Lica unit2 ppt converted

LINEAR INTEGRATED CIRCUITS &APPLICATIONS
UNIT ii
CHARACTERISTICS OF OPAMP
Prepared by,
E.ELAKKIA & M.PERARASI,
ASSISTANT PROFESSOR/EEE,
R.M.K.ENGINEERING COLLEGE
20 March 2021 1
E.ELAKKIA/AP/EEE/RMKEC

SYLLABUS
Ideal OP-AMP characteristics, DC characteristics, AC
characteristics, differential amplifier; frequency response of
OP-AMP; Basic applications of op-amp – Inverting and Non-
inverting Amplifiers, summer, differentiator and integrator-
V/I & I/V converters.
20 March 2021 2

AGENDA
• Ideal OP-AMP characteristics
• DC characteristics
• AC characteristics
• Differential amplifier
• Frequency response of OP-AMP
• Basic applications of op-amp - Inverting and Non-inverting
Amplifiers, summer, differentiator and integrator-V/I & I/V
converters.
20 March 2021 3

20 March 2021 E.ELAKKIA/AP/EEE/RMKEC 4

Operational Amplifier- OP AMP
◼ Originally an op-amp was an electronic circuit that
could carry out mathematical operations of addition,
subtraction, differentiation and integration.
Hence the word “operational”
◼ Op-amp is used to amplify DC and AC signals
20 March 2021 5

OP-AMP SYMBOL
Op-amps have five basic terminals:
1. Two input terminals:
Inverting terminal
Non Inverting Terminal
2. One output terminal
3. Two power supply terminals. (+VE and –VE
Supply)
20 March 2021 6

OP-AMP PACKAGES
1.TLC271 Programmable Op Amp
2.LMC660CN Quad CMOS Operational Amplifier
3. 741 General Purpose Op Amp
4. LM3875 56Watt Power Amplifier
5. TDA2003A 10Watt Car Audio Power Amplifier

20 March 2021 9

• The terminals V+ and V− are used for the input and they
control the output, usually as an amplified signal on Vout.
• The op-amp is built using several resistors and other
components called transistors. All these transistors and
resistors are packed inside the very small package you can
see in Figure 15(b).
• A dot and a dent on top of the package are generally used to
identify the orientation of the package and therefore the pin
number.
20 March 2021 10

Characteristics of Operational Amplifier
o Very high differential gain
o High input impedance
o Low output impedance
o Provide voltage changes (amplitude and polarity)
o Used in oscillator, filter and instrumentation
o Accumulate a very high gain by multiple stages
20 March 2021 11

EQUIVALENT CIRCUIT OF OPAMP

IDEAL CHARACTERISTICS OF OP-AMP
• Infinite voltage gain A.
• Infinite input resistance Ri, so that almost any signal source can drive it and
there is no loading of the proceeding stage.
• Zero output resistance Ro, so that the output can drive an infinite number of
other devices.
• Zero output voltage, when input voltage is zero.
• Infinite bandwidth, so that any frequency signals from o to ∞ HZ can be
amplified with out attenuation.
• Infinite common mode rejection ratio, so that the output common mode noise
voltage is zero.
• Infinite slew rate, so that output voltage changes occur simultaneously with
input voltage changes.
20 March 2021 14

Characteristics of Ideal Op-Amp
◼ Infinite input impedance (about 2 Mohm)
◼ Low output impedance (about 200 ohm)
◼ Very large voltage gain at low frequency
Thus, small changes in voltages can be amplified by
using an op-amp
◼ Infinite bandwidth (all frequencies are amplified by
same factor
◼ No slew rate – no delay between change in i/p and
changes in o/p
20 March 2021 15

Op Amp Characteristics Explained
◼ Infinite input impedance
no current flows into inputs
◼ Infinite voltage gain
a voltage difference at the two inputs is magnified to
a very large extent
in practice, voltage gain ~ 200000
means difference between + terminal and −
terminal is amplified by 200,000!

