ME2Tel CondicionamentoSinais

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Condicionamento de Sinais

Medidas Eléctricas II

Prof. Ricardo Queirós

Curso de Engenharia Electrotécnica: Electrónica e Telecomunicações Departamento de Electrónica e Electrotecnia

Faculdade de Engenharia Universidade Agostinho Neto

2

◦

Semestre 2017

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Conteúdo

1

_{Amplificadores Operacionais: Revisão}

2

_{Amplificador de Instrumentação}

3

Filtros

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Conteúdo

1

_{Amplificadores Operacionais: Revisão}

2

Amplificador de Instrumentação

3

Filtros

(4)

Conteúdo

1

_{Amplificadores Operacionais: Revisão}

2

Amplificador de Instrumentação

3

Filtros

(5)

Amplificadores Operacionais: Revisão

Introdução aos Amp-Ops

O termo “operacional” origina do facto de que estes

amplificadores eram principalmente usados para realizar

operações matemáticas (adição, subtracção, integração,

diferenciação);

Símbolo e terminais:

Entrada Inversora Entrada Não - Inversora +V -V

INTRODUCTION TOOPERATIONAL AMPLIFIERS ◆ 603

Inverting input – + – + Noninverting input Output +V –V

(b) Symbol with dc supply connections (a) Symbol

Typical packages. Pin 1 is indicated by a notch or dot on dual in-line (DIP) and surface-mount technology (SMT) packages, as shown.

(c) 1 8 DIP 1 20 SMT 1 8 SMT !_{FIGURE 12–1}

Op-amp symbols and packages.

The standard

operational amplifier (op-amp)

symbol is shown in Figure 12–1(a). It

has two input terminals, the inverting (!) input and the noninverting (") input, and one

output terminal. Most op-amps operate with two dc supply voltages, one positive and the

other negative, as shown in Figure 12–1(b), although some have a single dc supply.

Usually these dc voltage terminals are left off the schematic symbol for simplicity but are

understood to be there. Some typical op-amp IC packages are shown in Figure 12–1(c).

12–1

I

NTRODUCTION TO

O

PERATIONAL

A

MPLIFIERS

Early operational amplifiers (op-amps) were used primarily to perform mathematical

operations such as addition, subtraction, integration, and differentiation—thus the

term operational. These early devices were constructed with vacuum tubes and

worked with high voltages. Today’s op-amps are linear integrated circuits (ICs) that

use relatively low dc supply voltages and are reliable and inexpensive.

After completing this section, you should be able to

❏

Describe the basic operational amplifier and its characteristics

◆

Identify the schematic symbol and IC package terminals

❏

Discuss the ideal op-amp

❏

Discuss the practical op-amp

◆

Draw the internal block diagram

The Ideal Op-Amp

To illustrate what an op-amp is, let’s consider its ideal characteristics. A practical op-amp,

of course, falls short of these ideal standards, but it is much easier to understand and

ana-lyze the device from an ideal point of view.

First, the ideal op-amp has infinite voltage gain and infinite bandwidth. Also, it has an

infinite input impedance

(open) so that it does not load the driving source. Finally, it has a

zero output impedance.

Op-amp characteristics are illustrated in Figure 12–2(a). The input

voltage, V

in

, appears between the two input terminals, and the output voltage is A

v

V

in

, as

indicated by the internal voltage source symbol. The concept of infinite input impedance is

The operational amplifier concept originated around 1947. It was proposed that such a device would form an extremely useful analog building block. The first

commercial op-amps used vacuum tubes, but it was not until the introduction of the integrated circuit did the op-amp start to fulfill its true potential. In 1964, the first integrated circuit op-amp, designated the 702, was developed by Fairchild Semiconductor. This was later followed by the 709 and eventually the 741, which has become an industry standard.

H I S T O R Y N O T E

Amp-Op Ideal

Ganho de tensão (A

v

) infinito;

Largura de banda infinita;

Impedância de entrada infinita

(reduzir o efeito de carga na fonte);

Impedância de saída igual a zero;

V

s

=

A

v

V

e

;

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Folha de dados - Amp-Op 741

640 ◆ THEOPERATIONALAMPLIFIER

Absolute Maximum Ratings (TA = 25°C)

Parameter Symbol Value Unit

Supply Voltage VCC ±18 V

Differential Input Voltage VI(DIFF) 30 V

Input Voltage VI ±15 V

Output Short Circuit Duration - Indefinite

Power Dissipation PD 500 mW

Operating Temperature Range KA741

KA741I

TOPR 0 ~ + 70 -40 ~ +85

°C Storage Temperature Range TSTG -65 ~ + 150 °C

Electrical Characteristics

(VCC= 15V, VEE = - 15V. TA = 25° C, unless otherwise specified)

Parameter Symbol Conditions

KA741/KA741I Unit Min. Typ. Max.

