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LT1175-5 データシートの表示(PDF) - Linear Technology

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LT1175-5 Datasheet PDF : 38 Pages
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Applications Information
10
LTC1966
C = 0.1µF C = 0.22µF C = 0.47µF C = 1.0µF C = 2.2µF C = 4.7µF C = 10µF C = 22µF C = 47µF C = 100µF
1
0.1
0.01
10
1
0.1
1
10
SETTLING TIME (SEC)
Figure 19. Settling Time with Buffered Post Filter
100
1066 F14
C = 0.1µF C = 0.22µF C = 0.47µF C = 1.0µF C = 2.2µF C = 4.7µF C = 10µF C = 22µF C = 47µF C = 100µF
0.1
0.01
0.1
1
10
SETTLING TIME (SEC)
Figure 20. Settling Time with DC Accurate Post Filter
100
1066 F20
Although the settling times for the post filtered configu-
rations shown on Figures 19 and 20 are not that much
different from those with a single capacitor, the point of
using a post filter is that the settling times are far better
for a given level peak error. The filters dramatically reduce
the low frequency averaging ripple with far less impact
on settling time.
Crest Factor and AC + DC Waveforms
In the preceding discussion, the waveform was assumed
to be AC-coupled, with a modest crest factor. Both as-
sumptions ease the requirements for the averaging
capacitor. With an AC-coupled sine wave, the calculation
engine squares the input, so the averaging filter that
follows is required to filter twice the input frequency,
making its job easier. But with a sinewave that includes
DC offset, the square of the input has frequency content
at the input frequency and the filter must average out
that lower frequency. So with AC + DC waveforms, the
required value for CAVE should be based on half of the
lowest input frequency, using the same design curves
presented in Figures 6, 8, 17 and 18.
Crest factor, which is the peak to RMS ratio of a dynamic
signal, also effects the required CAVE value. With a higher
crest factor, more of the energy in the signal is concen-
trated into a smaller portion of the waveform, and the
averaging has to ride out the long lull in signal activity.
For busy waveforms, such as a sum of sine waves, ECG
traces or SCR chopped sine waves, the required value for
CAVE should be based on the lowest fundamental input
frequency divided as such:
fDESIGN = fINPUT(MIN)
3 CF 2
1966fb
21

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