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TNETX3150 数据表(PDF) 42 Page - Texas Instruments

部件名 TNETX3150
功能描述  ADDRESS-LOOKUP DEVICE
PDF  78 Pages
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制造商  TI [Texas Instruments]
网页  http://www.ti.com
标志 TI - Texas Instruments

TNETX3150 数据表(HTML) 42 Page - Texas Instruments

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TNETX15AE
ADDRESS-LOOKUP DEVICE
SPWS041A – AUGUST 1997 – REVISED OCTOBER 1997
42
POST OFFICE BOX 655303
DALLAS, TEXAS 75265
PRINCIPLES OF OPERATION
a graphical example of the lookup algorithm (continued)
Table
Table
Table
Table
Table
Table
Table
Figure 9. Address Distribution Table Hierarchy (Worst Case)
Another side effect of this algorithm is that the table order is already sorted (smallest to largest).
The choice of N (the number of bits to consider each cycle) must be an integer and is a balance of:
1.
The size of each table = 2N words; larger means more wasted space per lightly used table, and smaller
means less wasted space per lightly used table.
2.
The clocks required to process a whole address (in this case 48 IEEE addresses) = 48
÷ N rounded up to
the next integer; considering more bits at a time decodes 48 bits in less time (considering fewer at a time
takes more time). The upper limit is 10 or 11 clocks.
SRAM data storage
lookup table SRAM allocation
The TNETX15AE uses a 5-bit version of the lookup algorithm described in the previous section.
When a 5-bit version of the lookup algorithm is used, each address requires ten tables to store a 48-bit value.
Tables 0 through 9 are used to match the 48 bits involved in the lookup. Table 9 (shown in Figure 10) contains
the actual data describing an address.
The external SRAM must be a minimum of eight bits wide, so address descriptor information (see
findport and
f
indnodeage register definitions) can be stored in four locations as bytes. Starting at 16K, the external SRAM
must get wider so that a pointer of installed word width (when five bits are appended) gives a pointer with enough
bits to address any memory location. Both address descriptor blocks and pointer tables must be contiguous,
and both are allocated dynamically. The table is full when the next structure being allocated does not fit. Several
addresses may have to be deleted on a table-full condition to make room for a new address.
An allocated (but unused) table entry is set to 0 since zero cannot be a valid address for a table leaf.



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