PHY305F
Electronics Laboratory I
Section 5
The Analogue Package: Op-Amps
and Their Applications
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Introduction
• Operational amplifiers (op-amps) are general purpose amplifier
circuits designed to perform mathematical operations.
• They are commonly used in modern analog electronics.
• They are built from several transistor stages and are widely
available as integrated circuits, usually as 8-pin dual in-line
packages (DIP).
• Op-amps are linear amplifiers for which Vout ∝ Vin. Typically, the
DC open-loop voltage gain is 103 to 106. Because this gain is so
high, feedback is usually needed to get a particular transfer
function and to control the stability.
• Low-performance IC op-amps operate from DC to about 20 kHz.
• High-performance IC op-amps operate up to about 50 kHz.
• The 741 op-amp is particularly popular - used in Experiment 5.
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Ideal Amplifiers
• Amplification of low-level electrical signals is an important function
of electronic instrumentation. e.g., conversion of low-voltage
signal from a CD player to a high voltage than can drive speakers.
• For a linear amplifier, the output signal is directly proportional to
the amplitude of the input signal. This is true of all op-amps.
• The simplest model for an amplifier has a signal, vS(t), amplified
by a constant factor A which is called the gain of the amplifier.
The ideal load voltage is then:
v L ( t ) = Av S ( t )
• It is common to model such circuits using Thevenin’s Theorem:
→ the Thevenin equivalent source circuit (with vS and RS) is seen at
the input port of the amplifier
→ the load, RL, is the Thevenin equivalent resistance seen at the
output port of the amplifier
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
A Voltage Amplifier and Its Model
• The amplifier acts as an
RS
_+ v (t) Gain,
A RL
S
+
vL (t)
–
equivalent load wrt the
source and as an equivalent
source wrt the load.
• The model below replaces
the gain “black box” with an
equivalent circuit.
Source Amplifier Load
RS
+
v
_
+ S v in
–
(from Rizzoni Figures 12.2 and 12.3)
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Rout
Avin
Rin
_
+
RL
+
vL
–
Equivalent Circuit of Ideal Amplifier - 1
• The characteristics of the equivalent circuit used to represent the
amplifier can be derived as follows.
• The input circuit has equivalent resistance
Rin, so the input voltage is:
v in =
Rin
vS
RS + Rin
• The equivalent input voltage seen by the amplifier is then
amplified by gain A, represented by controlled voltage source Avin.
• The controlled source is in series with resistance
vL =
Rout, (the internal or output resistance of the
amplifier), so the voltage presented to the load is:
RL
Av in
Rout + RL
• Substituting to get the load voltage as an amplification of the
source voltage:


Rin
RL
 v S
v L =  A
 RS + Rin R out + RL 
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Equivalent Circuit of Ideal Amplifier - 2
• The amplification now depends on both the source and load
resistances, and on the input and output resistances of the
amplifier. So a given amplifier would behave differently with
different loads or sources. How can this be changed?
• Note, if Rin is very large, then: v in ≈ v S .
• Similarly, if Rout is very small, then: v L ≈ Av in .
• Thus, if Rin → ∞ and Rout → 0, then the ideal amplifier will magnify
the source voltage by the gain A:
v L = Av S
which is what we want for our “black box” amplifier.
• So practical voltage amplifiers will ideally have a very large input
impedance and a very small output impedance.
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Operational Amplifiers
• Operational amplifiers are integrated circuits, which means that
they consist of a large collection of electrical and electronic
circuits integrated on a silicon wafer.
• They can perform various operations (e.g., addition, filtering,
integration) using the properties of ideal amplifiers and of ideal
circuit elements.
• IC op-amps are now widely used in place of electronic circuits
involving many individual components.
• We will focus on the use of simple circuit models for op-amps,
avoiding detailed consideration of the internal workings.
• Advanced IC technology lets us treat op-amps as ideal devices
for most applications.
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Open-Loop Mode
• An ideal op-amp behaves as a difference amplifier, which is a
device that amplifies the difference between two input voltages.
So the op-amp output is: v out = A V ( OL ) v + − v −
→ Input
(
)
v+
is the noninverting input (or terminal) and input v- is the
inverting input (or terminal) for which the output signal will be 180°
out of phase with the input applied to this terminal. Both are
needed for amplification of + and - signals without special biasing.
→ Amplification factor AV(OL) is the open-loop voltage gain (without
feedback), which is typically 105 to 107.
• The large AV(OL), combined with the near-infinite input resistance
and near-zero output resistance, allows op-amp circuits to be
designed to perform as near-ideal voltage or current amplifiers.
• To analyze the performance of an op-amp circuit, only one
assumption is needed: the current into the amplifier input, iin = 0.
This is justified by the large input resistance and large AV(OL).
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Op-Amp Model Symbols, and Circuit
i in
v+
Positive power supply
+
+
v in
–
v–
R
A
Inverting input
R out
in
_
+
V(OL)
+
v
out
–
v in
–
V S+
Output
_
Noninverting
VS
input Negative power
supply
Op-amp model
Simplified circuit symbol
V+
S
Inverting
input in -
Output
Offset null
Inverting input
Noninverting input
_
V
_
S
VS
Noninverting
input in +
IC op-amp diagram
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
1
2
3
4
8
7
6
5
No connection
V+
S
Output
Offset null
Integrated circuit operational
amplifier (IC op-amp)
(from Rizzoni Figure 12.4)
Inverting Amplifiers - 1
• Inverting amplifiers are simple op-amps for which the input signal
to be amplified is connected to the inverting terminal and the
noninverting terminal is grounded.
• By choosing the ratio of two resistors, the gain can be arbitrarily
chosen. We will now show this.
RF
RS
+
_
iS
vS
(from Rizzoni Figure 12.5)
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
iF
–
v
iin
v+
–
+
+
v out
–
Inverting Amplifiers - 2
• Apply KCL to the inverting input node: iS + iF = iin
where iF is the feedback current which flows back to the inverting
terminal from the output.
• Apply Ohm’s Law to find the currents:
v out − v −
vS − v−
iin = 0 (our assumption )
iF =
iS =
RF
RS
• Voltage at the noninverting input is connected to ground: v + = 0
• The open-loop model for the op-amp requires that:
v
v out = A V ( OL ) (v + − v − ) = A V ( OL ) v − so v − = out
A V ( OL )
• We can now find the closed-loop gain, vout/vS, called this because
the presence of a feedback connection between the output and
the input makes a closed loop.
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Inverting Amplifiers - 3
• Since iS = −iF we have:
vS
v out
v
v out
+
= − out −
RS A V ( OL )RS
RF A V ( OL )RF
 1
1
1 

