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IBM 5150 - 10/27/82 BIOS - Problem of less than 4 banks of RAM

None of IBM's documentation on the IBM 5150 indicates that all 4 banks of motherboard RAM need to be populated. However, it is observed that with the 10/27/82 revision BIOS fitted to either the 16KB-64KB or 64KB-256KB versions of the 5150 motherboard, if all 4 motherboard RAM banks are not populated (and the motherboard switches set accordingly), then the power-on self test (POST) will miscalculate the amount of RAM.

This 10/27/82 revision BIOS requirement for 4 banks is due to two software bugs in the 10/27/82 revision BIOS. Details are below.

Note that this 10/27/82 revision BIOS requirement for 4 banks, if not met, does not prevent the motherboard's POST from starting and displaying something on-screen.

See note 1 at the bottom of this page for an example of misbehavior seen if the subject requirement is not met.

Bug #1 - In detail

This is a bug in the 10/27/82 revision of the 5150 BIOS, and this particular bug affects only the 64KB-256KB version of the 5150 motherboard.

The bug appears to come about because much code was copied from the early BIOS, as used on 16KB-64KB motherboards (up to 4 banks of 16 KB RAM), and inadequate testing done to verify that the code worked fully (repeat: fully) on the 64KB-256KB motherboard (up to 4 banks of 64 KB RAM).

Reference

BIOS source listing in the APR84 edition of the 5150 Technical Reference

Symptoms

When less than 4 banks of motherboard RAM are indicated by the switches in SW1/SW2, the bug results in the POST of the BIOS reporting much less RAM than it should.

Motherboard banks indicated via SW1	RAM in expansion cards	SW2 setting	RAM total reported by BIOS	Comment
0 (64 KB)	N/A	64 KB	16 KB	BAD - Should be 64 KB (1 x 64 KB bank)
0/1 (128 KB)	N/A	128 KB	32 KB	BAD - Should be 128 KB (2 x 64 KB bank)
0/1/2 (192 KB)	N/A	192 KB	48 KB	BAD - Should be 192 KB (3 x 64 KB bank)
0/1/2/3 (256 KB)	0 KB	256 KB	256 KB	GOOD (4 x 64 KB bank)
0/1/2/3 (256 KB)	64 KB	320 KB	320 KB	GOOD ([4 x 64 KB bank] + 64 KB)
0/1/2/3 (256 KB)	128 KB	384 KB	384 KB	GOOD ([4 x 64 KB bank] + 128 KB)

When this problem happens, the amount of BIOS reported RAM may be too little for certain versions of DOS. See here.

Where did I get the 'RAM total reported by BIOS' figures above?

The POST of the 5150 BIOS stores the total amount of RAM (in KB) that it believes is fitted, into the BIOS Data Area (BDA), specifically the word (little endian) at 0040:0013.
One way to read that amount is to:
1. Get into BASIC by removing any hard drive, and then boot without a boot floppy, then
2. In BASIC, enter the following:
def seg = &h40
print peek(&h14)*256 + peek(&h13)

Where in the code is the bug?

The block of code causing the problem is on page 5-42 of the reference, specifically lines 934 to 951. That block calculates the total KB amount of RAM (motherboard + expansion cards) and then stores the amount in the BDA.

You can see that the block I'm referring to is divided in to 2 sections: "DETERMINE RAM SIZE ON PLANAR BOARD" and "DETERMINE I0 CHANNEL RAM SIZE". Those descriptions are correct for a 16KB-64KB motherboard, but not a 64KB-256KB motherboard.

DETERMINE RAM SIZE ON PLANAR BOARD

Lines 935 to 941. This section of the block examines switches 3 and 4 on SW1. Irrespective of whether the 10/27/82 BIOS is on a 16KB-64KB motherboard or on a 64KB-256KB motherboard, the output of that section is always the same:
SW1-3 on / SW1-4 on (i.e. bank 0) - results in register BX containing 16 (0010h).
SW1-3 off / SW1-4 on (i.e. banks 0/1) - results in register BX containing 32 (0020h).
SW1-3 on / SW1-4 off (i.e. banks 0/1/2) - results in register BX containing 48 (0030h).
SW1-3 off / SW1-4 off (i.e. banks 0/1/2/3) - results in register BX containing 64 (0040h).