Op Amp Characteristics Explained
◼ Infinite bandwidth
In practice, bandwidth limited to few MHz range
slew rate limited to 0.5–20 V/s
20 March 2021 E.ELAKKIA/AP/EEE/RMKEC

BLOCK DIAGRAM OF OPAMP
20 March 2021 18

OPEN LOOP OPERATION OF OP-AMP

Operational Amplifier Without Feedback
◼ The op-amp can be regarded as a device which
generates an voltage Vo given by:
Vo = A (V2 – V1)
A is called as the gain of the amplifier.
V1 is the voltage applied at the inverting input,
V2 is the voltage applied at the non-inverting input,

Variation of Gain with Frequency
◼ The value of gain A depends on the frequency of the i/p
signal and is very high at low frequencies.
◼ At DC, (f = 0 Hz), gain A is about 105.
◼ But the gain decreases with frequency.

Variation of Output voltage with V1
◼ Vo = A (V2 –V1)
When V2 = 0, Vo = -AV1
◼ So, the output voltage is out of phase with the input voltage
applied to the inverting input.
◼ That is why it is called the “inverting” input

Variation of Output voltage with V2
◼ Vo = A (V2 – V1)
When V1 = 0, Vo =AV2
◼ So, the output voltage is in phase with the input voltage
applied to the non-inverting input.
◼ That is why it is called the “non-inverting” input

Variation of Output with Input Voltages
◼ Vo = A (V2 – V1)
If V2 > V1, Vo is positive
If V2 < V1, Vo is negative
If V2 = V1, Vo is zero

Consequences of Ideal characteristics
◼ Infinite input resistance means the current into the
inverting input is zero:
i- = 0
◼ Infinite gain means the difference between V1 and V2 is
zero:
V2 – V1 = 0

The Basic Inverting Amplifier
Resistor used to control amplification
R2
V in
–
+ +
–
V out
R 1
+
–
I1
I2

How to Calculate the Gain
◼ For an Inverting amplifier: Gain =
-R2 / R1
Example : if R2 is 100 kilo-ohm and R1 is 10 kilo-ohm,
Gain = -100 / 10 = -10
If the input voltage is 0.5V then the output voltage would be Vin x Gain:
Vout = 0.5V X -10 = -5V

Inverting Amplifier
◼ The i/p voltage to be amplified is fed to the inverting i/p
◼ A fraction of the o/p signal is fed back to the op-amp through the
inverting i/p.
◼ R2 is the feedback resistance in this circuit
◼ Since we have used the inverting i/p, the o/p is out of phase with the
i/p signal.
◼ This process is called negative feedback.

Inverting Amplifier
◼ It is called negative feedback because the overall gain of the amplifier
reduces.
◼ So why use negative feedback if gain is reduced?
The gain is constant over a wide range of input frequencies
and input voltages.
Stability is greater
Amplification is linear – i.e. distortion of o/p is less
Gain is independent of the characteristics of op amp.

Solving the Amplifier Circuit
Apply KCL at the inverting input:
R2
i1 + i2 + i-=0
–
R 1
i1
20
20
Mukesh N. Tekwani
August 27, 2009
i-
i2

KCL
R 1 R 1
i1 =
v i n
i − = 0
− v − =
v i n
2 2
2
R R
=
vo u t − v −
=
vo u t
i

Solve for Vo
Amplifier gain:
R1 R2
vin
= −
vout
R1
vin
R2
A= −
vout
A =
Thus, Gain of an op-
amp depends only
on the two
resistances and not
on the op- amp
characteristics

Assumptions made in deriving gain equation
◼ Each input draws zero current from the signal source.
Typically, i/p current is 1A
That is, input impedances are infinite
◼ The i/ps are both at the same potential if the op-amp is
not saturated.