Input Offset Voltage VIO

RS≤ 10 KΩ - 2.0 6.0 mV

RS≤ 50 Ω - -

-Input Offset Voltage Adjustment Range V

IO(R) VCC = ±20 V - ±15 - mV Input Offset Current IIO - - 20 200 nA Input Bias Current IBIAS - - 80 500 nA Input Resistance (Note1) RI VCC = ±20 V 0.3 2.0 - MΩ Input Voltage Range VI(R) - ±12 ±13 - V Large Signal Voltage Gain GV

RL≥ 2 KΩ VCC = ± 20 V, VO(P-P) = ± 15 V - - -V/mV VCC = ± 15 V, VO(P-P) = ±10 V 20 200 -Output Short Circuit Current ISC - - 25 - mA

Output Voltage Swing VO(P-P)

VCC = ±20V RL≥ 10 KΩ - - -V RL≥ 2 KΩ - - -VCC = ±15V RL≥ 10 KΩ ±12 ±14 -RL≥ 2 KΩ ±10 ±13 -Common Mode Rejection Ratio CMRR

RS≤ 10 KΩ, VCM=±12V 70 90 -dB RS≤ 50 Ω, VCM=±12V - - -Power Supply Rejection Ratio PSRR

VCC = ±15V to VCC = ±15V RS≤ 50 Ω - - -dB VCC = ±15V to VCC = ±15V RS≤ 10 KΩ 77 96

-Transient Rise Time TR

Unity Gain

- 0.3 - µs

Response Overshoot OS - 10 - %

Bandwidth BW - - - - MHz

Slew Rate SR Unity Gain - 0.5 - V/µs

Supply Current ICC RL=∞Ω - 1.5 2.8 mA Power Consumption PC VCC = ±20V - - -mW VCC = ±15V - 50 85 Electrical Characteristics

(VCC=±15V, unless otherwise specified)

The following specification apply over the range of 0°C ≤ TA≤ +70 °C for the KA741; and the -40°C ≤ TA≤ +85 °C

for the KA741I

Parameter Symbol Conditions

KA741/KA741I Unit Min. Typ. Max.

Input Offset Voltage VIO

RS≤ 50 Ω - -

-mV RS≤ 10 KΩ - - 7.5 Input Offset Voltage Drift ∆ VIO/∆ T - - - µV/ °C Input Offset Current IIO - - - 300 nA Input Offset Current Drift ∆ IIO/∆ T - - - nA/°C Input Bias Current IBIAS - - - 0.8 µA Input Resistance (Note1) RI VCC=±20 V - - - MΩ Input Voltage Range VI(R) - ±12

±12 ±10

±13 - V Output Voltage Swing VO(P-P)

VCC = ± 20 V - - -V - - -VCC = ±15 V ±14 -±13 -Output Short Circuit Current ISC - 10 - 40 mA Common Mode Rejection Ratio CMRR

RS≤ 10 KΩ, VCM = ±12 V RS≥ 10 KΩ RS≥ 10 KΩ RS≥ 2 KΩ RS≥ 2 KΩ RS≤ 50 Ω, VCM = ±12 V 70 90 -dB - -

-Power Supply Rejection Ratio PSRR VCC=±20 V

to±5 V

RS≤ 50 Ω - - -dB RS≤ 10 KΩ 77 96

-Large Signal Voltage Gain GV RS≥ 2 KΩ

VCC = ±20V, VO(P-P) = ±15V - - -V/mV VCC = ±15V, VO(P.P)=±10V 15 - -VCC = ±15V, VO(P-P) = ±2V - -

www.fairchildsemi.com

Rev. 1.0.1

Features

• Short circuit protection • Excellent temperature stability • Internal frequency compensation • High Input voltage range • Null of offset

Description

The KA741 series are general purpose operational amplifiers. It is intended for a wide range of analog applications. The high gain and wide range of operating voltage provide superior performance in intergrator, summing amplifier, and general feedback applications.

8-DIP

8-SOP

1 1

Internal Block Diagram

KA741

Single Operational Amplifier

!FIGURE 12–48

Partial datasheet for the KA741 op amp. Copyright Fairchild Semiconductor Corporation. Used by permission.

5. Using the datasheet, assign pin numbers to the op-amp in Figure 12–47. 6. Determine the maximum power consumption of the op-amp with the

supply voltages.

