any variation in output

voltage. The voltage that we get from full wave rectifier fluctuates between 0 V and Vpeak, and

hence it contains AC components. These AC components needs to be filtered out so as to obtain

DC voltage. The filter circuit reduces the variation in magnitude (ripples) of the rectifier output.



However, the output from the filter is still not a pure DC but filters removes the AC component

in the voltage to a considerable extent. This increases the average DC value of the output

voltage.

The output of the filter is given to the voltage regulator circuit for stabilizing the magnitude of

the DC output voltage against the variations caused by changes in the load current and input

voltage.

Characteristics of a power supply

There are various factors that determine the quality of the power supply like the load voltage,

load current, voltage regulation, source regulation, output impedance, ripple rejection, and so

on. Some of the characteristics are briefly explained below:

Load Regulation – The load regulation or load effect is the change in regulated output

voltage when the load current changes from minimum to maximum value.

퐿표푎푑푅푒푔푢푙푎푡푖표푛 = 푉 − 푉

푉 – Load Voltage at no load

푉 – Load voltage at full load.

From the above equation we can understand that when 푉 occurs the load resistance is

infinite, that is, the out terminals are open circuited. 푉 occurs when the load resistance

is of the minimum value where voltage regulation is lost.

%퐿표푎푑푅푒푔푢푙푎푡푖표푛 = 푉 − 푉 /푉 ∗ 100

Minimum Load Resistance – The load resistance at which a power supply delivers its full-

load rated current at rated voltage is referred to as minimum load resistance.

푀푖푛푖푚푢푚퐿표푎푑푅푒푠푖푠푡푎푛푐푒 = 푉 퐼⁄

The value of 퐼 , full load current should never increase than that mentioned in the data

sheet of the power supply.



Source/Line Regulation – In the block diagram, the input line voltage has a nominal value of

230 Volts but in practice, here are considerable variations in ac supply mains voltage. Since this

ac supply mains voltage is the input to the ordinary power supply, the filtered output of the

bridge rectifier is almost directly proportional to the ac mains voltage.

The source regulation is defined as the change in regulated output voltage for a specified rage

of lie voltage.

Output Impedance – A regulated power supply is a very stiff dc voltage source. This means

that the output resistance is very small. Even though the external load resistance is varied,

almost no change is seen in the load voltage. An ideal voltage source has an output impedance

of zero.

Ripple Rejection – Voltage regulators stabilize the output voltage against variations in input

voltage. Ripple is equivalent to a periodic variation in the input voltage. Thus, a voltage

regulator attenuates the ripple that comes in with the unregulated input voltage. Since a

voltage regulator uses negative feedback, the distortion is reduced by the same factor as the

gain.

Rectifiers:

Diode is a unidirectional device. It provides high resistance in one direction and a low

resistance in the other direction. Therefore it can be used as a rectifier for converting AC to

DC.There are two types of rectifiers.

Half wave rectifier

Full wave rectifier

Half wave rectifier:

Half wave rectifier is the simplest form of the rectifier. The circuit arrangement is shown

in fig(a).the primary side is connected to the power mains(230 V,50 Hz).An AC voltage

induces across the secondary winding, which is equal to less than or greater than the



primary voltage depending upon the turns ratio of the transformer. This will be the input of

the rectifier.

When the transformer secondary voltage is on its positive half cycle, the diode is

forward biased and current flows through the load RL, developing a voltage across it. When

the negative voltage is on its negative half cycle, the diode is reverse biased and no current

flows through the RL.Thus there will be no voltage across the output terminal during

negative half cycle. The fig (b) shows the input and output wave forms of a half wave

rectifier. Here only positive voltages are obtained at the output and all negative half cycles

are suppressed. Thus we get a pulsating DC at the output.



Peak Inverse Voltage (PIV)

In the rectifier circuit, during negative half cycle of the secondary voltage, the diode is

reverse biased. As there is no voltage across the load resistor RL during this half cycle. The

whole secondary voltage will come across the diode. When the secondary voltage reaches

its maximum value Vm in the negative half cycle, the voltage across the diode in reverse bias

is also maximum. This maximum voltage is known as peak inverse voltage (PIV).It is the

maximum voltage the diode must withstand during the reverse bias half cycle of the input.

Therefore the diode is used in the rectifier circuit must have the higher PIV rating than this

voltage.

Disadvantages:

Low output because only one half cycle delivers the output. AC components are more in the output.

Performance of Half wave Rectifier:

DC Output:

The DC output voltage or a current of the rectifier is the average value of the output voltage or current.

Ie,

The average value of a sine wave over one complete cycle is zero because it is symmetrical. But in the case of a half wave rectifier output, it consists of only alternate cycles as shown in output wave form. Hence there will be some DC current flowing through the RL.

Let,

Alternating voltage appearing across the secondary winding of the transformer V = V Sinωt

Current flowing through the load resistor i = I Sinωt

Diode forward resistance r = 0

From wave form, it is clear that current flowing through the load only in the period 0 to 휋.

i.e., i = I Sinωt ∶ for0 ≤ ωt ≤ π

I = Areaunderonecompletecycle

Timeperiod



i = 0 ∶ forπ ≤ ωt ≤ 2π Here area = ∫ I Sinωtd(ωt)

= I [−Cosωt]

= 2 I

Time period = 2 π

Now, I = 2I /2π

I = I /π

Voltage across RL V = I R

= I /πR

While diode forward resistance r is taken into account

I = V (R + r )⁄

DC voltage across RL can be written as

V = V R /π(R + r )

= V π(1 + r /R )⁄

Since, r ≪R

RMS Value:

Root mean square (RMS) is the effective value of the current flowing through the load and is given by

In the case of half wave rectifier,

V = V π⁄

퐼 =퐴푟푒푎푢푛푑푒푟푡ℎ푒푠푞푢푎푟푒푑푟푒푐푡푖푓푖푒푑푤푎푣푒표푣푒푟푎푐푦푐푙푒

푇푖푚푒푝푒푟푖표푑

I = 1

2π i

π

d(ωt)

=1

2π I Sin ωtd(ωt)

=I2π

(1− Cos2ωt)

2d(ωt)



Hence,

This is the rms value of the total current which include DC value and AC components. In the output of the rectifier, the instantaneous value of AC fluctuation is the difference of the instantaneous total value and the DC value.

Instantaneous AC value is given by

i = i − I

Here the rms value of AC component is given by,

i = I − I

Ripple Factor:

The purpose of the rectifier is to convert AC voltage to DC. But not type of rectifier converts AC to pure DC. It produces pulsating DC. This residual pulsation is called ripple. Ripple factor indicates the effectiveness of the rectifier in converting AC to perfect DC. It is the ratio of the ripple voltage to the DC voltage.

