Slide_the-operational-amplifier.pdf

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Electronic Devices, 9th edition
Thomas L. Floyd
The Operational Amplifier
(Op-Amp)

Thomas L. Floyd
Chapter Goals
• Develop understanding of linear amplification
concepts such as:
– Voltage gain, current gain, and power gain
– Gain conversion to decibel representation
– Input and output resistances
– Biasing for linear amplification
– Distortion in amplifiers
– Two-port representations of amplifiers
– Understand behavior and characteristics of ideal differential
and op amps.
– Demonstrate circuit analysis techniques for ideal op amps.
– Characterize inverting, non-inverting amplifiers.

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Amplification
Introduction
A complex periodic signal can be represented as the sum of many
individual sine waves. We consider only one component with
amplitude Vs = 1 mV and frequency ωs with 0 phase (signal is
used as reference):
Amplifier output is sinusoidal with same frequency but different
amplitude Vo and phase θ:

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Amplification
Introduction (cont.)
Amplifier output power is:
Here, we desire PO = 100 W with RL = 8 Ω and Vs = 1 mV
Output power also requires output current which is:
Input current is given by
Output phase is zero because circuit is purely resistive.

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Amplification
Voltage Gain & Current Gain
• Voltage Gain:
Magnitude and phase of voltage gain are given by
and
For our example,
• Current Gain:
Magnitude of current gain is given by

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Amplification
Power Gain
• Power Gain:
For our example,

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Amplification
Expressing Gain in Decibels (dB)
The logarithmic decibel or dB scale compresses the huge numeric range of
gains encountered in real systems.

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Amplification
Expressing Gain in dB - Example
For our example:

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Mismatched Source and Load
Resistances
 In introductory circuit theory, the maximum power transfer
theorem is usually discussed.
 Maximum power transfer occurs when the source and load
resistances are matched (equal in value).
 In most amplifier applications, however, the opposite situation is
desired.
 A completely mismatched condition is used at both the input and
output ports of the amplifier.

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Mismatched Source and Load
Resistances
If Rin >> Rs and Rout<< RL, then
In an ideal voltage amplifier,
and Rout = 0
For the voltage amplifier shown

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Distortion in Amplifiers
• In this graph,
different gains for
positive and
negative values of
the input cause
distortion in the
output.
• Total Harmonic
Distortion (THD) is
a measure of signal
distortion that
compares
undesired harmonic
content of a signal
to the desired
component.

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Total Harmonic Distortion
dc desired
output
2nd harmonic
distortion
3rd harmonic
distortion
Numerator = rms amplitude of distortion terms
Denominator = desired component

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Differential Amplifier
Basic Model
vo= A vid
A = open-circuit voltage gain
vid = (v+- v-) = differential input
signal voltage
Rid = amplifier input resistance
Ro = amplifier output
resistance
An ideal differential amplifier produces an
output that depends on the voltage
difference between its two input terminals.
Signal developed at amplifier output is in
phase with the voltage applied at + input
(non-inverting) terminal and 180o out of
phase with that applied at - input (inverting)
terminal.

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Differential Amplifier Model
Impact of Source and Load
RL = load resistance
RS = Thevenin equivalent
resistance of signal source
vs = Thevenin equivalent voltage
of signal source
•Op amp circuits are mostly dc-coupled amplifiers. Signals vo and vs may have
a dc component representing a dc shift of the input away from the Q-point.
•Op-amp amplifies both dc and ac components.

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Example including Source and Load Resistances
• Problem: Calculate voltage gain for an amplifier
• Given Data: A = 100, Rid = 100kΩ, Ro = 100Ω, RS = 10kΩ, RL =
1000Ω

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Example including Source and Load Resistances
• Analysis:
• An ideal amplifier’s output depends only on the input voltage
difference and not on the source and load resistances. This can
be achieved by using a fully mismatched resistance condition
(Rid >> RS or infinite Rid, and Ro << RL or zero Ro ). Then:
• A = open-loop gain (maximum voltage gain available from the
device)

