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```EMT 212/4
ANALOG ELECTRONIC II
CHAPTER 2:
OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
Content
1.
2.
Op-amp Application

Introduction

Inverting Amplifier

Non-inverting Amplifier

Voltage Follower / Buffer Amplifier

Summing Amplifier

Differencing Amplifier

Integrator

Differentiator

Comparator

Summary
Frequency Response
Op-amp Application

Introduction
Op-amps are used in many different applications.
We will discuss the operation of the fundamental opamp applications. Keep in mind that the basic
operation and characteristics of the op-amps do not
change — the only thing that changes is how we use
them
Inverting Amplifier



Circuit consists of an op-amp and three resistors.
The positive (+) input to the op-amp is grounded through
R2.
The negative (-) input is connected to the input signal (via
R1) and also to the feedback signal from the output (via RF).
Inverting Amplifier
V
V+


Assume that amplifier operates in its linearly
amplifying region.
For an ideal op-amp, the difference between the
input voltages V+ and V to the op-amp is very small,
essentially zero; V   V   0
V  V 
Inverting Amplifier
V
V+

Hence;
Vin  V  Vin
iin 

R1
R1
Inverting Amplifier

The op-amp input resistance is large, so the current
into the +ve and –ve op-amp inputs terminal will be
small, essentially zero
V   Vout  iin RF

V  Vout
Vin  V 

RF
R1
V 0
 Vout
Vin

RF
R1
Vout
RF
Av 

Vin
R1
Inverting Amplifier

Currents and voltages in the inverting op-amp
Inverting Amplifier - Example

Design an inverting amplifier with a specified
voltage gain.
Specification: Design the circuit such that the voltage
gain is Av = -5. Assume the op-amp is driven by an
ideal sinusoidal source, vs = 0.1sin wt (V), that can
supply a maximum current of 5 µA. Assume that
frequency w is low so that any frequency effects
can be neglected.
Non-Inverting Amplifier



Circuit consists of an
op-amp and three
resistors.
The negative (-) input
to the op-amp is
grounded through R1
and also to the
feedback signal from
the output (via RF).
The positive (+) input
is connected to the
input signal.
Non-Inverting Amplifier



Input current to opamp is very small.
No signal voltage is
created across R2 and
hence V   vin
V   V  so it follows
that;
V   vin
Non-Inverting Amplifier

I   0 so
RF and R1
carry the same
current. Hence vout is
related to V through
a voltage-divider
relationship
R1
V 
vout
R1  RF
R1
vin 
vout
R1  RF

vout
RF
Av 
 1
vin
R1
Non-Inverting Amplifier

The output has the same
polarity as the input,


a positive input signal produces
a positive output signal.
The ratio of R1 and RF
determines the gain.

When a voltage is applied to
the amplifier, the output
voltage increases rapidly and
will continue to rise until the
voltage across R1 reaches the
input voltage. Thus negligible
input current will flow into the
amplifier, and the gain
depends only on R1 and RF
Non-Inverting Amplifier

The input resistance to
the non-inverting
amplifier is very high
because the input
current to the
amplifier is also the
input current to the
op-amp, I+, which must
be extremely small.
Non-Inverting Amplifier
v1  v2
v1
i1  
R1
v1  vo
i2 
R2
i1  i2
v1  vo
v1


R1
R2
vout
R2
Av 
 1
vin
R1
Voltage Follower / Buffer Amplifier



This “buffer” is used to control
impedance levels in the circuit
– it isolates part of the overall
(measurement) circuit from the
output (driver).
The input impedance to the buffer is very high and its
output impedance is low.
The output voltage from a source with high output
impedance can, via the buffer, supply signal to one or
more loads that have a low impedance.
Voltage Follower / Buffer Amplifier



High input impedance.
Low output impedance.
Voltage gain = 1
Vout  Vin
Vout
Av 
1
Vin
Summing Amplifier



The inverting amplifier
can accept two or more
inputs and produce a
weighted sum.
Using the same
reasoning as with the
inverting amplifier, that
V ≈ 0.
The sum of the currents
through R1, R2,…,Rn is:
Vn
V1 V2
iin  
 ... 
R1 R2
Rn
Summing Amplifier

itself to draw iin
through Rf (iin = if).
if
iin
Vout  iin R f
Rf
Rf 
 Rf

 V1
 V2
 ...  VN
R2
RN 
 R1

The output will thus be the sum of V1,V2,…,Vn,
weighted by the gain factors, Rf/R1 , Rf/R2 …..,
Rf/Rn respectively.
Summing Amplifier

Special Cases for this Circuit:
1. If R1 = R2 =……= R then:
Vout  
Rf
R1
VIN1  VIN 2  .....  VINn 
if
iin
Summing Amplifier
2. If R1 = R2 = … = R and VIN1, VIN2, … are either
0V (digital “0”) or 5V (digital “1”) then the output
voltage is now proportional to the number of (digital)
1’s input.
if
iin
Summing Amplifier - Application

Digital to Analog Converter
- binary-weighted resistor DAC
Summing Amplifier - Application

Digital to Analog Converter
Differencing Amplifier

This circuit produces an output which is proportional to
the difference between the two inputs
vout 
Rf
R1
v1  v 2 
Differencing Amplifier