Problem 1:
In this circuit, we want a gain of ten. If R1 is 5 K ohm, what is
the value you need to use for R0?
Give your answer in ohms.
50,000 ohm

Problem 2:
In this circuit, you have it set up for a gain of -10. The input voltage is 0.24v.
What is the output voltage?
Gain = - Vo / Vi Vo = Gain x Vi Vo
= (-10) x 0.24 Vo = -2.4 V

Problem 3:
In this circuit, Ro and R1 values are shown. The input signal is also shown. Sketch the
o/p signal.
10 K ohm
2.7 K ohm

Problem 3:
• Gain A = Ro / R1
So, A = - 10 K / 2.7 K = -3.7
▪ Amplitude of i/p signal is 4 V
▪ So max o/p voltage is Vo = A x Vin
▪ Vo = 3.7 x 4 = 14.8 V
▪ But power supply is only +9V
▪ So 9V is the max o/p the amplifier can provide.

Problem 3:
• Amplifier is saturated
• It will remain saturated as long as size of i/p voltage is
greater than 9V / 3.7 = 2.4 V
•That is why we observe that the o/p gets clipped as
soon as the i/p rises above 2.4 V

Non inverting amplifier

DC Characteristics of op-amp
Current is taken from the source into the op-amp inputs respond
differently to current and voltage due to mismatch in transistor.
DC output voltages are,
1. Input bias current
2. Input offset current
3. Input offset voltage
4. Thermal drift

1. Input bias current
• The op-amp‘s input is differential amplifier, which may be
made of BJT or FET. In an ideal op-amp, we assumed that no
current is drawn from the input terminals.
• The base currents entering into the inverting and non-
inverting terminals (IB
-& IB
+ respectively).
• Even though both the transistors are identical, IB
- and IB
+ are
not exactly equal due to internal imbalance between the two
inputs.
• Manufacturers specify the input bias current IB

In application where the signal levels are measured in mV, this is totally unacceptable. This can be compensated. Where a
compensation resistor Rcomp has been added between the non-inverting input terminal and ground as shown in the figure
below.

• Current IB
+ flowing through the compensating resistor Rcomp,
then by KVL we
• -V1+0+V2-Vo = 0 (or)
• Vo = V2 – V1 ——>(3)
• By selecting proper value of Rcomp, V2 can be cancelled with
V1 and the Vo = 0. The value of Rcomp is derived a
• V1 =IB+*Rcomp (or)
• IB
+ = V1/Rcomp ——>(4)
• The node =a‘ is at voltage (-V1). Because the voltage at the
non-inverting input terminal is (-V1). So with Vi = 0 we get,

• I1 = V1/R1 ——>(5)
• I2 = V2/Rf ——>(6)
• For compensation, Vo should equal to zero (Vo = 0, Vi = 0). i.e.
from equation (3) V2 = V1. So that,
• I2 = V1/Rf ——>(7)
• KCL at node =a‘ gives,
• IB
-= I2 + I1
• IB
- = RfR1
• Assume IB
- = IB
+ and using equation (4) & (8) we get
• Rcomp = R1 + Rf
• Rcomp = R1 || Rf --->(9)
• i.e. to compensate for bias current, the compensating resistor,
Rcomp combination of resistor R1 and Rf.

2. Input offset current
• Bias current compensation will work if both bias currents IB+ and
IB- are equal.
• Since the input transistor cannot be made identical. There will
always be some small difference between IB
+ and IB
-. This
difference is called the offset curren
• |Ios| = IB
+ -IB
-
• Offset current Ios for BJT op-amp is 200nA and for FET op-amp is
10pA. Even with bias current compensation, offset current will
produce an output voltage when Vi = 0.

the offset current can be minimized by keeping feedback resistance small.
• Unfortunately to obtain high input impedance, R1 must be kept large.
• R1 large, the feedback resistor Rf must also be high. So as to obtain
reasonable gain.
• The T-feedback network is a good solution. This will allow large feedback
resistance, while keeping the resistance to ground low (in dotted line).
• The T-network provides a feedback signal as if the network were a single
feedback resistor.
• By T to Π conversion,
• To design T- network first pick Rt<<Rf/2

