429

## 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 *

*out*

*F* *v*

*v* (16.1-1a)

430
2
1
1
*R*
*R*
*R*
(16.1-1b)

And since *v _{out}*

*A*(

_{V}*v*

_{}

*v*

_{})

*A*(

_{V}*v*

_{S}*v*)

_{F}*A*(

_{V}*v*

_{S}*v*) Collecting terms in

_{out}*vout and vS we have the result*

*V*

*S*

*out*

A

v

v

1

1

(16.1-2a)Assuming *AV → large (i.e. nearly ideal) this equation will simplify to *

1

*S*

*out*

v

v

=*AIDEAL*=

*ANI*(16.1-2b)

for which with equation (16.1-1b) the transfer gain (for the non-inverting topology) is then

1
2
1
2
1 _{1}
*R*
*R*
*R*
*R*
*R*
*v*
*v*
*S*
*out* _{} _{} _{} _{= }
*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

*V*
*S*
*out*
*A*
*v*
*v*
1
1
1 _{ } _{(16.1-4a) }

Using equation (16.1-2b) this can be written as

*V*
*NI*
*S*
*out*
*A*
*A*
*v*
*v*

1 1_{}

*V*

*IDEAL*

*S*

*out*

*A*

*A*

*v*

*v*

1 1 (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.

0 )

( 1 2 2 1

*G* *G* *v* *G* *v* *G*

*v* _{out}* _{S}* (16.1-6a)

and since *v* *v* *vout*/*AV* and *v+ = 0, *

then *v*_{} *v _{out}* /

*A*and (

_{V}*G*

_{1}

*G*

_{2})

*v*

*G*

_{2}

*v*

*G*

_{1}0

*A*

*v*

*S*

*out*

*V*

*out*

_{(16.1-6b) }

431
2
1
2
2
1 )
(
*G*
*G*
*v*
*v*
*A*
*v*
*G*
*G*
*G*
*S*
*out*
*V*
*out*
(16.1-6c)

The first conductance ratio in (16.1-6c) is the same as the reciprocal of equation (16.1-1b), i.e.

1
2
1
2
1
2
*R*
*R*
*R*
*G*
*G*
*G* _{}
1

Even though this may be a ‘so what’, it puts equation (16.1-6c) in the form

2
1
1
1
*G*
*G*
*v*
*A*
*v* _{S}*V*
*out*

(16.1-6d)If we should let *AV*→ large (i.e. like that for the ideal opamp) equation (16.1-6d) becomes

1
2
2
1
*R*
*R*
*G*
*G*
*v*
*v*
*S*
*out* _{}_{} _{}_{} _{(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

*V*

*S*

*out*

A

R

R

v

v

1

1

1 2 *V*

*IDEAL*

*A*

*A* 1 1 (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 :

**Table 16.1-1.** Summary of the *R1,R2 (voltage divider) opamp topologies. The last equation (the ‘Where’ *
one) is how the feedback factor is acquired.

Non-inverting topology (input at *v+): *

1
2
1
*R*
*R*
*A _{IDEAL}* =

*ANI*Inverting topology (input to

*v-*thru R1):

1 2

### R R AIDEAL

If gain *AV is finite (and large) then: * _{}
*V*
*IDEAL*
*S*
*out*
*A*
*A*
*v*
*v*
1
1
Where: 1 *A _{NI}*

432

Table 16.1-1 is a benchmark that implies a universal recipe for all opamp topologies. This recipe is reflected by table16.1-2.

**Table 16.1-2.** Revised recipe for finding *vout/vS for any and all opamp topologies. *

The feedback factor is needed for almost all of the non-ideal constraints on an opamp topology. So an analysis of the non-inverting topology is one of the essentials.

## 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).
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 reached. *
4. Execute an (additional) analysis of the equivalent non-inverting topology to find
feedback factor

5. Qualify the result of part 3 by means of _{}

*V*
*IDEAL*
*S*
*out*
*A*
*A*
*v*
*v*
1
1

433

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. *

1 0

/

1

s

A

A

*V*

*V*

(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

*V*
*S*
*out*
*A*
*v*
*v* 1
1

which is better served if expressed as

*V*

_{V}

*VF*

*A*

*A*

*A* 1 (16.2-2)

If equation (16.2-1) is applied to equation (16.2-2) a little more informative form results

1 (1 / )

) / 1 ( 1 0 1 0 *s*

*A*

*s*

*A*

*A*

*V*

*V*

*VF*

_{}

_{}

1 0

0 ) / 1 (

_{V}*V*

*A*

*s*

*A*

which can be written as

0 1

0 / 1 *A*

*s*

*A*

*A*

*V*

*V*

*VF*

_{}

_{}

0 1 0 0 1 / 1 1 1

_{V}

_{V}*V*

*A*

*s*

*A*

*A* or in simplified form,

*F*

*VF*

*VF*

*s*

*A*

*A*1 0 / 1 (16.2-3)