= V /V

Or 훾 = I /I

= I − I /I

= I I⁄ − 1

=I

2π × 2 ωt − sin

2ωt2

tπ0

Irms = Im / 2

Vrms = Vm / 2

후 = 퐑퐢퐩퐩퐥퐞퐯퐨퐥퐭퐚퐠퐞퐃퐂퐯퐨퐥퐭퐚퐠퐞

=rmsvalueoftheACcomponent

DCcomponent



= (퐼 /2)/(퐼 /휋) − 1

= (1.57) − 1

= 1.21

Efficiency:

Efficiency of the rectifier is a measure of conversion of AC power to the useful DC output power. It is expressed as the ratio of DC output power to AC input power.

i.e., 휂 = 푃 푃⁄

Where 푃 is the power delivered to load and 푃 is the AC input power from the secondary winding of the transformer.

Here 푃 = 퐼 × 푅

= (퐼 휋⁄ ) × 푅

푃 = 퐼 (푅 + 푟 )

= (퐼 2⁄ ) (푅 + 푟 )

Therefore,

휂 =(퐼 휋⁄ ) × 푅

(퐼 2⁄ ) (푅 + 푟 )

= . 406푅(푅 + 푟 )

= . 406(1 + 푟 /푅 )

휂 = .406표푟40.6%

Transformer Utilization Factor:

For designing a power supply using transformer, transformer rating must be known. In order to

calculate the transformer utilization factor, the power supplied by the transformer must be

arranged in terms of the d.c power.

푉 × 퐼 = 푃



TUF is defined as the ratio of d.c output power to a.c power supplied to it by the secondary

winding,

i.e,푇푈퐹 = (푟푎푡푒푑)

As the rating of the transformer is different than the actual power delivered by the

secondary,the TUF is entirely different quantity than efficiency.

In case of a single diode half wave rectifier,the rated voltage of the secondary=Vm/√2 but the

actual r.m.s value of the current flowing through the secondary winding =Im/2 and not Im/√2.

Thus

푇푈퐹 =(퐼 휋⁄ ) × 푅푉√2

× 퐼2

=2√2휋

퐼 푅푉

푉 = 퐼 (푟 + 푅 )

푇푈퐹 =2√2휋

퐼 푅퐼 (푟 + 푅 )

Hence,

푇푈퐹 =2√2휋 = .287

Regulation of Half Wave Rectifier

Regulation is defined as the variation of output DC voltage with the changes in the output DC

current.

%푅푒푔푢푙푎푡푖표푛 =푉 − 푉

푉 × 100%

Where, 푉 = no load voltage.

푉 = full load voltage.

An ideal power supply should have a 0% regulation. The output voltage should not vary with

load current IL. For a half wave rectifier, the percentage regulation is given by,



%푅푒푔푢푙푎푡푖표푛 =

푉휋 − 푉

휋 − 푉휋푅 푅 + 푅

푉휋 − 푉

휋푅 푅 + 푅

Where,

푉 =푉휋

푉 = 푉 − 퐼

%푅푒푔푢푙푎푡푖표푛 =푉휋푉휋 ⎣⎢⎢⎢⎡ 1− 1 +

푅 + 푅푅

1 −푅 + 푅푅 ⎦

⎥⎥⎥⎤

%푅푒푔푢푙푎푡푖표푛 = −푅 + 푅푅

Form Factor:

Form factor is the ratio of rms value to the average value.

퐹표푟푚푓푎푐푡표푟 =푟푚푠푣푎푙푢푒

푎푣푒푟푎푔푒푣푎푙푢푒

푓표푟푚푓푎푐푡표푟 = 푉 2⁄푉 휋⁄

= 휋 2 = 1.57⁄

Peak Factor:

Peak factor is the ratio of peak value to rms value.

푃푒푎푘푓푎푐푡표푟 =푝푒푎푘푣푎푙푢푒푟푚푠푣푎푙푢푒

푓표푟푚푓푎푐푡표푟 = 푉푉 2⁄

= 2



Full Wave Rectifiers:

In a full wave rectifier, current flows through the load during both half cycles of the

input AC supply. Thus in full wave rectifier, alternate half cycle of the input AC are inverted to

get a unidirectional output current. Full wave rectifiers are of two types,

Centre tap rectifier

Bridge rectifier

Centre Tap Rectifier:

The circuit arrangement of a centre-tap full wave rectifier is shown in fig(a).during

positive half cycle of the secondary voltage, the diode D1 is forward biased and D2 is reverse

biased. The current flows through D1, RL and upper half of the secondary winding. During

negative half cycle of the secondary voltage, current flows through D2, RL and lower half of the

secondary winding. Here current through RL during both half cycle of input AC in the same

direction. Therefore the output voltage taken across RL will be DC. The input and output wave

forms of this rectifier are shown in fig (b).



Peak Inverse Voltage (PIV)

In the centre tap full wave rectifier, during positive half cycle of the secondary voltage

diode D1 conducts and when secondary voltage attains its maximum value Vm, a voltage equal

to Vm will develop across RL with the polarity marked in fig(c).The diode D2 at that instant is

reverse biased and the voltage across it will be the sum of the voltages developed across the

lower winding of the secondary (Vm) and the voltage developed across RL=Vm. Therefore peak

inverse voltage across the diode in this case will be Vm+Vm=2Vm.

i.e, PIV=2 Vm

Advantages:

High efficiency

Low ripple factor



Disadvantages:

Requires centre tap rectifier with secondary voltage twice as required for half wave or

bridge rectifier.

Diodes of high PIV (2V) rating are to be used.

Bridge Rectifier:

The circuit arrangement of the bridge rectifier is show in fig (a).It contains four diodes,

but avoids the need for a centre-taped transformer. During the positive half cycle of the

secondary voltage, diode D2 and D3 will be forward biased. At the same time diodes D1 and D4

are reverse biased. Therefore diodes D2 and D3 conduct and current flows through D2, RL, D3

and transformer secondary. During the negative half cycle of the secondary voltage diodes D1

and D4 will be forward biased and D2 and D3 will be reverse biased. Therefore the current flows

through D1, RL ,D4 and transformer secondary. Here in both cases the current flow through load

resistor RL is in the same direction. Hence DC voltage is obtained at the output. The wave forms

are shown in fig (b).