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Operational Amplifers
Op-amp is an electronic device that amplify the difference of voltage
at its two inputs.
–
+
+V
–V
Very high gain dc coupled amplifiers with differential inputs.
One of the inputs is called the inverting input (−); the other is called the
non-inverting input. Usually there is a single output.
Most op-amps operate from plus and minus supply voltages, which
may or may not be shown on the schematic symbol.
1
20
1
8
DIP
1
8
DIP SMT
1
8
SMT

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The Ideal Op-Amp
Ideally, op-amps have characteristics (used in circuit analysis):
 Infinite voltage gain
 Infinite input impedance (does not load the driving sources)
 Zero output impedance (drive any load)
 Infinite bandwidth (flat magnitude response, zero phase shift)
 Zero input offset voltage.
–
+
Zin = ‘
Vin Vout
Zout =0
AvVin
Av = ‘
The ideal op-amp has characteristics that
simplify analysis of op-amp circuits.
The concept of infinite input impedance is
particularly a valuable analysis tool for
several op-amp configurations.

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The Practical Op-Amp
Real op-amps differ from the ideal model in various respects.
In addition to finite gain, bandwidth, and input impedance,
they have other limitations.
–
+
Zin
Vin Vout
Zout
AvVin
 Finite open loop gain.
 Finite input impedance.
 Non-zero output impedance.
 Input current.
 Input offset voltage.
 Temperature effects.

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Internal Block Diagram of an Op-Amp
Internally, the typical op-amp has a differential input, a
voltage amplifier, and a push-pull output. Recall from the
discussion in Section 6-7 of the text that the differential
amplifier amplifies the difference in the two inputs.
Differential
amplifier
input stage
Voltage
amplifier(s)
gain stage
Push-pull
amplifier
output
stage
Vin Vout
+
–

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Input Signal modes
 As the input stage of an op-amp is a differential amplifier,
there are two input modes possible: differential mode
and common mode.
 In differential mode any one of the two scenarios can
occur.
 Either one input is applied to one input while the other
input is grounded (single-ended).
 Or opposite polarity signals are applied to the inputs
(double-ended).

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Differential Mode Operation
Vin
–
+
Vout
Vin
–
+
Vout
Single-ended differential amplifier
Double-ended differential amplifier

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Common Mode Operation
 The same input tend to cancel
each other and the output is
zero.
 This is called common-mode
rejection.
Vin
Vin
–
+
Vout
 In common mode, two signals voltages of the same
amplitude, frequency and phase are applied to the two
inputs.
 This is useful to reject unwanted signal that appears to both
inputs. It is cancelled and does not appear at the output

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Common-Mode Rejection Ratio
The ability of an amplifier to amplify differential signals and
reject common-mode signals is called the common-mode
rejection ratio (CMRR).
CMRR is defined as ol
cm
A
A
=
CMRR
where Aol is the open-loop
differential-gain and Acm is
the common-mode gain.
CMRR can also be expressed in decibels as 20log ol
cm
A
A
 
=  
 
CMRR
Acm is zero in ideal op-amp and much less than 1
is practical op-amps.
Aol ranges up to 200,000 (106dB)
CMRR = 100,000 means that desired signal is amplified 100,000 times
more than un wanted noise signal.
Op-Amp parametrs

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Common-Mode Rejection Ratio
Example
What is CMRR in decibels for a typical 741C op-amp?
The typical open-loop differential gain for the 741C is 200,000 and the
typical common-mode gain is 6.3.
90 dB
20log ol
cm
A
A
 
=  
 
CMRR
200,000
20log
6.3
= =
(The minimum specified CMRR is 70 dB.)

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Maximum Output Voltage Swing
VO(p-p): The maximum output voltage swing is determined
by the op-amp and the power supply voltages
With no input signal, the output of an op-amp is ideally 0 V.
This is called the quiescent output voltage.
When an input signal is applied, the ideal limits of the peak-
to-peak output signal are ± VCC .
In practice, however, this ideal can be approached but never
reached (varies with load resistance).

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Input Offset Voltage
 Ideal op-amp produces zero output voltage if the
differential input is zero
 But practical op-amp produces a non-zero output voltage
when there is no differential input applied. This output
voltage is termed VOUT(error).
 It is due to unavoidable mismatches in the differential
stage of the op amp.
 The amount of differential input voltage required
between the inputs to force the output to zero volts is the
input offset voltage VOS .
 Typical value of VOS is about 2mV.