434

In this equation *AVF0 is the zero frequency gain with feedback. And is *

0

0 0 1

_{V}*V*

*VF*

*A*

*A*

*A* (16.2-4a)

Now *1F is the frequency corner (bandwidth) for the feedback topology and is given by *

)

1

(

_{0}1 1

*F*

A

*V*

(16.2-4b)Note that the product of (16.2-4a) and (16.2-4b) will be
*T*

*V*
*F*

*VF* *A*

*A* _{0}

_{1}

_{0}

_{1}

(16.2-5)This result is not unexpected since unity gain frequency and all frequencies associated with equation (16.2-4b) will fall on the linear slope of the roll-off.

### Summary:

**Table 16.2-1.** Frequency response of opamp topologies (with passive feedback networks).

The analysis shows that the frequency response of a passive opamp topology (no reactive components) is
defined entirely by the feedback factor . So it is the same for either the non-inverting or the inverting
topology and in simplest form is given as *f1F = ** fT . *

Frequency response of the non-ideal opamp is specified (1) zero frequency gain *AV0*
and (2) by *fT*= unity-gain frequency. *fT is *also called the gain-bandwidth product.
Bandwidth *f1F of a circuit with feedback = fT /AVF0 (by equation 16.2-5) and *

435

**EXAMPLE 16.2-1:** A double-T topology is constructed with an opamp that has *AV0 =* 50dB and *fT = *
1.0MHz . Determine the gain and bandwidth of this circuit.

**SOLUTION:** Transfer gain is found by nodal analysis, beginning with the feedback node *v-* .
node *v- *: *vS*(0.01.005)*v*1(.005)0 where

v

v

*S*(virtual connection of inputs)

*S*
*v*
*v*_{1} 3
node *v1 *: *v*1(0.005.005.005)*vS*(.005)*v*2(.005)0
0
)
3
(
3 1 2
*v* *v _{S}*

*v*

_{S}*v*

v

_{2}

8

v

*node*

_{S}*v2*:

*v*2(0.005.005.005)

*v*1(.005)

*vout*(.005)0 0 ) 3 ( ) 8 ( 3 2 1

*v* *v _{S}*

*v*

*v*

_{S}*v*

_{out}*v*21

_{out}*v*(same as

_{S}*ANI =*21) (for which = 1/21). Thus the (non-inverting) gain

*AVF0*=

**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
*non-inverting* 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

436

**Figure 16.3-1.** Inside look at the opamp, with *Rin enhanced by feedback. *

From the figure _{iF}_{I}_{F}_{in}_{I}_{V}_{I}_{in}

*in*
*S*

i

v

A

v

i

v

v

R

i

v

/

)

(

/

)

(

∴

_{V}

_{in}

_{V}

*in*

*I*

*iF*

A

R

A

i

v

R

1

1

(16.3-1)The multiplication 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.

437

*in*

*V*

*in*

*V*

*iF*

*R*

*s*

*A*

*R*

*A*

*Z*

_{} 1 0 / 1 1 1

*R*

_{i}*Z*(16.3-2a)

_{eq}where *Zeq is the frequency dependent part and is of form *

)

/

1

(

_{1}0

s

A

R

Z

*i*

*V*

*eq*

. Rewritten as an admittance 0 1 / 1 1*V*

*i*

*eq*

*eq*

*RA*

*s*

*Y*

*Z*

G

_{eq}

sC

*(16.3-2b) for which*

_{eq}*V*

*i*

*eq*

*eq*

*A*

*R*

*G*

*R* 1

_{0}(16.3-3a) and

*i*

*T*

*i*

*V*

*eq*

*R*

*R*

*A*

*C* 1 1 1 0

*i*

*FR*1 1 (16.3-3b)

where the (gain-bandwidth) definition *T = AV0 *

×

*1*has been invoked as well as

*1F =*

*T*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 circuit is a non-inverting topology. By nodal analysis (and by inspection) *ANI = *
5.0V/V and so * =* 1/*ANI = 0.2. *

From equation (16.3-3a), *R _{eq}*

*A*

_{V}_{0}

*R*0.2400100

_{i}*k*

**= 8.0M**

and from equation (16.3-3b),

*k*
*MHz*
*R*
*C*
*i*
*T*
*eq*
100
)
2
2
(
2
.
0
1
1
*k*
*MHz* 100
2
2
.
0
1
2
1
_{.}_{04} _{10}12
1
16
.
0
**= 4.0pF**

438

Revisiting the analysis of *ZIF, Zeq, Yeq, Req*and* Ceq, the input to the opamp should be expected to be *
equivalent to an *RC* circuit of the topology shown by figure 16.3.4.

