Operational Amplifiers
Operational Amplifier was originally used for D.C. Amplifiers which
perform mathematical operations such as summation, subtraction,
integration, and differentiation in analog computers.
Op-Amp is a high gain amplifier. Output is taken from only one
terminal. Vin  small, Vo  high. So Vo/ Vin = high
Used to amplify both A.C and D.C signals.
Its applications are,
A.C. to D.C. conversion,
Analog-to-Digital and Digital-to-Analog conversion,
Signal conditioning,
Active filters, and
Many
other applications.
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Operational Amplifiers
The Ideal Operational Amplifier:
It has two input terminals 1(V1) and
2(V2) and output terminal 3(V0).
Since the Op-amp is an active device, it requires a D.C. power supply
for its operation. As shown in figure a positive supply +VCC is
connected to terminal 4 and a negative supply –VEE is connected to
terminal 5 (Gnd ???).
The Op-amp is also known as a differential amplifier. This means that
the Op-amp, senses the difference between the input signals applied at
terminals
1 and 2| Website
and amplifies
the difference by an amount A.
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Operational Amplifiers
The output Vo is therefore given by
Vo = A(V2 – V1)
Where V1 and V2 are the input signals and the constant ‘A’ is known as
the open-loop gain (There is no feedback loop connecting the output
terminal to the input terminals).
Properties of an ideal Op-Amp:
1. Infinite Input Impedance: The ideal Op-amp does not draw any
current from the voltage sources connected to its input terminals.
This implies that the input impedance of an Op-amp is infinity.
2. Zero Output Impedance: The voltage at the output terminal is
independent of the current drawn from it i.e. output impedance is
zero.
Hence the Op-amp can drive an infinite number of devices.
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Operational Amplifiers
3. Infinite Bandwidth: This implies that the amplifier can amplify any
frequency from zero to infinity without attenuation. In other
words, the ideal Op-amp will amplify signals of any frequency
with equal gain.
4. Infinite Voltage Gain: The open-loop voltage gain of an ideal Opamp is very large, i.e., infinity.
5. Perfect Balance: The output voltage is zero when equal voltages
are present at the two input terminals.
6. Infinite CMRR : This means that the output common-mode noise
voltage is zero.
7. Infinite Slew Rate: Slew rate indicates the rapidity with which the
output of an Op-amp changes in response to the changes in input
frequency. (how fast the output of op-amp is going to respond for
any change in input)
8. Temperature:
The characteristics do not change with temperature.
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Operational Amplifiers
In the figure, two signals V1 & V2
are fed to the input of an
operational amplifier. One of the
properties of an ideal Op-amp is
that no current should enter the
terminals 1, & 2, i.e., il and i2 are
both zero as shown in figure.
It is observed that in the expression for the output V0, the input signal
at terminal 1 appears with a negative sign attached to it. On the other
hand, the input signal at terminal 2 appears without alteration in its
sign.
Terminal 1 is therefore known as the Inverting Terminal while
Terminal
2 is called
theforNon-inverting Terminal.
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Operational Amplifiers
Saturable Property of an Op-Amp:
‘A’ is a constant for a particular Operational Amplifier and is known as
the open-loop gain since there is no feedback loop connecting the
output terminal to the input terminal, which is the basic circuit of an
op-amp.
This op-amp senses the difference between the input signals and
amplifies the difference between the input signals.
The output Vo is given by
Vo = A*(V2 - V1) ---(i)
Hence,
if V1= 0, Vo = A*V2 ----------(ii)
and
if V2= 0, Vo = - A*V1 --------(iii)
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Operational Amplifiers
The open-loop voltage gain of an ideal op-amp is infinity; then,
substituting A = ∞ in eq’s (ii) & (iii) above, the output Vo of the opamp should range from +∞ to -∞.
However, in practice, Vo is limited by the magnitudes of the power
supply voltages. If the supply voltage are ± 15V, (Terminals 4 & 5)
V0 would be about ±10 V.
When the output attains this level, it does not increase any further
even if the input voltages are increased. This op-amp is now said to
be saturated.