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The input bias current is the
average of the two dc currents
required to bias the differential
amplifier
Ideally, input bias current is zero. 1 2
BIAS
2
I I
I
+
=
Input Bias Current (IBIAS )

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 The input impedance of an op-amp is
specified in two ways:
 Differential input impedance and
common-mode input impedance.
 Differential input impedance, ZIN(d),
is the total resistance between
inverting and noninverting input.
 Common-mode input impedance,
ZIN(c), is the resistance between each
input and ground.
Input Impedance
ZIN(d)
–
+
ZIN(cm)
–
+

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The offset voltage developed by the input offset current is
𝑉𝑉𝑂𝑂𝑂𝑂 = 𝐼𝐼𝑂𝑂𝑂𝑂𝑅𝑅𝑖𝑖𝑖𝑖
The output error volt is
𝑉𝑉𝑂𝑂𝑈𝑈𝑈𝑈(𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒) = 𝐴𝐴𝑣𝑣𝐼𝐼𝑂𝑂𝑂𝑂𝑅𝑅𝑖𝑖𝑖𝑖
Input offset Current IOS
Ideally, the two input bias currents are equal, and thus their
difference is zero
The input offset current is
the difference of the input
bias currents 𝐼𝐼𝑜𝑜𝑜𝑜 = 𝐼𝐼1 − 𝐼𝐼2

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Zout : The output impedance is the resistance viewed from the
output of the circuit.
Output Impedance
Zout
–
+

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Slew Rate
The slew rate is the maximum
rate of change of the output
voltage in response to a step input
voltage (V/µs)
Slew rate is measured with an op-
amp connected as shown and is
given as
The slew rate is dependent upon
the high-frequency response of
the amplifier stages within the
op-amp.

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Example
Determine the slew rate for the
output response to a step input
shown.
Since this response is not
ideal, the limits are taken
at the 90% points.

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Negative Feedback
Negative feedback is the process of returning a portion of
the output signal of an amplifier to the input with a phase
angle that is opposite to the input signal.
Vout
+
–
Vin
Vf
Internal inversion makes Vf
180° out of phase with Vin.
Negative
feedback
circuit
The advantage of negative
feedback is that precise
values of amplifier gain
can be set. In addition,
bandwidth and input and
output impedances can be
controlled.

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Importance of Negative Feedback
 The usefulness of an op-
amp operated without
negative feedback is
generally limited to
comparator applications
 With negative feedback the gain of op-amp (called close-
loop gain Acl) can be reduced and controlled so that an
op-amp can function as a linear amplifier.
 The inherent open-loop voltage gain of a typical op-amp is
very high. Therefore, an extremely small input voltage (even
the input offset voltage) drives the op-amp into saturation.

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Op-Amps with Negative Feedback
 An op-amp can be connected using negative feedback to
stabilize the gain and increase frequency response.
 This close-loop gain (Acl ) is usually much less than the open-
loop gain (Aol).
 The close-loop voltage gain is the voltage of an op-amp with
external feedback.
 The amplifier circuit consists of an op-amp and an external
negative feedback circuit.
 The feedback from the output is connected to the inverting input
of the op-amp.
 The negative feedback is determined and controlled by external
components.

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Noninverting Amplifier
A noninverting amplifier is a configuration in which the
signal is on the noninverting input and a portion of the
output is returned to the inverting input.
The feedback circuit is formed by input resistance Ri and
feedback resistance Rf .
Rf
Ri
Vf
Vin
+
–
Feedback
circuit
Vout
This feedback creates a
voltage divider circuit
which reduces Vout and
connects the reduced
voltage Vf to the
inverting input and can
be expressed as:

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The differential input is
amplified by the open-
loop gain and produces
the output voltage as:
The attenuation, B, of
the feedback circuit is

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(NI) 1
f
cl
i
R
A
R
= +
Substituting BVout for Vf , we get
The overall voltage gain of the amplifier can be expressed
as
The product AolB is typically much greater than 1, so the
equation simplifies to
Which means