**Figure 16.3-4.** Equivalent input topology of the opamp with feedback.
The *RC* time constant of this network is approximately = *RinCeq. *

So for example 16.3-1, = 100k

× 4pF =

400ns, or a frequency corner*f =*(1/2)

×

(1/) = 400 kHz. This result is the same as*f1F =*

*fT.*

In like manner as the resistance at the input *vS, resistance at the output vout*is also defined by feedback.
The internal equivalent circuit that feeds the output node is shown by figure 16.3-5.

**Figure 16.3-5. **Internal equivalent circuit and output resistance topology with feedback.
Resistance into the opamp at *vout*is *Rout = vout/io *

As represented by the figure the current into *vout*will be *iO = (vout – AV *

×v

*I)/Ro .*

So output conductance *Gout = 1/Rout will be *

*out*

*o*

*I*

*v*

*out*

*out*

*out*

v

R

v

A

v

v

i

G

0

(16.3-4)439

For evaluation of output resistance *vS = 0 is required. So for v-* = *vF = ** vout *
then *vI = vS – vF = –** vout and equation (16.3-4) becomes*

*Gout = Go*(*vout + **AV vout)/vout* = Go (*1 + **AV) * (16.3-5)
By equation (16.2-1) *AV is frequency dependent and so conductance Gout*→ *Yout is of the form *

*eq*
*o*
*v*
*o*
*o*
*out* *G* *G* *A* *G* *Y*
*Y*

(16.3-6)for which *Yeq contains the frequency character of AV. Rewriting it as Zeq = 1/Yeq and making use of *
equation (16.2-1) gives a form with separated equivalent components.

1
0
0
0
1 1
/
1
*V*
*o*
*V*
*o*
*V*
*o*
*eq*
*A*
*G*
*s*
*A*
*G*
*A*
*G*
*s*
*Z* *Req* *sLeq*

(

16.3-7)

## For which the two equivalents

R

*eq*

and

L

*eq*

are then

0
0
1
*V*
*o*
*V*
*o*
*eq*
*A*
*R*
*A*
*G*
*R*

(

16.3-8a)

*T*

*o*

*V*

*o*

*eq*

*G*

*A*

*G*

*L* 1 1 1 0

(

16.3-8b)

since*AV0*

×

*1 =*

*T.*

Courtesy of this analysis the output node of an opamp with feedback will then be equivalent to the topology of figure 16.3-6.

440

**EXAMPLE 16.3-2:** The circuit shown uses an opamp with zero frequency gain *AV0 = 52 dB, *
gain-bandwidth product GB = 2MHz, and output resistance *Ro = 100. Determine output equivalent *
impedance terms *Req and Leq. *

**SOLUTION:** The circuit shown is an inverting topology. (By inspection) the equivalent non-inverting
topology will have *ANI = 8.0. Therefore ** =* 1/*ANI = 1/8. *

∴
400
8
/
1
100
0
*V*
*o*
*eq*
*A*
*R*
*R*
**= 2.0** *f* *MHz*
*R*
*L*
*T*
*o*
*eq*
2
8
/
1
100
16
.
0
2
1
**= 64****H **

If the backend of the opamp is assessed in terms of time constants, then for the values of example #16.3-2

= *Leq/Ro = *64H* /*100* = *0.64s and corresponds to frequency corner *f = *(1/2) ×(1/) = 250 kHz.
Note that this value is the same as *f1F*= *fT.* (= 0.125

×

2MHz)**Summary table:** Effect of the feedback loop on input and output impedance.

** Table 16.3-1. **Input/output equivalent impedance/admittance terms.
Input equivalent impedance *Zin*: *R _{eq}*

*A*

_{V}_{0}

*R*

_{in}*in*
*T*
*eq*
*R*
*f*
*C*
1
2
1

Output equivalent admittance *Yout : *

0
*V*
*o*
*eq*
*A*
*R*
*R*
*T*
*o*
*eq*
*f*
*R*
*L*
2
1

441

## 16.4 COMMON MODE SIGNAL REJECTION (CMRR)

As emphasized by its circuit symbol, an opamp is a differential amplifier. A diffamp is a topology that is
characterized by a large differential gain *Adiff, and a common-mode gain ACM that usually is fairly small. *
The ability to reject common-mode signal is defined by the common-mode rejection ratio (CMRR)

*CM*
*diff*
*A*
*A*

*CMRR* (16.4-1)

The opamp construct has a font end of the form of an operational transconductance amplifier (OTA) which is either an emitter- or source- coupled pair. For most OTAs the ECP or SCP transistor pair is a matched pair of transistors with stiff active load to create a relatively high gain amplifier. The shared current tail for the transistor pair is also fairly stiff. With those caveats, the CMRR for an opamp is typically very large, on the order of 100dB.