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Operational Amplifiers
Op-Amp as Comparator:
A comparator is a device which compares a signal voltage with a
reference voltage. An Op-Amp comparator is an open loop Op-Amp.
The reference voltage is applied to one of its input terminals, and the
signal to be compared is applied to the other input terminal.
Depending upon which of the two voltages is greater, the output
voltage is held at the positive or negative saturation voltage.
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Operational Amplifiers
A time-varying signal Vin is applied to the non-inverting terminal
through a resistor R, and the reference voltage Vref of 1 V is applied to
the inverting terminal through resistor R1.
When Vin < Vref the voltage at the inverting (-) terminal is greater than
the voltage at the non-inverting (+) terminal and hence Vo = -Vsat it
being approximately equal to -VEE.
When Vin > Vref the voltage at the non-inverting (+) terminal is greater
than the voltage at the inverting (-) terminal and hence Vo = + Vsat it
being approximately equal to +Vcc Therefore, when Vin crosses Vref the
output voltage V0 changes instantaneously from one saturation level
i.e. from +Vcc to -VEE or from -VEE to +Vcc to other and waveforms are
shown in the figure.
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Operational Amplifiers
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Operational Amplifiers
The Inverting Op-Amp Circuit:
An Inverting Op-Amp circuit is one whose output is out of phase by
180o with respect to the input.
The point G is called the ‘virtual ground’. since G is not directly
connected to the ground, no current from G flows to the ground. Also,
the voltage drop across R1 makes the potential V at G to be nearly
zero. This is the reason G is referred to as Virtual Ground (not the true
ground)
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Operational Amplifiers
Because of the infinite input impedance of the Op-Amp, no current
enters the Op-Amp. Hence the same current flows through the
feedback resistor Rf.
where Vo is the output voltage
The ratio Vo/Vi is the closed loop voltage gain VCL of the circuit. As
there is a feedback loop, the gain is known as the closed loop gain. We
see that it differs from A, the open-loop gain and depends only on the
values of the external resistances R1and Rf.
The minus sign (-) indicates that the output is inverted with respect to
the input. Thus the above circuit represents an inverting amplifier.
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Operational Amplifiers
The Non-Inverting Op-Amp Circuit:
A non-inverting Op-Amp circuit is one whose output is in phase with
the input. Figure shows an Op-Amp circuit with a small input Vi
applied to the non-inverting terminal. The inverting terminal is
connected through a resistor R1 to the ground. Rf is the feedback
resistor.
Because of infinite gain of Op-Amp, practically no voltage drop exists
between the input terminals. Hence the potential at G is also taken to
be Vi. Further, assuming no current flows through the Op-amp, the
same
current ‘i’ flows
R1and Rf. Thus we have
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Operational Amplifiers
Since the gain is positive, no phase change occurs for the signal.
Thus the above circuit represents a non-inverting amplifier.
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Operational Amplifiers
Properties of an Practical Op-Amp:
1. Input Impedance: This is the ratio of the input voltage change to
input current change, measured at one input terminal. Input
impedance may be in the range of 10 KΩ to 100 MΩ.
2. Output Impedance: This is the ratio of the output voltage change to
output current change. Output impedance could be in the range of
tens to hundreds of ohms. (75 Ω for µA-742 IC).
3. Gain: Though an ideal Op-amp has infinite gain, a practical Opamp has a gain of the order of 5 x 104.
4. Common Mode Rejection Ratio (CMRR): The relative sensitivity
of an Op-amp to a difference signal as compared to a common
mode signal is called the common-mode rejection ratio, and is
given by
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Operational Amplifiers
 AD 
CMRR(log)  20log10  
 AC 
For commercial Op-amps the CMRR lies in the range 60 to 100 dB.
5. Output Voltage Swing: This is the peak output voltage with respect
to zero available at the output without distortion. It is a function of
the supply voltage. Thus, in an amplifier operating between supply
voltages +6 and -6 V, the peak-to-peak undistorted output swing
might be from -4 V to +4V (in practice the allowable voltage swing
is not always symmetrical).