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(NI) 1
f
cl
i
R
A
R
= +
Determine the gain of the noninverting amplifier shown.
Rf
82 kΩ
Vin +
–
Vout
Ri
3.3 kΩ
82 k
1
3.3 k
Ω
= +
Ω
= 25.8

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Voltage Follower
A special case of the inverting amplifier is when Rf =0 and
Ri = ∞. This forms a voltage follower or unity gain buffer
with a gain of 1.
Rf
82 kΩ
Vin +
–
Vout
Ri
3.3 kΩ
This configuration offers very
high input impedance and its very
low output impedance.
These features make it a
nearly ideal buffer amplifier for
interfacing high-impedance
sources and low-impedance loads
Vin +
–
Vout
It produces an excellent circuit for isolating one circuit stage from another,
which avoids "loading" effects.

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Inverting Amplifier
We have: and
Since Iin = If , then
The overall gain is

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Example
Determine the gain of the inverting amplifier shown.
82 k
3.3 k
Ω
= −
Ω
= −24.8
–
+
Rf
Vout
Ri
Vin
82 kΩ
3.3 kΩ
(I)
f
cl
i
R
A
R
= −
The minus sign
indicates inversion.
Inverting Amplifier

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Oxford University Publishing
Microelectronic Circuits by Adel S. Sedra and Kenneth C. Smith (0195323033)
2.2.1.
Closed-Loop Gain
• question: how will we…
– step #4: define vOut in terms of
current flowing across R2
– step #5: substitute vin / R1 for
i1.
Analysis of the inverting configuration. The circled numbers indicate
the order of the analysis steps.
closed-loop
gain
G = -R2/R1

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Effect of Finite Open-Loop Gain
• Q: How does the gain expression change if open
loop gain (A) is not assumed to be infinite?
– A: One must employ analysis similar to the
previous, result is presented below…

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
if then the previous
gain expression is
2 1
2
yielded
2
1
1
/
1 ( / )
1
A
G
A
Out
A
In
v R R
G
R R
v
A
R
R
=∞
=∞
<∞
−
= =
+
 
+ 

≠ −


ideal gain
non-ideal gain
Collecting terms, the closed-loop
gain G is found as

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• Q: Under what condition can G = -R2 / R1 be
employed over the more complex expression?
– A: If 1 + (R2/R1) << A, then simpler expression
may be used.
2 2 2 1
2 1
1 1
if then el
/
1
1 ( /
1
se
)
A A
R R R R
A G G
R R
R R
A
=∞ <∞
−
+ << = − =
+
 
+  
 
ideal gain non-ideal gain

Thomas L. Floyd
Oxford University Publishing
Microelectronic Circuits by Adel S. Sedra and

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From the Figure we get
Impedances of the Noninverting Amplifier
Input Impedance
Substituting IinZin = Vd , where Zin is open loop input impedance
( )
(NI) 1
in ol in
Z A B Z
= +
The input impedance of the noninverting amplifier with negative
feedback is much greater than the internal open-loop input
impedance of the op-amp.

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Referring to the shown Figure,
after some mathematical
manipulations (Floyd 620) it
can be shown that the output
impedance of a noninverting
(NI) amplifier can be given as
Impedances of the Noninverting Amplifier
Output Impedance
The output impedance of the noninverting amplifier with
negative feedback is much less than the internal open-loop input
impedance of the op-amp.
( )
(NI)
1
out
out
ol
Z
Z
A B
=
+

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Since voltage-follower is a special case on noninverting
amplifier, the same formulas are used with B = 1, therfore
Impedances of the Voltage-Follower
The input impedance of voltage-follower is very high, it can be
seen that it is extremely high as compared to the noninverting
amplifier as the feedback attenuation B = 1.
The same is true for the output impedance where the factor of
is removed so it becomes even less than the output impedance of
the noninverting amplifier.

