What information does BIOS load into RAM?

Question

I know that, on booting, BIOS loads the first sector (512 bytes) of a pre-defined device drive on memory 0x7c00 and then jump to that address.

So, memory from 0x7c00 to

Answer 1

The x86 Real Mode Memory Map is as follows:

 - 0x00000000 - 0x000003FF - Real Mode Interrupt Vector Table
 - 0x00000400 - 0x000004FF - BIOS Data Area
 - 0x00000500 - 0x00007BFF - Unused
 - 0x00007C00 - 0x00007DFF - Our Bootloader
 - 0x00007E00 - 0x0009FFFF - Unused
 - 0x000A0000 - 0x000BFFFF - Video RAM (VRAM) Memory
 - 0x000B0000 - 0x000B7777 - Monochrome Video Memory
 - 0x000B8000 - 0x000BFFFF - Color Video Memory
 - 0x000C0000 - 0x000C7FFF - Video ROM BIOS
 - 0x000C8000 - 0x000EFFFF - BIOS Shadow Area
 - 0x000F0000 - 0x000FFFFF - System BIOS

In my real mode programming I usually stick from 0x00007E00 - 0x0009FFFF (Not all of it).. I use segment:offset addressing to use the memory.. To go from a 1 Stage bootloader to a Kernel or a Bootloader 2nd Stage, I use:

; bootloader.s

BITS  16
ORG   0x7C00

  CLI
  JMP 0xE000      ; Can also be JMP 0x7C00:200
  HLT

TIMES 510 - ($-$$) DB 0
DW 0xAA55

--

; Something.s

BITS  16
ORG   0x7E00      ; Can also be ORG   0x7C00:200

; Code goes here for your purposes.. whether it be a 2nd stage
; bootloader or your 16bit kernel..

CLI
HLT

If you are going into Protected mode, you will still need a stub as shown above.. In Something.s you can program in your protected mode routines (GDT, A20, Set video mode, etc..)

To explain about the memory location at 0x7C00 (Bootloader Entry Point), 0x7C00 - 0x7DFF is where you place your bootloader (the bootloader.s above). You place it there because the BIOS jumps to that location after doing its routines. The bootloader must be exactly 512 bytes in size (notice the TIMES directive). From there, your code can be any size (as long as it fits in the memory map), and you will be able to work on the OS fully.

If you DO go into 32Bit Protected Mode, you will be able to use ANYTHING about the 1MiB mark.