The circuit is linear so we can look at the output due
to each input individually and then add them
(superposition theorem)
Differencing Amplifier

Set v1 to zero. The output due to v2 is the same as
the inverting amplifier, so
v out  2  
Rf
R1
v2
Differencing Amplifier

The signal to the non-inverting output, is reduced by
the voltage divider:
v in 
Rf
R1  R f
v1
Differencing Amplifier

The output due to this is then that for a non-inverting
amplifier:
vout 1
Rf

  1 
R1


v in

Differencing Amplifier
v in 
Rf
R1  R f
Rf

vout 1   1 
R1

v1

v in

Rf

v out 1   1 
R1

 Rf 
v1
vout 1  
 R1 
 R f

R R
f
 1

v1


Differencing Amplifier
 Rf
vout 1  
 R1



v1

Thus the output is:
v out  2  
vout  vout 1  vout 2 
Rf
R1
Rf
R1
v2
v1  v2 
Thus the amplifier subtracts the inputs and amplifies
their difference.
Integrator


The basic integrator is easily identified by the
capacitor in the feedback loop.
A constant input voltage yields a ramp output. The
input resistor and the capacitor form an RC circuit.
Integrator


The slope of the ramp is determined by the RC time
constant.
The integrator can be used to change a square wave
input into a triangular wave output.
Integrator

The capacitive impedance:
1
1
Xc 

jC sC
Integrator

The input current: I  Vin   Vout   Vout   sCVout
Ri
Xc
1 /sC
Integrator
Vout
1

Vin
sCRi
Integrator

Thus the output in time domain: 1
Vout
1

Vin  
Vindt

jRi C
Ri C
Differentiator
V in
0
t0
t1

+
t2
V in
0
+

t0 - t1
V out
0
t0
t1
V out
t2
t
Differentiator



The differentiator does the opposite of the integrator
in that it takes a sloping input and provides an output
that is proportional to the rate of change of the input.
Note the capacitor is in the input circuit.
The output voltage can be determined by the formula
below:
Vout
dVin
R

Vin   RC
1
dt
jC
Comparator



The comparator is an op-amp circuit that compares
two input voltages and produces an output indicating
the relationship between them.
The inputs can be two signals (such as two sine waves)
or a signal and a fixed dc reference voltage.
Comparators are most commonly used in digital
applications.
Comparator


Digital circuits respond to rectangular or square
waves, rather than sine waves.
These waveforms are made up of alternating (high
and low) dc levels and the transitions between them.
Transitions
"High" dc level
"Low" dc level
Comparator

Example:
Assume that the digital system is designed to perform
a specific function when a sine wave input reaches a
value of 10 V
Comparator
V ref
V
+

Variable
voltage
source
Comparator
Digital
system

V

Comparator

With nonzero-level
+V
detection the voltage
divider or zener
R
diode sets the
reference voltage atV Z
which the op-amp
turns goes to the
maximum voltage
level.

V out
+
V in
(c) Zener diode sets reference voltage
Comparator
V REF
V in 0
t
+V out (max)
V out
0
V out (max)
t
Comparator Waveforms
Comparator


Remember that the comparator is configured in open-loop,
making the gain very high. This is open-loop configuration.
This makes the comparator very susceptible to unwanted
signals (noise) that could cause the output to arbitrarily switch
states.
Comparator


If the level of the pulse must be less than the output of a
saturated op-amp, a zener-diode can be used to limit the
output to a particular voltage. This is called output bounding.
Either positive, negative, or both halves of the output signal can
be bounded by use of one or two zener diodes respectively
D1 D2
0
+V Z 2 + 0.7 V
V in R i
0
V Z 2  0.7 V
Comparator - Application

Over-Temperature Sensing circuit
Comparator - Application

Analog to Digital Converter
Summary






The summing amplifier’s output is the sum of the inputs.
An averaging amplifier yields an output that is the
average of all the inputs.
The scaling adder has inputs of different weight with each
contributing more or less to the input.
Integrators change a constant voltage input to a sloped
output.
Differentiators change a sloping input into a step voltage
proportional to the rate of change.
The op-amp comparator’s output changes state when the
input voltage exceeds the reference voltage.
Frequency Response of Op-amp


The “frequency response” of any circuit is the
magnitude of the gain in decibels (dB) as a function
of the frequency of the input signal.
The decibel is a common unit of measurement for the
relative magnitude of two power levels. The
expression for such a ratio of power is:
Power level in dB = 10log10(P1/P2)

Note: A decibel is one-tenth of a "Bel", a seldom-used
unit named for Alexander Graham Bell, inventor of
the telephone.
Frequency Response of Op-amp


The voltage or current gain of an amplifier expressed in dB is
20 log10|A|, where A = Vout/Vin.
The frequency response of an op-amp has a low-pass
characteristic (passing low-frequency signals, attenuating highfrequency signals).
Gain (log scale)
A
-3 dB point
fc
Freq (Hz)
Frequency Response of Op-amp




The bandwidth is the frequency at which the power of
the output signal is reduced to half that of the
maximum output power.
This occurs when the power gain A drops by 3 dB. In
Figure shown before, the bandwidth is fc Hz.
For all op-amps, the Gain*Bandwidth product is a
constant.
Hence, if the gain of an op-amp is decreased, its
operational bandwidth increases proportionally.
```
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