3. Input offset voltage
• Inspite of the use of the above compensating techniques, it is
found that the output voltage may still not be zero with zero
input voltage [Vo ≠ 0 with Vi = 0]. This is due to unavoidable
imbalances inside the op-amp and one may have to apply a
small voltage at the input terminal to make output (Vo) = 0.
• This voltage is called input offset voltage Vos. This is the
voltage required to be applied at the input for making output
voltage to zero (Vo = 0).

Total output offset voltage:
• The total output offset voltage VOT could be either more or less than the
offset voltage produced at the output due to input bias current (IB) or
input offset voltage alone(Vos).
• This is because IB and Vos could be either positive or negative with
respect to ground. Therefore the maximum offset voltage at the output
of an inverting and non-inverting amplifier (figure b, c) without any
compensation technique used is given by many op-amp provide offset
compensation pins to nullify the offset voltage.
• 10K potentiometer is placed across offset null pins 1&5. The wipes
connected to the negative supply at pin 4.
• The position of the wipes is adjusted to nullify the offset voltage

Balancing circuit:
Inverting amplifier:
Non-inverting amplifier:

4. Thermal drift:
• Bias current, offset current, and offset voltage change with
temperature.
• A circuit carefully nulled at 25ºC may not remain. So when
the temperature rises to 35ºC. This is called drift.
• Offset current drift is expressed in nA/ºC.
• These indicate the change in offset for each degree Celsius
change in temperature.

AC CHARACTERISTICS
• FREQUENCY RESPONSE
• STABILITY OF OPAMP
• FREQUENCY COMPENSATION

HIGH FREQUENCY MODEL

MAGNITUDE &PHASE CHARCTERISTICS

VOLTAGE TRANSFER FUCTIONIN S DOMAIN

TRANSFER FUNCTION OF OPAMP WITH THREE
BREAK FREQUENCY

STABILITY OF AN OPAMP

FREQUENCY COMPENSATION TECHNIQUES
• EXTERNAL COMPENSATION
Dominant pole
Pole zero compensation
• INTERNAL COMPENSATION

Dominant pole compensation

Pole zero compensation

Open loop gain Vs frequency for pole zero
compensation

Transfer function of pole zero compenstion

Comparison of dominant pole and pole zero
compensation

Internal miller compensation
• Another effective compensation technique is the miller
compensation technique and it is an in-loop compensation
technique where a simple capacitor is used with or without
load isolation resistor (Nulling resistor). That means a
capacitor is connected in the feedback loop to compensate
the op-amp frequency response.
• The miller compensation circuit is shown below. In this
technique, a capacitor is connected to the feedback with a
resistor across the output.

Internal miller compensation

DIFFERENTIAL AMPLIFIER

Difference mode and common mode Gain

Determine the output voltage
=12 V

Scaling Adder
• A different weight can be assigned to each input
of a summing amplifier by simply adjusting the
values of the input resistors. The output voltage
can be expressed as
• VOUT=((Rf/R1)VIN1+(Rf/R2)VIN2+…+(Rf/Rn)VINn)

Problem
Determine the weight of each input voltage for the
scaling adder in below Figure and find the output
voltage.

Averaging Amplifier
• A summing amplifier can be made to produce the
mathematical average of the input voltages. The
amplifier has a gain of Rf/R, where R is the value of
each input resistor. The general expression for the
output of an averaging amplifier is
• VOUT=-(Rf /R)(VIN1+VIN2+…+VINn)
• Averaging is done by setting the ratio Rf/R equal to
the reciprocal of the number of inputs (n); that is ,
Rf/R=1/n.

Problem
Show that the amplifier in produces an output whose magnitude is the
mathematical average of the input voltages.