7. To what typical voltage can the output swing with ;15 Vsupply voltages? ;15 V Prof. Ricardo Queirós (UAN 2017) Medidas Eléctricas II: Elec. Tel. 4 / 30

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Amplificador Diferencial: Introdução

A configuração mais usual para efeitos de medida é a do

amplificador diferencial;

R _E V _s1 R _C1 R _C2 +V _cc V _s2 V _e1 V _e2 - V _EE V _s1 V _s2 V _e1 V _e2

Tipicamente, um amplificador operacional consiste em dois ou

mais amplificadores diferenciais;

Estes amplificadores operam tipicamente em três modos:

1) Diferencial (single-ended), 2) Diferencial (double-ended) e

3) Modo-Comum;

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Amplificador Diferencial: Introdução

A configuração mais usual para efeitos de medida é a do

amplificador diferencial;

Tipicamente, um amplificador operacional consiste em dois ou

mais amplificadores diferenciais;

Estes amplificadores operam tipicamente em três modos:

1) Diferencial (single-ended), 2) Diferencial (double-ended) e

3) Modo-Comum;

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Amplificador Diferencial: Introdução

A configuração mais usual para efeitos de medida é a do

amplificador diferencial;

Tipicamente, um amplificador operacional consiste em dois ou

mais amplificadores diferenciais;

Estes amplificadores operam tipicamente em três modos:

1) Diferencial (single-ended), 2) Diferencial (double-ended) e

3) Modo-Comum;

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Amplificador Diferencial: Modos de operação

Diferencial (single ended) : Uma das entrada é ligada à massa,

enquanto que o sinal a amplificar é ligado à outra entrada. O sinal

amplificado e invertido aparece no colector do transístor no qual o

sinal é aplicado. No colector do outro transístor aparece um sinal

amplificado mas não invertido;

Diferencial (double ended) : Dois sinais em oposição de fase são

aplicados às entradas. A saída do transístor 1 terá o dobro da

amplitude da sua entrada, invertida. O mesmo se aplica para o

transístor 2;

Modo-Comum : Dois sinais em fase, com a mesma amplitude e

frequência são aplicados às entradas. As saídas dos

transístores 1 e 2 serão iguais a zero.

Rejeição de Modo Comum

Se um sinal indesejado (ex: ruído) é comum às duas entradas, não

aparecerá na saída (não distorce o sinal desejado).

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Factor de Rejeição de Modo Comum

Sinais desejados: Aparecem numa entrada apenas ou nas duas

entradas mas em oposição da fase;

Sinais indesejados: Aparecem com a mesma polaridade e são

cancelados pelo amplificador diferencial;

Idealmente, o ganho do amplificador diferencial é muito elevado

para sinais desejados e zero para sinais de modo comum;

O factor de rejeição de modo comum (CMRR) é uma indicação da

habilidade de um amplificador diferencial em rejeitar sinais de

modo comum: (A

d

- ganho modo diferencial ou seja ganho de

malha aberta e A

c

- ganho de modo comum)

CMRR =

A

d

A

c

Um sinal desejado é amplificado CMRR vezes mais do que o

ruído.

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CMRR: Exemplo

O ganho de malha aberta de um amplificador operacional é o

ganho de tensão interno do dispositivo e é determinado pelo

quociente entre a tensão de saída e a tensão de entrada quando

não existem componentes externos. Ou seja, depende apenas da

estrutura interna do dispositivo.

CMRR - Exemplo

Um amp-op tem um ganho de tensão diferencial de malha aberta de

85000 e um ganho de modo comum de 0.25. Determine o CMRR e

expresse-a em decibéis.

CMRR = 85000/0.25 = 340000

CMRR = 20log(340000) = 110.6 dB

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Realimentação Negativa

NEGATIVEFEEDBACK ◆ 613

Negative feedback is illustrated in Figure 12–14. The inverting

input effectively

makes the feedback signal 180° out of phase with the input signal.

(

-)

1. Distinguish between single-ended and double-ended differential mode.

2. Define common-mode rejection.

3. For a given value of open-loop differential gain, does a higher common-mode gain

result in a higher or lower CMRR?

4. List at least ten op-amp parameters.

5. How is slew rate measured?

SECTION 12–2 CHECKUP

12–3

N

EGATIVE

F

EEDBACK

Negative feedback is one of the most useful concepts in electronics, particularly in

op-amp applications.

Negative feedback

is the process whereby a portion of the

output voltage of an amplifier is returned to the input with a phase angle that

opposes (or subtracts from) the input signal.