The CMRR primarily affects differential amplifier constructs such
as that shown by figure 16.4-1. In this figure the signal that
appears at the *v+ node is common to both inputs and so an error *
signal will be passed to the output

**Figure 16.4-1. **Simple diffamp.
Symmetry is important and this circuit will reject common signals as long as *R4/R3 = R2/R1 *(*= *)*. *The
symmetry criterion plays a role in other diffamp constructs such as the IA shown by figure 16.4-2.

**Figure 16.4-2a.** 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.

442

**Figure 16.4-2b.** 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.
A diffamp circuit with less symmetry, as represented by figure 16.4-3, will not necessarily be successful
in suppression of the common signal.

**Figure 16.4-3. ** 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 poorly rejected. A CMRR
assessment using Fourier analysis gives a circuit CMRR of only 39dB.

From these figures it should be emphasized that simulation is an analytical necessity for the behavior of circuits with constraints.

The CMRR for an opamp is assessed by a circuit construct of the form shown by figure 16.4-4. Since

_{}

_{}

_{}

_{}

2*A* *v* *v* *A* *v* *v*

*v _{O}*

_{V}

_{CM}and *v* *v _{O}*

*v*

_{CM}*v*

*vCM*

**Figure 16.4-4:** construct for CMRR evaluation

Dividing by *AV and collecting terms gives * _{O}

*V*
*CM*
*V*
*CM*
*V*
*CM*
*v*
*A*
*A*
*A*
*v*
*A*
*A*
2
1
1
Since 1
2
1 _{}
*V*
*CM*
*V* *A*
*A*
*A* then *A* *CMRR*
*A*
*v*
*vo*
*V*
*CM*
*CM*
1
(16.4-2)

443

The reciprocal of the transfer ratio *vO/vCM identified by equation (16.4-2) for several opamps is shown by *
figure 16.4-5 for several opamps. The y-axis is in dB and the x-axis = frequency. The one with the lowest
CMRR of 70dB corresponds to the general purpose LM324 opamp used in figures 16.4-2 and 16.4-3..

### Figure 14.4-5: CMRR vs frequency for the LM324, uA741C, and the LF411

## 16.5 SLEW-RATE

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

*dt*
*dV _{out}*

_{= constant }≡ SR

*C*

*EE*

*C*

*I* (16.5-1)

Current *IEE*is that of the 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 an amplitude distortion limit on the frequency since as the output cannot swing
large voltages at higher frequencies. For*V _{out}*(

*t*)

*V*sin(

_{a}*t*)this limit manifests itself as

444
and thus *V* cos( *t*)

*dt*
*dV*

*a*

*out* _{} _{which has a maximum slope at }
*t =* 0

Since the maximum amplitude is defined by the power supply rails, then the slew rate defines a maximum frequency at maximum output swing defined by

)
2
(
(max) *m*
*a*
*MAX*
*out*
*f*
*V*
*SR*
*dt*
*dV* _{}

where *fm = maximum power bandwidth, given by *

*S*
*m*
*V*
*SR*
*f*
2
1
(16.5-2)
where *VS = Va(max)* is (approximately) the supply rail voltage.

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 general- purpose 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 the *dV/dt* slope constraint of *SR =* 0.5V/us. Consequently the output ends up looking like a
triangular wave

.

**EXAMPLE 16.5-1:** An opamp with SR = 1.0V/us is supplied by voltage rails of

±

10V. (a) What is the maximum power bandwidth and(b) what is the highest frequency *f1 for an undistorted V1 of 2.0V? *

### SOLUTIONS: (a) fm (0.16)1.0(V s)/10V = 16kHz

445

## 16.6 OPAMP MACROMODELS

While it might be expected that the opamp parts in the (SPICE) library are a multiple-transistor construct,
that is neither a fact nor practical. The typical opamp circuit consists of 24-30 transistors, somewhat like
the form1_{ shown by figure 16.6-1. If such a circuit were invoked for each opamp the simulation would }

become swamped by an excess of invisible nodes and devices. The simulator would then slow to a crawl trying to iterate all of these nodes and interconnected transistor equations.

**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)

446

**Figure 16.6-2(a).** Macromodel for the LM324 pSPICE opamp.

**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

.