Peak inverse voltage (PIV)

When the secondary voltage is at its positive maximum value Vm diodes D2 and D3 are

forward biased and conduct. Conducting diodes have zero resistance and zero voltage drop

across them, the points A and C in fig (a) are in same potential. Similarly points B and D

.Therefore the reverse voltage coming across diodes D1 and D4 will be equal to the potential

difference between the points A and D will be equal to Vm.

Advantages:

Does not require centre tap rectifier.

Diodes of low PIV rating can be used.

Disadvantages:

Needs four diodes.



Performance of Full Wave Rectifier:

DC output:

In a full wave rectifier, it utilizes both half cycles of the input AC voltage. Therefore the DC voltage or average voltage available in the output of a full wave rectifier will be double the DC voltage available in the half wave rectifier.

i.e. 2× Vdc of the half wave rectifier

푉 = 2(푉 휋⁄ )

퐼 = 2(퐼 휋⁄ )

RMS Value:

When writing the rms value of the full wave rectifier both half cycles are to be considered. Hence the rms value of the current is given by,

Hence,

I = 1π i

π

d(ωt)

=1π I Sin ωtd(ωt)

=Iπ

(1− Cos2ωt)2

d(ωt)

=I2π

ωt− sin2ωt

2π0

Irms = Im / √2

Vrms = Vm / √2

=I π

2π



Ripple Factor:

훾 = I − I /I

= (I I⁄ ) − 1

= 퐼 /√2 /(2퐼 /휋) − 1

훾 = .482

Efficiency:

휂 = 푃 푃⁄

푃 = 퐼 × 푅

= (2퐼 휋⁄ ) × 푅

푃 = 퐼 (푅 + 푟 )

= 퐼 √2⁄ (푅 + 푟 )

휂 =(2 퐼 휋⁄ ) × 푅

퐼 √2⁄ (푅 + 푟 )

= . 812(1 + 푟 /푅 )

( Where 푅 ≫ 푟 )

휂 = .812표푟81.2%

TUF of Centre Tap Rectifier The two diode full-wave rectifier can be thought of equivalent to two half wave rectifiers feeding to a common load. Hence TUF with respect to two half wave secondary can be written as

TUF (secondary) = 2 TUF of half wave



= 2× .287 =.574

푇푈퐹(푃푟푖푚푎푟푦) =(2퐼 휋⁄ ) × 푅

푉√2

× 퐼√2

=8퐼휋

퐼 푅푉 퐼

=8휋 = .812

Thus,

푇푈퐹 = (푇푈퐹 + 푇푈퐹 )/2

=. 812+ .574

2 = .693

TUF of Bridge Rectifier

푇푈퐹 = 푃 푃⁄

푇푈퐹 =(2퐼 휋⁄ ) × 푅

푉√2

× 퐼√2

=8퐼휋

퐼 푅푉 퐼

=8휋 = .812

Regulation of Full Wave Rectifier

Regulation of output signal is defined as the variation of output DC voltage with the changes in the DC load current.

%푅푒푔푢푙푎푡푖표푛 =푉 − 푉

푉 × 100%

Where, 푉 = no load voltage.

푉 = full load voltage.

An ideal power supply should have a 0% regulation. The output voltage should not vary with load current IL. The % regulation is given by,



%푅푒푔푢푙푎푡푖표푛 =

2푉휋 − 2푉

휋 − 2푉휋푅 푅 + 푅

2푉휋 − 2푉

휋푅 푅 + 푅

Where,

푉 =2푉휋

푉 =2푉휋 − 퐼

%푅푒푔푢푙푎푡푖표푛 =2푉휋

2푉휋 ⎣

⎢⎢⎢⎡ 1− 1 +

푅 + 푅푅

1 −푅 + 푅푅 ⎦

⎥⎥⎥⎤

%푅푒푔푢푙푎푡푖표푛 = −푅 + 푅푅

Form Factor:

퐹표푟푚푓푎푐푡표푟 =푟푚푠푣푎푙푢푒

푎푣푒푟푎푔푒푣푎푙푢푒

푓표푟푚푓푎푐푡표푟 = (푉 √2)⁄(2푉 휋⁄ )

= 휋 2√2 = 1.11⁄

Peak Factor:

푃푒푎푘푓푎푐푡표푟 =푝푒푎푘푣푎푙푢푒푟푚푠푣푎푙푢푒

푃푒푎푘푓푎푐푡표푟 = 푉(푉 √2)⁄

= √2



Problems:

1. In a half wave rectifier the load resistance RL=1KΩ the forward resistance of the diode

rd=100Ω, input alternating voltage (Vm) =325 volt. Find (a) Peak value, average value and

rms value of the output current (b) Efficiency of the rectifier.

Solution

(a) Peak value

퐼 = 푉 (푟 + 푅 )⁄

= 325 (100 + 1000)⁄

퐼 = 295.45푚퐴

Average value

퐼 = 퐼 휋⁄

= 295.45 휋⁄

퐼 = 94.046푚퐴

Rms value

퐼 = 퐼 /2

= 295.45 2⁄

= 147.725푚퐴

(b) Efficiency

휂 = 푃 푃⁄

푃 = 퐼 × 푅

= (94.046 × 10 ) × 1000

푃 = 8.845푊

푃 = 퐼 (푅 + 푟 )

= (147.725 × 10 ) (100 + 1000)

= 24푊

휂 = 푃 푃⁄

= 8.845/24

휂 = .3685표푟36.85%



2. In a full wave centre tap rectifier, the load resistance used is of 2KΩ,the forward resistance

of each diode is 400Ω,the voltage across the half of the secondary winding is 240 Sin50t.Find

(a) Im, (b) Idc (c) Irms (d) ripple factor (e) PIV.

Solution

(a) 푰풎

퐼 = 푉 (푟 + 푅 )⁄

= 240 (400 + 2000)⁄

= .1퐴푚푝

(b) 푰풅풄

퐼 = 2 퐼 휋⁄

= (2 × .1) 휋⁄

= .636퐴푚푝 = 63.6푚퐴

(c) 푰풓풎풔

퐼 = 퐼 /√2

= . 1 √2⁄

= .070퐴푚푝 = 70.72푚퐴

(d) Ripple factor

훾 = (퐼 퐼⁄ ) − 1

= (. 07072/.0632) − 1

= .501표푟50.1%

(e) PIV

푃퐼푉 = 2푉 = 2 × 240 = 480 V

Filter Circuits:

The out put voltage of both half wave and full wave rectifier is pulsating and not pure DC.If

such a pulsating DC is given to any equipment, the working of the equipment will be disturbed.

