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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:
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Integrator (perform timing functions, generate linear ramps, triangular waves)
is
is
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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
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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)
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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.
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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
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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.
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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.
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.As the output rises with time, the op amp eventually saturates.
Effect of the op-amp input offset voltage VOS
VOS/R
VOS/R
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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
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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
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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
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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
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Show the output of integrator forinput pulse of 1-V height and 1ms
width (1KHz):
(R=10K ) and C = 10nF
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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
=
==
=
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-0.99V
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If the Integrator capacitor is shunted by a 1M resistor
how will the response be modified.
Opamp specified to saturate at 13 V
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RF=1M and C=10nF
mSt
ponentialexVetV
mS
VRRtV
tAs
t
o
Fo
1
)(51.9)1(100)(
10CRofconstanttime
awith100Vtoheadingllyexponentialbeoutput wil
100)(
,
10
F
=
==
=
=
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)(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
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-0.99V
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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
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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:
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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
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dt
dVRCV
R
V
dt
dVC
so
os
=
=
Ideal Differentiator Circuit:
0
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0
t
vVo = -C1R2 (v)/ (t)
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atordifferentiofntconstaTimeRC
CRV
V
CRj
Cj
R
Z
Z
V
V
s
o
F
s
o
=
=
===
1
1
0
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Frequency response of a
differentiator with a time-
constant CR.
.
)(01
)(1
aneouslyntinstaalmostrgeschacapacitor
valuergelaI
R
V
sayVinputstepfor
wire
=
=
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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.
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atordifferentipracticalofntconstaTimeCR
CRjV
VCR
for
CRjCRj
CjR
RZZ
VV
s
o
c
F
s
o
=
=
=>
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Antoniou Inductance simulation circuit
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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
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Theoretical Analysis
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