Analog Filters: Basics of OP AMP-RC Circuits

Analog Filters:Basics of OP AMP-RC Circuits

Stefano GregoriThe University of Texas at Dallas

Stefano Gregori Basics of OP AMP-RC Circuits 2

Introduction

So far we have considered the theory and basic methods of realizing filters that use passive elements (inductors and capacitors)

Another type of filters, the active filters, are in very common use They were originally motivated by the desire to realize inductorless

filters, because of the three passive RLC elements the inductor is the most non-ideal one (especially for low-frequency applications of filters in which inductors are too costly or bulky)

When low-cost, low-voltage solid-state devices became available, active filters became applicable over a much wider frequency range and competitive with passive ones

Now both types of filters have their appropriate applications


Active-RC filters

Active filters are usually designed without

regard to the load or source impedance; the terminating impedance may not affect the performance of the filter

it is possible to interconnect simple standard blocks to form complicated filters

are noisy, have limited dynamic ranges and are prone to instability

can be fabricated by integrated circuits

In this lesson we concentrate on active-RC filters. They make use of active devices as well as RC components.

Passive filters the terminating impedance is an

integral part of the filter: this is a restriction on the synthesis procedure and reduces the number of possible circuits

are less sensitive to element value variations

are generally produced in discrete or hybrid form


Operational Amplifier

In an ideal op-amp we assume:

input resistance Ri approaches infinity, thus i1 = 0

output resistance Ro approaches zero

amplifier gain A approaches infinity

Ri

Ro

A(e+-e-)

e+

e-

e2

i1

equivalent circuitsymbol

e+

e-e2


Inverting voltage amplifier

R1

R2

vinvout

i

i1

in )()(

Rtv

ti

)()()( in1

22out tv

RR

tiRtv

Example:

ftπVtv 2sin)( 0in

given

we have

R1 = 1 kΩR2 = 2 kΩV0 = 1 Vf = 1 MHz

ftπVRR

tv 2sin)( 01

2out

vin(t)

vout(t)


Weighted summer

k

kk R

tvti

)()(

)()( )( 1out titiRtv nf

R1

Rf

v1voutR2

v2

Rn

vn

n

k k

kf R

tvR

1

)(

n

nf R

tvR

tvR

)()(

1

1


Noninverting voltage amplifier

R1

R2

vin

vouti

i 1

in )()(

Rtv

ti

)( 1)( )( in1

221out tv

RR

tiRRtv

Example:

ftπVtv 2sin)( 0in

given

we have

R1 = 1 kΩR2 = 1 kΩV0 = 1 Vf = 1 MHz

ftπVRR

tv 2sin1)( 01

2out

vin(t)

vout(t)


Buffer amplifier

vin

vout

)()( inout tvtv


Inverting or Miller integrator

Rtv

ti)(

)( in

)(1)0()(0

inoutout t

dttvRC

vtv

R

C

vinvout

i dttdv

Cti)(

)( out

Example:

00

2sin

0)(

0in

tt

ftπVtv

0)0(out v

12cos2

)( 0out πft

fRCπV

tv

given

we have

vin(t)

vout(t)

R = 1 kΩC = 1 nFV0 = 1 Vf = 1 MHz

RCsV

V

inout


Inverting differentiator (1)

dttdv

RCtRitv)(

)()( inout

dttdv

Cti)(

)( in

Example:

00

2sin

0)(

0in

tt

ftπVtv

given

we have

R = 1 kΩC = 100 pFV0 = 1 Vf = 1 MHz

C

R

vinvout

i

vin(t)

vout(t)

00

2cos2

0)(

0out

tt

ftπfRCVπtv

Cs

inout VRCsV


Inverting differentiator (2)

vout(t) is a square waveform with:- vout max 2,068 V- vout min -2,068 V- frequency 500 kHz

R = 22 kΩC = 47 pFvin(t) is a triangular waveform with:

- vin max 2 V- vin min 0 V- frequency 500 kHz

vin(t)

vout(t)

C

R

vinvout

Cs Cs


Inverting lossy integrator

R1

R2

vinvout

C

in

21

out1

1 V

RsCR

V


Inverting weighted summing integrator

R1

v1voutR2

v2

Rn

vn

C

n

k k

kout R

VsC

V1

1


Subtractor

2

321

1031

1

0 VRRRRRR

VRR

Vout

R1

R0

v1vout

R2

v2

R3


Integrator and differentiator

integrator

differentiator

frequency behavior

R

C

vinvout

C

R

vinvout

integrator differentiator

R = 1 kΩC = 1 nF

2

1V fRCπ

A

fRCπA 2V vin(t) is a sinewave with frequency f.

