8/2/2019 Int & Diff

1/43

8/2/2019 Int & Diff

2/43

An Integrating circuit gives an output voltage proportional to

the Integral of the input signal

vo = k vidt

RC Circuit:

Voltage across a capacitor is 1/C iCdt.A current proportional to the input voltage is passed through C, voltage

across C will be integral of the input signal.

LR Circuit:

Current through inductor is 1/L vLdt.If this current is passed through a series resistor R, then the voltage

developed across R will be proportional to voltage vL

Output small in magnitude:

8/2/2019 Int & Diff

3/43

Integrator (perform timing functions, generate linear ramps, triangular waves)

is

is

8/2/2019 Int & Diff

4/43

Input Vs(t) is a time varying signal. The current is(t) will be Vs(t)/R.

t

Charge deposited on C is = is(t) dt0

t

vc

(t) = VC

+ (1/C) is

(t) dt ( VC

initial voltage on C (at t = 0))

0

v0(t) = -vc(t)

t

vo

(t) = -VC

- (1/CR) vi

(t) dt. Output proportional to time integral of input.

0 CR integrator time constant.

Integrator (perform timing functions, generate linear ramps, triangular waves)

is

is

8/2/2019 Int & Diff

5/43

Output of an Integrating circuit

proportional to the area under each half

cycle of the input waveform

(ex: designed to produce a triangularwaveform from a square wave input)

8/2/2019 Int & Diff

6/43

Vs

During positive half cycle of a square

wave input, Current through R is

constant.

Effectively all I flows through C,

capacitor charged linearly and output

is a negative going ramp.

8/2/2019 Int & Diff

7/43

Vs

t

V

-Vs

Vo

Vo

Since I is proportional

to the input peak

voltage vp and t isthe time duration ofthe input pulse,

output is directly

proportional to the

area under each half

cycle of input

8/2/2019 Int & Diff

8/43

Figure 2.39 (b) Frequency response of the integrator.

RCV

V

RCjV

V

Cj

VR

V

S

o

S

o

oS

1

1

100

=

=

=

As doubles, the magnitude is halved.

Integrator frequency int = 1/CR at = 0 integrator transfer functionis infinite. Any tiny dc component in the input signal will

theoretically produce an infinite output.

8/2/2019 Int & Diff

9/43

Source of error in practical integrators is usually due to

offset of the opamp.

Even with zero applied input signal, the op-amps input

offset voltage and bias current can cause a continuous

charging of the feedback capacitor.

Eventually, the opamps output will drift to either positive

or negative saturation.

8/2/2019 Int & Diff

10/43

.As the output rises with time, the op amp eventually saturates.

Effect of the op-amp input offset voltage VOS

VOS/R

VOS/R

8/2/2019 Int & Diff

11/43

Effect of the op-amp input bias and offset

currents

IOS=IB1-IB2

(IB2.R)/R = IB2

=-IB2R +(IOS/C)t

vo to ramp linearly with time until the opamp saturates

8/2/2019 Int & Diff

12/43

A large resistance RFconnected in parallel

withC

in order to providenegative feedback and

hence finite gain at dc.

But, time constant of

integrator remain RC

A Practical Inverting Integrator: (low pass filter)

10K

8/2/2019 Int & Diff

13/43

A large resistance RFconnected in parallel

withC

in order to providenegative feedback and

hence finite gain at dc.

But, time constant of

integrator remain RC

A Practical Inverting Integrator: (low pass filter)

1)(

)(

+=

jCR

RR

SVi

SV

F

F

o

jRCjCR

RR

SVi

SV

F

F

o

c

1

)(

)( =

>

10K

8/2/2019 Int & Diff

14/43

Basic Integrator response

Ideal response of Practical integrator

Actual response of Practical

integrator

RF/R1

Gain db

fa fbfa = 1/2RFC

fb= 1/2RC

|(Vo/vi)| = 1/2fRC

(RF/R)dB -3dB

Usually fa

8/2/2019 Int & Diff

15/43

Show the output of integrator forinput pulse of 1-V height and 1ms

width (1KHz):

(R=10K ) and C = 10nF

8/2/2019 Int & Diff

16/43

I=1V/10k=0.1mA and C=10nF

ideal integrator,

(Time constant of integrator=RC=0.1ms.)