DETERMINE I0 CHANNEL RAM SIZE

Lines 945 to 951.

The first thing that this bit of code does (at line 945) is fetches the "IO_RAM_SIZE" value was calculated (based on SW-2 setting) and stored earlier (page 3-37). IO_RAM_SIZE contains the amount of RAM past 64 KB. On a 16KB-64KB motherboard, that corresponds to the amount of IO RAM (RAM on IO expansion cards), and so on that motherboard, "IO_RAM_SIZE" is an accurate description. But on a 64KB-256KB motherboard, "IO_RAM_SIZE" is an incorrect description. It is best to think of "IO_RAM_SIZE" as RAM past 64 KB.

Putting it altogether (lines 935 to 951), we end up with:

Motherboard banks indicated via SW1	RAM in expansion cards	SW2 setting	BX register	BX = 0040h ?	IO_RAM_SIZE (KB RAM past 64)	Calculated RAM total (MEMORY_SIZE)	Comment
0 (64 KB)	N/A	64 KB	16 (0010h)	No	not relevant	16 + 0 = 16 KB ( BX + 0 , because BX not 0040h )	BAD - 48K less than actually exists
0/1 (128 KB)	N/A	128 KB	32 (0020h)	No	not relevant	32 + 0 = 32 KB ( BX + 0 , because BX not 0040h )	BAD - 96K less than actually exists
0/1/2 (192 KB)	N/A	192 KB	48 (0030h)	No	not relevant	48 + 0 = 48 KB ( BX + 0 , because BX not 0040h )	BAD - 144K less than actually exists
0/1/2/3 (256 KB)	0 KB	256 KB	64 (0040h)	Yes	192 (00C0h)	64 + 192 = 256 KB ( BX + IO_RAM_SIZE )	GOOD
0/1/2/3 (256 KB)	64 KB	320 KB	64 (0040h)	Yes	256 (0100h)	64 + 256 = 320 KB ( BX + IO_RAM_SIZE )	GOOD
0/1/2/3 (256 KB)	128 KB	384 KB	64 (0040h)	Yes	320 (0140h)	64 + 320 = 384 KB ( BX + IO_RAM_SIZE )	GOOD

Bug #2 - In detail

This is a software bug in the 10/27/82 revision of the 5150 BIOS.
This particular bug affects both the 16KB-64KB 5150 motherboard (if the 10/27/82 BIOS is fitted) and the 64KB-256KB 5150 motherboard.

Reference

BIOS source listing in the APR84 edition of the 5150 Technical Reference

Symptoms

Mostly, when less than 4 banks of motherboard RAM are indicated by the switches in SW1/SW2, a 201 error (RAM error) is seen on screen.

Why did I use "mostly"?
Here is an example. 64KB-256KB motherboard.
SW1 and SW2 set for 1 bank of RAM only (64K), and only 1 bank is physically fitted - 201 error seen.
SW1 and SW2 set for 1 bank of RAM only (64K), but 4 banks are physically fitted - 201 error not seen.

Why?

Because of a software bug, if less than 4 banks of motherboard RAM are indicated via SW1/SW2, then the portion of the POST that does a data test of the RAM thinks that there is a lot more RAM fitted than indicated by SW1/SW2.
Here are some examples for a 64KB-256KB motherboard:

Motherboard banks indicated via SW1	RAM in expansion cards	SW2 setting	RAM test routine thinks this much RAM exists	Comment
0 (64 KB)	N/A	64 KB	192 KB	BAD - 128K more than actually exists
0/1 (128 KB)	N/A	128 KB	384 KB	BAD - 256K more than actually exists
0/1/2 (192 KB)	N/A	192 KB	448 KB	BAD - 256K more than actually exists
0/1/2/3 (256 KB)	0 KB	256 KB	256 KB	GOOD
0/1/2/3 (256 KB)	64 KB	320 KB	320 KB	GOOD
0/1/2/3 (256 KB)	128 KB	384 KB	384 KB	GOOD

Where in the code is the bug?