This disturbance of ripples can be avoided by using filter circuits. Many types of passive filters

are in use such as,

Shunt capacitor filter.



Series inductor filter.

Chock input LC filter.

Π- Section Filter.

Shunt Capacitor Filter:

This type of filter consists of a large value capacitor connected across the load resistance

RL as shown in fig (a).This capacitor offers low reactance path to AC components and very high

impedance to DC.So that AC components in the rectifier output find low reactance path

through the capacitor and only a very small part flows through RL, producing a small ripple

voltage at the output as shown in fig (B).

When the diode conducts, the capacitor charges up to peak voltage Vm (point B in

figure).From this point onwards the rectifier voltage starts to fall, but the charged capacitor

tries to send current back through the diodes. Thus the diodes are reverse biased and become

open circuited. The capacitor then starts to discharge through the load. This action prevents the

load voltage from falling to zero during no conduction period of the diode. The capacitor

continues to discharge until the source voltage becomes more than the voltage across the

capacitor(at point C).Then the diode begins to conduct and the capacitor again charges to Vm

(portion CD in figure).During this time, the rectifier delivers the charging current as well as load

current. Thus the capacitor removes the AC components and also improves the output voltage.

The discharging rate of the capacitor depends upon the time constant C.RL. The ripple factor in

the capacitor filter is given by,

휸 = ퟏퟒ√ퟑ풇푪푹푳



Half Wave Rectifier with Capacitor Filter:

Full Wave Centre-tap Rectifier with Capacitor Filter:



Full Wave Bridge Rectifier with Capacitor Filter:



Ripple factor of the full wave rectifier with a capacitor filter

From the figure, it is obvious that discharging of the capacitor is given by,

푄 = 퐼 × 푇

The charge acquired by capacitor is,

푄 = 푉 × 퐶

We know that the charge acquired by the capacitor during the charging is equal to the charge

lost during discharge.

푄 = 푄



푉 × 퐶 = 퐼 × 푇

푉 × 퐶 = × 푇

푉 × 퐶 = × ---------------------(1)

For a full wave rectifier the capacitor charges and discharges the charge carries in both the half

cycles and the total time period is equal to charging and discharging of charges in positive cycle

plus charging and discharging of charges in negative cycle.

T = (T1 + T2) |positive cycle + (T1 + T2) |negative cycle

Since Tdischarge >> Tcharge , charging time Tcharge can be neglected.

Therefore, 푇 = -----------------------(2)

Substituting equation (2) in equation (1)

Vrp-p = ×

For the ripple waveform,

Vr rms = √

Vrp-p = 2√3 Vr rms ------------------------(3)

Substituting equation (3) in equation (2)

2√3 Vr rms = ×

Vr rms = √

But ripple factor is given by,

γ = = √

Advantages

Low cost

Small size and weight

Good characteristics

Can be connected for both HW and FW rectifiers

Improved d.c. output



Disadvantages

Capacitor draws more current

Series inductor filter

The working of series inductor filter depends on the inherent property of the inductor to oppose

any variation in current intend to take place. Figure below shows a series inductor filter connected at

the output of a FWR. Here the reactance of the inductor is more for ac components and it offers more

opposition to them. At the same time it provides no impedance for d.c. component. Therefore the

inductor blocks a.c. components in the output of the rectifier and allows only d.c. component to flow

through RL.

The action of an inductor depends upon the current through it and it requires current to flow at all time.

Therefore filter circuits consisting inductors can only be used together with full wave rectifiers.

The operation of the inductor filter depends on its property to oppose any change of current passing through it. To analyze this filter for full wave,

The Fourier series can be written

as

The dc component is .

Assuming the third and higher terms contribute little output, the output voltage is



The diode, choke and transformer resistances can be neglected since they are very small compared with RL. Therefore the dc component

of current

The impedance of series combination of L and RL at 2w is

Therefore for the ac component,

Therefore, the resulting current i is given by,

The ripple factor which can be defined as the ratio of the rms value of the ripple to the dc value of the wave, is

If , then a simplified expression for g is

Advantages

Sudden changes in current is smoothen out

Improved filtering action at high load currents across the inductor.



Disadvantages

Reduced output voltage due to the drop

Bulky and large in size

Note suite for HWR

LC filter

It is a combination of inductor and capacitor filter. Here an inductor is connected in series and a

capacitor is connected in parallel to the load as shown in fig 5.6. As discussed earlier, a series inductor

filter will reduce the ripple, when increasing the load current. But in case of a capacitor filter it is reverse

that when increasing current the ripple also increases. So a combination of these two filters would make

ripple independent of load current.

From Fourier series, the output voltage can be expressed as

The dc output voltage,



The ripple factor

Advantages

Reduced ripples at the output

Action is independent of load current

Disadvantages

Low output voltage

Bulky and large in size

Not suit to connect with HWR.

π – filter (Capacitor input filter) or CLC filter

This filter is basically a capacitor filter followed by an LC filter as shown in fig 5.8. Since its shape (C-LC)

is like the letter π it is called π – filter. It is also called capacitor input filter because the rectifier feeds

directly into the capacitor C1. Here the first capacitor C1 offers a low reactance to a.c. component of

rectifier output but provide more reactance to d.c components. Therefore most of the a.c. components

will bypass through C1 and the d.c. component flows through chock L. The chock offers very high

reactance to the a.c. component. Thus it blocks a.c. components while pass the d.c. The capacitor C2

bypasses any other a.c. component appears across the load and we get study d.c. output as shown

below.



Advantages

More output voltage

Ripple less output

Suitable to be used with both HWR and FWR

Disadvantages

Large in size and weight

High cost

Need of Regulated Power Supply

In an ordinary power supply, the voltage regulation is poor i.e. d.c. output voltage changes

appreciably with load current. Moreover, output voltage also changes due to variations in the

input a.c. voltage. This is due to the following reasons :

(i) In practice, there are considerable variations in a.c. line voltage caused by outside factors

beyond our control. This changes the d.c. output voltage. Most of the electronic circuits will

refuse to work satisfactorily on such output voltage fluctuations. This necessitates to use

regulated d.c. power supply.

(ii) The internal resistance of ordinary power supply is relatively large (> 30 Ω). Therefore,

output voltage is markedly affected by the amount of load current drawn from the supply. These

variations in d.c. voltage may cause erratic operation of electronic circuits. Therefore, regulated

d.c. power supply is the only solution in such situations.



