Neamen D. Microelectronics: Circuit Analysis and Design

Подождите немного. Документ загружается.

Sections 14.4 and 14.5 Offset Voltage and Input Bias Current: Total Effects

14.51 For the op-amp in Figure P14.51, the input offset voltage is

= 3

mV,

the average input bias current is

= 0.4 μ

A, and the offset bias current is

= 0.06 μ

A. (a) Determine the possible range in output voltage for

= 0

and

R = 0

. (b) Repeat part (a) for

= 0

and

R = 9.09



= 0.2

V and

R = 9.09



1058 Part 2 Analog Electronics

D14.52 Consider the op-amp circuit in Figure P14.52. (a) Find the value of R

needed for a

±10

mV offset voltage adjustment. (b) Determine R

to mini-

mize bias current effects. (Assume

 R

D14.53 For each op-amp in the circuit in Figure P14.38, the offset voltage is

= 10

mV and the input bias currents are

= I

= 2 μ

A. (a) Find

the worst-case output voltages

and

for

= 0

. (b) Design compen-

sation circuits to adjust both v

and v

to zero when

= 0

14.54 The op-amps in the circuit in Figure P14.49 have an offset voltage

= 2

mV, an average input bias current of

= 0.2 μ

A, and an offset

current of

= 0.02 μ

A. (a) For

= 0

and

= R

= 0

, determine the

possible range in output voltages

, and

. (b) Repeat part (a) for

= 8.33



and

= 10



14.55 Each op-amp in Figure P14.50 has an offset voltage of

= 2

mV, an

average input bias current of

= 500

nA, and an input offset current

100 nA. Determine the worst-case output voltage for each

circuit.

Section 14.6 Additional Nonideal Effects

14.56 For each op-amp in Figure P14.50, the input offset voltage is

= 2

T = 25

◦

and the input offset voltage temperature coefﬁcient is

TCv

= 6.7 μ

V/°C. Find the output voltage v

due to the input offset

voltage effects at: (a)

T = 25

◦

C and (b)

T = 50

◦

14.57 The input offset voltage in each op-amp in Figure P14.57 is

= 1

mV at

T = 25

◦

C and the input offset voltage coefﬁcient is TCv

= 3.3

◦

Find the worst-case output voltages

and

at: (a)

T = 25

◦

C and

(b)

T = 50

◦

10 kΩ

100 kΩ

–

Figure P14.51

100 kΩ

+15 V

–

15 V

= 10 kΩ

15 kΩ

–

Figure P14.52

nea80644_ch14_1009-1060.qxd 07/12/2009 3:45 Page 1058 pinnacle MHDQ-New:MHDQ134:MHDQ134-14:

14.58 For each op-amp in Figure P14.50, the input bias current is

= 500

T = 25

◦

C, the input offset current is

= 200

nA at

T = 25

◦

C, the

input bias current temperature coefﬁcient is 8 nA/

◦

C, and the input offset

current temperature coefﬁcient is 2 nA/

◦

C. (a) Find the output voltage due

to the average input bias currents at

T = 25

◦

C. (b) Find the worst-case out-

put voltage due to the input bias current and input offset current at

T = 25

◦

C. (c) Repeat parts (a) and (b) for

T = 50

◦

14.59 For each op-amp in Figure P14.57, the input bias current is

= 2 μ

T = 25

◦

C, the input offset current is

= 0.2 μ

A at

T = 25

◦

C, the

input bias current temperature coefﬁcient is 20 nA/

◦

C, and the input offset

current temperature coefﬁcient is 5 nA/

◦

C. (a) Find the worst-case output

voltages

and

due to the average input bias currents at

T = 25

◦

(b) Find the worst-case output voltages

and

due to the input bias

currents and input offset current at

T = 25

◦

C. (c) Repeat parts (a) and

(b) for

T = 50

◦

14.60 The op-amp in the difference ampliﬁer conﬁguration in Figure P14.60 is

ideal. (a) If the tolerance of each resistor is

±1.5%

, determine the minimum

value of

CMRR

. (b) Repeat part (a) if the tolerance of each resistor is

±3%

14.61 If the tolerance of each resistor in the difference ampliﬁer in Figure P14.60

±x%

, what is the maximum value of

if the minimum

CMRR

is (a) 50 dB

and (b) 75 dB.

