B.TECH FIRST YEAR
ACADEMIC YEAR: 2020-2021
COURSE NAME: BASIC ELECTRONICS
COURSE CODE
: EC 1001
LECTURE SERIES NO :
CREDITS
:
3
MODE OF DELIVERY : ONLINE (POWER POINT PRESENTATION)
FACULTY
:
DR. TEJPAL
EMAIL-ID
:
Tej.pal@Jaipur.manipal.edu
PROPOSED DATE OF DELIVERY:
SESSION OUTCOME
“UNDERSTAND THE BASIC
PRINCIPLE OF OPERATIONAL
AMPLIFIER”
ASSIGNMENT
QUIZ
MID TERM EXAMINATION –II
END TERM EXAMINATION
ASSESSMENT CRITERIA’S
PROGRAM
OUTCOMES
MAPPING WITH
CO3
[PO1]
ENGINEERING KNOWLEDGE: APPLY THE KNOWLEDGE
OF MATHEMATICS, SCIENCE, ENGINEERING
FUNDAMENTALS, AND AN ENGINEERING
SPECIALIZATION TO THE SOLUTION OF COMPLEX
ENGINEERING PROBLEMS.
OPERATIONAL AMPLIFIER (OP-AMP)
INTEGRATOR
𝑑𝑣𝑐
𝑖𝑐 = 𝐶
𝑑𝑡
𝑣𝑖𝑛 − 0
𝑑(0 − 𝑣𝑜 )
=𝐶
𝑅
𝑑𝑡
𝑡
0
𝑡
0
𝑑𝑣0
1
=−
𝑑𝑡
𝑅𝐶
𝑡
𝑣𝑖𝑛 𝑑𝑡
0
𝑣𝑖𝑛
𝑑𝑡 =
𝑅
𝑡
0
1
𝑣0 = −
𝑅𝐶
𝑑(−𝑣0 )
𝐶
𝑑𝑡
𝑑𝑡
𝑡
𝑣𝑖𝑛 𝑑𝑡
0
INTEGRATOR
 Integrator is a circuit whose output is
proportional to (negative) integral of the
input signal with respect to time
 Feedback is given through capacitor to
inverting terminal
 Since same current flows through R and C,
v in
dv o
 C
R
dt
t
1
vo 
v in dt

RC 0
INTEGRATOR
 Integrator is a circuit whose output is proportional to
(negative) integral of the input signal with respect to time
 Feedback is given through capacitor to inverting terminal
 Since same current flows through R and C,
vin
dvo
= −C
R
dt
−1
vo =
𝑅𝐶
t
vin dt
0
INTEGRATOR
OP-AMP INTEGRATOR EXAMPLE
 Given an input signal of 4V square wave
for 10 ms duration, what is the
integrator output versus time for the
integrator circuit at the left?
R. W. Knepper
SC412, slide 215

The current into the capacitor during the
square wave is constant at 4V/5Kohm =
0.8 mA

Using the integral expression from the
previous chart, the capacitor voltage will
increase linearly in time (1/R1C) 4t = 0.8t
V/ms during the square wave duration

The output will therefore reduce linearly
in time by – 0.8t V/ms during the pulse
duration, falling from 0 to –8 volts, as
shown in the figure at left

Since at 10 ms the output will be –8 V >
VNEG, the op-amp will not saturate during
the 10 ms input pulse
DIFFERENTIATOR
 Input signals are applied to the capacitor C. Capacitive reactance is the
important factor in the analysis of the operation of a differentiator.
Capacitive reactance is Xc = 1/2πfC. Capacitive reactance is inversely
proportional to the rate of change of input voltage applied to the
capacitor. At low frequency, the reactance of a capacitor is high and at
high frequency reactance is low. The capacitor blocks any DC content so
there is no current flow to the amplifier summing point, X resulting in
zero output voltage. The capacitor only allows AC type input voltage
changes to pass through and whose frequency is dependant on the rate of
change of the input signal.
 Therefore, at low frequencies and for slow changes in input voltage, the
gain, Rf / Xc, is low, while at higher frequencies and for fast changes the
gain is high, producing larger output voltages.
 If a constant DC voltage is applied as input, then the output voltage is
zero. If the input voltage changes from zero to negative, the voltage
output voltage is positive. If the applied input voltage changes from zero
to positive, the output voltage is negative. If a square wave input is
applied to a differentiator, then a spike waveform is obtained at the output.
DIFFERENTIATOR
 Differentiator is circuit whose output is proportional to
(negative) differential of input voltage with respect to time
 Input is given through capacitor, feedback given through
resistor to inv terminal
Since current through R and C are same,
𝑖𝑐 = 𝑖𝑓
𝑑(𝑣𝑖𝑛 − 0) (0 − 𝑣0 )
𝐶
=
𝑑𝑡
𝑅
dvin
vo
C
=−
dt
R
dvin
vo = −𝑅𝐶
dt