OP176_15 Datasheet, PDF (15/21 Pages) Analog Devices – Bipolar/JFET, Audio Operational Amplifier

High Pass Sections
Figure 44a illustrates the high-pass form of a 2-pole SK filter
using an OP176. For simplicity and practicality, capacitors C1
and C2 are set equal (“C”), and resistors R2 and R1 are
adjusted to a ratio, N, which provides the filter damping
coefficient, α, as per the design expressions. This high pass
design is begun with selection of standard capacitor values for
C1 and C2 and a calculation of N. The values for R1 and R2
are then determined from the following expressions:
R1 =
1
2π × FREQ × C × N
and
R2 = N × R1
IN
C1
0.01µF
C2
0.01µF
R2
22k
(22.508k)
R1
11k
(11.254k)
OP176
3
2
+V_S
7
6
4
–V_S
Z_COMP
OUT
GIVEN: α, FREQ
SET C1 = C2 = C
α= ²
N
=
1
Q
N= ⁴
α²
=
R2
R1
R1 =
1
2 π FREQ x C x N
R2 = N x R1
1 kHz BW SHOWN
Z_COMP (HIGH PASS)
IN (–)
OUTPUT
R2
C2 C1
R1
Figures 44a. Two-Pole Unity Gain HP/LP Active Filters
In this examples, circuit α (or 1/Q) is set equal to √2, providing
a Butterworth (maximally flat) characteristic. The filter corner
frequency is normalized to 1 kHz, with resistor values shown in
both rounded and (exact) form. Various other 2-pole response
shapes are possible with appropriate selection of α, and fre-
quency can be easily scaled, using inversely proportional R or
C values for a given α. The 22 V/µs slew rate of the OP176 will
support 20 V p-p outputs above 100 kHz with low distortion.
The frequency response resulting with this filter is shown as the
dotted HP portion of Figure 45.
OP176
Low Pass Sections
In the LP SK arrangement of Figure 44b, the R and C elements
are interchanged where the resistors are made equal. Here, the
ratio of C2/C1 (“M”) is used to set the filter α, as noted.
Otherwise, this filter is similar to the HP section, and the
resulting 1 kHz LP response is shown in Figure 45. The design
begins with a choice of a standard capacitor value for C1 and a
calculation of M. This then forces a value of “M × C1” for C2.
Then, the value for R1 and R2 (“R”) is calculated according to
the following equation:
R=
1
2π × FREQ × C1 × M
IN
R1
11k
(11.254k)
R2
11k
(11.254k)
C2
0.01µF
C1
0.02µF
OUT
GIVEN: α, FREQ
OP176
3
2
+V_S
7
6
4
–V_S
Z_COMP
α=
2
M
=
1
Q
M= ⁴
α²
=
C2
C1
CHOOSE C1
C2 = M x C1
R=
1
2 π FREQ x C1 x M
1 kHz BW SHOWN
Z_COMP (LOW PASS)
IN (–)
OUTPUT
C2
R2
R1
C1
Figures 44b. Two-Pole Unity Gain HP/LP Active Filters
10.000
LP
0.0
–10.00
–20.00
–30.00
–40.00
–50.00
–60.00
–70.00
20
HP
100
1k
10k
50k
FREQUENCY – Hz
Figure 45. Relative Frequency Response of 2-Pole, 1 kHz
Butterworth LP (Left) and HP (Right) Active Filters
REV. 0
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