Chapter 14 Nonideal Effects in Operational Ampliﬁer Circuits 1059

10 kΩ

50 kΩ

20 kΩ

50 kΩ

60 kΩ

–

Figure P14.57

10 kΩ

50 kΩ

–

Figure P14.60

COMPUTER SIMULATION PROBLEMS

14.62 Consider an inverting ampliﬁer such as shown in Figure 14.2. Bias a stan-

dard op-amp at

±5

V, and let

= 100



and

= 10



. Using a com-

puter simulation, plot

versus

over the range

−0.7 ≤ v

≤ 0.7

V. What

is the output saturation voltage?

nea80644_ch14_1009-1060.qxd 07/12/2009 3:45 Page 1059 pinnacle MHDQ-New:MHDQ134:MHDQ134-14:

14.63 Consider the simpliﬁed op-amp shown in Figure 14.11. Use standard tran-

sistors and take the output at the collector of

. Assume the bias current for

and

= 19 μA

and the bias current for

and

= 0.15

mA.

Let

= 30

pF. (a) Using a computer simulation, determine the slew rate of

the ampliﬁer. (b) Using a computer simulation, determine the small-signal

bandwidth for (i)

= 1 μ

V and (ii)

= 5 μ

V. Use an appropriate load.

14.64 The equivalent circuit of the all-CMOS MC14573 op-amp was given in

Figure 13.14. Using a computer simulation, determine the slew rate of the

op-amp assuming

= 12

pF. Use standard transistors.

14.65 A basic bipolar input diff-amp stage is shown in Figure 14.22. Use standard

transistors and other appropriate circuit parameters. Let

= v

= 0

(a) Plot

and

as a function of the wiper arm position

. (b) Plot the

collector voltage of

as a function of wiper arm position

DESIGN PROBLEMS

[Note: Each design should be veriﬁed with a computer analysis.]

*D14.66An ampliﬁer system, using op-amps, is to be designed to provide a low-

frequency voltage gain of 50 and a bandwidth of 20 kHz. The only available

op-amps have a low-frequency open-loop voltage gain of

3 × 10

and a

bandwidth of 10 Hz. Design an appropriate system.

*D14.67Consider the simpliﬁed op-amp in Figure 14.11. Neglect the emitter-fol-

lower output stage. Assume bias voltages of

= 3

V and

−

=−3V

. Let

the bias current for

and

= 0.1

mA. The total power dissipated

in the circuit is to be limited to 0.65 mW. Design the circuit such that the

slew rate is

s. Determine

for

and

, and ﬁnd the appropriate

value for

*D14.68Consider the op-amp circuit shown in Figure P14.12. Each op-amp has an

offset voltage of

= 2

mV. Design an offset voltage compensation

circuit. Assume bias voltages are limited to

±5

*D14.69Consider the op-amp circuit shown in Figure P14.12. Each op-amp has an

average input bias current of

= 1 μ

A and the offset bias current is

= 0.1 μ

A. Design an optimum bias-current compensation circuit. What

is the possible range of output voltage

for

= 0

1060 Part 2 Analog Electronics

nea80644_ch14_1009-1060.qxd 07/12/2009 3:45 Page 1060 pinnacle MHDQ-New:MHDQ134:MHDQ134-14:

Chapter

Applications and Design

of Integrated Circuits

1061

In Chapter 9, we introduced the ideal operational ampliﬁer and analyzed and designed

basic op-amp circuits. In this chapter, we consider additional applications and designs

of op-amp and comparator circuits that may be fabricated as integrated circuits.

A comparator is essentially an op-amp operated in an open-loop conﬁguration with

either a high or low saturated output signal.

Circuits to be considered include active ﬁlters, oscillators, Schmitt trigger cir-

cuits, integrated circuit power ampliﬁers, and voltage regulators.

A general goal of this chapter is to increase our skill at designing electronic

circuits to meet particular speciﬁcations and to perform particular functions.

