Datasheet

AD6620
–41–REV. A
The impulse response length of the CIC2 is given by
21
2
×−
( )
M
f
CIC
ADC
The composite impulse response length of all three stages is
NMM MM M
f
TAPS CIC CIC CIC CIC CIC
ADC
××+××−×+
52 52 2
431
The Algorithmic Latency is
NMM MM M
f
TAPS CIC CIC CIC CIC CIC
ADC
××+××−×+
×
52 52 2
431
2
Fixed Latency is the delay due to each register between the input
and the output of the AD6620. The latency is the count of each
register multiplied by the period of the clock that drives it. The
fixed latency of the AD6620 can be approximated by the follow-
ing expression:
10 7 7 5
25
tt M M M N t
CLK SAMP CIC CIC RCF TAPS CLK
++ + +
[]
[]
[]
+×
where:
t
CLK
is the high speed clock to the AD6620.
t
SAMP
is the data rate delivered to the AD6620.
Normally t
CLK
and t
SAMP
are the same unless a clock multiplier is
used such as with the AD6600’s 2× clock output.
Variable Latency is due to any differences between the asyn-
chronous edge of the SYNC pulses and the data rate. This
includes use of the internal synchronization options.
Based on the information on latency, the plots shown below
provide typical latency for a variety of different applications.
They were obtained by inserting a –F
S
dc step into the Input Data
Port of the AD6620. These are I channel step responses for the
input transient. The latency is defined as the output period times
number of output samples until the output reached approxi-
mately 50% of the step value.
OUTPUT SAMPLES
0.20
–0.40
–1.00
1213
FRACTION OF F
S
5791113151719
0.00
–0.20
–0.60
–0.80
23 25 27 29
61.44MHz
SAMPLE RATE
M
CIC2
= 16
M
CIC5
= 8
M
RCF
= 8
N
TAPS
= 256
AT 19 OUTPUT SAMPLES,
THE LATENCY WOULD
BE 0.32ms
EXPECTED LATENCY = 0.303ms
Figure 52. AMPS Example
OUTPUT SAMPLES
0.20
–0.40
1112
FRACTION OF F
S
345678910
0.00
–0.20
–0.60
–0.80
12 13 14 15
58.9824MHz
SAMPLE RATE
M
CIC2
= 2
M
CIC5
= 4
M
RCF
= 6
N
TAPS
= 48
AT 8 OUTPUT SAMPLES,
THE LATENCY WOULD
BE 6.51␮s
EXPECTED LATENCY = 6.31␮s
Figure 53. CDMA Example
OUTPUT SAMPLES
0.20
–0.40
1213
FRACTION OF F
S
5791113151719
0.00
–0.20
–0.60
–0.80
23 25 27 29
64.512MHz
SAMPLE RATE
M
CIC2
= 2
M
CIC5
= 14
M
RCF
= 3
N
TAPS
= 84
AT 19 OUTPUT SAMPLES,
THE LATENCY WOULD
BE 24.7␮s
EXPECTED LATENCY = 24.31␮s
Figure 54. PHS Example
OUTPUT SAMPLES
0.20
–0.40
1112
FRACTION OF F
S
345678910
0.00
–0.20
–0.60
–0.80
12 13 14 15
–1.00
65.0MHz
SAMPLE RATE
M
CIC2
= 2
M
CIC5
= 6
M
RCF
= 20
N
TAPS
= 240
AT 9 OUTPUT SAMPLES,
THE LATENCY WOULD
BE 33.23␮s
EXPECTED LATENCY = 31.2␮s
Figure 55. WB-GSM Example