Electric Circuits, 9th Edition P60 pot

15.9, we require the magnitude of the high-pass filter be unity at the cutoff frequency of the low-pass filter and the magnitude of the low-pass filter be unity at the cut-off frequency

566 Active Filter Circuits

Finally, we need to meet the passband gain

specification We can adjust the scaled values of

either R x or R 2 , because K = #2/^1 • If w e adjust

R 2 , we will change the cutoff frequency, because

UJ C - l/RiC Therefore, we can adjust the value of

R] to alter only the passband gain:

R t = R 2 /K = (15,915.5)/(5) = 3183.1 O

The final component values are

R { = 3183.1X1, /?2 = 15,915.5 O, C = 0.01 /xF

The transfer function of the filter is given by

-31,415.93

H ™ ~ s + 6283.185'

The Bode plot of the magnitude of this transfer

function is shown in Fig 15.8

/ A S S E S S M E N T PROBLEM

Objective 2—Be able to design filter drcuits starting

response and component values

15.3 What magnitude and frequency scale factors

will transform the prototype high-pass filter

into a high-pass filter with a 0.5 fxF capacitor

and a cutoff frequency of 10 kHz?

NOTE: Also try Chapter Problems 15.15 and 15.16

15.3 Op Amp Bandpass

and Bandreject Filters

We now turn to the analysis and design of op amp circuits that act as band-pass and bandreject filters While there is a wide variety of such op amp circuits, our initial approach is motivated by the Bode plot construction shown in Fig 15.9 We can see from the plot that the bandpass filter con-sists of three separate components:

1 A unity-gain low-pass filter whose cutoff frequency is <wc.2, the larger

of the two cutoff frequencies;

2 A unity-gain high-pass filter whose cutoff frequency is co c[ , the

smaller of the two cutoff frequencies; and

3 A gain component to provide the desired level of gain in the passband

20

J 5

10

5 °

-5

-10

-15

10 50 100 500 1000 5000 10,000

/(Hz)

Figure 15.8 • The Bode magnitude plot of the low-pass filter from

Example 15.4

\

\

\

\ _ -^

with a prototype and use scaling to achieve desired frequency

Answer: k f = 62,831.85, k m = 31.831

5 10 50 100 500 1000 5000 10,000

a) (rad/s)

Figure 15.9 A Constructing the Bode magnitude plot of a bandpass filter

These three components are cascaded in series They combine additively

in the Bode plot construction and so will combine multiplicatively in the

s domain It is important to note that this method of constructing a band-pass magnitude response assumes that the lower cutoff frequency (oo c] ) is smaller than the upper cutoff frequency (o) c2 ) The resulting filter is called

a broadband bandpass filter, because the band of frequencies passed is

wide The formal definition of a broadband filter requires the two cutoff frequencies to satisfy the equation

> 2

As illustrated by the Bode plot construction in Fig 15.9, we require the magnitude of the high-pass filter be unity at the cutoff frequency of the low-pass filter and the magnitude of the low-pass filter be unity at the cut-off frequency of the high-pass filter Then the bandpass filter will have the cutoff frequencies specified by the low-pass and high-pass filters We need

to determine the relationship between ft>t<1 and (oc2 that will satisfy the requirements illustrated in Fig 15.9

We can construct a circuit that provides each of the three compo-nents by cascading a low-pass op amp filter, a high-pass op amp filter, and an inverting amplifier (see Section 5.3), as shown in Fig 15.10(a)

Figure 15.10(a) is a form of illustration called a block diagram Each

block represents a component or subcircuit, and the output of one block is the input to the next, in the direction indicated We wish to

establish the relationship between (o c] and OJ C2 that will permit each circuit to be designed independently, without concern for the other sub-circuits in the cascade Then the design of the bandpass filter is reduced

to the design of a unity-gain order low-pass filter, a unity-gain first-order high-pass filter, and an inverting amplifier, each of which is a sim-ple circuit

Vi *• Low-pass filter High-pass filter Inverting amplifier

(a)

-"VVV *

RL

R„

•>

*—wv-C f

RH

Ri

•> •—"wv—•

^ V v

-(b)

Figure 15.10 • A cascaded op amp bandpass filter, (a) The block diagram, (b) The circuit

The transfer function of the cascaded bandpass filter is the product of the transfer functions of the three cascaded components:

w c2 —s 1^

R;

S + (O c2 / \S + (O c i

-Kto c2 s (s + m cl )(s + <o c2 )