After completing this section, you should be able to

❏

Explain negative feedback in op-amps

❏

Discuss why negative feedback is used

◆

Describe the effects of negative feedback on certain op-amp parameters

Vout

+ – Vin

Vf

Internal inversion makes Vf

180° out of phase with Vin.

Negative feedback circuit

!FIGURE 12–14

Illustration of negative feedback.

Why Use Negative Feedback?

As you can see in Table 12–1, the inherent open-loop voltage gain of a typical op-amp is

very high (usually greater than 100,000). Therefore, an extremely small input voltage

drives the op-amp into its saturated output states. In fact, even the input offset voltage of

the op-amp can drive it into saturation. For example, assume V

IN

! 1 mV and A

ol

!

100,000. Then,

Since the output level of an op-amp can never reach 100 V, it is driven deep into saturation

and the output is limited to its maximum output levels, as illustrated in Figure 12–15 for

both a positive and a negative input voltage of 1 mV.

The usefulness of an op-amp operated without negative feedback is generally limited to

comparator applications (to be studied in Chapter 13). With negative feedback, the

closed-loop voltage gain (A

cl

) can be reduced and controlled so that the op-amp can function as a

V

IN

A

ol

= (1 mV)(100,000) = 100 V

Porquê usar realimentação negativa?

O ganho de malha aberta de uma amp-op é tipicamente muito

elevado. Portanto, mesmo uma pequena tensão de entrada satura

(perto de V+ ou V- dependendo da polaridade da tensão de entrada) o

amp-op. Com realimentação consegue-se controlar o ganho, as

impedâncias de entrada e saída, e a largura de banda.

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Amplificador Não-Inversor

OP-AMPS WITHNEGATIVEFEEDBACK ◆ 615

Closed-Loop Voltage Gain, A

cl

The

closed-loop voltage gain

is the voltage gain of an op-amp with external feedback.

The amplifier configuration consists of the op-amp and an external negative feedback

cir-cuit that connects the output to the inverting input. The closed-loop voltage gain is

deter-mined by the external component values and can be precisely controlled by them.

Noninverting Amplifier

An op-amp connected in a closed-loop configuration as a

noninverting amplifier

with a

controlled amount of voltage gain is shown in Figure 12–16. The input signal is applied to

the noninverting (

+) input. The output is applied back to the inverting

input through

the feedback circuit (closed loop) formed by the input resistor R

i

and the feedback resistor

R

f

.

This creates negative feedback as follows. Resistors R

i

and R

f

form a voltage-divider

circuit, which reduces V

_out

and connects the reduced voltage V

_f

to the inverting input. The

feedback voltage is expressed as

V

f

= a

R

i

R

i

+ R

f

bV

out

(

-)

Rf R_i Vf Vin + – Feedback circuit V_out !_{FIGURE 12–16} Noninverting amplifier.

The difference of the input voltage, V

in

, and the feedback voltage, V

f

, is the differential

input to the op-amp, as shown in Figure 12–17. This differential voltage is amplified by the

open-loop voltage gain of the op-amp (A

ol

) and produces an output voltage expressed as

The attenuation, B, of the feedback circuit is

Substituting BV

out

for V

f

in the V

out

equation,

V

out

= A

ol

(V

in

- BV

out

)

B

=

_R

R

i i

+ R

f

V

out

= A

ol

(V

in

- V

f

)

R_f Ri V_f Vin + – Vout Vdiff = Vin – Vf !FIGURE 12–17 Differential input, Vin- Vf.

V

f

= (

R

i

R

i

+

R

f

)V

out

V

out

=

A

ol

(V

in

− V

f

)

B =

R

i

R

i

+

R

f

A

cl

=

V

out

v

in

=

A

ol

1 + A

ol

B

≈

1 B

Z

in(NI)

= (1 + A

ol

B)Z

in

Z

_out(NI)

=

Z

out

1 + A

ol

B

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Seguidor de Tensão

Voltage-Follower

Thevoltage-followerconfiguration is a special case of the noninverting amplifier where all of the output voltage is fed back to the inverting input by a straight connection, as shown in Figure 12–19. As you can see, the straight feedback connection has a voltage gain of 1 (which means there is no gain). The closed-loop voltage gain of a noninverting amplifier is as previously derived. Since B ! 1 for a voltage-follower, the closed-loop voltage gain of the voltage-follower is

1/B (-) Acl(VF) ! 1 Equation 12–9 – + Vout Vin !FIGURE 12–19 Op-amp voltage-follower. – + Rf Vout Ri Vin !FIGURE 12–20 Inverting amplifier.