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The input impedance of the inverting (I) amplifier is
Impedances of the Inverting Amplifier
This is because the input is in series with Ri and that is connected
to virtual ground so that is the only resistance seen by input.
Input Impedance

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As with the noninverting amplifier, the output impedance of an
inverting (I) amplifier is decreased by the negative feedback.
In fact the the expression is the same as for the noninverting case.
Impedances of the Inverting Amplifier
The output impedance of the both noninverting and inverting
amplifier is low. In fact, practically it can be considered zero..
Output Impedance

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Noninverting amplifier:
Impedances
( )
(NI) 1
in ol in
Z A B Z
= +
( )
(NI)
1
out
out
ol
Z
Z
A B
=
+
Generally, assumed to be ∞
Generally, assumed to be 0
(I)
in i
Z R
≅
( )
(I)
1
out
out
ol
Z
Z
A B
=
+
Generally, assumed to be Ri
Generally, assumed to be 0
Inverting amplifier:
Note that the output impedance has the same form for both amplifiers.

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Example

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Solution

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Determine the closed-loop gain of each amplifier in Figure.
(a) 11 (b) 101 (c) 47.8 (d) 23

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If a signal voltage of 10 mVrms is applied to each amplifier in
Figure, what are the output voltages and what is there phase
relationship with inputs?.
(a) Vout ≅ Vin = 10 mV, in phase (b) Vout = AclVin = – 10 mV, 180º out of
phase (c) Vout = 233 mV, in phase (d) Vout = – 100 mV, 180º out of phase

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Effect of Input Bias Current - Inverting Amplifier
An inverting amplifier with zero input voltage is shown.
Ideally, the current through Ri is zero because the input voltage
is zero and the voltage at the inverting terminal is zero.
The small input bias current, I1, is through Rf from the output
terminal. This produces an output error voltage I1Rf when it
should be zero

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A voltage-follower with zero input voltage and a source
resistance, Rs is shown.
In this case, an input bias current, I1, produces a drop across
Rs and creates an output voltage error -I1 Rs as shown.
Effect of Input Bias Current – Voltage Follower

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Consider a noninverting amplifier with zero input voltage.
The input bias current, I1, produces a voltage drop across
Rf and thus creates an output error voltage of I1Rf
Effect of Input Bias Current - Noninverting Amplifier

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Bias Current Compensation – Voltage Follower
The output error voltage due to bias currents in a voltage-
follower can be sufficiently reduced by adding a resistor,
Rf , equal to the source resistance, Rs , in the feedback
path.
The voltage drop created by I1 across the added resistor
subtracts from the output error voltage (if I1 = I2 ).

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Bias Current Compensation – Inverting and Noninverting
To compensate for the effect of bias current in Inverting
and Noninverting Amplifiers, a resistor Rc is added at the
noninverting terminal.
The compensating resistor value equals the parallel
combination of Ri and Rf.
The input current creates a voltage drop across Rc that
offsets the voltage across the combination of Ri and Rf ,
thus sufficiently reducing the output error voltage.

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–
+
Rf
Vout
Ri
Vin
Rc = Ri || Rf
–
+
Rf
Vout
Ri
Vin
Rc = Ri || Rf
Noninverting
amplifier
Inverting
amplifier
Bias Current Compensation – Inverting and Noninverting

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Effect of Input Offset Voltage
 The output of an op-amp should be zero when the
differential input is zero.
 Practically this is not the case. There is always an output
error voltage present whose range is typically in
microvolts to millivolts.
 This is due to the imbalances within the internal op-amp
transistors
 This output error voltage is aside from the one produced
by the input bias

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Input Offset Voltage Compensation
Most ICs provide a mean of compensating for offset voltage.
An external potentiometer to the offset null pins of IC package

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Bandwidth Limitations
 Frequency response of amplifiers is shown in a plot called Bode
Plot.
 In Bode plot, the frequency is on the horizontal axis and is in
logarithmic scale. It means that the frequency change is not
linear but ten-times. This ten-time change in frequency is called
a decade.
 The vertical axis shows the voltage gain in decibel (dB).
 The maximum gain on the plot is called the midrange gain.
 The point in the frequency response of amplifiers where the
gain is 3dB less than the midrange gain is called the critical
frequency.