6. Input Common-mode Voltage Swing: This is the maximum range
of input voltage that can be simultaneously applied to both inputs
without
causing
cut foroff or saturation of amplifier stages.
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Operational Amplifiers
7. Input Offset Current (Iio): This is the difference of the currents into
the input terminals with the output at zero volts.
8. Input Offset Voltage: Whenever both the input terminals of the Opamp are grounded, ideally, the output voltage should be zero.
However, in this condition, the practical Op-amp shows a small nonzero output voltage. To make this output voltage zero, a small
voltage in milli volts (1-4) is required to be applied to one of the
input terminals. Such a voltage makes the output exactly zero. This
d.c. voltage, which makes the output voltage zero, when the other
terminal is grounded is called input offset voltage denoted as Vio.
9. Input Bias Current: Input bias current can be defined as the current
flowing into each of the two input terminals when they are biased to
the
same voltage level, i.e., when the Op-amp is balanced.
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Operational Amplifiers
10. Power-supply Voltage-rejection ratio: A change in supply voltage
ΔVCC will produce a change ΔVo in the amplifier output. The
power-supply voltage-rejection ratio ΔVo/ΔVCC is usually specified
for the condition that the difference voltage input Vi = 0. Typically
this ratio is in the range 10-5 to 7 x 10-5.
11. Frequency Response: Op-amps are usually low-frequency devices.
The open-loop (without feed back) 3 dB bandwidth of generalpurpose Op-amps will be in the region of 100 Hz or less.
Bandwidths as low as 10Hz are possible.
12. Slew Rate: Slew Rate is one of the most important specifications
of an Op-amp because it limits the size of the output voltage at
higher frequencies.
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Operational Amplifiers
The charging current in a capacitor is given by i = C dV/dt
OR
i/C = dV/dt
The above equation says that the rate of change of voltage is equal to
the charging current divided by the capacitance.
The larger the charging current, the faster the capacitor charges. If the
charging current is limited to a maximum value, the rate of change of
voltage is also limited to a maximum value.
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Operational Amplifiers
The rate of change of voltage with respect to time is
1. If Imax = 80 µA and Cc = 40 pF the maximum rate of voltage change
is
This shows that the output voltage across the capacitor changes at a
maximum rate of 2V/µS. The voltage cannot change faster than this
unless Imax is increased or Cc is decreased.
Slew Rate is defined as the maximum rate of output voltage change.
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rate = Imax/CC
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Operational Amplifiers
Modes of Operation: (Ideal)
1. Single-ended mode: Single-ended input operation results when the
input signal is connected to one input with the other input
connected to ground.
Figure-1 shows the input is applied to the plus input (with minus input
at ground), which results in an output having the same polarity. as the
applied input signal. Figure-2 shows an input signal applied to the
minus input, the output then being opposite in phase to the applied
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Operational Amplifiers
2. Differential mode:
Op-Amp has the ability to greatly amplify signals that are opposite at
the two inputs, while only slightly amplifying signals that are common
to both inputs.
In differential mode, the two input signals are equal but have opposite
polarity at every instant of time. Thus referring to figure, Vi1 = -Vi2. So
Differential input VDM = (Vi1 - Vi2)
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Operational Amplifiers
3. Common Mode: In this case, the two input signals are identical both
in amplitude & phase at every instant of time, and the circuit is said to
be operating in the common mode. The input signals are called
common mode signals.
Therefore, VCM = ½ * (Vi1 + Vi2)
i.e. either Vi1 or Vi2 (or average of both).
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Operational Amplifiers
Since amplification of the opposite input signals is much greater than
that of the common input signals, the circuit provides a common mode
rejection as described by a numerical value called the Common-Mode
Rejection Ratio (CMRR).
The overall operation being to amplify the difference signal while
rejecting the common signal at the two inputs. Since noise (any
unwanted input signal) is generally common to both inputs, the
differential connection attenuates this unwanted input while providing
an amplified output of the difference signal applied to the inputs.
V1 = (+Vin) + N,
V2 = (-Vin) + N,
Vo = V1 – V2
= [(+Vin) + N – ((-Vin) + N)]
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+Vin
-Vin
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Operational Amplifiers
Here Common mode signals is ‘N’ i.e. Noise.
