Digitronic Digital Cam Switch Unit
Automationsanlagen GmbH CamCon DC50/51
12. Calculation the RAM storage-requirement for CamCon
The required RAM-main storage (not similar to the constand value - Camstorage or the EEPROM)
depends on 7 factors:
1. Standard consumption (approximately 100000 Byte)
2. Number of outputs ( 8 to 200 in steps of 8 outputs).
3. Cycle time (displayed in miliseconds).
4. Actual value/measuring system (displayed in impulses)
resolution
5. Maximal Speed-compensation ( 0 to 9999.9 in steps of 100 microseconds).
6. Mode of program selection (the double ammount of storage is required ).
( See also chapter "7.4.6.8. Setting the program selection
mode" on page 64).
7. Size of the EE-Prom storage ( EE-Prom - storage size in Byte for Cache ).
The RAM - storage requirement is calculated by the following formula:
storage requirement in Bytes = standard consumption +
Number of outputs * IActual value resolution.* ( 2 If program Mode not slow )
8
+
max. delay-time * 4
cycletime
+ EE-Prom size
Example 1: The camswitch with a resolution of 360°, an EE-Promstorage of 32kByte, 16 outputs, a
speed-compensation of 1000ms and a cycletime of 250µs needs:
Storage requirement in Bytes = 100000 +
16 * 360
8
+
1000 * 4
0.250
+ 32768
Storage requirement in Bytes = 100000 + 720 + 16000 + 32768
Storage requirement in Bytes = 149488 = ca. 150kByte
Example 2: The camswitch with a resolution of 8192°, an EE-Promstorage of 48kByte, 64 outputs,
a speed-compensation of 500ms and a cycletime of 250µs, needs:
Storage requirement in Bytes = 100000 +
64 * 8182
8
+
500 * 4
0.250
+ 49152
Storage requirement in Bytes = 100000 + 65536 + 8000 + 49152
Storage requirement in Bytes = 222688 = ca. 220kByte
Note: If the required RAM - storage requirement greater than the CamCons total ammount of
storage, you need to reduce the measuring system's resolution.
Attention: Changings in the storage-structure of the CamCon software the storage requirement can
alternate at different software-versions!
,Version from: Aug. 04 Page: 85
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