Electric Circuits, 9th Edition P60 pot

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
we
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.
/ASSESSMENT 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
<w
c
.
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 (o
c2
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 sub-
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 first-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—•
^Vv-
(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,
#BP -
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
) » w
c2
,
(15.11)
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,
GJ
C2
:
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
K. (15.14)
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
= 100w
cl
, 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 ka.
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
» w
tl
.
0.2
fxF
7958 ft
7958 ft °
2 MF
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 w
cl
, 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
» w
cl
. 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)
-a>
c
i
+
S + 0)

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 func-
=3
(15.19)
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
()
= Vw
cl
, o>
c2
:
\H(j<*o)\
=
*/
(;o>
0
)
2
+ 2(0
cl
(j(o
o
) + (o
ci
io
c2
R
A(M
2
+ (Wcl + W
c2
)(/ft>

0
) +
(O
cl
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 magnitude.
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
= 20w
cl
, 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 ka,
C
L
= 0.5 uF.
I
20
15
10
6.54
5
0
-5
-10
-15
-20
-25
-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 10
3
)(0.5 X 10
-6
)
= lOOrad/s.
We use the same approach to design the high-
pass filter, starting with the prototype 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
>:> w

ti 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:
(-n
n
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

-10
-20
-30
-40
-50
-60
-70
-80
^^
Sec
^\
:ond ordc
^JFi
\
\
\
Third order
Fou
rth
c
rst
\
\
X
)Rk
or
\
\
v
\

k
\
s
\
\
\
I"
\
\
N
\
)|
1 1 II 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 w
t
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 fre-
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'
l
_
(±y
/n
, + 1 \V2/
V2 = o)c
tl
+ 1,
1.
(15.22)
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:
a)
c4
2 - 1 = 0.435 rad/s.
(15.23)