PREVIEW

In this chapter, we will:

• Analyze and design active ﬁlters that transmit desired frequency components

of an input signal and attenuate undesired frequency components.

• Analyze and design oscillators that provide sinusoidal signals at speciﬁed

frequencies.

• Analyze and design various Schmitt trigger circuits.

• Analyze and design multivibrator circuits that provide signals with particular

waveforms.

• Analyze and design IC power ampliﬁers that usually consist of high-gain

small-signal ampliﬁers in cascade with an output stage.

• Analyze and design voltage regulators that establish a relatively constant dc

voltage generated from an ac signal source.

• As an application, design an active bandpass ﬁlter to meet a set of speciﬁcations.

nea80644_ch15_1061-1140.qxd 07/12/2009 3:58 Page 1061 pinnacle MHDQ-New:MHDQ134:MHDQ134-15:

1062 Part 2 Analog Electronics

15.1 ACTIVE FILTERS

Objective: • Analyze and design active ﬁlters that transmit desired

frequency components of an input signal and attenuate undesired fre-

quency components.

An important application of an op-amp is the active ﬁlter. The word ﬁlter refers to

the process of removing undesired portions of the frequency spectrum. The word

active implies the use of one or more active devices, usually an operational ampliﬁer,

in the ﬁlter circuit. As an example of the application of op-amps in the area of active

ﬁlters, we will discuss the Butterworth ﬁlter. There are many types or classiﬁcations

of ﬁlters. However, the objective here is to concentrate mainly on a single type

(Butterworth) in order to demonstrate the use of op-amps in ﬁlter design. Additional

types of ﬁlters are discussed in other references.

Two advantages of active ﬁlters over passive ﬁlters are:

1. The maximum gain or the maximum value of the transfer function may be

greater than unity.

2. The loading effect is minimal, which means that the output response of the ﬁlter

is essentially independent of the load driven by the ﬁlter.

Active Network Design

From our discussions of frequency response in Chapter 7, we know that RC networks

form ﬁlters. Figure 15.1(a) is a simple example of a coupling-capacitor circuit. The

voltage transfer function for this circuit is

T (s) =

(s)

R +

sRC

1 + sRC

(15.1)

The Bode plot of the voltage gain magnitude

|T ( jω)|

is shown in Figure 15.1(b). The

circuit is called a high-pass ﬁlter.

Figure 15.2(a) is another example of a simple RC network. Here, the voltage

transfer function is

T (s) =

(s)

+ R

1 + sRC

(15.2)

15.1.1

(a)

|T( jw)|

Stopband Passband

(b)

Figure 15.1 (a) Simple high-pass ﬁlter and (b) Bode plot of transfer function magnitude

nea80644_ch15_1061-1140.qxd 07/12/2009 3:58 Page 1062 pinnacle MHDQ-New:MHDQ134:MHDQ134-15:

Chapter 15 Applications and Design of Integrated Circuits 1063

(a)

|T( jw)|

StopbandPassband

(b)

Figure 15.2 (a) Simple low-pass ﬁlter and (b) Bode plot of transfer function magnitude

(a)

–

(b)

–

Figure 15.3 (a) High-pass ﬁlter with voltage follower and (b) low-pass ﬁlter with voltage

follower

The Bode plot of the voltage gain magnitude

|T ( jω)|

for this circuit is shown in

Figure 15.2(b). This circuit is called a low-pass ﬁlter.

Although these circuits both perform a basic ﬁltering function, they may suffer

from loading effects, substantially reducing the maximum gain from the unity value

shown in Figures 15.1(b) and 15.2(b). Also, the cutoff frequencies

and

may

change when a load is connected to the output. The loading effect can essentially be

eliminated by using a voltage follower as shown in Figure 15.3. In addition, a nonin-

verting ampliﬁer conﬁguration can be incorporated to increase the gain, as well as

eliminate the loading effects.