The most important features of the voltage-follower configuration are its very high input impedance and its very low output impedance. These features make it a nearly ideal buffer amplifier for interfacing high-impedance sources and low-impedance loads. This is discussed further in Section 12–5.

Inverting Amplifier

An op-amp connected as an inverting amplifierwith a controlled amount of voltage gain is shown in Figure 12–20. The input signal is applied through a series input resistor Rito

the inverting input. Also, the output is fed back through Rfto the same input. The

non-inverting (_{+) input is grounded.} (_-)

At this point, the ideal op-amp parameters mentioned earlier are useful in simplifying the analysis of this circuit. In particular, the concept of infinite input impedance is of great value. An infinite input impedance implies zero current at the inverting input. If there is zero current through the input impedance, then there must be no voltage drop between the inverting and noninverting inputs. This means that the voltage at the inverting input is zero because the noninverting (_{+) input is grounded. This zero voltage at the inverting input} terminal is referred to as virtual ground. This condition is illustrated in Figure 12–21(a).

Since there is no current at the inverting input, the current through Riand the current

through Rfare equal, as shown in Figure 12–21(b).

Iin = If

(-)

Z

in(NI)

= (1 + A

ol

)Z

in

Z

_out(NI)

=

Z

out

1 + A

ol

Caso especial do amplificador não-inversor;

Toda a tensão de saída é realimentada para a entrada inversora;

O ganho de malha fechada é 1/B, sendo B = 1 para o seguidor

de tensão. Portanto: A

cl

=

1 Apresenta muito alta impedância de entrada e muito baixa

impedância de saída. É adequado para funcionar como buffer, ou

seja, interface entre fontes de elevada impedância de saída e

cargas de baixa impedância;

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Seguidor de Tensão

Voltage-Follower

Inverting Amplifier

Iin = If

(-)

Z

in(NI)

= (1 + A

ol

)Z

in

Z

_out(NI)

=

Z

out

1 + A

ol

Caso especial do amplificador não-inversor;

Toda a tensão de saída é realimentada para a entrada inversora;

O ganho de malha fechada é 1/B, sendo B = 1 para o seguidor

de tensão. Portanto: A

cl

=

1 Apresenta muito alta impedância de entrada e muito baixa

impedância de saída. É adequado para funcionar como buffer, ou

seja, interface entre fontes de elevada impedância de saída e

cargas de baixa impedância;

(17)

Seguidor de Tensão

Voltage-Follower

Inverting Amplifier

Iin = If

(-)

Z

in(NI)

= (1 + A

ol

)Z

in

Z

_out(NI)

=

Z

out

1 + A

ol

Caso especial do amplificador não-inversor;

Toda a tensão de saída é realimentada para a entrada inversora;

O ganho de malha fechada é 1/B, sendo B = 1 para o seguidor

de tensão. Portanto: A

cl

=

1 Apresenta muito alta impedância de entrada e muito baixa

impedância de saída. É adequado para funcionar como buffer, ou

seja, interface entre fontes de elevada impedância de saída e

cargas de baixa impedância;

(18)

Seguidor de Tensão

Voltage-Follower

Inverting Amplifier

Iin = If

(-)

Z

in(NI)

= (1 + A

ol

)Z

in

Z

_out(NI)

=

Z

out

1 + A

ol

Caso especial do amplificador não-inversor;

Toda a tensão de saída é realimentada para a entrada inversora;

O ganho de malha fechada é 1/B, sendo B = 1 para o seguidor

de tensão. Portanto: A

cl

=

1 Apresenta muito alta impedância de entrada e muito baixa

impedância de saída. É adequado para funcionar como buffer, ou

seja, interface entre fontes de elevada impedância de saída e

cargas de baixa impedância;

(19)

Amplificador Inversor

618 ◆ T_HEO_PERATIONALA_MPLIFIER

The voltage across R

i

equals V

in

because the resistor is connected to virtual ground at the

inverting input of the op-amp. Therefore,

Also, the voltage across R

f

equals

because of virtual ground, and therefore,

Since

Rearranging the terms,

Of course, V

out

/V

in

is the overall gain of the inverting (I) amplifier.

V

out

V

in

=

-R

f

R

i

-V

out

R

f

=

V

in

R

i

I

f

= I

in

,

I

f

=

-V

_R

out f

-V

out

I

in

=

V

_R

in i

Equation 12–10 shows that the closed-loop voltage gain of the inverting amplifier

(A

cl(I)

) is the ratio of the feedback resistance (R

f

) to the input resistance (R

i

). The

closed-loop gain is independent of the op-amp’s internal open-closed-loop gain.