Differential mode signals are +Vin and –Vin.
So Common mode signals are cancelled out and differential mode
signals are amplified.
1. VDM = (Vi1 - Vi2)
Vo = AD VDM
2. VCM = ½ * (Vi1 + Vi2)
Vo = AC VCM
Since any signals applied to an op-amp in general have both in-phase
and out-of phase components, the resulting output can be expressed as
Total Vo = (ADVDM + ACVCM)
Where VDM = difference mode input voltage
VCM = common mode input voltage
AD = differential mode gain of the amplifier
AC = common-mode
gain of the amplifier
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Operational Amplifiers
Opposite Polarity Inputs:
If opposite polarity inputs applied to an op-amp are ideally opposite
signals, Vi1 = -Vi2 = Vs,
the resulting difference voltage is
VDM = Vi1 -Vi2 = Vs - (-Vs) = 2Vs
While the resulting common voltage is
VCM = ½ * (Vi1 + Vi2) = ½ * [Vs + (-Vs)] = 0
so that the resulting output voltage is
Vo = ADVDM + ACVCM = AD 2Vs + AC 0 = 2 ADVs
This shows that when the inputs are an ideal opposite signal (no
common element), the output is the differential gain times twice the
input signal applied to one of the inputs.
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Operational Amplifiers
Same Polarity Inputs:
If same polarity inputs applied to an op-amp Vi1 = Vi2 = Vs,
the resulting difference voltage is
VDM = Vi1 -Vi2 = Vs - (Vs) = 0
While the resulting common voltage is
VCM = ½ * (Vi1 + Vi2) = ½ * [Vs + Vs] = Vs
so that the resulting output voltage is
Vo = ADVDM + ACVCM = AD 0 + AC Vs = AcVs
This shows that when the inputs are ideal in-phase signals (no
difference signal), the output is the common-mode gain times the input
signal Vs, which shows that only common-mode operation occurs.
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Operational Amplifiers
Common-Mode Rejection:
The solutions above provide the relationships that can be used to
measure AD and AC in op-amp circuits.
To measure AD: Set Vi1 = -Vi2 = Vs = 0.5 V, so that
VD = (Vi1 - Vi2) = [0.5 V - (-0.5 V)] = 1V
VC = ½ * (Vi1 + Vi2) = ½ [0.5 V + (-0.5 V)] = 0V
Under these conditions the output voltage is
Vo = ADVD + ACVC = AD(1 V) + AC(0) = AD
Thus, setting the input voltages Vi1 = -Vi2 = 0.5V results in an output
voltage numerically equal to the value of AD.
(i.e.forAD = 2*Vs = 2*Vi1 or 2*Vi2)
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Operational Amplifiers
To measure AC: Set Vi1 = Vi2 = Vs = 1V, so that
VD = (Vi1 - Vi2) = [1V – 1V] = 0V
VC = ½ * (Vi1 + Vi2) = ½ [1V + 1V] = 1V
Under these conditions the output voltage is
Vo = ADVD + ACVC = AD(0 V) + AC(1 V) = AC
Thus, setting the input voltages Vi1 = Vi2 = 1V results in an output
voltage numerically equal to the value of AC. (i.e. AC = Vs = Vi1 or Vi2)
AD
CMRR 
AC
 AD 
CMRR(log)  20 log10  
AC 
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Operational Amplifiers
Op-Amp Applications:
1. Comparator:
2. Voltage follower:
The input voltage vi is applied to the non-inverting input, and the
inverting input is grounded through a resistor R. The input and output
are shorted, as shown in figure.
We can show that this Op-Amp gives an output voltage which is equal
to the input voltage, and is in phase with it.
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Operational Amplifiers
As the Op-Amp has practically infinite input impedance and zero
current through the virtual short at N, the potential at N is the same as
the potential at the non-inverting input. But the voltage (with respect
to ground) of this terminal (i.e., non-inverting input) is Vi
Vo = Vi
Hence the output voltage is equal to the input voltage. Also, it is in
phase with the input voltage. As the output voltage follows the input
voltage at all instants, this Op-Amp is called Voltage Follower.
