LINEAR INTEGRATED CIRCUITS

Course Code: 19ECE46

Module 4 : FILTERS AND IC REGULATORS

Richard Lincoln Paulraj 1
 Introduction
• Filters are circuits that pass only a certain range of signal
frequencies and attenuate unwanted frequencies. This
property of filters is also called “frequency selectivity”.
• There are two broad categories of filters:
✔An analog filter processes continuous-time signals
✔A digital filter processes discrete-time signals.
• Filters are usually classified according to the band of
frequencies they pass.
• Low pass filter
• High pass filter
• Band pass filter
• Notch filter ( Band Stop filter)

Richard Lincoln Paulraj 2

 Introduction Cont . .

• Passive Filters : Employ only passive components such as

resistors, capacitors and inductors.
• Active filers : Uses amplifiers , along with passive components
• Analysis of filter circuits can be extremely complex, but the
operation of most commonly used filters can be readily
understood, and simple design techniques are available.

Filters types and characteristics

• To represent filter characteristics, gain Vs frequency is plotted.

• Signal frequency (f) is plotted to logarithmic scale and voltage gain
( Av= Vo / Vi) in decibel scale

Richard Lincoln Paulraj 3

 Analog Filter Responses

H(f) H(f)

0 f 0 f
fc fc

Ideal “brick wall” filter Practical filter

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 Ideal Filters

Lowpass Filter Highpass Filter

M( )

Passband Stopband Stopband Passband

c c

Bandstop Filter Bandpass Filter

M( )

Passband Stopband Passband Stopband Passband Stopband

c c c c
1 2 1 2

Richard Lincoln Paulraj 5

 Passive Filters
• Made up of passive components - resistors, capacitors and
inductors
• No amplifying elements (transistors, op-amps, etc)
•
No signal gain
• to
1st order - design is simple (just use standard
• equations find resonant
2n order - complex frequency of the circuit)
equations
• dRequire no power supplies
• Buffer amplifiers might be required
●
Desirable to use inductors with high quality factors

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 Inductor - BIG PROBLEM!
• Physical size, and large inductance values are required.
• Tuning inductors to the required values is time-consuming
and expensive for larger quantities of filters.
• Often prohibitively expensive.
●
Difficult to implement at frequencies below 1 kHz.
●
Lossy

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 Active Filter

• No inductors
• Made up of op-amps, resistors and capacitors
• Provides arbitrary gain
• Generally easier to design
• High input impedance prevents excessive loading of the driving
source
• Low output impedance prevents the filter from being affected
by the load
• Easy to adjust over a wide frequency range without altering the
desired response

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Types of filters based on Fall-of rate
First order filter :
• 20 db/ decade fall of rate in transition
band
• Single pole filter
• Only one reactive component

Second order filter :

• 40 db/ decade fall of rate in transition
band
• Two pole filter
• Two reactive component

Third order filter :

• 60 db/ decade fall of rate in transition
band
• Three pole filter
• Three reactive component

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 Op Amp Advantages

• Advantages of active RC filters

include:
– Reduced size and weight
– Increased reliability and improved performance
– Simpler design than for passive filters and can realize a
wider range of functions as well as providing voltage gain
– In large quantities, the cost of an IC is less than its passive
counterpart

Richard Lincoln Paulraj 10

 Op Amp Disadvantages
• Active RC filters also have some disadvantages:
– limited bandwidth of active devices limits the highest
attainable pole frequency and therefore applications
– nearby
100 kHz (passive RLC filters can be used up to 500 MHz)
– require power supplies (unlike passive filters)
increased sensitivity to variations in circuit parameters
caused by environmental changes compared to passive
filters.
• For many applications, particularly in voice and data
communications, the economic and performance advantages of
active RC filters far outweigh their disadvantages.