Thus, the negative

feed-back stabilizes the voltage gain. The negative sign indicates inversion.

A

cl(I)

! "

R

f

R

i

Equation 12–10

Given the op-amp configuration in Figure 12–22, determine the value of R

f

required to

produce a closed-loop voltage gain of

-100.

EXAMPLE 12–4

– + Rf V_out Ri V_in 0 V Virtual ground (0 V) – + Rf V_out I_in V_in ₊ Ri I_f – _I 1 = 0 0 V + –

(b) Iin = If and current at the inverting input (I1) is 0. (a) Virtual ground

! FIGURE 12–21

Virtual ground concept and closed-loop voltage gain development for the inverting amplifier.

– + Rf Vout R_i Vin 2.2 k! ! _{FIGURE 12–22}

I

in

=

I

f

=>

V

in

R

i

=

−V

out

R

f

V

out

= −V

in

R

f

R

i

A

cl

= −

R

f

R

i

Z

in(I)

≈ R

i

Z

out(I)

=

Z

out

1 + A

ol

B

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Comparador

Amp-op que permite comparar duas tensões de entrada

produzindo sempre à saída um de dois estados, dependendo da

diferença das entradas;

Os amp-ops podem ser utilizados como comparador com ou sem

feedback. Contudo, o amp-op sem feedback é mais comum em

aplicações pouco críticas.

COMPARATORS ◆ 669

Nonzero-Level Detection

The zero-level detector in Figure 13–1 can be modified to detect positive and negative

volt-ages by connecting a fixed reference voltage source to the inverting

input, as shown in

Figure 13–2(a). A more practical arrangement is shown in Figure 13–2(b) using a voltage

divider to set the reference voltage, V

REF

, as follows:

where

_+V

is the positive op-amp dc supply voltage. The circuit in Figure 13–2(c) uses a

V

REF

=

_R

R

2 1

+ R

2

(

+V )

(

-)

– + Vout Vin (a) (b) Vin0 Vout 0 –Vout(max) +Vout(max) t t !FIGURE 13–1

The op-amp as a zero-level detector.

–

+

V_out

Vin

(a) Battery reference – + VREF – + V_out Vin

(c) Zener diode sets reference voltage

R +V VZ – + V_out Vin (b) Voltage-divider reference VREF R₁ +V R2 (d) Waveforms Vin 0 Vout –Vout(max) +Vout(max) V_REF 0 t t "_{FIGURE 13–2} Nonzero-level detectors.

zener diode to set the reference voltage (V

REF

!

V

Z

). As long as V

in

is less than V

REF

, the

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Comparador

Amp-op que permite comparar duas tensões de entrada

produzindo sempre à saída um de dois estados, dependendo da

diferença das entradas;

Os amp-ops podem ser utilizados como comparador com ou sem

feedback. Contudo, o amp-op sem feedback é mais comum em

aplicações pouco críticas.

COMPARATORS ◆ 669

Nonzero-Level Detection

The zero-level detector in Figure 13–1 can be modified to detect positive and negative

volt-ages by connecting a fixed reference voltage source to the inverting

input, as shown in

Figure 13–2(a). A more practical arrangement is shown in Figure 13–2(b) using a voltage

divider to set the reference voltage, V

REF

, as follows:

where

_+V

is the positive op-amp dc supply voltage. The circuit in Figure 13–2(c) uses a

V

REF

=

_R

R

2 1

+ R

2

(

+V )

(

-)

– + Vout Vin (a) (b) Vin0 Vout 0 –Vout(max) +Vout(max) t t !FIGURE 13–1

The op-amp as a zero-level detector.

–

+

V_out

Vin

(a) Battery reference – + VREF – + V_out Vin

(c) Zener diode sets reference voltage

R +V VZ – + V_out Vin (b) Voltage-divider reference VREF R₁ +V R2 (d) Waveforms Vin 0 Vout –Vout(max) +Vout(max) V_REF 0 t t "_{FIGURE 13–2} Nonzero-level detectors.

zener diode to set the reference voltage (V

REF

!

V

Z

). As long as V

in

is less than V

REF

, the

(22)

Integrador

Amp-op com um condensador no circuito de realimentação;

A carga Q num condensador é proporcional à corrente que o

carrega e ao tempo de carga (Q = I

c

t = CV

c

);

Assim: V

c

= (I

c

/C)t, representa a equação de uma recta;

Uma tensão de entrada constante produz uma rampa na saída.

