Topic: Hacking the E-Tech ADRT04, need small informations (IC)

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Post #1

DDave

5 Dec 2008, 14:43

Hello everyone, I've been checking out my router lately. It's an E-Tech ADRT04.
It is based on the Texas Instruments (tnetd7300agdw), so it's an AR7.
PROBLEM SOLVED

I have opened the case and looked at the IC's and there are two IC's where I'm not too sure how much memory it has.
Mira P2V64S40ETP-G6 59AE6VW :
64mb SDRAM? This confuses me.. That much?? On the official homepage there isn't any smaller SDRAM than 64mb!
http://www.deutron.com.tw/data_sheets/sdram/64mb_sd.pdf
MX29LV160CBTC-70G :
2mb FLASH ROM (16M-BIT [2Mx8/1Mx16] ): http://pdf1.alldatasheet.com/datasheet- … LV160.html

This is what confuses me..2mb ROM and 64mb RAM? xD
EDIT: 8mb RAM
Here are some informations...Might help someone..If you need anything else. Just reply.

# cat /proc/version
Linux version 2.4.17_mvl21-malta-mips_fp_le (root@localhost.localdomain) (gcc version 2.95.3 20010315 (release/MontaVista)) #1 Tue Jun 13 14:17:31 CST 2006
#
# cat /proc/meminfo
        total:    used:    free:  shared: buffers:  cached:
Mem:   6524928  6311936   212992        0   327680  1470464
Swap:        0        0        0
MemTotal:         6372 kB
MemFree:           208 kB
MemShared:           0 kB
Buffers:           320 kB
Cached:           1436 kB
SwapCached:          0 kB
Active:           2172 kB
Inactive:         1432 kB
HighTotal:           0 kB
HighFree:            0 kB
LowTotal:         6372 kB
LowFree:           208 kB
SwapTotal:           0 kB
SwapFree:            0 kB
#
# cat /proc/iomem
00000000-13ffffff : reserved
14000000-1401ffff : System RAM
14020000-147fffff : System RAM
  14020000-1419464f : Kernel code
  141a5300-141bdfff : Kernel data
a8612800-a8612fff : eth0
#
# cat /proc/mtd
dev:    size   erasesize  name
mtd0: 00169000 00010000 "mtd0"
mtd1: 00076f70 00010000 "mtd1"
mtd2: 00010000 00008000 "mtd2"
mtd3: 00010000 00010000 "mtd3"
mtd4: 001e0000 00010000 "mtd4"
#

Edit: Confirmed 2mb ROM
http://pdf1.alldatasheet.com/datasheet- … LV160.html

(Last edited by DDave on 6 Dec 2008, 00:13)

Post #2

lizby

6 Dec 2008, 00:06

64 megabits, not bytes--8 megabytes.

Post #3

DDave

6 Dec 2008, 00:12

lizby wrote:

64 megabits, not bytes--8 megabytes.

Thank you, with 2mb ROM and 8mb RAM it isn't very interesting for Openwrt..

Post #4

mazilo

6 Dec 2008, 01:36

DDave wrote:

lizby wrote:
64 megabits, not bytes--8 megabytes.
Thank you, with 2mb ROM and 8mb RAM it isn't very interesting for Openwrt..

2mb = 2mb * 1MByte / 8mb * 1024 KByte/MByte = 256KByte.
8mb = 8mb * 1MByte / 8mb = 1MByte

So, with 256KByte ROM and 1MByte RAM, that does not seem to be possible to run OpenWRT.

Post #5

DDave

6 Dec 2008, 09:19

mazilo wrote:

DDave wrote:
lizby wrote:
64 megabits, not bytes--8 megabytes.
Thank you, with 2mb ROM and 8mb RAM it isn't very interesting for Openwrt..
2mb = 2mb * 1MByte / 8mb * 1024 KByte/MByte = 256KByte.
8mb = 8mb * 1MByte / 8mb = 1MByte
So, with 256KByte ROM and 1MByte RAM, that does not seem to be possible to run OpenWRT.

Ouch, so it only has 256Kbytes ROM and 1Mbyte RAM...even worse...

Post #6

mazilo

6 Dec 2008, 19:10

DDave wrote:

mazilo wrote:
DDave wrote:
Thank you, with 2mb ROM and 8mb RAM it isn't very interesting for Openwrt..
2mb = 2mb * 1MByte / 8mb * 1024 KByte/MByte = 256KByte.
8mb = 8mb * 1MByte / 8mb = 1MByte
So, with 256KByte ROM and 1MByte RAM, that does not seem to be possible to run OpenWRT.
Ouch, so it only has 256Kbytes ROM and 1Mbyte RAM...even worse...

What I tried to say is a 1MB = 8mb, where B is a Byte and b is a bit.

Post #7

DDave

6 Dec 2008, 19:15

Duh sorry, didn't see you were calculating it. My bad..
I'll update the wiki page then.
Thanks.

The discussion might have continued from here.

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