micrologix memory size
- Thread starter bwiring
- Start date
As most PLC's still assume that memory is allocated in BYTES the answer here is that 1 WORD
= 2 BYTES; 1 DWORD = 4 BYTES.In the classical sense 1k = 1024 (2^10) and 4k is 4096 bytes.
Hence 4k would hold 2048 WORDS and 1024 DWORDS.
This will not give how many instructions the memory can hold as different instructions will occupy n bytes each.
Hope this helps
= 2 BYTES; 1 DWORD = 4 BYTES.In the classical sense 1k = 1024 (2^10) and 4k is 4096 bytes.
Hence 4k would hold 2048 WORDS and 1024 DWORDS.
This will not give how many instructions the memory can hold as different instructions will occupy n bytes each.
Hope this helps
Mickey
Lifetime Supporting Member
Look under controller properties. See picture below
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