∆V

out

∆t

= −

V

in

R

i

C

688 ◆ B_ASICO_P-A_MPC_IRCUITS – + Vin R Vout C !FIGURE 13–31 An op-amp integrator.

How a Capacitor Charges To understand how an integrator works, it is important to

re-view how a capacitor charges. Recall that the charge Q on a capacitor is proportional to the charging current (IC) and the time (t).

Also, in terms of the voltage, the charge on a capacitor is

From these two relationships, the capacitor voltage can be expressed as

This expression has the form of an equation for a straight line that begins at zero with a constant slope of IC/C. Remember from algebra that the general formula for a straight line

is y ! mx " b. In this case, y ! VC, m ! IC/C, x ! t, and b ! 0.

Recall that the capacitor voltage in a simple RC circuit is not linear but is exponential. This is because the charging current continuously decreases as the capacitor charges and causes the rate of change of the voltage to continuously decrease. The key thing about using an op-amp with an RC circuit to form an integrator is that the capacitor’s charging current is made constant, thus producing a straight-line (linear) voltage rather than an ex-ponential voltage. Now let’s see why this is true.

In Figure 13–32, the inverting input of the op-amp is at virtual ground (0 V), so the volt-age across Riequals Vin. Therefore, the input current is

Iin = Vin Ri VC = a IC C bt Q = CVC Q = ICt – + Vin Ri _V_out IC C + – Iin 0 V 0 A VC !FIGURE 13–32 Currents in an integrator.

If Vinis a constant voltage, then Iinis also a constant because the inverting input always

re-mains at 0 V, keeping a constant voltage across Ri. Because of the very high input

imped-ance of the op-amp, there is negligible current at the inverting input. This makes all of the input current go through the capacitor, as indicated in Figure 13–32, so

IC = Iin

INTEGRATORS ANDDIFFERENTIATORS ◆ 689

The Capacitor Voltage Since Iinis constant, so is IC. The constant ICcharges the

capaci-tor linearly and produces a linear voltage across C. The positive side of the capacicapaci-tor is held at 0 V by the virtual ground of the op-amp. The voltage on the negative side of the capacitor, which is the op-amp output voltage, decreases linearly from zero as the capacitor charges, as shown in Figure 13–33. This voltage, VC, is called a negative ramp and is the

consequence of a constant positive input.

The Output Voltage Voutis the same as the voltage on the negative side of the capacitor.

When a constant positive input voltage in the form of a step or pulse (a pulse has a constant amplitude when high) is applied, the output ramp decreases negatively until the op-amp saturates at its maximum negative level. This is indicated in Figure 13–34.

Rate of Change of the Output Voltage The rate at which the capacitor charges, and

therefore the slope of the output ramp, is set by the ratio IC/C, as you have seen. Since

IC! Vin/Ri, the rate of change or slope of the integrator’s output voltage is ¢Vout/¢t. – + Vin Ri 0 V + – VC 0 Constant IC !FIGURE 13–33

A linear ramp voltage is produced across the capacitor by the constant charging current. – + 0 Ri + – IC Vout Vin ₀ –Vmax 0 V C !FIGURE 13–34

A constant input voltage produces a ramp on the output of the integrator.

Integrators are especially useful in triangular-wave oscillators as you will see in Chapter 16. ¢Vout

¢t ! "

Vin R_iC

Equation 13–7

(a) Determine the rate of change of the output voltage in response to the input square

wave, as shown for the ideal integrator in Figure 13–35(a). The output voltage is initially zero. The pulse width is

(b) Describe the output and draw the waveform.

Solution (a) The rate of change of the output voltage during the time that the input is at +2.5 V (capacitor charging) is ¢Vout ¢t = -Vin RiC = -2.5V (10kÆ)(0.01mF) = -25kV/s = "25mV/Ms 100ms. EXAMPLE 13–10

(23)

Integrador

Amp-op com um condensador no circuito de realimentação;

A carga Q num condensador é proporcional à corrente que o

carrega e ao tempo de carga (Q = I

c

t = CV

c

);

Assim: V

c

= (I

c

/C)t, representa a equação de uma recta;

Uma tensão de entrada constante produz uma rampa na saída.

∆V

out

∆t

= −

V

in

R

i

C

IC = Iin

¢t ! "

Vin R_iC

Equation 13–7

(24)

Integrador

Amp-op com um condensador no circuito de realimentação;

A carga Q num condensador é proporcional à corrente que o

carrega e ao tempo de carga (Q = I

c

t = CV

c

);

Assim: V

c

= (I

c

/C)t, representa a equação de uma recta;

Uma tensão de entrada constante produz uma rampa na saída.

