Datasheet
8
LTC1264
1264fb
ODES OF OPERATIO
W
U
tors R
H
and R
L
to create a notch. This is shown in Figure
8. Mode 3a is more versatile than Mode 2 because the
notch frequency can be higher or lower than the center
frequency of the 2nd order section. The external op amp
of Figure 8 is not always required. When cascading the
sections of the LTC1264, the highpass and lowpass
outputs can be summed directly into the inverting input of
the next section.
Please refer to the Maximum Frequency of Operation
paragraph under Applications Information for a guide to
the use of capacitor C
C
.
Mode 2n
This mode extends the circuit topology of Mode 3a to
Mode 2 (Figure 9) where the highpass notch and lowpass
outputs are summed through two external resistors R
H
and R
L
to create a lowpass output with a notch higher in
frequency than the notch in Mode 2. This mode, shown in
Figure 8, is most useful in lowpass elliptic designs. When
cascading the sections of the LTC1264, the highpass
notch and lowpass outputs can be summed directly into
the inverting input of the next section.
Please refer to the Maximum Frequency of Operation
paragraph under Applications Information for a guide to
the use of capacitor C
C
.
Figure 7. Mode 2, 2nd Order Filter Providing Highpass
Notch, Bandpass and Lowpass Outputs
–
+
Σ
∫ ∫
AGND
R1
N
BP
LP
V
IN
1264 F07
+
–
S
1/4 LTC1264
R2
R3
C
C
R4
f
i
= ; f
O
= f
i
; f
n
= f
O
f
CLK
20
√
R2
R4
1 +
Q = 1.005
R3
R2
( )
√
R2
R4
1 +
R3
6.42•R4
( )
1
1 –
H
OHP
= – (AC GAIN, f > f
n
); H
OHPn
= –
R2
R1
R2
R1
1
R2
R4
1 +
( )
(DC GAIN, f < f
n
)
H
OBP
= –
R3
R1
R3
6.42•R4
( )
1
1 –
; H
OLP
= H
OHPn
Mode 3a
This is an extension of Mode 3 where the highpass and
lowpass output are summed through two external resis-
Figure 8. Mode 3a, 2nd Order Filter Providing a Highpass Notch or Lowpass Notch Output
–
+
Σ
∫ ∫
AGND
R1
HP
BP
LP
V
IN
1264 G08
+
–
S
R2
R3
R4
C
C
–
+
EXTERNAL OP AMP OR
INPUT OP AMP OF THE
LTC1264, SIDES A, B, C, D
HIGHPASS
OR LOWPASS
NOTCH OUTPUT
R
G
R
L
R
H
1/4 LTC1264
R2
R1
( )
R
G
R
H
()
f
i
= ; f
n
= f
i ;
f
O
= f
i
H
OHPn
(f = ∞) = ; H
OLPn
(f = 0) =
f
CLK
20
√
R
H
R
L
√
R2
R4
Q = 1.005
R3
R2
√
R2
R4
R3
6.42•R4
( )
1
1 –
( )
R4
R1
( )
R
G
R
L
()