Circuits, Devices, Networks, and Microelectronics
CHAPTER 16.
THE NON-IDEAL OPAMP
16.1 EFFECT OF FINITE GAIN OF THE OPAMP
In Chapter 6 on the ideal operational amplifier it is characterized as an artifact with infinite gain, infinite
input resistance and zero output resistance. These unrealistic but approachable properties give it the
highly desirable ‘nullator-norator’ behavior when used in the feedback mode. As such the opamp can
effectively disappear and the transfer function then magically becomes an active realization of the
feedback network. But the opamp certainly does not completely disappear. It is a construct, a topology
of nodes and branches consisting of transistors, resistances, and capacitances. These internal components
impose circuit performance qualifications on its use and application. And in order to effectively employ
it as a circuit component these qualifications need to be sorted out, identified, and passed along to the
circuit simulation macro that is used to represent and distinguish one opamp from another.
After transistors and transistor amplifiers have been introduced and analyzed, most of the constraints are
obvious. Transfer characteristics Rin, AV, and Rout are finite, not infinite or zero. The frequency response
is finite. And since diffamps are involved, the common-mode rejection concern is finite. And finally,
there is a constraint associated with large Vout swing, i.e. dVout/dt ≡ slew-rate = finite, typically on the
order of 1.0V/s. There are even a few others associated with offsets and power supply constraints, but
these constraints are the big four.
Reconsider the very first instance of the opamp feedback topology identified by chapter 6 in which the
feedback network is a R1, R2 voltage divider. This topology is indicated by figure 16.1-1 (which is the
same as figure 6.2-1)
Figure 16.1-1. The opamp with feedback via a voltage divider.
The fraction of vout fed back as vF to the inverting input (v-) is
v F    vout
(16.1-1a)
for which the feedback factor is
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
R1
R1  R2
(16.1-1b)
 AV  (v S  vout )
And since vout  AV  (v   v  )  AV  (v S  v F )
Collecting terms in vout and vS we have the result
vout
1
vS

1
  
AV




(16.1-2a)
Assuming AV → large (i.e. nearly ideal) this equation will simplify to
vout
1

vS

= AIDEAL
= ANI
(16.1-2b)
for which with equation (16.1-1b) the transfer gain (for the non-inverting topology) is then
v out R1  R2
R

 1 2
vS
R1
R1
= ANI
(16.1-3)
and is the same as that of the ideal. Otherwise, for Av = finite, equation (16.1-2a) should be rewritten as
vout 1

vS


1
1 
 AV



(16.1-4a)
Using equation (16.1-2b) this can be written as
vout
 ANI
vS

1
1 
 AV




1 

 AIDEAL 1 
 AV 
(16.1-5)
It so happens that equation (16.1-5) will have the same form for its sibling figure 16.1-1(b). And this
premise can be confirmed by nodal analysis at the feedback node vF (= inverting input v- ), i.e.
v (G1  G2 )  vout G2  v S G1  0
(16.1-6a)
and since v   v   vout / AV and v+ = 0,
then v  vout / AV and

vout
(G1  G2 )  vout G2  vS G1  0
AV
Equation (16.1-6b) can be rewritten as
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(16.1-6b)
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(G1  G2 ) vout
G
 vout  v S 1
G2
AV
G2
(16.1-6c)
The first conductance ratio in (16.1-6c) is the same as the reciprocal of equation (16.1-1b), i.e.
G2  G1 R1  R2

G2
R1

1

Even though this may be a ‘so what’, it puts equation (16.1-6c) in the form
 1

G
vout  
 1  v S 1
G2
 AV

(16.1-6d)
If we should let AV → large (i.e. like that for the ideal opamp) equation (16.1-6d) becomes
vout
G
R
 1  2
vS
G2
R1
(16.1-7)
And this result is (as expected) the ideal transfer gain for the inverting topology. Equation (16.2-6d) then
would be of the form
vout
R
 2
vS
R1