Types of Voltage Regulators

A device which maintains the output voltage of an ordinary power supply constant irrespective

of load variations or changes in input a.c. voltage is known as a voltage regulator. A voltage

regulator generally employs electronic devices to achieve this objective. There are basic two

types of voltage regulators viz., (i) shunt voltage regulator (ii) series voltage regulator.

Zener Voltage Regulator:

The function of regulator is to provide an output voltage V0 that is as constant as

possible in spite of the ripple in source voltage, Vs and the variations in the load current IL. Two

parameters used to measure the performance of voltage regulator are line regulation and load

regulation.

Line regulation is the ratio of change in output voltage to the change in source voltage.

퐿푖푛푒푟푒푔푢푙푎푡푖표푛 =∆푉∆푉

Load regulation is defined as the ratio of change in output voltage to the change in load

current.

퐿표푎푑푟푒푔푢푙푎푡푖표푛 =∆푉∆퐼

Working of Zener Voltage regulator

The figure shows the basic circuit arrangement for a zener voltage regulator. Here the

zener diode with a break down voltage Vz is connected in reverse bias across the load. A

resistor Rs is also connected in series. If a voltage Vs is applied to the circuit, until the voltage

across the load is less than the break down voltage Vz the zener diode does not conduct and no

current flows through the zener diode. At this time the same voltage Vs will appear across the

load. When the input voltage Vs is more than Vz, the zener diode will conduct and a current Iz

will flow through it. This excess current will increase the drop across Rs and limits the zener

current from exceeding the maximum rated value.Thus the current ‘I’ from the DC supply splits

to the load and to the zener diode.



At this time when the zener diode conducts the voltage across it is equal to its break

down voltage Vz and remains fairly constant even when the current through it vary

considerably. Thus a constant voltage is maintained across the load.

A figure of merit for a voltage regulator is called the percent regulation, and is defined as

% Regulation = ( ) ( )

( )× 100

where VL(nom) is the nominal value of the output voltage.

As the percent regulation approaches zero percent, the circuit approaches that of an ideal

voltage regulator.

Analysis and Design

We first determine the proper input series resistance Rs. This resistance limits the current

through the Zener diode and drops the excess voltage between Vs and VZ.

Rs = =

which assumes that the Zener resistance is zero for the ideal diode. Solving this equation for

the diode current. Iz, we get



Iz = − 퐼

where IL = Vz / RL , and the variables are the input voltage source VS and the load current IL.

For proper operation of this circuit, the diode must remain in the breakdown region and the

power dissipation in the diode must not exceed its rated value. In other words:

1. The current in the diode is a minimum, Iz(min), when the load current is a maximum,

IL(max), and the source voltage is a minimum, VS(min).

2. The current in the diode is a maximum, Iz(max), when the load current is a minimum,

IL(min), and the source voltage is a maximum, VPS(max).

Inserting these two specifications into the previous equation, we obtain

Rs = ( )

( ) ( )

and

Rs = ( )

( ) ( )

Equating these two expressions, we then obtain

[푉 ( ) − 푉 ][퐼 ( ) + 퐼 ( )] =[푉 ( ) − 푉 ][퐼 ( ) + 퐼 ( )]

Reasonably, we can assume that we know the range of input voltage, the range of output

load current, and the Zener voltage. The previous equation then contains two unknowns

Iz(min) and Iz(max). Further, as a minimum requirement, we can set the minimum Zener current



to be one-tenth the maximum Zener current, or IZ(min) =0.1 IZ(max). We can then solve for IZ(max),

using the previous equation, as follows:

Iz(max) = ( ) ( ) ( )[ ( ) ]

( ) . . ( )

Use the maximum current obtained from the above equation, we can determine the

maximum required power rating of the Zener diode. Then we can determine the required

value of the input resistance using one of the previous equations.

Limitations

A basic zener voltage regulator has the following draw backs :

(i) It has low efficiency for heavy load currents. It is because if the load current is large, there

will be considerable power loss in the series limiting resistance.

(ii) The output voltage slightly changes due to zener impedance as Vout = VZ + IZ ZZ. Changes in

load current produce changes in zener current. Consequently, the output voltage also changes.

Therefore, the use of this circuit is limited to only such applications where variations in load

current and input voltage are small.

Transistor Shunt Voltage Regulator- Block Diagram

The block diagram of a discrete transistor shunt voltage regulator is given below. As the

name says the voltage regulation is provided by shunting the current away from the load. The

control element shunts a part of the current that is produced as a result of the input unregulated

voltage that is given to the load. Thus the voltage is regulated across the load. Due to the change

in load, if there is a change in the output voltage, it will be corrected by giving a feedback signal

to the comparator circuit which compares with a reference voltage and gives the output control

signal to the control element to correct the magnitude of the signal required to shunt the current

away from the load.



If the output voltage increases, the shunt current increases and thus produces less load current

and maintains a regulated output voltage. If the output voltage reduces, the shunt current reduces

and thus produces more load current and maintains a regulated constant output voltage.

Transistor Shunt Voltage Regulator-Circuit Diagram

A shunt voltage regulator provides regulation by shunting current away from the load to regulate

the output voltage. Fig. below shows the circuit of shunt voltage regulator.

Operation

As there is a voltage drop in the series resistance Rseries the unregulated voltage is also decreased

along with it. The amount of voltage drop depends on the current supplied to the load Rload. The

value of the voltage across the load depends on the zener diode and the transistor base emitter

voltage Vbe.

Thus, the output voltage can be written as

Vout = Vz + Vbe

The output remains nearly a constant as the values of Vz and Vbe are nearly constant. This

condition is explained below.



When the supply voltage increases, the output voltage and base emitter voltage of transistor

increases and thus increases the base current Ib and therefore causes an increase in the collector

current Ic (Ic = β.Ib).

Thus, the supply voltage increases causing an increase in supply current, which inturn causes a

voltage drop across the series resistance Rseries and thereby decreasing the output voltage. This

decrease will be more than enough to compensate for the initial increase in output voltage. Thus,

the output remains nearly a constant. The working explained above happens in reverse if the

supply voltage decreases.

When the load resistance Rload increases,the ouput voltage Vout also increases.This increases the

voltage across Vbe alone as Vz is constant.So, zener current throught the base emitter junction of

the transistor increases. As base current Ib increases,the collector current Ic also increases.The

input current is the sum of base current Ib, collector current Ic and load current IL.

I = Ib + IC + ILoad

The increase in collector current Ic decreases the load current IL and hence output voltage Vout is

regulated. It happens in reverse if there is a decrease in load resistance.



Drawbacks.