3. Summing Amplifier: (Adder)
The three inputs to be added are connected to the inverting input as
shown. The non-inverting terminal is connected to the ground. This
op-amp summing amplifier is a feedback amplifier in which the
feedback
is provided through Rf. Point G is at virtual ground.
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Operational Amplifiers
Since G is at virtual ground,
if = il + i2 + i3 ---- (1)
where i1 i2 and i3 are the currents through input resistors R1, R2 and R3
respectively.
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Operational Amplifiers
Substituting these above values in eq (i) we get,
An averaging circuit is a special case of a summing amplifier.
Suppose R1 = R2 = R3 = R,
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Operational Amplifiers
4. Subtractor:
The subtraction of 2 input voltages is possible by using an Op-amp
circuit called the subtractor. Its operation is similar to that of a
summing amplifier.
Vo = (V2 – V1)
In order to determine the output of the subtractor, the principle of
superposition is used. (Analyze one input by grounding the other
input).
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Operational Amplifiers
First V1 will be analyzed by grounding V2.
So Op-Amp becomes inverting amplifier.
V01 = (-Rf/R1) * V1
If Rf = R1 then V01 = -V1
Second V2 will be analyzed by grounding V1.
So Op-Amp becomes Non-inverting amplifier.
At point ‘N’ the potential will be (Voltage divider theorem)
R2
VN 
 V2
R2  R f
if R f  R2
V2
VN 
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Operational Amplifiers
For Non-inverting amplifier,
V02 = [1 + (Rf/R2)] * VN
V02 = 2 VN = 2 (V2/2) = V2
So total Vo = V01 + V02
Vo = -V1 + V2
i.e.
Vo = (V2 – V1)
(Here Rf = R2)
(Subtractor)
The output voltage is equal to the voltage V2 applied to the Noninverting terminal minus the voltage V1 applied to the inverting
terminal. Hence the circuit is called “Subtractor”.
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Operational Amplifiers
5. Integrating Circuit: (Integration amplifier)
An integrator circuit is one whose output is the integral of the input. It
is obtained by using a basic inverting amplifier configuration if the
feedback resistor ‘Rf ’ is replaced by a capacitor ‘Cf ’.
The charge on the capacitor is Q = C * V
Q = C (0 - Vo)
where ‘0’ is the potential at G (virtual ground).
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Operational Amplifiers
Equate the two equations
But i = Vi/R, where ‘Vi’ is the input voltage and ‘R’ resistor through
which the input is fed.
Thus we see that the output voltage is the integral of the input voltage
and hence the circuit functions as an integrator.
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Operational Amplifiers
6. Differentiating Circuit:
A differentiator circuit is one whose output is the differential
coefficient of the input. Figure shows a differentiator circuit using an
Op-Amp. ‘G’ is the virtual ground.
The input voltage ‘Vi’ is applied to the inverting terminal of the OpAmp through the capacitor. The non-inverting terminal is connected
to the ground.
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Operational Amplifiers
If ‘q’ is the charge acquired by the capacitor then,
Vi = q/C
Differentiating equation w.r.t. ‘t’ we get
The charging current of the capacitor is
but i = dq/dt
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Operational Amplifiers
Equating these two equations
but
Substitute for dq/dt
Output = (Constant) x (Differential coefficient of the Input)
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Operational Amplifiers
1. An Inverting Amplifier has R1 = 20 KΩ and Rf = 100 KΩ. Find the
Output Voltage, the Input Resistance and the Input Current for an input
voltage of 1V.
2. Design an inverting amplifier with an input resistance of 30 KΩ and
a closed loop gain of -10.
ACL = -10, R1 = 30 KΩ
Rf = 300 KΩ
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Operational Amplifiers
3. An inverting amplifier of figure has a load resistor of 50 KΩ
connected to its output. If the input voltage is 0.5 V, find the load
current, output voltage and input current.