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 Categories of Filters
Low Pass Filters: High Pass Filters:
Pass all frequencies from dc Pass all frequencies that are
up to the upper cutoff above its lower cutoff
frequency. frequency
Av(
Av( dB
dB
)
)
-3 { -3
d
{
d
B
B

f f
f f1

Low-pass response High-pass response

Richard Lincoln Paulraj 12
 Categories of Filters
Band Pass Filters: Band Stop (Notch) Filters:
Pass only the frequencies Eliminate all signals within
that fall between its values the stop band while passing
of the lower and upper all frequencies outside this
cutoff frequencies. band.
A v(
Av( dB
dB )
)

-3 { -3
d
{
d
B
B

f f
f f f f
1 2 1 2

Band Pass Response Band Stop Response

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Filter Response Characteristics
Av
Different filter circuit design
provides different performance
characteristics.

Butterworth
Bessel
Chebyshev

f
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ACTIVE LOW PASS FILTER

Passive filter

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Low pass filter
• Here low frequency signal is
reproduced at output with very
small attenuation
• High frequency (Unwanted
noise) signal is fully attenuated

• Cut-off frequency is defined as

frequency at which gain is
attenuated by 3 dB from its
normal level in pass band

• Pass band and attenuation band

is as shown in diagram

• Here pass band is from f=0 to

f=fc and BW = fc – 0 = fc

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High pass filter
• Here High frequency signal is
reproduced at output with very small
attenuation

• Low frequency signal is fully

attenuated

• Cut-off frequency is defined as

frequency at which gain is attenuated
by 3 dB from its normal level in pass
band

• Pass band and attenuation band is as

shown in diagram

• Here pass band is from fc to upper

cut-off frequency of amplifier

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BASICS :
• Capacitive Reactance is inversely proportion to frequency
Xc = 1 / (2* pi * f * C )

• At low frequency capacitive reactance is very high

• At high frequency capacitive reactance is very small

• In Circuit below if Xc is very high ( at low frequency) then there is very little attenuation
of low frequency signals.

If Xc is very small (at high frequency ) then input voltage drop is

across R1

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First Order Low Pass Active Filter
• In Passive filter load will affect the filter performance

• Hence to isolate load we use active element voltage

follower and circuit becomes Active filter.

• Here Voltage gain of filter is given by

• At low frequency capacitive reactance is very

high therefore entire voltage drop is across
Xc1 . Therefore there is little attenuation of
signal

• At high frequency capacitive reactance is

very low therefore entire voltage drop is
across R1 . Therefore there is full attenuation
of signal
Richard Lincoln Paulraj 19
 • Gain versus frequency and phase versus
Frequency is as shown in figure

• At low frequencies, gain is constant with

Small insertion loss . It is called Pass band

• Attenuation band is divided into

transition band and stop band

• At cut-off frequency gain falls by 3-db

• In transition band gain is attenuated with fall rate of 20 dB/decade.

•In Stop band gain is attenuated

•Because the filter is an RC circuit, there is a signal waveform phase shift from i/p to o/p
as well as attenuation.

• Phase angle between input and output is 0 degree at low frequency

• Phase angle between input and output is -45 degree at cut-off frequency
• Phase angle between input and output is –90 degree at high frequency

• Cut-off frequency occurs when gain is at 3 dB or (0.707) this happens when Xc1 = R1
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therefore fc = 1/ ( 2 * pi * R1 * C1)
Design : 1) Select C1 very much greater than stray capacitance
Select C1 = 1000 pF

2) Then calculate R1 using equation

fc = 1/ ( 2 * pi * R1 * C1)

3) Use R2 as same as R1 i.e R2 = R1

Problem : Using 741 op-amp , design the first-order active low pass filter to have 1.2 KHz
cut-off frequency

Select C1 = 1000 pF

R1 = 1/ ( 2 * pi * fc * C1)

= 1/ ( 2 * pi * 1.2K * 1000p)

= 132.63 K ohms ( Use 120 K ohms)

R2 = R1 = 120 K ohms
For 120K ohms if we calculate C1 once again

C1 = 1/ ( 2 * pi * fc * R1)= 1/ ( 2 * pi * 1.2k * 120 K)= 1105 pF ( Use 1100 pF Standard

Value) Richard Lincoln Paulraj 21
First Order High Pass Active Filter
Here Voltage gain is :

•The first order active high pass filter is made up of a passive high-pass circuit (C1 & R1)

•The voltage follower functions as a buffer amplifier to isolate the passive filter circuit
from the load.