Applications Of Summing Amplifier
• Summing, averaging & scaling
• Providing DC offset
• Digital to Analog converter
• Audio mixer

ADDER-SUBTRACTOR
It is possible to perform addition and subtraction simultaneously with a
single op-amp using the circuit shown above.
The output voltage Vo can be obtained by using superposition theorem.
To find output voltage Vo1 due to V1 alone, make all other input voltages
V2, V3 and V4 equal to zero. The simplified circuit is shown in Fig.b.
This is the circuit of an inverting amplifier and its output voltage is,

4.3 (b) Op-amp adder- subtractor, (c) Simplifier
circuit for V2 = V3 = V4 = 0,
(d) Simplified circuit for V1 = V2 = V4 =0
Similarly V02 = -V2
Now, the output voltage
Vo3 due to the
voltage signal V3
applied at the (+)
input
alone
input
terminal can be found by setting
V1=V2=V4=0. The circuit now
becomes a non-inverting amplifier
as shown in Fig. 4.3 (d). The
voltage Va at the non- inverting
terminal is
Applying Thevinons theorem at
inverting terminal, we get
=R/2
=1/3 R
=V3

So, the output voltage Vo3 due to V3 aloneis
Similarly, it can be shown that the output
voltage V04: due to V4 alone is Vo4 = V4
(4.14)
Thus, the output voltage Vo due to all four input
voltages is given by

Difference between DA& Differentiator

Differentiator output

Differentiator Circuit
Figure 7-27 Basic
differentiator, (a) Circuit,
(b) Frequency response.
Figure 7-27b
Gain increases with frequency
fc = Unity GainBW
Practical
Figure 7-28
Ideal
fc = Unity GainBW

Since IB = 0,
ic = iF
Since v1= v2 = 0V,

fa is the frequency at which the gain is 0db and is given by
At fa gain is 0db
fC is the unity gain-bandwidth of the op-amp, and f is some relative operating
frequency.
Both the stability and the high-frequency noise problems can be corrected by the addition of two
components: R1 and CF, as shown in Figure 7-28(a), This circuit is a practical differentiator, the
frequency response of which is shownin Figure 7-27(b) by a dashed line. From frequency f to fb, the
gain increases at20 dB/decade. However, after fb the gain decreases at 20 dB/decade. This 40-dB/
decade change in gain is caused by the R1C1 and RFCF combinations. The gain- limiting frequency fb is
given by (7-29)
where R1C1= RFCF , help to reduce significantly the effect of high-frequency
input, amplifier noise, and offsets. Above all, it makes the circuit more stableby preventing the increase
in gain with frequency. Generally, the value of fb and in turn R1C1 and RFCF values should be selected
such that
fa< fb< fc (7-30)

But v1= v2 = 0 V
,because A is very large.Therefore,
Thus the output v0 is equal to the RFC times the negative instantaneous rate of change
of the input voltage vin with time. Since the differentiator performs the reverse of the
integrator's function, a cosine wave input will produce a sine wave output, or a
triangular input will produce a square wave output. However, the differentiator of
Figure 7-27(a) will not do this because it has some practical problems. The gain of the
circuit
(RF /XC1) increases with increase in frequency at a rate of 20 dB/decade. This makes the
circuit unstable. Also, the input impedance XC decreases with increase in frequency,
which makes the circuit very susceptible to high-frequency noise. When amplified,
this noise can completely override the differentiated output signal. The frequency
response of the basic differentiator is shown in Figure 7-27(b). In this figure, fa is the
frequency at which the gain is 0 dB and is given by

EXAMPLE 7-16
Design a differentiator to differentiate an input signal that varies in frequency from 10 Hz to about 1 kHz. If
a sine wave of 1 V peak at 1000 Hz is applied to the differentiator of part (a), draw its output waveform.
SOLUTION (a) To design a differentiator, we simply follow the steps outlined previously:

Problem
Determine the output voltage of the ideal op-amp differentiator in Figure
7-26 for the triangular-wave input shown.