-K(O c2 s

s 2 + {oi cX + <o c2 )s + co c ico c2

(15.10)

We notice right away that Eq 15.10 is not in the standard form for the transfer function of a bandpass filter discussed in Chapter 14, namely,

# B P - 13s

s 2 + f3s + a) 2 ,

In order to convert Eq 15.10 into the form of the standard transfer func-tion for a bandpass filter, we require that

When Eq 15.11 holds,

(x) c2 » U) c i

(a) c[ + 0) c2 ) » wc2,

( 1 5 1 1 )

and the transfer function for the cascaded bandpass filter in

Eq 15.10 becomes

H(s) Kco c2 s

S + (O c2 S + 0) c [OJ c2

Once we confirm that Eq 15.11 holds for the cutoff frequencies spec-ified for the desired bandpass filter, we can design each stage of the cas-caded circuit independently and meet the filter specifications We

15.3 Op Amp Bandpass and Bandreject Filters 569

compute the values of R L and C L in the low-pass filter to give us the

desired upper cutoff frequency, GJC2:

1

W , 9

RJC L '

(15.12)

We compute the values of R H and C u in the high-pass filter to give us the

desired lower cutoff frequency, a) c] :

1

U) cl =

RHCH

(15.13)

Now we compute the values of R; and Rf in the inverting amplifier

to provide the desired passband gain To do this, we consider the

magni-tude of the bandpass filter's transfer function, evaluated at the center

frequency, co () :

\H(j<«o)\ Ka) c2 (j(D 0 )

&c2

Recall from Chapter 5 that the gain of the inverting amplifier is

Rf/Ri- Therefore,

R<

\H(j*>„)\ = - /

Ri

(15.15)

Any choice of resistors that satisfies Eq 15.15 will produce the desired

passband gain

Example 15.5 illustrates the design process for the cascaded

band-pass filter

Example 15.5 Designing a Broadband Bandpass Op Amp Filter

Design a bandpass filter for a graphic equalizer to

provide an amplification of 2 within the band of

fre-quencies between 100 and 10,000 Hz Use 0.2 fx¥

capacitors

Solution

We can design each subcircuit in the cascade and

meet the specified cutoff frequency values only if

Eq 15.11 holds In this case, (o c2 = 100wcl, so we can

say that a) c2 » co cl

Begin with the low-pass stage From Eq 15.12,

1

<*c2

R, =

R,C iy-L 2TT( 10000),

1

[2TT( 10000)] (0.2 X 10"6)

son

Next, we turn to the high-pass stage From Eq 15.13,

1

°M RuC

H^H

Ru —

= 277(100),

1 [27r(100)](0.2 x 10"(1)

% 7958 O

Finally, we need the gain stage From Eq 15.15, we see there are two unknowns, so one of the resistors can be selected arbitrarily Let's select a 1 kH

resis-tor for R[ Then, from Eq 15.15,

R f = 2(1000)

= 2000 a = 2 k a

570 Active Filter Circuits

The resulting circuit is shown in Fig 15.11 We

leave to you to verify that the magnitude of this

cir-cuit's transfer function is reduced by 1/V2 at both

cutoff frequencies, verifying the validity of the

assumption co c2 » wtl

0.2 fxF

7958 ft

7958 ft ° 2 M F

I "WV 1 £ —

Figure 15.11 A The cascaded op amp bandpass filter designed in Example 15.5

We can use a component approach to the design of op amp bandreject filters too, as illustrated in Fig 15.12 Like the bandpass filter, the band-reject filter consists of three separate components There are important differences, however:

1 The unity-gain low-pass filter has a cutoff frequency of wcl, which is the smaller of the two cutoff frequencies

2 The unity-gain high-pass filter has a cutoff frequency of a) c2 , which

is the larger of the two cutoff frequencies

3 The gain component provides the desired level of gain in the passbands

-40

5 10 50 100 500 1000 5000 10,000

(o (rad/s)

Figure 15.12 • Constructing the Bode magnitude plot of a bandreject filter

The most important difference is that these three components cannot

be cascaded in series, because they do not combine additively on the Bode plot Instead, we use a parallel connection and a summing amplifier, as shown both in block diagram form and as a circuit in Fig 15.13 Again, it is assumed that the two cutoff frequencies are widely separated, so that the

resulting design is a broadband bandreject filter, and io c2 » wcl Then each component of the parallel design can be created independently, and the cutoff frequency specifications will be satisfied The transfer function

of the resulting circuit is the sum of the low-pass and high-pass filter trans-fer functions From Fig 15.13(b),