1
1 
 AV




1 

 AIDEAL 1 
 AV 
(16.1-8)
Notice that this is the same form as equation (16.1-5). How about that?
So we have a (readjusted) summary for the R1, R2 opamp topologies :
Non-inverting topology (input at v+):
Inverting topology (input to v- thru R1):
AIDEAL  1 
R2
R1
AIDEAL  
R2
R1
vout
 AIDEAL
vS
If gain AV is finite (and large) then:
Where:
= ANI

1 
1 

 AV 
  1 ANI
Table 16.1-1. Summary of the R1,R2 (voltage divider) opamp topologies. The last equation (the ‘Where’
one) is the means for ascertaining feedback factor .
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It should be noted that all opamp topologies need to include analysis of the non-inverting topology for
determination of the value of feedback factor .
Summary and recipe for any and all opamp topologies
1. Assume that the inputs are ‘virtually’ connected (nullator context).
2. Execute a nodal analysis at feedback node vF (= v-)
3. Continue with a nodal analysis through the feedback network until vout is included.
4. Execute a ‘discovery’ analysis of the equivalent non-inverting topology (ANI) to find feedback
factor
5. Qualify the result of part 3 with reality by means of
vout
 AIDEAL
vS

1 
1 
 
 AV 
Table 16.1-2. Revised standard analysis (recipe) for transfer gain of (any and all) opamp topologies.
16.2 FINITE FREQUENCY RESPONSE OF THE OPAMP
The frequency response of the real opamp has the same character as any other real circuits due to the time
constants inherent to all circuits and devices. The principal distinction is that opamp circuits also must
answer to feedback, and so must be specifically compensated so that the frequency response will be of the
form of a single pole (single time constant) low-pass response, as illustrated by figure 16.2-1. STC
(single-times constant) response is necessary and essential for feedback stability.
Figure 16.2-1. Frequency response of the LM324 opamp (SPICE simulation).
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Figure 16.2-1 is a SPICE rendition of Bode magnitude plot for the LM324 opamp. The cursors mark the
specific characteristics of this response. For this opamp |AV(f= low)| = 51.9dB. At high frequencies f
rolls off to 0.895 MHz at |AV(f)| = 0dB. These benchmarks are called the zero frequency gain AV0 and the
unity-gain frequency. The unity-gain frequency fT is also called the gain-bandwidth product (GB).
Note that the frequency response for the opamp is not identified by the corner (= f1 ). For the LM324 the
corner (by inspection) is approximately = 2.3kHz. If you do the math you will find that this value
corresponds almost exactly to f1 = fT/|AV0|.
Otherwise the opamp will have frequency response (for f < fT ) of the form.
AV 
AV 0
1  s / 1
(16.2-1)
where 1 = 2f1 and AV0 is the zero frequency (voltage) gain.
Using the same mathematics as before, i.e. equation (16.2-2a) then
vout
1
vS

1 
  

AV 

which is better served if expressed as
AV
1  AV 
AVF 
(16.2-2)
If equation (16.2-1) is applied to equation (16.2-2) a little more informative form results
AVF 
AV 0 (1  s / 1 )
1   AV 0 (1  s / 1 )