(RC=0.1 ms fc= 1.591 KHz)

)(1010

11.01.0

1)(

.1

)()1(

1

0

0

linearVnF

msmAdtmA

CtV

dti

C

tV

ms

o

t

o

=

==

=

8/2/2019 Int & Diff

17/43

-0.99V

8/2/2019 Int & Diff

18/43

If the Integrator capacitor is shunted by a 1M resistor

how will the response be modified.

Opamp specified to saturate at 13 V

8/2/2019 Int & Diff

19/43

RF=1M and C=10nF

mSt

ponentialexVetV

mS

VRRtV

tAs

t

o

Fo

1

)(51.9)1(100)(

10CRofconstanttime

awith100Vtoheadingllyexponentialbeoutput wil

100)(

,

10

F

=

==

=

=

8/2/2019 Int & Diff

20/43

)(99.0)1(1)(

1)(

,

1.0 ponentialexVetV

VR

RtV

tAs

t

o

Fo

==

=

RF=10K And C=10nF (i.e. RFC=0.1 ms)

Correct values of RF not chosen

8/2/2019 Int & Diff

21/43

-0.99V

8/2/2019 Int & Diff

22/43

IC

tF

o

IC

t

o

CCF

IC

Ve

R

RtV

or

VdtR

V

CtV

VR

RtV

+=

+=

=

)1()()2(

.1

)()2(

)()()1(

0 2

1

2

Practical integrator with initial condition:

Step configuration S2 S1

1st Inverting Amplifier

charges capacitor C with

IC

ON OFF

2nd Ideal integrator with

initial condition

OFF ON

2nd Practical Integrator with

initial condition

ON ON

8/2/2019 Int & Diff

23/43

An Differentiating circuit gives an output voltage proportional tothe time rate of change of the input signal

vo = k (dvi/dt)

RC Circuit:

Current through a capacitance is C dv/dt where v is the voltage

across it. If this current is allowed to flow through a resistance R then the

voltage drop across R will be RC dv/dt.

Output small in magnitude:

8/2/2019 Int & Diff

24/43

A differentiating circuit

- output amplitude proportional to rate of

change of an input voltage.

(ex: designed to respond to a triangular and

rectangular input waveforms)

Interchanging the location of the resistor and

capacitor of the integrator circuit

8/2/2019 Int & Diff

25/43

dt

dVRCV

R

V

dt

dVC

so

os

=

=

Ideal Differentiator Circuit:

0

8/2/2019 Int & Diff

26/43

0

t

vVo = -C1R2 (v)/ (t)

8/2/2019 Int & Diff

27/43

atordifferentiofntconstaTimeRC

CRV

V

CRj

Cj

R

Z

Z

V

V

s

o

F

s

o

=

=

===

1

1

0

8/2/2019 Int & Diff

28/43

Frequency response of a

differentiator with a time-

constant CR.

.

)(01

)(1

aneouslyntinstaalmostrgeschacapacitor

valuergelaI

R

V

sayVinputstepfor

wire

=

=

8/2/2019 Int & Diff

29/43

Practical Differentiator Circuit :

In the basic circuit output voltage of the differentiator increases withfrequency, making the circuit susceptible to high frequency noise.

Practical circuit a resistor Rs is placed in series with input capacitor to

decrease the high frequency gain to the ratio RF/Rs

The circuit functions as a differentiator only for frequencies higher than

fc = 1/2 R1 C. For frequencies higher than fc, the circuit approaches thebehavior of an inverting amplifier.

8/2/2019 Int & Diff

30/43

atordifferentipracticalofntconstaTimeCR

CRjV

VCR

for

CRjCRj

CjR

RZZ

VV

s

o

c

F

s

o

=

=

=>

8/2/2019 Int & Diff

40/43

Antoniou Inductance simulation circuit

8/2/2019 Int & Diff

41/43

The Basic Simulation

Circuit Analysis begins with the source V1 at node 1

Theoretical analysis leads to the input

impedance of Zin=V1/I1=sC4R1R3R5/R2

This is the the same impedance of an inductor

where L= C4R1R3R5/R2

8/2/2019 Int & Diff

42/43

Theoretical Analysis

8/2/2019 Int & Diff

43/43