The block of code causing the problem is on page 5-37 of the reference, specifically lines 525 to 539, which is the majority of the section named "DETERMINE I0 CHANNEL RAM SIZE". The block is calculating something known as "IO_RAM_SIZE", and as explained in BUG #1, IO_RAM_SIZE is the amount of RAM past 64 KB.

This block is a modified form of the one in the earlier BIOS. It is modified because this BIOS now reads switches #1 to #5 on SW2. The earlier revisions of BIOS only went as high as switch #4. In modifying the code, the programmer has introduced a bug. The bug results in wrong values of IO_RAM_SIZE being calculated when less than 4 banks of motherboard RAM are indicated via SW1/SW2.

Here are some examples for a 64KB-256KB motherboard:

Motherboard banks indicated via SW1	RAM in expansion cards	SW2 setting	IO_RAM_SIZE (KB RAM past 64)	RAM test routine thinks this much RAM exists	Comment
0 (64 KB)	N/A	64 KB	128 KB	128 + 64 = 192 KB	BAD
0/1 (128 KB)	N/A	128 KB	320 KB	320 + 64 = 384 KB	BAD
0/1/2 (192 KB)	N/A	192 KB	384 KB	384 + 64 = 448 KB	BAD
0/1/2/3 (256 KB)	0 KB	256 KB	192 KB	192 + 64 = 256 KB	GOOD
0/1/2/3 (256 KB)	64 KB	320 KB	256 KB	256 + 64 = 320 KB	GOOD
0/1/2/3 (256 KB)	128 KB	384 KB	320 KB	320 + 64 = 384 KB	GOOD

By the way. It is relatively easy for you to see the IO_RAM_SIZE figure for yourself. That is because the POST of the 5150 BIOS stores the IO_RAM_SIZE figure that it calculated into the BIOS Data Area (BDA), specifically the word (little endian) at 0040:0015.
One way to read that amount is to:
1. Get into BASIC by removing any hard drive, and then boot without a boot floppy, then
2. In BASIC, enter the following:
def seg = &h40
print peek(&h16)*256 + peek(&h15)

Where did the programmer go wrong?

If you examine the block of code (lines 525 to 539) carefully, you will discover that the block, by accident no doubt, requires that the upper nibble in register AH be zero before the block is executed.
A non-zero nibble corrupts the calculated value of IO_RAM_SIZE

So what's before the block that affects the upper nibble in register AH?
Another block, lines 511 to 521, in which register AH will end up as:

Motherboard banks indicated via SW1	AH register	Non-zero upper nibble?
0	64 (40h)	Yes
0/1	128 (80h)	Yes
0/1/2	192 (C0h)	Yes
0/1/2/3	0 (00h)	No

Note 1

For example, I have a working IBM 5150 motherboard of type 64B-256KB (i.e. fitted with 10/27/82 BIOS, and 4 banks of RAM).
There are no RAM cards fitted.
If I change the RAM related switches on the motherboard to what follows, I will see a '4098 201' RAM error at power-up.
SW1: 3=OFF, 4=ON (indicating two banks)
SW2: 1=ON, 2=OFF, 3=ON, 4=ON, 5=ON (indicating a total of 128K conventional RAM)

In this particular case, bug #2 is taking effect, making the POST believe that there is a total of 384 KB of RAM.
1. The POST successfully tests up to 256 KB (the four populated banks on the 64B-256KB motherboard), then
2. When the POST tests the next address, it fails (no RAM there) and displays the '4098 201' RAM error.