A shunt voltage regulator has the following drawbacks :

i) A large portion of the total current through RS flows through transistor rather than to the load.

ii) There is considerable power loss in RS. iii) There are problems of overvoltage protection in this circuit.

For these reasons, a series voltage regulator is preferred over the shunt voltage regulator.

Series Voltage Regulator

The series voltage regulator or series pass voltage regulator uses a regulating element placed in

series with the load. By changing the resistance of the series element, the voltage dropped across

it can be varied to ensure that the voltage across the load remains constant.

The output voltage is sampled by a circuit that provides a feedback voltage to be compared to a

reference voltage.

1. If output voltage increases, the comparator circuit provides a control signal to cause the series

control element to decrease the output voltage-thereby maintaining the output voltage.

2. If the output voltage decreases, the comparator circuit provides a control signal to cause the

series control element to increase the amount of the output voltage.

The advantage of the series voltage regulator is that the amount of current drawn is effectively

that used by the load, although some will be consumed by any circuitry associated with the

regulator. Unlike the shunt regulator, the series regulator does not draw the full current even

when the load does not require any current. As a result the series regulator is considerably more

efficient.



Transistor Series Voltage Regulator

Figure below shows a simple series voltage regulator using a transistor and zener diode.

The circuit is called a series voltage regulator because the load current passes through the series

transistor Q1 . The unregulated d.c. supply is fed to the input terminals and the regulated output is

obtained across the load. The zener diode provides the reference voltage.

Operation

When the input supply voltage Vin increases the output voltage Vout also increases. This

increase in Vout will cause a reduced voltage of the transistor base emitter voltage Vbe as the

zener voltage Vz is constant. This reduction in VBE causes a decrease in the level of conduction

which will further increase the collector-emitter resistance of the transistor and thus causing an

increase in the transistor collector-emitter voltage and all of this causes the output voltage Vout to

reduce. Thus, the output voltage remains constant. The operation is similar when the input

supply voltage decreases.

The next condition would be the effect of the output load change in regard to the output

voltage. Let us consider a case where the current is increased by the decrease in load resistance

RL. This causes a decrease in the value of output voltage and thus causes the transistor base

emitter voltage to increase. This causes the collector emitter resistance value to decrease due to

an increase in the conduction level of the transistor. This causes the input current to increase

slightly and thus compensates for the decrease in the load resistance RL.

The biggest advantage of this circuit is that the changes in the zener current are reduced by a

factor β and thus the zener effect is greatly reduced and a much more stabilized output is

obtained.



Limitations

(i) Although the changes in zener current are much reduced, yet the output is not

absolutely constant. It is because both VBE and VZ decrease with the increase in room

temperature.

(ii) The output voltage cannot be changed easily as no such means is provided.

Series Feedback Voltage Regulator

Figure below shows the circuit of series feedback voltage regulator. It employs principles

of negative feedback to hold the output voltage almost constant despite changes in line voltage

and load current. The transistor Q2 is called a pass transistor because the entire load current

passes through it. The sample and adjust circuit is the voltage divider that consists of R1 and R2.

The voltage divider samples the output voltage and delivers a negative feedback voltage to the

base of Q1. The feedback voltage VF controls the collector current of Q1.

Operation.

The unregulated d.c. supply is fed to the voltage regulator. The circuit maintains constant output

voltage irrespective of the variations in load or input voltage. Here is how the circuit operates.

(i) Suppose the output voltage increases due to any reason. This causes an increase in

voltage across KL (i.e., R2 ) as it is a part of the output circuit. This in turn means that

more VF is fed back to the base of transistor Q1; producing a large collector current of



Q1. Most of this collector current flows through R3 and causes the base voltage of Q2

to decrease. This results in less output voltage i.e., increase in voltage is offset. Thus

output voltage remains constant.

(ii) Similarly, if output voltage tries to decrease, the feedback voltage VF also decreases.

This reduces the current through Q1 and R3. This means more base voltage at Q2 and

more output voltage. Consequently, the output voltage remains at the original level.

Short-Circuit Protection

The main drawback of any series regulator is that the pass transistor can be destroyed by

excessive load current if the load is accidentally shorted. To avoid such an eventuality, a current

limiting circuit is added to a series regulator as shown in Figure below. A current limiting circuit

consists of a transistor (Q3) and a series resistor (R4) that is connected between base and

emitter terminals.

(i) When the load current is normal, the voltage across R4 (= voltage across base-emitter of Q3)

is small and Q3 is off. Under this condition, the circuit works as described earlier.



(ii) If load current becomes excessive, the voltage across R4 becomes large enough to turn on

Q3. The collector current of Q3 flows through R3, thereby decreasing the base voltage of Q2. The

decrease in base voltage of Q2 reduces the conduction of pass transistor (i.e., Q2), preventing

any further increase in load current. Thus, the load current for the circuit is limited to about 700

mA.

Clippers

There are a variety of diode networks called clippers that have the ability to “clip” off a

portion of the input signal without distorting the remaining part of the alternating waveform.

Depending on the orientation of the diode, the positive or negative region of the input signal is

“clipped” off. Often, dc battery is also used to fix the clipping level. The input waveform can be

clipped at different levels by simply changing the battery voltage and by interchanging the

position of various elements.

Clipper circuits are used in radars and digital computers etc. when it is desired to

remove signal voltages above or below a specified voltage level. Another application is in radio

receivers for communication circuits where noise pulses that rise well above the signal

amplitude are clipped down to the desired level.

There are two general categories of clippers: series and parallel. The series configuration

is defined as one where the diode is in series with the load, while the parallel variety has the

diode in a branch parallel to the load. Here we assume that the diode is ideal. This simply

means that the turn-on voltage of the diode is any voltage greater than 0 V, and the diode

resistance is 0 Ω.

Series Clippers

When the diode is connected in series with the load to form a clipper, the

configuration is called series clippers. Series clippers can be used to clip either the positive or

negative portions of the input signal.



Series Negative Clipper

During the positive half cycle of the input signal, the diode is turned on and it acts as

closed switch. Hence no voltage drops across the diode. All the input voltage appears across the

load, and hence the ouptut signal voltage is same as input voltage.

During negative half cycle of the input voltage, the diode is reverse biased and acts as open

switch. In such a case, no current flows through the diode and the load, and hence the output

voltage across the load is zero.

Fig: Series Negative Clipper

Biased Series Negative Clippers

If we want to change/adjust the clipping level of AC voltage, then external biasing

voltage must be used. The figure given below shows a biased (series) clipper. A biasing voltage

is connected in series with the diode as shown in the figure.



Fig: Biased Series Negative Clipper

When the input signal is positive, it tries to establish the current through the diode.