447

**Figure 16.6-3(a).** Macromodel for the LF411 pSPICE opamp. The input is a nJFET
source-coupled pair and it distinguishes its performance characteristics

.

## Figure 16.6-3(b).

pSPICE node list for LF411 macromodel

Macromodels are a necessity for packaged circuits. The forms indicated by figures 16.6-2 and 16.6-3 are the ones that exist in the pspice library.

448

The somewhat simpler cousin, the ideal opamp macromodel is a VVT (voltage-voltage transducer) with large gain factor, as indicated by figure 16.6-4. It is an equivalent to a nullator-norator element.

**Figure 16.6-4.** Macromodel for ideal opamp. Since there are no internal nodes and the part (the VVT) is
linear, this model does not add any additional overhead to the simulation.

An option that is one step closer to a real opamp is one that adds an an

R

*in and an*

R

*out to figure 16.6-4.*

**Figure 16.6-5.** Enhanced macromodel for ideal opamp. *Rin*= 1 M and *Rout = 50. *

STC frequency character can be added to the model with a topology as represented by figure 16.6-6

449

The macromodel of figure 16.6-6 does not include performance constraints such as CMRR and slew rate. They 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. Power supply nodes must be added to the circuit symbol. In the macromodels represented by figures 16.6-2 and 16.6-3, the differential pair front end is the signature of a more mature construct. A simplified version is shown by figure 16.6-7.

**Figure 16.6-7.** In this case there are two stages with the input stage defined by the differential
pair M1 and M2 and the next stage 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 critical 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 10HZ to 100Hz if it is
to match the compensated frequency profile for the opamp, the capacitance must be large. This

requirement is accomplished by a Miller multiplication effect across the capacitance by means of the CCT. The multiplying factor is approximately that of the CCT and gives result

2
1
1
2
1
*kC*
*R*
*f*
*O*
(16.6-1)

Other relationships that relate to the enhanced macros of figure 16.6-7 are

dt

dV

*= constant = SR 2*

_{out}C

I

_{SS}

(16.6-2)450

with transfer gain _{m}_{D}_{a}_{O}

*I*
*O*
*kR*
*G*
*R*
*g*
*v*
*v*
_{1} (16.6-3)

where *gm1 is the transconductance of the (balanced) input FETs. The CMRR is self-evident as *
*Gcm*

*Ga*

*CMRR* (16.6-4)

A simulation of figure 16.6-7 with the MOSFET 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* = 1000A/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 two more nodes must be included with the part, namely those of the power
supply. Then constraints and implications associated with the power supply need to be modeled. Figure
16.6-7 does not do so. Power supply 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.

As a footnote to the macromodel constructs, the value given to transconductance gain *Ga* is always set as

*Ga = 1/Rc* (*= 1/Rd *for FET) (16.6-5)

This choice assures that equation (16.6-2) for the macromodel SR value is valid and consistent with the definition of slew rate.

451

## SUMMARY AND PORTFOLIO

Feedback factor *1/TNI * where *TNI * = gain for non-inverting topology, ideal opamp

*V*
*IDEAL*
*A*
*T*
*ideal*
*non*
*T*
1
1
)

( where *AV* = finite gain

Gain-bandwidth:
1
3
*f _{T}*

*f*

_{dB}*GB*3dB corner

*f*3

*dB*

*fT*Slew rate:

*C*

*I*

*dt*

*dV*

*MAX*

*out*

*m*

*S*

*a*

*f*

*V*

*f*

*V*

*SR* 2

*fm*≡ full-power bandwidth

Finite *Rin *and finite Rout:

**Input equivalent impedances: ** *R _{eq}*

*A*0

_{V}*R*

_{i}

*i*
*T*
*eq*
*R*
*f*
*C*
1
2
1
**Output equivalent admittances: **

0
*V*
*o*
*eq*
*A*
*R*
*R*

*T*

*o*

*eq*

*f*

*R*

*L* 2 1 Macromodels:

*SR*

*ieec*2

*CMRR*

*ga*

*gcm*

452

**APPENDIX 15-1.** Simulations of selected opamps
(1) Open-loop transfer characteristics

* * *AV0* (741C) = 106 dB
f*T = 913 kHz *
*AV0* (LM324) = 50 dB
f*T = 955 kHz *
*AV0* (LF411) = 112 dB
f*T = 7.0 MHz *
(1) Common-mode rejection ratio

*CMRR*
*A*
*A*
*v*
*v*
*V*
*CM*
*CM*
*O* 1
CMRR (LM324) = 70 dB
CMRR (uA741) = 90 dB
CMRR (LF411) = 106 dB