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–20 dB/decade roll-off
Unity-gain frequency (fT)
Critical frequency
10
1 100 1k 10k 100k 1M
f (Hz)
106
100
75
50
25
0
Aol (dB)
Midrange

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 An open-loop response curve (Bode plot) for a certain op-amp
is shown.
 The differential open-loop gain Aol of an op amp is not
infinite; rather, it is finite and decreases with frequency.
 Note that although the gain is quite high at dc and low
frequencies, it starts to fall off at a rather low frequency.
 The process of modifying the open-loop gain is termed
frequency compensation, and its purpose is to ensure that op-
amp circuits will be stable (as opposed to oscillatory).
 These are units that have a network (usually a single
capacitor) included within the same IC chip whose function is
to cause the op-amp gain to have the single-time-constant
low-pass response shown.

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Gain-Versus-Frequency Analysis
The RC lag (low-pass) circuits within an
op-amp are responsible for the roll-off in
gain as the frequency increases, just as for
the discrete amplifiers. The attenuation of
an RC lag circuit shown is expressed as:
The critical frequency of an RC circuit is:

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Gain-Versus-Frequency Analysis
If an op-amp is represented by a voltage gain element with a
gain of Aol(mid) plus a single RC lag circuit, as shown, it is
known as a compensated op-amp. The total open-loop gain of
the op-amp is the product of the midrange open-loop gain,
Aol(mid), and the attenuation of the RC circuit as:

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Phase Shift
An RC circuit causes a propagation delay from input to
output, thus creating a phase shift between the input signal
and the output signal. An RC lag circuit such as found in an
op-amp stage causes the output signal voltage to lag the input.
The phase shift, θ, is given by:

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Overall Frequency Response
 Most op-amps have a constant roll-off of -20 dB/decade
above its critical frequency. The more complex IC
operational amplifier may consist of two or more
cascaded amplifier stages.
 The gain of each stage is frequency dependent and rolls
off at above its critical frequency. Therefore, the total
response of an op-amp is a composite of the individual
responses of the internal stages. dB gains are added and
phase lags of the stages are added as shown in next slide
for a three stage op-amp.

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closed-loop frequency response
 Op-amps are usually used in a closed-loop configuration
with negative feedback in order to achieve precise
control of gain and bandwidth.
 Midrange gain of an op-amp is reduced by negative
feedback as we have already seen in previous sections.
 Now we will see its effects on bandwidth.

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Effect of Negative Feedback on Bandwidth
 The closed-loop critical frequency of an op-amp is given by:
𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐) = 𝑓𝑓𝑐𝑐 𝑜𝑜𝑙𝑙 (1 + 𝐵𝐵𝐴𝐴𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚 )
Where B is the feedback attenuation of the closed-loop op-amp.
 The above expression shows that the closed-loop critical
frequency, 𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐), is higher than the open-loop critical
frequency 𝑓𝑓𝑐𝑐 𝑜𝑜𝑜𝑜 by a factor of (1 + 𝐵𝐵𝐴𝐴𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚 )
 Since 𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐) equals bandwidth therefore the closed-loop
bandwidth, 𝐵𝐵𝐵𝐵𝑐𝑐𝑐𝑐, is also increased:
𝐵𝐵𝐵𝐵𝑐𝑐𝑐𝑐 = 𝐵𝐵𝐵𝐵𝑜𝑜𝑙𝑙(1 + 𝐵𝐵𝐴𝐴𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚 )

Thomas L. Floyd
Summary
The Figure shows the concept of closed-loop response. When the open-
loop gain is reduced due to negative feedback, the bandwidth is
increased.
This means that you can
achieve a higher BW by
accepting less gain.
Av
f
Aol(mid)
0 fc(ol) fc(cl)
Acl(mid)
Closed-loop gain
Open-loop gain

Thomas L. Floyd
Gain-Bandwidth Product
 An increase in the closed-loop gain causes a decrease in the
bandwidth and vice versa, such that the product of gain and
bandwidth is constant.
 If 𝐴𝐴𝑐𝑐𝑐𝑐 is the gain of an op-amp with 𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐) bandwidth then:
𝐴𝐴𝑐𝑐𝑐𝑐𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐) = 𝐴𝐴𝑜𝑜𝑙𝑙𝑓𝑓𝑐𝑐(𝑜𝑜𝑙𝑙)
 The equation, Acl fc(cl) = Aol fc(ol) shows that the product of the
gain and bandwidth are constant.
 The gain-bandwidth product is always equal to the frequency
at which the op-amp’s open-loop gain is unity or 0 dB (unity
gain bandwidth, 𝑓𝑓𝑇𝑇)
𝑓𝑓𝑇𝑇 = 𝐴𝐴𝑐𝑐𝑐𝑐𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐)

