SAK-C164CI-8EM Infineon Technologies, SAK-C164CI-8EM Datasheet - Page 56

16BIT MCU, 64K OTP, 4K RAM, CAN, 20MHZ

SAK-C164CI-8EM

Manufacturer Part Number
SAK-C164CI-8EM
Description
16BIT MCU, 64K OTP, 4K RAM, CAN, 20MHZ
Manufacturer
Infineon Technologies
Datasheet

Specifications of SAK-C164CI-8EM

No. Of I/o's
59
Ram Memory Size
2KB
Cpu Speed
20MHz
No. Of Timers
5
Digital Ic Case Style
MQFP
Supply Voltage
RoHS Compliant
Core Size
16bit
Program Memory Size
64KB
Oscillator Type
External, Internal
Controller Family/series
C164CI
Rohs Compliant
Yes

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Due to this adaptation to the input clock the frequency of
it is locked to
duration of individual TCLs.
The timings listed in the AC Characteristics that refer to TCLs therefore must be
calculated using the minimum TCL that is possible under the respective circumstances.
The actual minimum value for TCL depends on the jitter of the PLL. As the PLL is
constantly adjusting its output frequency so it corresponds to the applied input frequency
(crystal or oscillator) the relative deviation for periods of more than one TCL is lower than
for one single TCL (see formula and
For a period of
deviation D
where
So for a period of 3 TCLs @ 25 MHz (i.e.
and (3TCL)
This is especially important for bus cycles using waitstates and e.g. for the operation of
timers, serial interfaces, etc. For all slower operations and longer periods (e.g. pulse train
generation or measurement, lower baudrates, etc.) the deviation caused by the PLL jitter
is neglectible.
Note: For all periods longer than 40 TCL the N = 40 value can be used (see
Figure 12
Data Sheet
Max. jitter
(
N
N
±26.5
× TCL)
±30
±20
±10
= number of consecutive TCLs and 1 ≤
±1
D
ns
N
N
min
:
1
Approximated Maximum Accumulated PLL Jitter
f
= 3TCL
min
OSC
This approximated formula is valid for
1
N
< –
=
. The slight variation causes a jitter of
× TCL the minimum value is computed using the corresponding
N
N
< –
NOM
× TCL
40 and 10 MHz
10
- 1.288 ns = 58.7 ns (@
NOM
- D
< –
Figure
N
f
CPU
; D
20
N
< –
N
52
25 MHz.
[ns] = ±(13.3 +
= 3): D
12).
N
3
≤ 40.
f
CPU
= (13.3 +
30
= 25 MHz).
f
CPU
N
f
CPU
× 6.3)/
3
is constantly adjusted so
× 6.3)/25 = 1.288 ns,
which also effects the
f
CPU
40
[MHz],
C164CL/SL
V2.0, 2001-05
C164CI/SI
Figure
MCD04455
10 MHz
16 MHz
20 MHz
25 MHz
12).
N

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