17
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LT1994
Differential 1st Order Lowpass Filter
Maximum –3dB frequency (f
3dB
) 2MHz
Stopband attenuation: –6dB at 2 • f
3dB
and 14dB at 5 • f
3dB
Example: The specifi ed –3dB frequency is 1MHz Gain = 4
1. Using f
3dB
= 1000kHz, C11
abs
= 400pF
2. Nearest standard 5% value to 400pF is 390pF and C11
= C12 = 390pF
3. Using f
3dB
= 1000kHz, C11 = 390pF and Gain = 4, R21
= R22 = 412Ω and R11 = R12 = 102Ω (nearest 1%
value)
Differential 2nd Order Butterworth Lowpass Filter
Maximum –3dB frequency (f
3dB
) 1MHz
Stopband attenuation: –12dB at 2 • f
3dB
and –28dB at 5 • f
3dB
TYPICAL APPLICATIONS
+
–
–
+
V
+
0.1µF
3
7
6
8
1
2
5
4
R21
C11
R12
R11
R22
C12
0.1µF
V
IN
–
V
IN
+
LT1994
V
OUT
–
V
OUT
+
1994 TA03
Component Calculation:
R11 = R12, R21 = R22
f MHz and Gain
MHz
f
dB
dB
3
3
2
2
≤≤
1. Calculate an absolute value for C11 (C11
abs
) using a
specifi ed –3dB frequency
C
f
C inpFandf inkHz
abs
dB
abs dB
11
410
11
5
3
3
=
•
()
2. Select a standard 5% capacitor value nearest the absolute
value for C11
3. Calculate R11 and R21 using the standard 5% C11
value, f
3dB
and desired gain
R11 and R21 equations (C11 in pF and f
3dB
in kHz)
R
Cf
R
R
Gain
dB
21
159 2 10
11
11
21
6
3
=
=
.•
•
+
–
–
+
V
+
0.1µF
3
7
6
8
1
2
5
4
R32
R31
C21
R12
R11
0.1µF
C11
V
IN
–
V
IN
+
LT1994
V
OUT
–
V
OUT
+
1994 TA04
R22
R21
C22
Component Calculation:
R11 = R12, R21 = R22, R31 = R32, C21 = C22,
C11 = 10 • C21, R1 = R11, R2 = R21, R3 = R31,
C2 = C21 and C1 = C11
1. Calculate an absolute value for C2 (C2
abs
) using a
specifi ed –3dB frequency
C
f
C inpFandf inkHz Note
abs
dB
abs dB
2
410
2
5
3
3
=
•
()(22)
2. Select a standard 5% capacitor value nearest the absolute
value for C2 (C1 = 10 • C2)