Figure 7-28 Practical differentiator, (a) Circuit, (b) Sine wave input and resulting cosine wave output, (c) Square
wave input and resulting spike output.

The frequency response of the basic integrator is shown in Figure 7-24. In
this figure, fb, is the frequency at whichthe gain is 0 dB and is given by
Both the stability and the low-frequency roll-off problems can be corrected by the
addition of a resistor RF as shown in the practical integrator of Figure 7-25. The
frequency response of the practical integrator is shown in Figure 7-24 by a dashed
line. In this figure, f is some relative operating frequency, and for frequencies up to fa
the gain RF/Ri is constant. However, after fa, the gain decreases at a rate of 20
dB/decade. In other words, between fa, and fb, the circuit of Figure 7-25 acts as
an integrator. The gain-limiting frequency fa is givenby
Frequency Response

If the op-amp was ideal, an integrator as shown in above Figure would require just one resistor, R, and one
capacitor, C, and the relation between the output and input voltages would be given by
fa < fb. If fa = fb/10, then RF = 10R1. The input signal will be integrated properly, if time period T of the input
signal
=fa =fb
The Integrating range is in between fa & fb

Frequency Response of an Ideal & Lossy Integrator
=fa =fb
Integrating Range
20db/decade

Comparison Between Integrator &
Differentiator.
The process of integration involves the accumulation
of signal over time and hence sudden changes in the
signal are suppressed. Therefore an effective smoothing
of signal is achieved and hence, integration can be
viewed as low-pass filtering.
The process of differentiation involves identification
of sudden changes in the input signal. Constant and
slowly changing signals are supressed by a differentiator.
It can be viewed as high-pass filtering.

Voltage to Current Converter (TransconductanceAmplifier)
In many applications, one may have to convert a voltage signal to a proportional output current.
For this, there are two types of circuits possible.
•V-I Converter with floating load
•V-I Converter with grounded load
Figure 4.8 (a) shows a voltage to current converter in which load ZL is
floating. Since voltage at node 'a' is vi therefore,
Fig. 4.8 Voltage to current converter
Vi
(b) Grounded load
iL = vi /R
(a) Floating load
= iLR1
iL proportional to Vi
is controlled by R1
Vi

That is the input voltage vi is converted into an output current of vi /R1, it may be seen that
the same current flows through the signal source and load and,
therefore, signal source should be capable of providing this load current.
A voltage-to-current converter with grounded load is shown in Fig. 4.8 (b). Let vi
be the voltage at node 'a'. Writing KVL, we get
Since the op-amp is used in non-inverting mode, the gain of the circuit is
1 + R/R= 2. The output voltage is,
vo = 2v1 = vi + vo – iLR Vi = iLR
iL = vi /R

As the input impedance of a non-inverting amplifier is very high, this circuit has the advantage of drawing very
little current from the source.
A voltage to current converter is used for low voltage dc and ac voltmeter, LED and zenar diode tester.

Current to Voltage Converter (Trans-resistance Amplifier) Photocell,
photodiode and photovoltaic cell give an output current that is
proportional to an incident radiant energy or light. The current through these devices
can be converted to voltage by using a current-to-voltage converter and thereby the
amount of light or radiant energy incident on the photo-device
canbe measured.
Figure 4.9. shows an op-amp used as I to V converter. Since the (-) input terminal is
at virtual ground, current ii flows through the feedback resistor Rf. Thus the output
voltage v0 = - iiRf. It may be pointed out that the lowest current that this circuit can
measure will depend upon the bias current of the op-amp. This means that for 741
(bias current is 3 nA) can be used to detect lower currents. The resistor Rf is
sometimes shunted with a capacitor Cf to reduce high frequency noise and the
possibility of oscillations.

Inverting Current to Voltage Converter

Non-inverting Current to Voltage Converter.

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