However, the biasing voltage V opposes the flow of current and tries to keep the diode in “off”

state. As soon as the signal voltage rises above voltage V, the diode starts to conduct and the

current flows through the load. The output appears across the load as soon as the current flows

though it. Maximum value of output voltage is Vm – V.

Series Positive Clipper

Positive clippers are used to clip positive portions of the input signal and allow the

negative portions of the signal to pass through .That means,during the positive half cycle of

the input signal, diode is reverse biased and acts as open switch. In such a case, no current

flows through the diode and the load, and hence the output voltage across the load is zero.



Fig: Series Positive Clipper

During negative half cycle of the input voltage, the diode is turned on and it acts as

closed switch. Hence no voltage drops across the diode. All the input voltage appears across the

load, and hence the ouptut signal voltage is same as input voltage.

Biased Series Positive Clippers

When the input signal is positive, the diode is reverse biased, it is considered to be in

“off” state and it acts as an open switch. As the signal voltage goes positive, it aids the battery

voltage in reverse biasing the diode and consequently the reverse bias voltage across the diode

increases. For the complete positive half cycle, the diode stays in reverse bias and acts as open

switch.

Now consider the instant when the signal voltage goes negative. As soon as the signal

voltage goes negative, it starts to oppose the battery voltage and tries to forward bias the

diode. However at this point, the signal voltage is not sufficient to forward bias the diode. Once



the signal voltage is greater than the battery voltage (we assume Vm to be greater than battery

voltage V) it forward biases the diode. The diode is now in the “on” state and output voltage is

developed acoss the load. Minimum value of output voltage is -(Vm – V).

Fig: Biased Series Positive Clipper

In all the above discussions, note that the diode is considered to be ideal one. In a practical

diode, the breakdown voltage will exist (0.7 V for silicon and 0.3 V for Germanium).

Parallel Clippers

In parallel or shunt clipping circuits, the diode is connected in parallel to the output. In this

discussions, the diode is considered to be a practical one. In a practical diode, the breakdown

voltage will exist (0.7 V for silicon and 0.3 V for Germanium).



Parallel Positive Clipper

In this diode clipping circuit, the diode is forward biased during the positive half cycle of the

sinusoidal input waveform. For the diode to become forward biased, it must have the input

voltage magnitude greater than +0.7 volts (0.3 volts for a germanium diode).

When this happens the diodes begins to conduct and all positive half cycles are bypassed

through the diode to the ground. Thus the output voltage which is taken across the diode can

never exceed 0.7 volts during the positive half cycle. During the negative half cycle, the diode is

reverse biased and does not conduct. As a result input voltage will appears at the output. In

this way positive clipper passes only negative going half cycle of the input to the output.

Fig: Parallel Positive Clipper

Parallel Negative Clipper

The diode is forward biased during the negative half cycle of the sinusoidal waveform and limits

or clips it to -0.7 volts while allowing the positive half cycle to pass unaltered when reverse

biased. As the diode limits the negative half cycle of the input voltage it is therefore called a

negative clipper circuit.



Fig:Parallel Negative Clipper

Combinational Clipper

If we connected two diodes in inverse parallel as shown, then both the positive and negative

half cycles would be clipped as diode D1 clips the negative half cycle of the sinusoidal input

waveform while diode D2 clips the positive half cycle.

For ideal diodes the output waveform above would be zero. However, due to the forward bias

voltage drop across the diodes the actual clipping point occurs at +0.7 volts and -0.7 volts

respectively. But we can increase this ±0.7V threshold to any value we want up to the

maximum value, (Vm) of the sinusoidal waveform either by connecting together more diodes in

series creating multiples of 0.7 volts, or by adding a voltage bias to the diodes.



Fig: Combinational clipper

Biased Diode Clipping Circuits

To produce diode clipping circuits for voltage waveforms at different levels, a bias

voltage, V is added in series with the diode as shown. The voltage across the series combination

must be greater than V + 0.7V before the diode becomes sufficiently forward biased to conduct.

For example, if the V level is set at 4.0 volts, then the sinusoidal voltage at the diode’s anode

terminal must be greater than 4.0 + 0.7 = 4.7 volts for it to become forward biased.

Biased Positive Parallel Clippers

Here the positive terminal of the external bias is connected to the cathode of the diode.

Therefore the diode is reverse biased until the voltage across it is greater than V+.7V (here the

voltage level is 4+.7= 4.7 Volt).When the input signal voltage level is greater than 4.7 volt, the

diode becomes forward biased and begin to conduct. Any voltage levels above 4.7 volt are

clipped off.



Fig: Biased Positive Parallel Clipper

Biased Negative Parallel Clippers

Here the negative terminal of the external bias is connected to the anode of the diode.

Therefore the diode is reverse biased until the voltage across it is less than –V-.7V (here the

voltage level is -4-.7= -4.7 Volt).When the input signal voltage level is less than 4.7 volt, the

diode becomes forward biased and begin to conduct. Any voltage levels below -4.7 volt are

clipped off.

Fig: Biased Negative Parallel clipper



Biased Combinational Clippers

During the positive half cycle of the input signal,first branch is effective and others

remain open. When the voltage of the positive half cycle reaches +4.7 V, diode D1 conducts and

limits the waveform at +4.7 V. Diode D2 does not conduct during this period. During the

negative half cycle of the input, second branch will effective and first branch is open. When the

voltage of the negative half cycle reaches -4.7 V, diode D2 conducts and limits the waveform at -

4.7 V. Diode D1 does not conduct during this period. Therefore, all positive voltages above +4.7

V and negative voltages below –4.7 V are automatically clipped.

Fig: Biased combinational Clipper

The advantage of biased diode clipping circuits is that it prevents the output signal from

exceeding preset voltage limits for both half cycles of the input waveform, which could be an

input from a noisy sensor or the positive and negative supply rails of a power supply.

If the diode clipping levels are set too low or the input waveform is too great then the

elimination of both waveform peaks could end up with a square-wave shaped waveform.



Drawbacks of Series and Parallel Diode Clippers

In series clippers, when the diode is in ‘OFF’ position, there will be no transmission of input

signal to output. But in case of high frequency signals transmission occurs through diode

capacitance which is undesirable. This is the drawback of using diode as a series element in

such clippers.

In shunt clippers, when diode is in the ‘off condition, transmission of input signal should

take place to output. But in case of high frequency input signals, diode capacitance affects

the circuit operation adversely and the signal gets attenuated (that is, it passes through

diode capacitance to ground).