AV 0
(1  s / 1 )  AV 0 
which can be written as
AVF 
AV 0
1  AV 0  s / 1 

AV 0
1

1  AV 0 1  s / 1 1  AV 0 
or in simplified form,
AVF 
AVF 0
1  s /  1F
(16.2-3)
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In this equation AVF0 is the zero frequency gain with feedback. And is
AVF 0 
AV 0
1  AV 0 
(16.2-4a)
Now 1F is the frequency corner (bandwidth) for the feedback topology and is given by
1F  1 (1  AV 0 )
(16.2-4b)
Note that the product of (16.2-4a) and (16.2-4b) will be
AVF 01F  AV 01  GB   T
(16.2-5)
This is a nice clean result. It is not unexpected since unity gain frequency and all frequencies associated
with equation (16.2-4b) lie on the linear slope of the roll-off.
Summary:
Frequency response of the non-ideal opamp is specified by AV0 = zero frequency gain and by fT = unitygain frequency (a.k.a. gain-bandwidth product).
The bandwidth of a circuit with feedback is
or better yet,
f1F = fT /AVF0 (by equation 16.2-5)
f1F =  fT since AVF0 is approximately = 1/
(by equation 16.2-4b)
Table 16.2-1. Frequency response of opamp circuits.
Frequency response of opamp circuits is defined by feedback , and that factoid is contained by feedback
factor. Therefore be aware that the corner for a inverting topology is defined by its non-inverting
sibling, and is always f1F =  fT .
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EXAMPLE 16.2-1: A T-network constructed using an opamp with AV0 = 50dB and fT = 1.0MHz .
Determine the gain and bandwidth of this circuit.
Transfer gain is found by nodal analysis, beginning with the feedback node v- .
node v- : v S (0.01  .005)  v1 (.005)  0 where v   v S (virtual connection of inputs)
 v1  3v S
node v1 : v1 (0.005  .005  .005)  v S (.005)  v 2 (.005)  0
 3  (v1  3v S )  v S  v2  0  v 2  8v S
node v2 : v 2 (0.005  .005  .005)  v1 (.005)  v out (.005)  0
 3  (v 2  8v S )  (v1  3v S )  vout  0
(for which  = 1/21). Thus the (non-inverting) gain AVF0
 vout  21v S
(same as ANI = 21)
= 21
with frequency corner (bandwidth) at f1F =fT = 1.0MHz/21 = 47.6 kHz
Note that if the configuration in this example been of the inverting option it is still necessary to determine
ANI to determine the bandwidth. Emphasis is repeated, that equation (16.2-5) relates only to the noninverting gain.
16.3 FINITE INPUT AND OUTPUT RESISTANCES OF THE OPAMP
Another qualification of the opamp is that the input resistance Rin is finite. If not large, Rin will load down
the source vS and compromise the ‘buffer’ nature of the opamp input. Fortunately feedback enhances
input resistance and so even a modest Rin will become large. This aspect is easy enough to ascertain and
may be determined by an expanded view of the opamp, as represented by figure 16.3-1
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Circuits, Devices, Networks, and Microelectronics
Figure 16.3-1. Inside look at the opamp, with Rin enhanced by feedback.
vS
 RiF  (v I  v F ) / iin  (v I  AV v I ) / iin
iin
From the figure
∴
RiF 
vI
 1  AV   Rin  1  AV 
iin
(16.3-1)
Needless to say the enhancement factor (1+ AV) can be huge since AV is expected to be large and  is not
likely to be small.
However the enhancement of input resistance must be qualified because AV is also frequency dependent,
as defined by equation (16.2-1). So the input resistance becomes an element ZiF with frequency character
due to the feedback loop, as represented by figure 16.3-2.
Figure 16.3-2. Effect of feedback on input resistance.
The nature of this dependence can be resolved by applying equation (16.2-1) to (16.3-1), for which
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
AV 0 
Z iF  Rin  1  AV   Rin  1  
  Ri  Z eq
1  s / 1 

where Zeq is the frequency dependent part and is of form
(16.3-2a)
Z eq  Ri AV 0 (1  s / 1 ) .
Its characteristics are more separable if expressed as an admittance:
Yeq 
1  s / 1
1

 Geq  sCeq
Z eq
Ri AV 0
for which
and
C eq 
Req 
(16.3-2b)
1
 AV 0 Ri
Geq
1
AV 01 Ri

1
T Ri
(16.3-3a)

1
(16.3-3b)
1 F R i
where we have made use of the (gain-bandwidth) definition T = AV01
Typical values of Req and Ceq are represented by example 16.3-1
EXAMPLE 16.3-1: Evaluate the equivalent input impedance terms for the equal-resistance T-network
driven by an opamp for which AV0 = 52dB, Ri = 100k, and fT = 2MHz.
SOLUTION: The ideal transfer gain is of type non-inverting. By nodal analysis ANI = 5.0V/V.
Therefore  = 1/ANI = 0.2.
And from equation (16.3-3a),
Req  AV 0 Ri  0.2  400  100k
and from equation (16.3-3b),
Ceq 

