LTC1562 View Datasheet(PDF) - Linear Technology
Part Name
Description
MFG CO.
'LTC1562' PDF : 28 Pages View PDF
LTC1562
APPLICATIONS INFORMATION
tolerance (by a factor of 2 incrementally), but it also
implies that R2 has a wider range than fO. (RQ and RIN also
tend to scale with R2.) At high fO these resistors fall below
5k, heavily loading the outputs of the LTC1562 and leading
to increased THD and other effects. At the other extreme,
a lower fO limit of 10kHz reflects an arbitrary upper
resistor limit of 1MΩ. The LTC1562’s MOS input circuitry
can accommodate higher resistor values than this, but
junction leakage current from the input protection cir-
cuitry may cause DC errors.
The 2nd order transfer functions HLP(s), HBP(s) and
HHP(s) (below) are all inverting so that, for example, at DC
the lowpass gain is – HL. If two such sections are cas-
caded, these phase inversions cancel. Thus, the filter in the
application schematic on the first page of this data sheet
is a dual DC preserving, noninverting, rail-to-rail lowpass
filter, approximating two “straight wires with frequency
selectivity.”
Figure 4 shows further details of 2nd order lowpass,
bandpass and highpass responses. Configurations to
obtain these responses appear in the next three sections.
Basic Lowpass
When ZIN of Figure 3 is a resistor of value RIN, a standard
2nd order lowpass transfer function results from VIN to V2
(Figure 5):
( ) V2(s)
VIN(s)
=
HLP(s)
=
s2
+
–HLωO2
ωO / Q s + ωO2
The DC gain magnitude is HL = R2/RIN. (Note that the
transfer function includes a sign inversion.) Parameters
ωO (= 2πfO) and Q are set by R2 and RQ as above. For a 2nd
order lowpass response the gain magnitude becomes QHL
RIN
VIN
RQ R2
VOUT
INV V1 V2
2nd ORDER
1/4 LTC1562
1562 F05
Figure 5. Basic Lowpass Configuration
HB
0.707 HB
BANDPASS RESPONSE
HP
HL
0.707 HL
LOWPASS RESPONSE
HP
HH
0.707 HH
HIGHPASS RESPONSE
fL fO fH
f (LOG SCALE)
fP fC
f (LOG SCALE)
fC
fP
f (LOG SCALE)
Q
=
fO
fH –
fL ; fO
=
fL fH
fL
=
fO
–1
2Q
+
1
2Q
2
+
1
fH
=
fO
1
2Q
+
1 2
2Q
+
1
fC = fO
1–
1
2Q2
+
1–
1
2Q2
2
+
1
fP = fO
1– 1
2Q2
HP
=
HL
1
Q
1
1–
1
4Q2
fC
=
fO
1–
1
2Q2
+
1–
1
2Q2
2
+
1
–1
fP = fO
1–
1
2Q2
–1
HP
=
HH
1
Q
1
1–
1
4Q2
Figure 4. Characteristics of Standard 2nd Order Filter Responses
8
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