Clampers

The clamping network is one that will “clamp” a signal to a different dc level. The

network must have a capacitor, a diode, and a resistive element, but it can also employ an

independent dc supply to introduce an additional shift. The magnitude of R and C must be

chosen such that the time constant τ= RC is large enough to ensure that the voltage across the

capacitor does not discharge significantly during the interval the diode is non conducting.

Clamping circuits are often used in TV receivers as dc restorers. The incoming composite

video signal is normally processed through capacitively-coupled amplifiers which eliminate the

dc component thereby losing the black and white reference levels and the blanking level. These

reference levels have to be restored before applying the video signal to the picture tube.

Note that voltage drop across the diode (.7 volt) is considered for representing the

output wave forms of the following clamper circuits and it is avoided in the discussions.

Positive Clamper

Let the input signal be Vm sint. During the negative half cycle of the input sine wave,

the diode conducts and capacitor charges to Vm (peak value of the input signal) with positive

polarity at the right side of the capacitor.Here we take the Vm as 5 volt. During the positive half

cycle of the input signal, the capacitor can’t discharge since the diode does not conduct. Thus



the capacitor acts as a DC source of Vm volts connected in series with the input signal source.

Then the output voltage can be expressed as Vo= Vm+Vm sint.

Fig: Positive Clamper

Negative Clamper

During the positive half cycle of the input sine wave, the diode conducts and capacitor

charges to Vm with negative polarity at the right side of the capacitor. During the negative half

cycle of the input signal, the capacitor can’t discharge since the diode does not conduct. Thus

the capacitor acts as a DC source of -Vm volts connected in series with the input signal source.

Then the output voltage can be expressed as Vo= -Vm+Vm sint.

Fig: Negative Clamper



Biased Positive Clamper

Here we take a 3 volt battery as DC source connected in series with the diode in such a

way that positive terminal of the battery is connected to the anode terminal of the diode.

During the negative half cycle of the input sine wave, the diode conducts and capacitor charges

through diode and the DC source till (Vm +3) volts with positive polarity at the right side of the

capacitor. The charging of the capacitor is limited to (Vm +3) volts due to the presence of the DC

source. The output is then represented as Vo= (Vm +3)+ Vm sint.

Fig: Biased Positive Clamper

Biased Negative Clamper

Here we take a 3 volt battery as DC source connected in series with the diode in such a

way that negative terminal of the battery is connected to the cathode terminal of the diode.

During the positive half cycle of the input sine wave, the diode conducts and capacitor charges

through diode and the DC source till (Vm +3) volts with negative polarity at the right side of the

capacitor. The charging of the capacitor is extended up to (-Vm-3) volts due to the presence of

the DC source. The output is then represented as Vo= -Vm -3+ Vm sint = -(Vm +3)+ Vm sint.



Fig: Biased Negative Clamper

High Pass RC Circuit as a Differentiator

The High pass RC circuit is also known as a differentiator. The name high pass is so called

because the circuit blocks the low frequencies and allows high frequencies to pass through it. It

is due to reason that reactance of the capacitor decreases with the increasing frequency. As a

result of this, at very high frequencies, the capacitor acts as a short circuit and virtually all the

input appears at the output.

A circuit in which the output voltage is directly proportional to the derivative of the input

voltage is called a differentiating circuit. Mathematically, the output voltage is given by:

Output ∝ d/dt (input)



Vout ∝ dvin/dt

or Vout = RC d/dt Vin

Where RC is a constant of proportionality.

A differentiating circuit is a simple RC series circuit with output taken across the resistor R. The

circuit is designed in such a way that output is proportional to the derivative of the input. Thus

if a d.c. or constant input is applied to such a circuit, the output will be zero. It is because the

derivative of the constant is zero.

In order to achieve a good differentiation, the following two conditions should be satisfied. The time constant RC of the circuit should be smaller than the time period of the input

signal. The value of a capacitive reactance X should be 10 or more time larger then R at the

operating frequency. Let Vin , be the input alternating voltage and let i be The resulting alternating current.

The charge q on the capacitor at any instant is:

q = CV

i = dq/dt

i = d/dt (CVin )

i = C d/dt (Vin )

Since the capacitive reactance is very larger than R, the input voltage can be consider

equal to the capacitor voltage without any error,, i.e Vc = Vin ,

output voltage is given by:

Vout = iR

Or Vout =(C d/dt (Vin ))R

Vout =RC d/dt Vin

where RC is a constant, and Hence

Output ∝ d/dt (input)

The output waveform from a differentiating circuit depends upon the time constant and

the shape of the input signal.



Fig: Response of RC Differentiator to a Square Wave Input

Applications:

Some important applications of a differentiating circuit are given as under:

To generate a square wave from a triangular wave input.

To generate a step from a ramp input.

To generate a series of narrow pulses called spikes from the rectangular or square

waveform. The pips are used as trigger pulses or synchronization pulses in circuits used

in television and cathode ray oscilloscopes.

Low Pass RC Circuit as Integrator

Low pass circuits also known as an integrator. The name low pass circuit is designated because

of the fact that the circuit pass low frequencies but attenuates high frequencies.

A circuit in which the output voltage is directly proportional to the integral of the input voltage

is called an integrating circuit. Mathematically, the output voltage is given by:

output ∝ ∫ (input) Vout ∝ ∫Vin dt

or

Vout =(1/ RC) ∫Vin dt



Where RC is a constant of proportionality.

In order to achieve a good integration, the following conditions must be satisfied.

The time constant RC of the circuit should be very large as compared to the time period

of the input signal.

The value of R should be 10 or more times larger than X,.

Let Vin be the input alternating voltage and let i be the resulting alternating current. Since R is

very large as compared to capacitive reactance XC of the capacitor. It may be assume that

voltage across R is equal to the input voltage, i.e.

VR = Vin

Now

i = VR /R = Vin /R

The charge q on the capacitor at any instant is:

Q =∫i dt

Output voltage is given by:

Vout = = ∫ 푖푑푡

Vout = ∫ 푑푡

Vout = ∫푉 푑푡

Vout ∝∫푉푖푛푑푡

Output ∝ ∫(푖푛푝푢푡)

The output waveform from an integrating circuit depends upon the time constant and the

shape of the input signal.



Applications:

Some important applications of an integrating circuit are given as under:

To perform mathematical integration in analogue computers.

To generate a triangular wave from a square wave.

To generate a sawtooth wave from a rectangular wave.

To trigger the electronics devices.

Analysis of L, C, LC & π filters. · 2015-08-19 · Differentiator and Integrator circuits...

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