+
+
• Rearranging: v S = − v out 

R
R
A
R
R
A
F
S
V
(
OL
)
F
S
V
(
OL
)


• Thus, if the open-loop gain of the amplifier is large enough, the
second two terms on the RHS are negligible wrt 1/(RF/RS).
• With AV(OL) ~ 105 to 107, the closed-loop gain of an inverting
amplifier is:
v out = −
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
RF
vS
RS
Inverting Amplifiers - 4
The analysis can also be further simplified, as follows.
−
• The inverting terminal voltage is: v =
v out
A V ( OL )
• So, as AV(OL) → ∞, v - will become very small (~µV). This leads to
the assumption that for an inverting amplifier,
• One consequence of this assumption is: v − ≈ 0.
The effect of the feedback connection from output to inverting
input is to force the voltage at the inverting input to be equal to
the voltage at the noninverting input.
• Equivalent to saying: v − ≈ v + for an op-amp with negative
feedback.
Analysis of ideal op-amps with negative feedback can now be
simplified with two assumptions: (1) iin = 0, and (2) v - = v+ .
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Op-Amp Feedback
• Why design an op-amp with very large open-loop gain and negative
feedback, rather than with a smaller gain and no feedback?
→ The large open-loop gain provides flexibility in the design of amplifiers
with almost arbitrary gain (using RS and RF).
→ Negative feedback is what allows the use of this flexibility.
• Consider an open-loop amplifier.
→ The large AV(OL) will amplify any input voltage difference.
→ When a negative feedback is added, vout causes current iF to flow
through the feedback resistor to satisfy KCL at the inverting node.
∆v = v + − v − = 0 then vout will continue to ensure that KCL is
satisfied at the inverting node, i.e., iF = iS .
If there is now a small imbalance in ∆v at the op-amp input, then vout
will increase by AV(OL) ∆v.
So a small current (AV(OL) ∆v/RF) will flow from output to input through
the feedback resistor, reducing ∆v to 0, restoring the original balance.
∴Negative feedback is a self-balancing mechanism that allows the
amplifier to preserve 0 potential difference between its input terminals.
→ If
→
→
→
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Op-Amp Summing Amplifiers - 1
• The op-amp summer or summing amplifier is based on the
inverting amplifier and is used to add signal sources.
• Advantage: the summation is independent of the load and source
impedances, so sources with different internal impedances do not
interact with each other.
R F
R S
1
vS
1
+
_
iF
i1
v out
R S2
vS
2
+
_
i2
R S
N
(from Rizzoni Figure 12.7)
vS
N
+
_
iN
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Op-Amp Summing Amplifiers - 2
• Apply KCL at the inverting node: the sum of the N source currents
and the feedback current must equal zero, so:
i1 + i2 + i3 + ... + iN = −iF
v
• Each source current is: in = Sn n = 1, 2, ..., N
R Sn
• The feedback current is: i = v out
F
RF
• Combining these:
N
v Sn
∑R
n=1
=−
Sn
v out
RF
N
RF
v Sn
n=1 R Sn
or v out = −∑
• So the output is the weighted sum of the N input signal sources,
with the weighting factor for each source equal to the ratio of
feedback resistance to the source resistance.
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Noninverting Amplifiers - 1
• Noninverting amplifiers (or voltage followers) avoid the negative
gain (i.e., phase inversion) of inverting amplifiers.
• They are op-amps for which the input signal is connected to the
noninverting terminal and the inverting terminal is grounded.
• We can calculate the gain as we did for the inverting amplifier.
RF
i
RS
i
R
+
_
S
i
in
v–
v+
i in
v
S
F
–
+
+
v
out
–
(from Rizzoni Figure 12.8)
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Noninverting Amplifiers - 2
• Apply KCL at the inverting node: iF = iS + iin ≈ iS
v out − v −
v−
• With:
iF =
and iS =
RF
RS
• Because iin = 0, the voltage drop across the source resistance R
+
is also 0, so: v = v S .
−
+
• Using this and v − ≈ v + gives: v = v = v S
v out − v S v S
=
• We can show that: iF = iS and
RF
RS
• The closed loop gain of a noninverting amplifier is thus:
v out
R
= 1+ F
vS
RS
• Note that this gain is always positive and ≥ 1.
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Differential Amplifiers - 1
• Differential amplifiers are combinations of inverting and
noninverting amplifiers that are often used when the difference
between two signals needs to be amplified.
• They have two input sources, v1 and v2, which may or may not be
independent.
R2
i2
R1
–
+
i1
+
_
+
v1
v2
v out
R1
+
_
–
R2
(from Rizzoni Figure 12.10)
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Differential Amplifiers - 2
One approach to analyzing differential amplifiers:
(1) Compute noninverting- and inverting-terminal voltages, v+ and v -.
(2) Equate the noninverting and inverting voltages:
v - = v+ .
(3) Apply KCL at the inverting node, where i2 = -i1.
R
•
2
Using our assumption iin = 0, the noninverting- v + =
v2
R
+
R2
1
terminal voltage is (using voltage divider):
•
Given v - = v+ , we have currents:
•
Given i2 = -i1, we can solve for vout:
i1 =
v1 − v +
R1
i2 =
v out − v +
R2
 1

