Refresh some Olimex LPC1766STK networking configurations
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@ -132,51 +132,8 @@
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*
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* 16384 <= ntx * (pktsize + 8 + 4) + nrx * (pktsize + 8 + 8)
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*
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* If ntx == nrx and pktsize == 424, then you could have
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* ntx = nrx = 18.
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*
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* An example with all of the details:
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*
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* NTXDESC=18 NRXDESC=18 CONFIG_NET_EMACRAM_SIZE=16Kb CONFIG_NET_ETH_MTU=420:
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* LPC17_TXDESCTAB_SIZE = 18*8 = 144
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* LPC17_TXSTATTAB_SIZE = 18*4 = 72
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* LPC17_TXTAB_SIZE = 216
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*
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* LPC17_RXDESCTAB_SIZE = 16*8 = 144
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* LPC17_RXSTATTAB_SIZE = 16*8 = 144
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* LPC17_TXTAB_SIZE = 288
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*
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* LPC17_DESCTAB_SIZE = 504
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* LPC17_DESC_BASE = LPC17_SRAM_BANK0 + 0x00004000 - 504
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* = LPC17_SRAM_BANK0 + 0x00003e08
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* LPC17_TXDESC_BASE = LPC17_SRAM_BANK0 + 0x00003e08
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* LPC17_TXSTAT_BASE = LPC17_SRAM_BANK0 + 0x00003e98
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* LPC17_RXDESC_BASE = LPC17_SRAM_BANK0 + 0x00003ee0
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* LPC17_RXSTAT_BASE = LPC17_SRAM_BANK0 + 0x00003f70
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*
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* LPC17_PKTMEM_BASE = LPC17_SRAM_BANK0
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* LPC17_PKTMEM_SIZE = 0x00004000-504 = 0x00003e40
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* LPC17_PKTMEM_END = LPC17_SRAM_BANK0 + 0x00003e08
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* LPC17_MAXPACKET_SIZE = ((420 + 3 + 2) & ~3) = 424
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* LPC17_NTXPKTS = 18
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* LPC17_NRXPKTS = 18
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* LPC17_TXBUFFER_SIZE = 18 * 424 = 0x00001dd0
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* LPC17_RXBUFFER_SIZE = 18 * 424 = 0x00001dd0
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* LPC17_BUFFER_SIZE = 0x00003ba0
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* LPC17_BUFFER_BASE = LPC17_SRAM_BANK0
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* LPC17_TXBUFFER_BASE = LPC17_SRAM_BANK0
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* LPC17_RXBUFFER_BASE = LPC17_SRAM_BANK0 + 0x00001dd0
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* LPC17_BUFFER_END = LPC17_SRAM_BANK0 + 0x00003ba0
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*
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* Then the check LPC17_BUFFER_END < LPC17_PKTMEM_END passes. The amount of
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* unused memory is small: 0x00003e08-0x00003ba0 or about 616 bytes -- not
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* enough for two more packets.
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*
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* [It is also possible, with some effort, to reclaim any unused
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* SRAM for the use in the heap. But that has not yet been pursued.]
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* If ntx == nrx and pktsize == 590, then you could have ntx = nrx = 13. In this
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* case, you would need only 15,704 bytes of EMAC RAM (but be careful with alignment!).
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*/
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#define LPC17_TXDESCTAB_SIZE (CONFIG_NET_NTXDESC*LPC17_TXDESC_SIZE)
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