1
 T Ri

1
0.2  (2  2MHz)  100 k
1
1
1

 0.16 
2 0.2  2MHz  100 k
.04  1012
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= 8.0M
= 4.0pF
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The signal source vS and whatever circuit is associated with it should operate as if the input to the opamp
is of the equivalent circuit represented by figure 16.3.4.
Figure 16.3-4. Equivalent input circuit.
The RC time constant of this network is approximately = RinCeq. So for example 16.3-1, = 400ns, or a
frequency corner at f = (1/2) x (1/) = 400 kHz.
Notice that this result is the same as f1F = fT.
In like manner the output resistance is affected by the feedback loop. If we look into the output of the
opamp, we would see an equivalent circuit of the form represented by figure 16.3-5.
Figure 16.3-5. Assessment of output resistance for the opamp as consequence of the feedback loop.
Output resistance is identified as Rout = vout/io . Therefore
Gout  i0 / vout 
(vout  Av v I )
vout
Ro
(16.3-4)
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In the assessment of output resistance it is necessary that there is no signal at the input, i.e. vS = 0. And
since there is a feedback voltage vF =  vout at the (v-) input as indicated by figure 16.3-5, then vI = - vout
and equation (16.3-4) will give
Gout  Go [vout  Av vout )] vout
 Go (1  AV )
(16.3-5)
i.e. the output conductance is enhanced by feedback factor (1+AV)..
Once again we acknowledge that opamp gain AV is frequency dependent, as defined by equation (16.2-1).
Consequently Gout will then be an admittance = Yout of the form
Yout  Go  Go Av  Go  Yeq
(16.3-6)
where Yeq is a consequence of the frequency dependence of AV. If we look at its inverse Zeq = 1/Yeq and
make use of equation (16.2-1) then
Z eq 
1  s / 1
1
s
 Req  sLeq


Go AV 0 Go AV 0 Go AV 01
(16.3-7)
The result is equivalent to a couple of pseudo-components Req and Leq, of the form
Req 
R
1
 o
Go AV 0 AV 0
(16.3-8a)
Leq 
1
1

Go AV 01 Go T
(16.3-8b)
in which use was made of AV01 = T.
The equivalent output will then be of the form shown by figure 16.3-6.
Figure 16.3-6. Equivalent output circuit according to equations (16.3-6) and (16.3-7)
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EXAMPLE 16.3-2: The circuit construct indicated uses an opamp with zero frequency gain AV0 = 52
dB, gain-bandwidth product GB = 2MHz, and output resistance Ro = 100. Determine output equivalent
terms Req and Leq.
Solution: The ideal transfer gain is of an inverting topology. For the associated non-inverting topology
we have, by inspection ANI = 8.0. Therefore  = 1/ANI = 1/8.
∴ Req 
Ro
100

AV 0 1 / 8  400
Leq 
= 2.0
1 Ro
100
 0.16 
2 f T
1 / 8  2MHz
= 64H
If we were to assess the backend of the opamp in terms of time constants, then = Leq/Ro which = 0.64s
for the values of example #16.3-2 and corresponds to frequency corner f = (1/2) x (1/) = 250 kHz.
Notice that this value is the same as f1F = fT.
Summary table: Effect of the feedback loop on input and output impedance.
Input equivalent impedances:
Req  AV 0 Rin
Output equivalent admittances:
Req 
Ro
AV 0
Table 16.3-1. Input/output impedance/admittance equivalences.
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C eq 
1
1
2 f T Rin
Leq 
1 Ro
2 f T
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16.4 INCOMPLETE REJECTION OF COMMON SIGNALS (CMRR)
One of the purposes in life of the opamp is to serve as a driver for differential amplifier topologies such as
the IA (instrumentation amplifier) and other differential amplifier constructs. A diffamp is supposed to
amplify the difference between inputs and reject signals that are common to both.
Real opamp circuits are not completely symmetrical and consequently a small common-mode signal will
be passed. We identify the ratio of differential gain to common-mode gain as a property of the opamp
called the common-mode rejection ratio (CMRR). It typically is on the order of 100dB and its measure
rolls off at high frequencies. The CMRR has nothing to do with the social life of electrical engineers.
CMRR 
Adiff
(16.4-1)
ACM
The common-mode signal shows up as a distortion of the output. If the common-mode signal is a carrier
signal, it will not be completely rejected and show up as a Fourier component at the output.
Since the common-mode signal is spurious then it will be enhanced or suppressed by the feedback
network in a different manner than other signals, since
vout (common)  vocm  ACM
(v  v )
2
(16.4-2)
where ACM is small.
If the opamp is effective in suppressing common-mode signals, why worry? For some instances we have
to do so, since the common mode signals can be enhanced by the feedback network. For example
consider the basic differential amplifier construct of figure 16.6-1. It will reject signals just fine as long
as R4/R3 =  (= R2/R1). But what if R4/R3 = where  is a small error due to component variation?
Then its CMRR is approximately
CMRRcircuit
 