H(s) S + 0)-a>ci c \ S + 0)+ c2

R f{u) cX {s + o) c2 ) + s(s + o>cl)

R \ (S + a> cl )(s + (O C2 ) Rffs 2 + 2co c iS + o) C ]C0 c2

R t \ (s + o) cl )(s + (o c2 ) (15.16)

Using the same rationale as for the cascaded bandpass filter, the two

cutoff frequencies for the transfer function in Eq 15.16 are a> cl and (o c2

only if oi cl » m c \ Then the cutoff frequencies are given by the equations

1

Wcl

&cl

RLCL

1

Rfl^H

(15.17)

(15.18)

Vi

Low-pass filter

High-pass filter

Summing amplifier

(b) Figure 15.13 A A parallel op amp bandreject filter, (a) The block diagram, (b) The circuit

572 Active Filter Circuits

In the two passbands (as s tion is Rf/Ri Therefore,

0 and s —» oo), the gain of the transfer

As with the design of the cascaded bandpass filter, we have six unknowns and three equations Typically we choose a commercially available capacitor

value for C L and C H Then Eqs 15.17 and 15.18 permit us to calculate R L

and R H to meet the specified cutoff frequencies Finally, we choose a value

for either Rf or R { and then use Eq 15.19 to compute the other resistance Note the magnitude of the transfer function in Eq 15.16 at the center

frequency, co () = Vwc l, o>c2:

\H(j<*o)\ =

* / (;o> 0 ) 2 + 2(0 cl (j(o o ) + (o ci io c2

R A(M 2 + (Wcl + Wc2)(/ft>0) + (Ocl CO c2

Rf 2d), c\

Rl 0) c \ + (x) c 2

R f 20),!

fy ft>c2 '

(15.20)

If (o c2 » a> ch then \H(j(0 o )\ <5C 2Rf/Rj (as <o c i/(o c2 <5C 1), so the

magni-tude at the center frequency is much smaller than the passband magnimagni-tude Thus the bandreject filter successfully rejects frequencies near the center frequency, again confirming our assumption that the parallel implementa-tion is meant for broadband bandreject designs

Example 15.6 illustrates the design process for the parallel band-reject filter

Example 15.6 Designing a Broadband Bandreject Op Amp Filter

Design a circuit based on the parallel bandreject op

amp filter in Fig 15.13(b) The Bode magnitude

response of this filter is shown in Fig 15.14 Use

0.5 fxF capacitors in your design

Solution

From the Bode magnitude plot in Fig 15.14, we see

that the bandreject filter has cutoff frequencies of

100 rad/s and 2000 rad/s and a gain of 3 in the

pass-bands Thus, (o c2 = 20wcl, so we make the

assump-tion that o) c2 » (o cl Begin with the prototype

low-pass filter and use scaling to meet the

specifica-tions for cutoff frequency and capacitor value The

frequency scale factor kt is 100, which shifts the

cut-off frequency from 1 rad/s to 100 rad/s The

magni-tude scale factor k m is 20,000, which permits the use

of a 0.5 /xF capacitor Using these scale factors

results in the following scaled component values:

R L = 20 k a ,

C = 0.5 u F

I

20

15

10 6.54

5

0

- 5

- 1 0

- 1 5

- 2 0

- 2 5 -30

1 \

V

1

/ /

1

» c 2

it.- 50 100 500 1000

o) (rad/s)

5000 10,000

Figure 15.14 A The Bode magnitude plot for the circuit to be designed

in Example 15.6

15.4 Higher Order Op Amp Filters 573

The resulting cutoff frequency of the low-pass filter

component is

1

ft>d =

RLCI

1 (20 X 103)(0.5 X 10-6)

= lOOrad/s

We use the same approach to design the

high-pass filter, starting with the prototype high-high-pass op

amp filter Here, the frequency scale factor is

kf = 2000, and the magnitude scale factor is

k m = 1000, resulting in the following scaled

com-ponent values:

R H = 1 kO,

C H = 0.5 /x¥

Finally, because the cutoff frequencies are

widely separated, we can use the ratio Rf/Ri to

establish the desired passband gain of 3 Let's choose

Ri = 1 kft, as we are already using that resistance for R H Then R f = 3 kft, and K = R f /Ri =

3000/1000 = 3 The resulting parallel op amp band-reject filter circuit is shown in Fig 15.15

