# CHAPTER 16. THE NON-IDEAL OPAMP

N/A
N/A
Protected

Share "CHAPTER 16. THE NON-IDEAL OPAMP"

Copied!
24
0
0

Full text

(1)

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 Voutswing, 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 vFto the inverting input (v-) is

out

F v

v  (16.1-1a)

(2)

430 2 1 1 R R R    (16.1-1b)

And since voutAV (vv) AV (vSvF) AV (vS vout) Collecting terms in 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 vout /AV and  (G1G2)v G2v G1 0 A v S out V out (16.1-6b)

(3)

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 AIDEAL   = 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 ANI

(4)

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 

(5)

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 AV0and 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 sA 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)
(6)

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 1F

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

10

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

(7)

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)v1(.005)0 where

v

v

S (virtual connection of inputs)

S v v1 3  node v1 : v1(0.005.005.005)vS(.005)v2(.005)0 0 ) 3 ( 3 1    2   v vS vS v

v

2

8

v

S node v2 : v2(0.005.005.005)v1(.005)vout(.005)0 0 ) 3 ( ) 8 ( 3 2   1   

v vS v vS voutvout 21vS (same as 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 Rinis 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

(8)

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.

(9)

437

in V in V iF R s A R A Z           1 0 / 1 1 1

RiZeq (16.3-2a)

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

eq (16.3-2b) for which 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), Req

AV0Ri 0.2400100k= 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 1012 1 16 . 0    = 4.0pF

(10)

438

Revisiting the analysis of ZIF, Zeq, Yeq, Reqand 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 voutis 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 voutis Rout = vout/io

As represented by the figure the current into voutwill 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)
(11)

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 GoutYout 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     ReqsLeq

(

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.

(12)

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= 2.0f MHz R L T o eq 2 8 / 1 100 16 . 0 2 1       = 64H

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

= Leq/Ro = 64H /100 = 0.64s 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: Req  AV0Rin

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 

(13)

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.

(14)

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

vO V CM

and v vOvCM 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)

(15)

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..

## 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 dVout = constant ≡ SR C EE C I  (16.5-1)

Current IEEis 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. ForVout(t)Vasin(t)this limit manifests itself as

(16)

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?

(17)

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)

(18)

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

.

(19)

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.

(20)

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

(21)

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

out = constant = SR 2

C

I

SS

(16.6-2)
(22)

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 = 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 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.

(23)

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    fT f dB GB 3dB corner f3dB fT Slew rate: C I dt dV MAX outm S a f V f V SR     2 fm ≡ full-power bandwidth

Finite Rin and finite Rout:

Input equivalent impedances: ReqAV0Ri

 

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: SRieec2 CMRRga gcm
(24)

452

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

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

Figure

+7

References

Related documents

With 90 members in the Bayer EWC, the select committee is crucially important for the information and consultation function of the EWC, during the annual plenary meetings, between

The input signal is a 16 digits binary vector, but it can not be all “1” or “0” there has to be some variation in the signal, to get an output that it is possible to count,

Online community: A group of people using social media tools and sites on the Internet OpenID: Is a single sign-on system that allows Internet users to log on to many different.

KEC also operates three world-class tower testing facilities in India, including its facility in Nagpur, which is capable of testing towers up to 1,200 kV making it the

Quality: We measure quality (Q in our formal model) by observing the average number of citations received by a scientist for all the papers he or she published in a given

Public Health Nurses are known Public Health Nurses are known leaders in the county, region and leaders in the county, region and. the state as they the state as they

Accessing elements of a structure Initialization of structure data Using pointers to access structures

Digital to Analog Converter DAC0_OUT2 / OPAMP output channel number 2.