(16.4.3)
And is usually much small than that for the opamp. So component accuracy is essential for differential
amplifier topologies, as is symmetry. For the instrumentation amplifier (IA), represented by (pSPICE)
figures 16.4-1, the CMRR for this topology is almost invariably excellent.
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Figure 16.4-1a. pSPICE schematic of IA (Instrumentation amplifier) constructed with the general
purpose LM324 opamp. Input signal = 10kHz carrier signal with one input carrying a 2kHz signal.
Figure 16.4-1b. The differential output appears as a 2kHz signal with the 10kHz carrier almost
completely rejected. Fourier analysis confirms that the CMRR for this circuit is approximately 90dB.
On the other hand, a circuit with less symmetry, as represented by figure 16.4-2, will not necessarily be
successful in suppression of the common signal that the non-ideal opamp does not succeed in rejecting.
Figure 16.4-2. pSPICE schematic of alternative differential amplifier topology using the general purpose
LM324 opamp. Input signal = 10kHz carrier signal with one input carrying a 2kHz signal, identical
inputs to that of figure 16.2-2 (for the IA). In this case the carrier signal is not well rejected. A CMRR
assessment using Fourier analysis gives a circuit CMRR of only 39dB.
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16.5 SLEW-RATE LIMITS ON THE OUTPUT
The frequency response limitation of the opamp as identified by section 16.2 is not unexpected, inasmuch
as all circuits have internal time constants that define their ability to respond to frequencies. It is less
apparent that there are other time constraints that are not related to the familiar RC time constants.
The non-ideal opamp has one such. It relates to the fact that non-ideal opamps are compensated with an
internal (compensation) capacitance = CC that pulls the dominant pole down to a level under which the
frequency response will be dominated by a single-pole response. This internal capacitance must be
charged and discharged not only by the normal time constants, but also by the differential pair of
transistors when they are fully tilted by a large voltage swing. This constraint manifests itself in terms of
an inability of the output to make large swings quickly, i.e. it cannot ‘slew’ the output voltage quickly and
dVout
= constant ≡ SR
dt

I
CC
(16.5-1)
Current I is the internal current source that drives the coupled (differential) pair. The slew-rate (SR) is
typically on the order of 1.0V/us for a general purpose opamp. High-performance opamps such as those
deployed within integrated circuit designs can be made with slew rates on the order of 50V/us.
The slew rate also places a distortion limit on the frequency, inasmuch as the output cannot swing large
voltages at higher frequencies, as represented by the fact that
Vout (t )  Va sin( t )
and thus
dVout
 Va cos(t ) which has a maximum slope at t = 0
dt
Since the maximum amplitude is defined by the power supply rails, then the slew rate defines a maximum
frequency at maximum output swing which we enticingly define by
dVout
dt
 SR  Va (max)  (2f m )
MAX
where fm = maximum power bandwidth, given by
fm 
1 SR

2 VS
(16.5-2)
where VS = Va(max) is (approximately) the supply rail voltage.
EXAMPLE 16.5-1: An opamp with SR = 1.0V/us is supplied by voltage rails of ± 10V.
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(a) What is the maximum power bandwidth and
(b) what is the highest frequency f1 for an undistorted V1 of 2.0V?
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Example 16.5-1 points out that a reduction of the output amplitude accommodates higher frequency
without distortion by the slew rate.
The effect of slew rate on the output signal is represented by figure 16.5-1, for which an LM324 generalpurpose opamp attempts to swing a large signal. The SR for the LM324 is about 0.5V/us.
Figure 16.5-1. Simulation of LM324 which is intending to swing an output signal from –5 to +5 but
afflicted by into the dV/dt constraint of SR = 0.5V/us. Consequently the output ends up looking like a
triangular wave.
16.6 OPAMP MACROMODELS
While it might be presupposed that the opamp parts available with the (SPICE) simulator are little more
than a multi-transistor construct packed into a triangular box, that is neither fact nor practical. In fact if
the actual transistor circuit of the opamp were used, simulation of circuits that use opamps would find
their little wiry images overloaded by a huge number of invisible nodes and transistors hiding inside the
opamp icon(s).. The typical opamp circuit consists of 24-30 transistors, somewhat like the form1 shown
by figure 16.6-1. The hapless simulator would grind itself to death trying to iterate all of these nodes and
the transistor equations interconnected thereto.
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Figure 16.6-1. Schematic of the ICL8741 opamp
So instead we devise a much simpler ‘macromodel’ that is little more than a transistor coupled-pair on the
front end followed by a few ideal dependent sources with parameters as necessary to effect the transfer
characteristics. Diodes are included to characterize the voltage/current limits of the opamp. A typical
opamp macro is represented by figure 16.6-2 (which in this case is for the LM324 opamp)
Figure 16.6-2(a). Macromodel for the LM324 pSPICE opamp.
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Figure 16.6-2(b). pSPICE node list for LM324 macromodel.
For opamp macromodels the construct is similar but the components may change according to the
technology. The LF411 part, for example is a jFET opamp, as reflected by its macromodel .
Figure 16.6-3(a). Macromodel for the LF411 pSPICE opamp. Note that the input is a nJFET
source-coupled pair, and this distinguishes its performance characteristics.
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Figure 16.6-3(b). pSPICE node list for LF411 macromodel
Macromodels are a necessity of packaged circuits. The forms indicated by figures 16.6-2 and 16.6-3 are
ones that exist in the pspice simulation package. In order to relate to these circuits we should consider
their somewhat simpler cousins.
For example the ideal opamp macromodel is merely a VVT (voltage-voltage transducer) with large gain
factor, as indicated by figure 16.6-4. This is almost equivalent to a nullator-norator element.
Figure 16.6-4. Macromodel for ideal opamp. Note that there are no internal nodes and the part (the
VVT) is linear (as well as ideal). So this model will not add any overhead to the simulation. The use of
the opamp symbol as a circuit icon is then mostly a feel good option.
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If we should like to be a little less ideal and more like a real opamp, it is simple enough to add an Rin and
an Rout to the above model as represented by figure 16.6-5.
Figure 16.6-5. Enhanced macromodel for ideal opamp, with finite Rin = 1 M and finite Rout = 50.
Of course this macromodel is still deficient as an equivalent to a circuit such as 16.6-1, since it totally
lacks any frequency character. But that is also easy enough to add, as represented by figure 16.6-6
Figure 16.6-6. The macromodel of figure 16.6-6 does not include the more critical limitations
associated with an ideal opamp, namely the CMRR and the slew rate. Those can only be accommodated
if a differential pair is included. The use of a differential pair is a major upgrade, since by default, a
power supply must also be assumed. In the (more realistic) macromodels, represented by figures 16.6-2
and 16.6-3, the differential pair front end is a signature of the more mature construct. A simplified
version is represented by figure 16.6-7.
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Circuits, Devices, Networks, and Microelectronics
Figure 16.6-7. In this case we have two stages, with the input stage being defined by the
differential pair M1 and M2 and the next stage being defined by the two VCTs G1 and G2. Slew rate is
defined by the current source and C2. Capacitance C2 also defines the frequency corner. The two VCTs
define (1) the differential gain and (2) the common-mode gain, with relative magnitudes as represented by
Ga and Gcm.
The CCT (denoted by F2) plays a role in both the signal gain and the frequency corner. For the sake of
the slew rate, which is an amplitude swing, capacitance C2 must be small, on the order of pF (see equation
16.5-1). For the sake of the roll-off corner which must be on the order of 10 – 100Hz if it is to match the
compensated frequency profile for the opamp, the capacitance must be large. This is accomplished by a
Miller multiplication effect across the capacitance accomplished by means of the CCT. The multiplying
factor is approximately that of the CCT and gives result
f1 
1
1