Now let's check our assumption that

°°c2 > : > wti by calculating the actual gain at the specified cutoff frequencies We do this by making

the substitutions s = /277-(100) and s = /2-77-(2000)

into the transfer function for the parallel bandreject filter, Eq 15.16 and calculating the resulting magni-tude We leave it to the reader to verify that the magnitude at the specified cutoff frequencies is 2.024, which is less than the magnitude of 3/V2 = 2.12 that we expect Therefore, our reject-ing band is somewhat wider than specified in the problem statement

Figure 15.15 • The resulting bandreject filter circuit designed in Example 15.6

NOTE: Assess your understanding of this material by trying Chapter Problems 15.30 and 15.31

15.4 Higher Order Op Amp Filters

You have probably noticed that all of the filter circuits we have

exam-ined so far, both passive and active, are nonideal Remember from

Chapter 14 that an ideal filter has a discontinuity at the point of cutoff,

which sharply divides the passband and the stopband Although we

can-not hope to construct a circuit with a discontinuous frequency response,

we can construct circuits with a sharper, yet still continuous, transition at

the cutoff frequency

Cascading Identical Filters

How can we obtain a sharper transition between the passband and the stopband? One approach is suggested by the Bode magnitude plots in Fig 15.16 This figure shows the Bode magnitude plots of a cascade of identical prototype low-pass filters and includes plots of just one filter, two

in cascade, three in cascade, and four in cascade It is obvious that as more filters are added to the cascade, the transition from the passband to the stopband becomes sharper The rules for constructing Bode plots (from Appendix E) tell us that with one filter, the transition occurs with an asymptotic slope of 20 decibels per decade (dB/dec) Because circuits in cascade are additive on a Bode magnitude plot, a cascade with two filters has a transition with an asymptotic slope of 20 + 20 = 40 dB/dec; for three filters, the asymptotic slope is 60 dB/dec, and for four filters, it is

80 dB/dec as seen in Fig 15.16

In general, an ^-element cascade of identical low-pass filters will transi-tion from the passband to the stopband with a slope of 20« dB/dec Both the block diagram and the circuit diagram for such a cascade are shown in

Fig 15.17 It is easy to compute the transfer function for a cascade of n

pro-totype low-pass filters—we just multiply the individual transfer functions:

(-nn

The order of a filter is determined by the number of poles in its trans-fer function From Eq 15.21, we see that a cascade of first-order low-pass

filters yields a higher order filter In fact, a cascade of n first-order filters produces an /zth-order filter, having n poles in its transfer function and a

final slope of 20« dB/dec in the transition band

3

i

20

10

0

—J

- 1 0

- 2 0

- 3 0

- 4 0

- 5 0

- 6 0

- 7 0

- 8 0

^^

Sec

^ \

:ond ordc

^JFi

\

\

\ Third order Fou rth c

rst

\

\

X

)Rk

or

\

\

v

\

k

\

s

\

\

\

I"

\

\

N

\

) | 1 1 I I 1 IM

0.1 0.5 1 5 10

a) (rad/s)

Figure 15.16 A The Bode magnitude plot of a cascade of identical

prototype first-order filters

15.4 Higher Order Op Amp Filters 575

Low-pass filter Low-pass filter Low-pass filter

(a)

(b)

Figure 15.17 • A cascade of identical unity-gain low-pass filters, (a) The block diagram, (b) The circuit

There is an important issue yet to be resolved, as you will see if you

look closely at Fig 15.16 As the order of the low-pass filter is increased by

adding prototype low-pass filters to the cascade, the cutoff frequency also

changes For example, in a cascade of two first-order low-pass filters, the

magnitude of the resulting second-order filter at o>c is - 6 dB, so the cutoff

frequency of the second-order filter is not wt In fact, the cutoff frequency

is less than oo c

As long as we are able to calculate the cutoff frequency of the higher

order filters formed in the cascade of first-order filters, we can use

quency scaling to calculate component values that move the cutoff

fre-quency to its specified location If we start with a cascade of n prototype

low-pass filters, we can compute the cutoff frequency for the resulting

Azth-order low-pass filter We do so by solving for the value of m cn that results

in 1//(/0))1 = 1/V2:

H{s) =

\H(jO>cn)\ =

1

(-1)*

(s + ir

i

(M-„ + 1)"

l

l

(V^~TT)" V2'

, + 1 \ V 2 /

V 2 = o)c tl + 1,

To demonstrate the use of Eq 15.22, let's compute the cutoff

fre-quency of a fourth-order unity-gain low-pass filter constructed from a

cas-cade of four prototype low-pass filters: