COM20019IP datasheet(40/81 Pages) SMSC

40 / 81 page

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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

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

COM20019IP датащи(PDF) 40 Page - SMSC Corporation

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