Thomas L. Floyd
Example
Determine the bandwidth of each of the amplifiers shown.
Both op-amps have an open-loop gain of 100 dB and a unity-
gain bandwidth (fT ) of 3 MHz.

Thomas L. Floyd
Selected Key Terms
Operational
amplifier
Differential
mode
Common mode
A type of amplifier that has very high voltage
gain, very high input impedance, very low
output impedance and good rejection of
common-mode signals.
A mode of op-amp operation in which two
opposite-polarity signals voltages are applied to
the two inputs (double-ended) or in which a
signal is applied to one input and ground to the
other input (single-ended).
A condition characterized by the presence of
the same signal on both inputs

Thomas L. Floyd
Selected Key Terms
Open-loop
voltage gain
Negative
feedback
Closed-loop
voltage gain
Gain-
bandwidth
product
The voltage gain of an op-amp without external
feedback.
The process of returning a portion of the output
signal to the input of an amplifier such that it is
out of phase with the input.
The voltage gain of an op-amp with external
feedback.
A constant parameter which is always equal to
the frequency at which the op-amp’s open-loop
gain is unity (1).

Thomas L. Floyd
Quiz
1. The ideal op-amp has
a. zero input impedance and zero output impedance
b. zero input impedance and infinite output impedance
c. infinite input impedance and zero output impedance
d. infinite input impedance and infinite output
impedance

Thomas L. Floyd
Quiz
2. The type of signal represented in the figure is a
a. single-ended common-mode signal
b. single-ended differential signal
c. double-ended common-mode signal
d. double-ended differential signal
Vin
–
+
Vout

Thomas L. Floyd
Quiz
3. CMRR can be expressed in
a. amps
b. volts
c. ohms
d. none of the above

Thomas L. Floyd
Quiz
4. The difference in the two dc currents required to bias the
differential amplifier in an op-amp is called the
a. differential bias current
b. input offset current
c. input bias current

Thomas L. Floyd
Quiz
5. To measure the slew rate of an op-amp, the input signal is a
a. pulse
b. triangle wave
c. sine wave

Thomas L. Floyd
Quiz
6. The input impedance of a noninverting amplifier is
a. nearly 0 ohms
b. approximately equal to Ri
c. approximately equal to Rf
d. extremely large

Thomas L. Floyd
Quiz
7. The noninverting amplifier has a gain of 11. Assume that
Vin = 1.0 V. The approximate value of Vf is
a. 0 V
b. 100 mV
c. 1.0 V
d. 11 V
Rf
10 kΩ
V
V
in
f
+
–
Vout
Ri
1.0 kΩ

Thomas L. Floyd
Quiz
8. The inverting amplifier has a gain of −10. Assume that
Vin = 1.0 V. The approximate value of the voltage at the
inverting terminal of the op-amp is
a. 0 V
b. 100 mV
c. 1.0 V
d. 10 V
–
+
Rf
Vout
Ri
Vin
1.0 kΩ
10 kΩ

Thomas L. Floyd
Quiz
9. To compensate for bias current, the value of Rc should
be equal to
a. Ri
b. Rf
c. Ri||Rf
d. Ri + Rf
–
+
Rf
Vout
Ri
Vin
Rc

Thomas L. Floyd
Quiz
10. Given a noninverting amplifier with a gain of 10 and a
gain-bandwidth product of 1.0 MHz, the expected high
critical frequency is
a. 100 Hz
b. 1.0 kHz
c. 10 kHz
d. 100 kHz

Thomas L. Floyd
Quiz
Answers:
1. c
2. d
3. d
4. b
5. a
6. d
7. c
8. a
9. c
10. d

Slide_the-operational-amplifier.pdf

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