Figure shows how circuit gain AV changes with the frequency f

AV is the ratio between the amplitude of the output sinewave vout(t) and the amplitude of the input sinewave vin(t)


Low-pass and high-pass circuits

low-passvoutlpvin

R

C

C

vin vouthp

R

low-pass circuit

high-pass circuit

frequency behavior

high-pass

R = 1 kΩC = 1 nF


Inverting first-order section

v1v2

R1

C1

R2

C2

Z1

Z2

v1v2

2

1

1

2

1

2 YY

ZZ

VV

11

22

11

1

2

1

2

CsRCsR

RR

VV

inverting lossing integrator


Noninverting first-order section

2

1

1

2

1

2 1 1YY

ZZ

VV

11

122

11

1

2

1

2

CsRCsR

RR

VV

noninverting lossing integrator

Z1

Z2

v1

v2

v1

v2

R1

C1

R2

C2


Finite-gain single op-amp configuration

V1

RCthreeport V2

V3

i2

i3i1

Many second-order or biquadratic filter circuits use a combination of a grounded RC threeport and an op-amp

3332321313

3232221212

3132121111

VyVyVyIVyVyVyI

VyVyVyI

μEEI

/0

23

3

μyy

yEE

3332

31

1

2


Infinite-gain single op-amp configuration

32

31

1

2

yy

EE

V1

RCthreeport V2

V3


Gain reduction

V1

NV2

Z

V1

NV2

Z1

Z2

V1'

To reduce the gain to α times its original value (α < 1) we make

21

2

1

1

ZZZ

αVV

Z

ZZZZ

21

21and

solving for Z1 and Z2, we get

αZZ 1 Z

αZ

11

2and


Gain enhancement

A simple scheme is to increase the amplifier gain and decrease the feedback of the same amount

V1

RCthreeport V2

V3 K

1/KV2/K

μyy

KyEE

3332

31

1

2


RC-CR transformation (1)

is applicable to a network N that contains resistors, capacitors, and dimensionless controlled sources

conductance of Gi [S] → capacitance of Gi [F]

capacitance of Cj [F] → conductance of Cj [S] the corresponding network functions with the dimension of the

impedance must satisfy

sZ

ssZ 11)(

the corresponding network functions with the dimension of the admittance must satisfy

ssYsY 1)(

the corresponding network functions that are dimensionless must satisfy

sHsH 1)(


RC-CR transformation (2)

67212)( 2

1

2

ssVV

sH

1 F

v1' v2'2

1/2 F

3

2

v1 v2

12

2

1/3 F

1/2 F

)12(672)(

2

1

111

ssss

IV

sZ

27612)( 2

2

1

2

ss

sVV

sH

)2(276)(

2

1

111

ss

ssIV

sZ

N N’


Sallen-Key filters

lowpass filter

highpass filter

frequency behavior

C

vin

vout

C

R

R

C

vin

vout

C

RR

lowpass highpass

bandpass

R = 1 kΩC = 1 nF


2 4 6 8 10

0.2

0.4

0.6

0.8

1

2 4 6 8 10

0.2

0.4

0.6

0.8

1

2 4 6 8 10

0.2

0.4

0.6

0.8

1

2 4 6 8 10

0.2

0.4

0.6

0.8

1

2 4 6 8 10

0.2

0.4

0.6

0.8

1

Types of biquadratic filters

012

0

bsbsGb

012

2

bsbsGs

012

1

bsbssGb

01

20

22

bsbsasa

012

012

bsbsbsbs

G

lowpass highpass bandpass bandreject allpass

Analog Filters: Basics of OP AMP-RC Circuits

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