參數(shù)資料
型號: LMF60CMJ-100
廠商: NATIONAL SEMICONDUCTOR CORP
元件分類: 模擬濾波器
英文描述: LMF60 High Performance 6th-Order Switched Capacitor Butterworth Lowpass Filter
中文描述: SWITCHED CAPACITOR FILTER, BUTTERWORTH, LOWPASS, CDIP14
封裝: CERAMIC, DIP-14
文件頁數(shù): 14/20頁
文件大?。?/td> 391K
代理商: LMF60CMJ-100
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
e
log (10
0.1AMin
b
1)
b
log(10
0.1AMax
b
1)
2 log (f
s
/f
b
)
where n is the order of the filter, A
Min
is the minimum stop-
band attenuation (in dB) desired at frequency f
s
, and A
Max
is
the passband ripple or attenuation (in dB) at frequency f
b
. If
the result of this equation is greater than 6, then more than
a single LMF60 is required.
(1)
The attenuation at any frequency can be found by the fol-
lowing equation:
Attn(f)
e
10 log
[
1
a
(10
0.1AMax
b
1) (f/f
b
)
2n
]
dB
where n
e
6 (the order of the filter).
(2)
2.1 A LOWPASS DESIGN EXAMPLE
Suppose the amplitude response specification inFigure 8 is
given. Can the LMF60 be used The order of the Butter-
worth approximation will have to be determined using eq. 1:
A
Min
e
30 dB, A
Max
e
1.0 dB, f
s
e
2 kHz, and f
b
e
1 kHz
n
e
log(10
3
b
1)
b
log(10
0.1
b
1)
2 log(2)
Since n can only take on integer values, n
e
6. Therefore
the LMF60 can be used. In general, if n is 6 or less a single
LMF60 stage can be utilized.
e
5.96
Likewise, the attenuation at f
s
can be found using equation
2 with the above values and n
e
6 giving:
Atten (2 kHz)
e
10 log
[
1
a
(10
0.1
b
1) (2/1)
12
]
e
30.26 dB
This result also meets the design specification given in Fig-
ure 8 again verifying that a single LMF60 section will be
adequate.
TL/H/9294–21
FIGURE 8. Design Example Magnitude Response
Specification Where the Response of the Filter Design
Must Fall Within the Shaded Area of the Specification
f
c
e
f
b
Since the LMF60’s cutoff freqency f
C
, which corresponds to
a gain attenuation of
b
3.01 dB, was not specified in this
example it needs to be calculated. Solving equation 2 where
f
e
f
C
as follows:
10
0.1(3.01 dB)
b
1)
(10
0.1AMax
b
1)
10
0.301
b
1
1/(2n)
e
1
10
0.1
b
1
1/12
e
1.119 kHz
where f
C
e
f
CLK
/50 or f
CLK
/100.
To implement this example for the LMF60-50 the clock fre-
quency will have to be set to f
CLK
e
50(1.119 kHz)
e
55.95 kHz or for the LMF60-100 f
CLK
e
100(1.119 kHz)
e
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
e
log (10
0.05 Amin
b
1)
b
log(10
0.05 AMax
b
1)
2 log (f
s
/f
b
)
Attn(f)
e
10 log
[
1
a
(10
0.05 AMax
b
1) (f/f
b
)
2n
]
dB
where n
e
6 (the order of each filter).
Equation 3 will determine whether the order of the filter is
adequate (n
s
6) while equation 4 can determine if the
required stopband attenuation is met and what actual cutoff
frequency (f
C
) is required to obtain the particular frequency
response desired. The design procedure would be identical
to the one shown in Section 2.1.
(3)
(4)
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
e
f
n
e
0.742 f
C
. The attenua-
tion at this frequency is 0.12 dB which must be compensat-
ed for by making R
1
e
1.014
c
R
2
.
Since R
1
does not equal R
2
there will be a gain inequality
above and below the notch frequency. At frequencies below
the notch frequency (f
m
f
n
), the signal through the filter
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
a
6 dB. For f
n
f
n
, the signal at the output of the
filter is greatly attenuated thus only the input signal will ap-
pear at the output of the Op-Amp. With R
3
e
R
1
e
1.014
R
2
the overall gain is 0.986 or
b
0.12 dB at frequencies
above the notch.
http://www.national.com
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