∆V

out

∆t

= −

V

in

R

i

C

IC = Iin

¢t ! "

Vin R_iC

Equation 13–7

(25)

Integrador

Amp-op com um condensador no circuito de realimentação;

A carga Q num condensador é proporcional à corrente que o

carrega e ao tempo de carga (Q = I

c

t = CV

c

);

Assim: V

c

= (I

c

/C)t, representa a equação de uma recta;

Uma tensão de entrada constante produz uma rampa na saída.

∆V

out

∆t

= −

V

in

R

i

C

IC = Iin

¢t ! "

Vin R_iC

Equation 13–7

(26)

Diferenciador

Amp-op com um condensador como elemento de entrada;

A saída é proporcional à taxa de variação da tensão de entrada.

V

out

= −

V

in

t

R

f

C

692 ◆ BASICOP-AMPCIRCUITS

Since the current at the inverting input is negligible, I

R

I

C

. Both currents are constant

because the slope of the capacitor voltage (V

C

/t) is constant. The output voltage is also

con-stant and equal to the voltage across R

f

because one side of the feedback resistor is always

0 V (virtual ground).

V

out

= I

R

f

= I

C

R

f

=

– + V_out – + t1– t2 0 V + – t0– t1 0t0 t1 t2 Vin 0 t0 t1 t2 t C !FIGURE 13–39

Output of a differentiator with a series of positive and negative ramps (triangle wave) on the input.

Equation 13–8

The output is negative when the input is a positive-going ramp and positive when the input

is a negative-going ramp, as illustrated in Figure 13–39. During the positive slope of the

input, the capacitor is charging from the input source and the constant current through the

feedback resistor is in the direction shown. During the negative slope of the input, the

cur-rent is in the opposite direction because the capacitor is discharging.

V

out

! "

a

V

C

t b

R

f

C

Notice in Equation 13–8 that the term V

C

/t is the slope of the input. If the slope

increases, V

out

increases. If the slope decreases, V

out

decreases. The output voltage is

proportional to the slope (rate of change) of the input. The constant of proportionality is

the time constant, R

f

C

.

Determine the output voltage of the ideal op-amp differentiator in Figure 13–40 for the

triangular-wave input shown.

EXAMPLE 13–11

– + V_out V_in 0 15 s t 2.2 k! Rf 5 s 10 s +5 V –5 V 0.001 F C µ µ µ µ !FIGURE 13–40

Solution

Starting at t ! 0, the input voltage is a positive-going ramp ranging from

(a "10 V change) in

Then it changes to a negative-going ramp ranging from

(a

change) in

The time constant is

R

f

C

= (2.2

k

Æ)(0.001

mF)

= 2.2

ms

5 ms.

-10

V

+5

V

to

-5

V

5 ms.

-5

V

to

+5

V

(27)

Diferenciador

Amp-op com um condensador como elemento de entrada;

A saída é proporcional à taxa de variação da tensão de entrada.

V

out

= −

V

in

t

R

f

C

692 ◆ BASICOP-AMPCIRCUITS

Since the current at the inverting input is negligible, I

R

I

C

. Both currents are constant

because the slope of the capacitor voltage (V

C

/t) is constant. The output voltage is also

con-stant and equal to the voltage across R

f

because one side of the feedback resistor is always

0 V (virtual ground).

V

out

= I

R

f

= I

C

R

f

=

– + V_out – + t1– t2 0 V + – t0– t1 0t0 t1 t2 Vin 0 t0 t1 t2 t C !FIGURE 13–39

Output of a differentiator with a series of positive and negative ramps (triangle wave) on the input.

Equation 13–8

The output is negative when the input is a positive-going ramp and positive when the input

is a negative-going ramp, as illustrated in Figure 13–39. During the positive slope of the

input, the capacitor is charging from the input source and the constant current through the

feedback resistor is in the direction shown. During the negative slope of the input, the

cur-rent is in the opposite direction because the capacitor is discharging.

V

out

! "

a

V

C

t b

R

f

C

Notice in Equation 13–8 that the term V

C

/t is the slope of the input. If the slope

increases, V

out

increases. If the slope decreases, V

out

decreases. The output voltage is

proportional to the slope (rate of change) of the input. The constant of proportionality is

the time constant, R

f

C

.

Determine the output voltage of the ideal op-amp differentiator in Figure 13–40 for the

triangular-wave input shown.

EXAMPLE 13–11

– + V_out V_in 0 15 s t 2.2 k! Rf 5 s 10 s +5 V –5 V 0.001 F C µ µ µ µ !FIGURE 13–40