lmf60 National Semiconductor Corporation, lmf60 Datasheet - Page 14

• Manufacturer Part Number: lmf60
• Description: High Performance 6th-order Switched Capacitor Butterworth Lowpass Filter
• Manufacturer: National Semiconductor Corporation

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ure 8 again verifying that a single LMF60 section will be
2 0 Designing with the LMF60
Given any lowpass filter specification two equations will
come in handy in trying to determine whether the LMF60 will
do the job The first equation determines the order of the
lowpass filter required
n
where n is the order of the filter A
band attenuation (in dB) desired at frequency f
the passband ripple or attenuation (in dB) at frequency f
the result of this equation is greater than 6 then more than
a single LMF60 is required
The attenuation at any frequency can be found by the fol-
lowing equation
Attn(f)
where n
2 1 A LOWPASS DESIGN EXAMPLE
Suppose the amplitude response specification in Figure 8 is
given Can the LMF60 be used The order of the Butter-
worth approximation will have to be determined using eq 1
Since n can only take on integer values n
the LMF60 can be used In general if n is 6 or less a single
LMF60 stage can be utilized
Likewise the attenuation at f
2 with the above values and n
This result also meets the design specification given in Fig-
adequate
Since the LMF60’s cutoff freqency f
a gain attenuation of
example it needs to be calculated Solving equation 2 where
f
where f
A
Specification Where the Response of the Filter Design
Must Fall Within the Shaded Area of the Specification
e
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e
Min
Atten (2 kHz)
FIGURE 8 Design Example Magnitude Response
f
log (10
C
e
e
as follows
C
n
30 dB A
e
e
10 log 1
e
0 1A Min
6 (the order of the filter)
f
CLK
log(10
f
c
e
e
e
e
e
Max
50 or f
a
f
1
1 119 kHz
10 log 1
30 26 dB
b
b
3
2 log (f

b
(10
e
1)
10
10
b
10
(10
1)
CLK
1 0 dB f
b
0 1A Max
2 log(2)
3 01 dB was not specified in this
0 301
0 1(3 01 dB)
0 1
b
s
0 1A Max
log(10
a
f
100
b
log(10
s
b
b
)
can be found using equation
(10
e
1
s
b
1
0 1A Max
e
0 1
6 giving
J
Min
b
1) (f f
0 1
C
b
1 12
2 kHz and f
b
1)
which corresponds to
is the minimum stop-
b
1)
1) (2 1)
(
b
1)
b
1 (2n)
)
2n
e
e
1)
s
dB
5 96
6 Therefore
12
and A
TL H 9294 – 21
b
e
1 kHz
Max
b
(1)
(2)
is
If
14
To implement this example for the LMF60-50 the clock fre-
quency will have to be set to f
55 95 kHz or for the LMF60-100 f
111 9 kHz
2 2 CASCADING LMF60s
In the case where a steeper stopband attenuation rate is
required two LMF60’s can be cascaded (Figure 9) yielding a
12th order slope of 72 dB per octave Because the LMF60
is a Butterworth filter and therefore has no ripple in its pass-
band when LMF60’s are cascaded the resulting filter also
has no ripple in its passband Likewise the DC and pass-
band gains will remain at 1V V The resulting response is
shown in Figure 10
In determining whether the cascaded LMF60’s will yield a
filter that will meet a particular amplitude response specifi-
cation as above equations 3 and 4 can be used shown
below
n
Attn(f)
where n
Equation 3 will determine whether the order of the filter is
adequate (n
required stopband attenuation is met and what actual cutoff
frequency (f
response desired The design procedure would be identical
to the one shown in Section 2 1
2 3 IMPLEMENTING A ‘‘NOTCH’’ FILTER WITH THE
LMF60
A ‘‘notch’’ filter with 60 dB of attenuation can be obtained by
using one of the Op-Amps available in the LMF60 and three
external resistors The circuit and amplitude response are
shown in Figure 11
The frequency where the ‘‘notch’’ will occur is equal to the
frequency at which the output signal of the LMF60 will have
the same magnitude but be 180 degrees out of phase with
its input signal For a sixth order Butterworth filter 180
phase shift occurs where f
tion at this frequency is 0 12 dB which must be compensat-
ed for by making R
Since R
above and below the notch frequency At frequencies below
the notch frequency (f m f
has a gain of one and is non-inverting Summing this with
the input signal through the Op-Amp yields an overall gain
of two or
filter is greatly attenuated thus only the input signal will ap-
pear at the output of the Op-Amp With R
R
above the notch
2
e
the overall gain is 0 986 or
log (10
e
1
e
10 log 1
a
does not equal R
0 05 A min
6 (the order of each filter)
C
6 dB For f n f
) is required to obtain the particular frequency
s
6) while equation 4 can determine if the
a
1
b
(10
e
2 log (f
1)
1 014
0 05 A Max
b
n
2
e
s
n
log(10
) the signal through the filter
there will be a gain inequality
the signal at the output of the
f
c
b
f
n
)
CLK
R
b
CLK
e
b
0 05 A Max
2
0 12 dB at frequencies
0 742 f
1) (f f
e
e
100(1 119 kHz)
50(1 119 kHz)
3
b
)
C
e
2n
b
The attenua-
R
1)
dB
1
e
1 014
(3)
(4)
e
e