•At low frequency capacitive reactance is very high therefore entire voltage drop is
across Xc1 . Therefore signal is fully attenuated

• At high frequency capacitive reactance(Xc1) is very low therefore entire voltage drop is
across R1 . Therefore there is zero ( minimum) input signal attenuation.
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 • Gain versus frequency and phase versus
Frequency is as shown in figure

• At high frequency gain is constant with

Small insertion loss It is called pass band

• At cut-off frequency (Xc1 = R1) gain falls

by 3-db across C1

• The voltage gain falls of at the rate of 20 dB/decade as the frequency is decreased from fc

• Phase angle between input and output is +90 degree (Phase lead) at low frequency

• Phase angle between input and output is +45 degree at cut-off frequency

• Phase angle between input and output is 0 degree at high frequency (Pass band)

•The bandwidth for a HPF is calculated as BW = (f2-fc). f2 is the op-amp unity gain frequency.

• Cut-off frequency occurs when gain is at 3 dB or (0.707) this happens when R1 = XC1
therefore fc = 1/ ( 2 * pi * R1 * C1) Richard Lincoln Paulraj 23
 Design : 1) Select C1 very much greater than stray capacitance
Select C1 = 1000 pF

2) Then calculate R1 using equation

fc = 1/ ( 2 * pi * R1 * C1)

3) Use R2 = R1

Problem : Using 741 op-amp , design the first-order active high pass filter to have 10 KHz
cut-off frequency

Select C1 = 1000 pF

R1 = 1/ ( 2 * pi * fc * C1)

= 1/ ( 2 * pi * 10 K * 1000p)

= 15.9 K ohms ( Use 15 KΩ standard

value)

Richard Lincoln Paulraj 24

Second Order Low Pass Active Filter ( Sallen – Key filter)
Second order low pass filter has fall off rate 40
dB/decade

The circuit is known as Sallen-Key filter and

also as a voltage controlled voltage source
(VCVS) filter.

At low frequencies Xc1 and Xc2 are very high

there fore input signal are reproduced at the
output as it is ( Entire voltage appear across
Xc1 and Xc2 is almost open circuit).

At high frequencies Xc1 and Xc2 are very small

therefore voltage drop across Xc1 is very small
or zero and C2 shorts Vo ( small voltage or
nearly zero ) to junction between R1 and R2
• Because of combination C1 R2 section and feedback from C2 to junction of R1 and R2
We get fall-off rate as 40 db/decade
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 • Gain versus frequency and phase versus
Frequency is as shown in figure

• At low frequencies, Xc1 & Xc2 are very

much larger than R1 & R2, gain is constant
with Small insertion loss It is called pass
band

• At cut-off frequency gain falls by 3-db

• In transition band gain is attenuated with fall rate of 40 dB/decade. Produced by C1 & R2 , C2
and R1||R2

• Phase angle between input and output is 0 degree at low frequency

• Phase angle between input and output is -90 degree at cut-off frequency

• Phase angle between input and output is –180 degree at high frequency (Above fc)

• Cut-off frequency is given by

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Design : 1) Select C1 very much greater than stray capacitance
Select C1 = 1000 pF

2) Use R1 = R2 = 0.5 (R1 + R2)

3) C2 =2 C1

4) Find R2 or R1 using

It is like Xc1 = R2 at fc
5) R3 = R1 + R2
Problem : Using 741 op-amp , design the second-order active low pass filter to have 12
KHz cut-off frequency . Use the selected components to calculate actual cut-off frequency

Richard Lincoln Paulraj 27

Second Order High Pass Active Filter
Second order high pass Sallen-Key filter has
fall off rate 40 dB/decade

At low frequencies Xc1 and Xc2 are very high

than R1 and R2, therefore input signal are
totally attenuated ( Entire voltage appear
across Xc1 and Xc2 therefore voltage drop
across R2 is small)

At high frequencies Xc1 and Xc2 are very small

than R1 and R2, behaves as short circuit
passing Vi unattenuated to voltage follower .

• Because of the combination of C2 R2 section and feedback via R1 & C1||C2 junction
We get fall-off rate as 40 db/decade

Richard Lincoln Paulraj 28

 • Gain versus frequency and phase versus
Frequency is as shown in figure

• At high frequency gain is constant with

Small insertion loss It is called pass band

• At cut-off frequency gain falls by 3-db

• In transition band gain is attenuated with fall rate of 40 dB/decade.

•In Stop band gain is total Attenuated

• Phase angle between input and output is +180 degree at low frequency

• Phase angle between input and output is +90 degree at cut-off frequency

• Phase angle between input and output is 0 degree at high frequency

• Cut-off frequency is
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Design : 1) Select C1 very much greater than stray capacitance
Select C1 = 1000 pF

2) Use R1 = 0.5 R2

3) C2 = C1

4) Select R2 by using equation

5) R3 = R2

Problem : Using 741 op-amp , design the second-order active high pass filter to have 6
KHz cut-off frequency . Use the selected components to calculate actual cut-off frequency

Richard Lincoln Paulraj 30

BAND-PASS FILTERS
MULTISTAGE BAND PASS FILTER

A Band-pass filter can be constructed simply by connecting low-pass and high-pass filters in
cascade as shown in fig above

For example, a low pass filter with fc = 100 KHz is cascaded with a high – pass circuit that
has fc = 10 kHz as shown above

The low-pass circuit will pass all frequencies up to 100 kHz , while the high-pass circuit will
block all frequencies below 10 kHz

Consequently the combination given a filter band from 10 KHz to 100 kHz.

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SINGLE – STAGE BAND-PASS
FILTER
The capacitors are selected to have Xc2 large enough to
be neglected at low frequencies

Xc1 is selected small enough to be neglected at high

frequencies
At Low frequencies
Xc2 is selected large enough to be neglected at
low frequencies

Then circuit is inverting amplifier with gain

At signal frequencies in the pass band of the

circuit, Xc1 becomes very smaller than R1 then
gain becomes this is mid-band gain

If Xc1 = R1 then gain reduces to below 3-dB from

midband frequency gain . Therefore Xc1 = R1 at f1
where f1 is Lower cut-off frequency
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At High frequencies

Xc1 is selected small enough (than R1 & C1) to be neglected at high frequencies
then the Equivalent circuit is as shown below
The circuit acts as inverting amplifier with gain

At frequency in pass band Xc2 is much larger than R2

therefore at mid band gain R2 / R1

When Xc2 = R2 gain reduces by 3dB from mid band There fore Xc2 = R2 at f2
In pass band circuit behaves as inverting amplifier and as a low-pass filter for high
frequencies and high pass filter at low frequencies
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Design :
1) Select lowest capacitance C2 much larger than stray capacitance
C2 = 1000 pf

2) Find R2 using Xc2 = R2 at f2

f2 = 1/ ( 2 π C2 Xc2) = 1 / (2 π C2 R2)
there fore R2 = 1 / (2 π C2 f2)

3) Then find R1 using Av in mid band gain Av = (R2/R1)

4) Find C1 using Xc1 = R1 at f1

f1 = 1/ ( 2 π C1 Xc1) = 1 / (2 π C1 R1)
there fore R1 = 1 / (2 π C1 f1)

5) Select R3 = R2

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Problem : Design a single – stage band pass filter to have unity voltage gain and a pass band is
From 300 Hz to 30 kHz

Here f1 = 300 Hz and f2 = 30 kHz

Select C2 = 1000 pF and

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 BANDWIDTH

Wide band and narrow band filters is as shown in figure

In both cases BW = f2 – f1

The circuit Q factor is figure of merit for a filter circuit. It defines the selectivity of filter in
passing center frequency f0 and rejecting other frequencies

Q factor is given by Q = fo/BW

In diagram b we can see that filter with a Q of 10 has a much narrower band width than a filter
with Q equal to 1.
Narrow band filters have Q greater than 5 and wide band filters have Q less than 5

The centre frequency of filter is determined by

Richard Lincoln Paulraj 36
Problem: A single stage band pass filter has following components : R1 = R2 = 7.5 k ohms
R3 = 6.8 k ohms , C1 = 8200 pf and C2 = 750 pf . Determine the circuit BW , centre
frequency and Q factor
F1 = 1 / (2 π C1 R1) = 1 / ( 2* π* 8200p *7.5 K) = 2.58 KHz
F2 = 1 / (2 π C2 R2) = 1 / ( 2* π* 750p *7.5 K) = 28.29 KHz
BW = F2 – F1 = 28.29 kHz – 2.58 kHz = 25.7 kHz
Centre frequency
= 8.5 kHz
Quality factor Q = fo/(BW) = 8.5 K / (25.7 K) = 0.33
Richard Lincoln Paulraj 37
 Notch filter or Band limiter or Band Stop filter
Notch filter is inverse of band-pass filter , its
function is to block a band of signal frequencies.

Diagram below shows how a notch filter can be

constructed using low-pass and high-pass filters

Here both circuits are connected in parallel and

outputs are applied to summing circuit

Another method of creating a notch filter is to

sum the output of band pass filter with its own
Input signal

In this case the band pass filter should have gain

equal to one and it should be inverting.

During pass band of band pass circuit , filter

output (VBP)equals to –vi and so the two inputs to
summing circuit cancel giving Zero output voltage
Above and below the pass band of band pass circuit ,VBP is negligible therefore output of
summing circuit is vi, and so combination has a notch frequency response.
Richard Lincoln Paulraj 38
A SINGLE STAGE NOTCH FILTER
Here Bandpass filter can be converted into a
Single-stage notch filter by addition of resistor R4

If R4 is equal to R1 & R2 and R3 is equal to R2 then the

circuit functions as difference amplifier with one side
producing inverted Vi in range f1 and f2 ( Pass band of
band pass filter) .

Other side produce Vi in all range of frequencies

Output of difference amplifier is (R2/R1) (Vi – Vf)

As R2= R1 therefore Vo = (Vi – Vf)

Where Vf is output of band pass filter

In pass band of notch filter , band pass will be in stop band therefore Vo = ( Vi – 0) = Vi

In stop band of notch filter , band pass will be in pass band therefore Vo = ( Vi – Vi) = 0
Richard Lincoln Paulraj 39
Design :

Follow design procedure as same as Band pass filter

1) Select lowest capacitance C2 much larger than stray capacitance

C2 = 1000 pf

2) Find R2 using Xc2 = R2 at f2

f2 = 1/ ( 2 π C2 Xc2) = 1 / (2 π C2 R2)
therefore R2 = 1 / (2 π C2 f2)

3) Then find R1 Av = (R2/R1)=1 there fore R2 = R1

4) Find C1 using Xc1 = R1 at f1

f1 = 1/ ( 2 π C1 Xc1) = 1 / (2 π C1 R1)
therefore R1 = 1 / (2 π C1 f1)

5) Select R3 = R2 and R4 = R1

Richard Lincoln Paulraj 40

Design : A single – stage notch filter is to be designed to have a stop band ranging from
200 Hz to 20 kHz . Determine suitable components values .

Richard Lincoln Paulraj 41