1
R2
v out = R 2 − v 1 +
v2 +
v2 
R1 + R 2
R1 (R1 + R 2 ) 
 R1
•
The differential amplifier closed-loop gain is ∴ v = R 2 (v − v )
2
1
out
R1
(Note: input voltage difference is amplified.)
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Differential Amplifiers - An Example
• Electrocardiograms (EKGs) measure the difference v1 - v2
between the potentials at two electrodes on a patient’s chest.
• Electrical equipment powered by 60-Hz, 100-VAC line current
causes interference in the electrode leads which act as antennas.
• So the electrodes measure these inputs to the EKG amplifier
(interference signal is approx. same for identical electrodes):
Lead 1 : v1( t ) + v n ( t ) = v1( t ) + Vn cos(377 t + φn )
Lead 2 : v 2 ( t ) + v n ( t ) = v 2 ( t ) + Vn cos(377 t + φn )
• Now, the output voltage from the differential amplifier is:
R
R
v out = 2 [(v1 + v n ( t )) − (v 2 + v n ( t ))] = 2 [v1 − v 2 ]
R1
R1
• So the differential amplifier eliminates the noise term (commonmode signal) and amplifies the desired EKG waveform
(differential-mode signal). This is a highly desirable feature!
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
Instrumentation Amplifiers (IA)
• Instrumentation amplifiers (IAs) are often used to provide
impedance isolation between bridge transducers and the
differential amplifier stage, by amplifying v1 and v2 separately.
• The circuit shown is often packaged as a single IC, given the
widespread application of the IA (e.g., AD625). This allows
precise matching of R1 and R2.
• It can be shown (try it!),
v out
R  2R 2 
AV =
= F 1 +

v1 − v 2 R 
R1  v 2
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
RF
R
v
1
that the closed-loop
voltage gain of the
instrumentation amplifier is:
R2
R1
Vout
R2
R
RF
(from Rizzoni Figure 12.14)
Practical Op-Amp Design Considerations
Op-amps allow sophisticated circuits to be designed in few simple
steps using selected resistors, IF the circuit components are chosen
based on several criteria.
(1) Use standard resistor values, e.g., with 5% tolerance rating.
(2) Ensure that the load current is reasonable - do not select very
small resistor values, e.g., typically try to use resistors > 100 Ω.
(3) Avoid stray capacitance - do not select very large resistor values
that can couple unwanted signals into the circuit through “capacitive
coupling”, e.g., typically try to use resistors < 1 MΩ.
(4) Use precision designs where required. Certain applications may
require amplifier gain to be a very accurate value, e.g., requiring
precision resistors having 1% tolerance.
PHY305F - Electronics Laboratory, Fall Term (K. Strong)
The 741 Op-Amp
• The LM741 is a general-purpose, low-cost op-amp.
• It is available as a metal-can package, dual-in-line or
SO package (DIP), and two ceramic dual-in-line
packages.
• It offers high gain, a wide range of operating voltages,
good temperature stability, overload protection on the
input and output.
• It is used in many analogue applications,
V+
V-
including integrators, summing amplifiers,
and general feedback applications.
• The dual-in-line package is the most common
and has eight pins: two for the voltage
supplies (v+, v -), two inputs (inverting and
noninverting), one output, two offset null
connections, and a no-connection pin (NC).
PHY305F - Electronics Laboratory, Fall Term (K. Strong)