(i) Input current through R1 is,
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Operational Amplifiers
(ii) Output voltage
i = (0-Vo)/Rf
Vo = -iRf
Vo = 5V
(iii) Load current
4. Determine the closed loop voltage gain of the Op-Amp circuit
shown in figure. Also calculate the amplitude of the output voltage if
Vi = 0.2 sinωt. (Check the question)
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Operational Amplifiers
The figure is that of an inverting amplifier
= -5 (negative sign indicates phase shift i.e. inverting configuration)
Vi =Vm sin ωt, = 0.2 sin ωt
here Vm = 0.2
While determining the amplitude of the output voltage, it is not
required to consider the negative sign of the gain.
Vo = 5*0.2 = 1 V
5. Design an amplifier to have a gain of + 10 using a single Op-amp.
As the
gain is positive, the amplifier is a non-inverting amplifier.
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Operational Amplifiers
As both R1 and Rf are external resistors, they can be
selected arbitrarily.
Let R1 = 30 KΩ
Rf = 270 KΩ
then,
6. For the circuit shown in figure determine the output voltage.
The
circuit of figure
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Operational Amplifiers
7. Find the output voltage of a three-input adder circuit in which
R1= R2 = R3 = 4 KΩ and the feedback resistance Rf = 6 KΩ .
Given V1 = -4V, V2 = - 2V and V3 = 3V
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Operational Amplifiers
8. Design a scaling adder circuit using an Op-Amp, to give the output.
Vo = -(3Vl + 4V2 + 5V3)
For an inverting summing amplifier,
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Operational Amplifiers
Let us arbitrarily choose Rf =120KΩ
Comparing the above equation with the one given in the problem, we
have,
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Operational Amplifiers
P – 462
P – 478
P – 496
P – 503
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M.V. Rao
M.V. Rao
M.V. Rao
M.V. Rao
Q.6(b)
Q.6(b)
Q.6(a)
Q.6(c)
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Operational Amplifiers
1. Differentiate ideal op-amp and practical op-amp. Write at least 6-8 differences.
2. Explain (a) Integrating Amplifier (b) Differentiating Amplifier
3. Find the output voltage of a three-input adder circuit in which R1= R2 = R3 = 4 KΩ
and the feedback resistance Rf = 6 KΩ. Given V1 = -4V, V2 = - 2V and V3 = 3V
4. Design a scaling adder circuit using an Op-Amp, to give the output.
Vo = -(3Vl + 4V2 + 5V3)
5. The gain of voltage follower is ---------A) Unity
B) Zero
C) Infinity
D) None
6. The maximum rate at which amplifier output can change in volts per microsecond
(V/µs) is called
A) Over rate
B) Slew rate
C) Under rate
D) None
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Operational Amplifiers
1. Differentiate ideal op-amp and practical op-amp. Write at least 6-8
differences.
2. Explain (a) Inverting Amplifier (b) Non-Inverting Amplifier
3. Find the output voltage of a three-input adder circuit in which R1=
R2 = R3 = 4 KΩ and the feedback resistance Rf = 6 KΩ. Given V1
= -4V, V2 = - 2V and V3 = 3V
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Operational Amplifiers
Q. Draw the non-inverting voltage amplifier circuit using an op-amp
and show that the closed-loop voltage gain is given by
AVf 
AV
(1  AV  )
Where AV = open-loop voltage gain of an op-amp, β = feedback factor.
A non-inverting Op-Amp circuit is one whose output is in phase with
the input. Figure shows an Op-Amp circuit with a small input Vi
applied to the non-inverting terminal. The inverting terminal is
connected through a resistor R1 to the ground. Rf is the feedback
resistor.
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Operational Amplifiers
Because of infinite gain of Op-Amp, practically no voltage drop exists
between the input terminals. Hence the potential at G is also taken to
be Vi Further, assuming no current flows through the Op-amp, the
same current ‘i’ flows through R1 and Rf.
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Operational Amplifiers
Since the gain is positive, no phase change occurs for the signal. The
expression in eq-(2) was derived on the assumption of infinite openloop amplifier gain AV.
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Operational Amplifiers
For a finite value of Av, from figure, the differential voltage amplified
by the op-amp is
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