2 RO  kC2
(16.6-1)
Other relationships that relate to the enhanced macros of figure 16.6-7 are
dVout
dt
= constant = SR 
I SS
C2
(16.6-2)
with transfer gain
vO
 g m1 RD  Ga  kRO
vI
(16.6-3)
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where gm1 is the transconductance of the (balanced) input FETs. The CMRR is self-evident as
CMRR  Ga Gcm
(16.6-4)
A simulation of figure 16.6-7 with the transistor modeled as (LEVEL=1, TOX=13.8n, UO=400,
GAMMA = 0.5, PHI = 0.75) and W/L = 20.0u/1.0u is shown by figure 16.6-8.
Figure 16.6-8. Simulation of figure 16.6-7 with the MOSFETs defined such that the conduction
coefficient is K = 1000A/V2. As indicated by the cursors the gain is 84.71dB and fT = 3.396MHz.
Comparison of figures 16.6-3 and 16.6-7 emphasizes that the introduction of real components (i.e. the
coupled pair) means that we have to accommodate two more nodes, namely those of the power supply.
And then we also need to model the constraints implied by a power supply. Figure 16.6-7 does not do so.
Constraints require an entirely different level of complexity and must be accomplished by a framework of
diode components and polyfunctions, as indicated by figure 16.6-2 and 16.6-3. The model is still flawed
in that there is considerable input offset Voff associated with the MOSFETS.
As a footnote to the macromodel constructs, the value give to transconductance gain element Ga is
always
Ga = 1/Rc
(= 1/Rd for FET)
(16.6-5)
This choice assures that equation (16.6-2) for the macromodel value of slew rate is valid and consistent
with the definition of slew rate.
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SUMMARY AND PORTFOLIO
  1/TNI
Feedback factor
where TNI = gain for non-inverting topology, ideal opamp

1 

T (non  ideal )  TIDEAL 1 

AV 

GB  f T  f 3dB 
Gain-bandwidth:
Slew rate:
where AV = finite gain
dVout
dt

MAX
I
C
1
3dB corner

 SR  Va  2f  VS  f m
f 3dB  f T
fm ≡ full-power bandwidth
Finite Rin and finite Rout:
Input equivalent impedances:
Req  AV 0 Ri
Output equivalent admittances:
Req 
Macromodels:
SR  iee c2
Ro
AV 0
CMRR  ga gcm
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C eq 
1
1
2 f T Ri
Leq 
1 Ro
2 f T 
Circuits, Devices, Networks, and Microelectronics
APPENDIX 15-1. Simulations of selected opamps
(1) Open-loop transfer characteristics
AV0 (741C) = 106 dB
fT = 913 kHz
AV0 (LM324) = 50 dB
fT = 955 kHz
AV0 (LF411) = 112 dB
fT = 7.0 MHz
(1) Common-mode rejection ratio
vO  AV v   v    ACM v   v   2
v  vCM
v  vO  vCM
 vO  AV  vO   ACM vO 2  vCM
or

vO 1  AV  ACM 2  ACM vCM 
vO 1  AV   AV  vO   ACM vO  vCM 
∴ vO  AV   ACM vCM 

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vO
A
1
 CM 
vCM
AV
CMRR