COM20019 Datasheet, PDF (40/81 Pages) SMSC Corporation – Low Cost ARCNET (ANSI 878.1) Controller with 2K x 8 On-Board RAM

512 bytes is still guaranteed to be free because
Command Chaining only requires two pages for
transmit and two for receive (in this case, two 256
byte pages for transmit and two 512 byte pages
for receive, leaving 512 bytes free). Please note
that it is the responsibility of software to reserve
512 bytes for each receive page if the device is
configured to handle long packets. The
COM20019 does not check page boundaries
during reception. If the device is configured to
handle only short packets, then both transmit and
receive pages may be allocated as 256 bytes
long, freeing at least 1KByte at any given time.
Even if the Command Chaining operation is being
used, 1KByte is still guaranteed to be free
because Command Chaining only requires two
pages for transmit and two for receive (in this
case, a total of four 256 byte pages, leaving 1K
free).
The general rule which may be applied to
determine where in RAM a page begins is as
follows:
Address = (nn x 512) + (f x 256).
Transmit Sequence
During a transmit sequence, the microcontroller
selects a 256 or 512 byte segment of the RAM
buffer and writes into it. The appropriate buffer
size is specified in the "Define Configuration"
command. When long packets are enabled, the
COM20019 interprets the packet as either a long
or short packet, depending on whether the buffer
address 2 contains a zero or non-zero value. The
format of the buffer is shown in Figure 8. Address
0 contains the Source Identifier (SID); Address 1
contains the Destination Identifier (DID); Address
2 (COUNT) contains, for short packets, the value
256-N, where N represents the number of
information bytes in the message, or for long
packets, the value 0, indicating that it is indeed a
long packet. In the latter case, Address 3
(COUNT) would contain the value 512-N, where N
represents the number of information bytes in
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