These two ﬁlter circuits are called one-pole ﬁlters; the slope of the voltage gain

magnitude curve outside the passband is 6 dB/octave or 20 dB/decade. This charac-

teristic is called the rolloff. The rolloff becomes sharper or steeper with higher-order

ﬁlters and is usually one of the speciﬁcations given for active ﬁlters.

Two other categories of ﬁlters are bandpass and band-reject. The desired ideal

frequency characteristics are shown in Figure 15.4.

(a)

|T( jw )|

(b)

|T( jw )|

Figure 15.4 Ideal frequency characteristics: (a) bandpass ﬁlter and (b) band-reject ﬁlter

nea80644_ch15_1061-1140.qxd 07/12/2009 3:58 Page 1063 pinnacle MHDQ-New:MHDQ134:MHDQ134-15:

1064 Part 2 Analog Electronics

–

Figure 15.5 General two-pole active ﬁlter

General Two-Pole Active Filter

Consider Figure 15.5 with admittances

through

and an ideal voltage follower.

We will derive the transfer function for the general network and will then apply spe-

ciﬁc admittances to obtain particular ﬁlter characteristics.

15.1.2

A KCL equation at node

yields

− V

= (V

− V

+(V

− V

(15.3)

A KCL equation at node

produces

− V

= V

(15.4)

From the voltage follower characteristics, we have

= V

. Therefore, Equa-

tion (15.4) becomes

= V



+ Y



= V



+ Y



(15.5)

Substituting Equation (15.5) into (15.3) and again noting that

= V

, we have

+ V

+ Y

) = V

+ Y

)

= V



+ Y



+ Y

)

(15.6)

Multiplying Equation (15.6) by

and rearranging terms, we get the following

expression for the transfer function:

T (s) =

(s)

+ Y

)

(15.7)

To obtain a low-pass ﬁlter, both

and

must be conductances, allowing the

signal to pass into the voltage follower at low frequencies. If element

is a capaci-

tor, then the output rolls off at high frequencies.

To produce a two-pole function, element

must also be a capacitor. On the

other hand, if elements

and

are capacitors, then the signal will be blocked at

low frequencies but will be passed into the voltage follower at high frequencies, re-

sulting in a high-pass ﬁlter. Therefore, admittances

and

must both be conduc-

tances to produce a two-pole high-pass transfer function.

nea80644_ch15_1061-1140.qxd 07/12/2009 3:58 Page 1064 pinnacle MHDQ-New:MHDQ134:MHDQ134-15:

Chapter 15 Applications and Design of Integrated Circuits 1065

–

Figure 15.6 General two-pole low-pass ﬁlter

Two-Pole Low-Pass Butterworth Filter

To form a low-pass ﬁlter, we set

= G

= 1/R

= G

= 1/R

= sC

, and

= sC

, as shown in Figure 15.6. The transfer function, from Equation (15.7),

becomes

T (s) =

(s)

+sC

+ G

+sC

)

(15.8)

15.1.3

At zero frequency,

s = jω = 0

and the transfer function is

T (s = 0) =

= 1

(15.9)

In the high-frequency limit,

s = jω →∞

and the transfer function approaches zero.

This circuit therefore acts as a low-pass ﬁlter.

A Butterworth ﬁlter is a maximally ﬂat magnitude ﬁlter. The transfer func-

tion is designed such that the magnitude of the transfer function is as ﬂat as possible

within the passband of the ﬁlter. This objective is achieved by taking the derivatives

of the transfer function with respect to frequency and setting as many as possible

equal to zero at the center of the passband, which is at zero frequency for the low-

pass ﬁlter.

Let

= G

≡ G = 1/R

. The transfer function is then

T (s) =

+sC



+sC



1 + sRC

(2 + sRC

)

(15.10)

We deﬁne time constants at

= RC

and

= RC

. If we then set

s = jω

,we

obtain

T ( jω) =

1 + jωτ

(2 + jωτ

)

(1 − ω

) + j(2ωτ

)

(15.11)

The magnitude of the transfer function is therefore

|T ( jω)|=[(1 − ω

)

+(2ωτ

)

]

−1/2

(15.12)

nea80644_ch15_1061-1140.qxd 07/12/2009 3:58 Page 1065 pinnacle MHDQ-New:MHDQ134:MHDQ134-15:

