Config-Register

Answered Question

I have been going through the config-register and I know the following to be true:



0x210X = 9600 baud

0x212X = 19200 baud

0x312X = 57600 baud


0xXXX0 = ROM

0xXXX1 = Bootstrap

0xXXX2 = Normal


The problem is the ignore value of 0x2142.


My question is "Does anyone know why it is 0x2142?"  If 2 is normal NVRAM then does that mean that the prefix of 0x214X is the ignore value?  What is special about it?  I am just really curious about how this value was selected to be the ignore value.  If anyone knows of any resources about config-register values and how they relate please let me know.

Correct Answer by Phillip Remaker about 6 years 10 months ago

0xXX2X = bit 5 set = 19200

0xX8XX = bit 11 set = 4800

0x1XXX = bit 12 set = 1200


See also http://www.cisco.com/en/US/docs/routers/access/1800/1841/software/configuration/guide/b_creg.html, Table 4.


When all bits are cleared (0) it is the default rate of 9600.

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In that document I see the following:

5,11,12

0x0020, 0x0800, 0x1000

  • Console line speed



However it says that 0x2902 has a baud rate of 4800.  So I broke up the hex into the following:


0x0100 = no break

0x0800 = speed/baud rate

0x2000 = boots rom on failure

0x0002 = normal boot

-----------

0x2902


This tells me that 0x0800 is a 4800 baud rate, but that doesn't match what I was expecting.


The only explaination I can think of is the following:


if speed = !

     speed = 9600

elseif speed = 0x0800

     speed = 4800



or something very close to this.  Without me crunching the numbers to verify does anyone know what baud rates you get from 0x0020, 0x0800, and 0x1000?

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