1066 Part 2 Analog Electronics

For a maximally ﬂat ﬁlter (that is, a ﬁlter with a minimum rate of change), which

deﬁnes a Butterworth ﬁlter, we set

d|T |

dω



ω=0

= 0

(15.13)

Taking the derivative, we ﬁnd

d|T |

dω

=−

[(1 − ω

)

+(2ωτ

)

]

−3/2



−4ωτ

(1 − ω

) + 8ωτ



(15.14)

Setting the derivative equal to zero at

ω = 0

yields

d|T |

dω



ω=0



−4ωτ

(1 − ω

) + 8ωτ



= 4ωτ

[−τ

(1 − ω

) + 2τ

]

(15.15)

Equation (15.15) is satisﬁed when

2τ

= τ

,or

= 2C

(15.16)

For this condition, the transfer magnitude is, from Equation (15.12),

|T |=

[1 + 4(ωτ

)

]

1/2

(15.17)

The 3 dB, or cutoff, frequency occurs when

|T |=1/

√

, or when

4(ω

3dB

)

= 1

We then ﬁnd that

3dB

= 2π f

3dB

√

2RC

(15.18)

In general, we can write the cutoff frequency in the form

3dB

(15.19)

Finally, comparing Equations (15.19), (15.18), and (15.16) yields

= 0.707C

(15.20(a))

and

= 1.414C

(15.20(b))

The two-pole low-pass Butterworth ﬁlter is shown in Figure 15.7(a). The Bode

plot of the transfer function magnitude is shown in Figure 15.7(b). From Equa-

tion (15.17), the magnitude of the voltage transfer function for the two-pole low-pass

Butterworth ﬁlter can be written as

|T |=



1 +



3dB



(15.21)

Equation (15.15) shows that the derivative of the voltage transfer function mag-

nitude at

ω = 0

is zero even without setting

2τ

= τ

. However, the added condition

2τ

= τ

produces the maximally ﬂat transfer characteristics of the Butterworth

ﬁlter.

nea80644_ch15_1061-1140.qxd 07/12/2009 3:58 Page 1066 pinnacle MHDQ-New:MHDQ134:MHDQ134-15:

Chapter 15 Applications and Design of Integrated Circuits 1067

(a)

|T( jw)|

3dB

2pRC

–12 dB/octave

–40 dB/decade

(b)

= 1.414C

= 0.707C

–

Figure 15.7 (a) Two-pole low-pass Butterworth ﬁlter and (b) Bode plot, transfer function

magnitude

DESIGN EXAMPLE 15.1

Objective: Design a two-pole low-pass Butterworth ﬁlter for an audio ampliﬁer

application.

Speciﬁcations: The circuit with the conﬁguration shown in Figure 15.7(a) is to be

designed such that the bandwidth is 20 kHz.

Choices: An ideal op-amp is available and standard-valued resistors and capacitors

must be used.

Solution: From Equation (15.19), we have

3dB

2π RC

RC =

2π f

3dB

2π(20 ×10

)

= 7.96 ×10

−6

If we let

R = 100 k

, then

C = 79.6pF

, which means that

= 1.414C = 113 pF

and

= 0.707C = 56.3pF

Trade-offs: Standard-valued 100 k



resistors can be used. Standard-valued

120 pF and

= 56 pF

capacitors can be used. For these elements, a bandwidth of

20.1 kHz is obtained.

Comment: These resistance and capacitance values are generally too large to be fab-

ricated conveniently on an IC. Instead, discrete resistors and capacitors, in conjunc-

tion with the IC op-amp, would need to be used.

EXERCISE PROBLEM

Ex 15.1: Design a two-pole low-pass Butterworth ﬁlter with a bandwidth of

25 kHz. The largest capacitor value to be used is 50 pF. (Ans. Set

= 50

pF,

then

= 25

pF,

R = 180

)

nea80644_ch15_1061-1140.qxd 07/12/2009 3:58 Page 1067 pinnacle MHDQ-New:MHDQ134:MHDQ134-15: