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From the blog: > However, the 8-bit bus enabled cheaper computer hardware. I wonder if that can be expanded on. An old (now departed) friend once said to me t

From the blog: > However, the 8-bit bus enabled cheaper computer hardware. I wonder if that can be expanded on. An old (now departed) friend once said to me that this was a mirage, because the only thing that the narrower bus really bought them was that it made the 16kB configuration possible, which no-one actually bought (the minimum for using floppies was 32kB!). He claimed that the narrower bus didn't actually make the configurations with more RAM cheaper because it made them require more support chips. Is there any truth to this?

gary_0 3y ago
It's mentioned at the bottom of this article[0] that Intel had an obsession with using the smallest packaging and pin count possible, while competitors did not see any need to be that conservative. I wonder if that has something to do with it. [0] https://www.righto.com/2023/07/8086-pins.html https://www.righto.com/2023/07/8086-pins.html
- phire 3y ago
  The 8088 and 8086 actually have the same number of pins and almost identical pinouts, despite the the difference in data bus width. This is because the data and address are multiplexed onto the same pins, because of intel's obsession with lowest pincount (each pin does add quite a bit of cost to packaging). On the 8086, the bottom 16 address pins (of 20) are multiplexed with data, on the 8088, only the bottom 8 address pins are multiplexed. So you had to demultiplex the bus no matter which chip you chose. I think the cost savings mostly come from being able to configure systems with only 8 DRAM chips per bank (instead of 16 DRAM chips per bank with the 8086). A 16 bit bus also requires that your ROM chips are in pairs, and it probably increases the motherboard routing complexity. And a small bit of extra decoding logic. A 16 bit bus would have also required IBM to skip straight over the 8 bit ISA standard with it's smaller sockets and forced up the complexity of all PC expansion cards (which would also require double the ROM chips, double the RAM chips and extra logic) Edit: And now that I think about it, the fact that the upper bit of the address bus aren't multiplexed might actually allow you to simplify the DRAM row/column addressing logic... But only if you put the column into the upper bits of the address.
kens 3y ago
According to Dave Bradley, one of the creators of the IBM PC, "We chose the 8088 [over the 8086] because of its 8-bit data bus. The smaller bus saved money in the areas of RAM, ROM, and logic for the simple system." Another thing is that even if almost nobody bought the minimal RAM configuration, having a low-cost configuration can be very important from a marketing standpoint. (By the way, it's kind of amazing that the base RAM for an IBM PC was just 16 kilobytes, and now 16 gigabytes is a base RAM configuration.) https://www.tech-insider.org/personal-computers/research/1990/09.html https://www.tech-insider.org/personal-computers/research/199...
adrian_b 3y ago
The cost saving was more a marketing trick. Most of the savings were not from memory but from the reuse of the existing 8-bit peripherals without additional hardware and without software changes. On a 16-bit bus, you could connect an 8-bit peripheral to one of its halves, but then the internal registers that previously were at consecutive addresses now were spread at multiples of 2. When doing only 8-bit transfers, you could rewrite the software drivers to use the modified addresses. However you could not do 16-bit transfers, because the bytes were no longer in the same word. This could be fixed with additional buffers, decoders and registers, to convert 16-bit transfers into pairs of 8-bit transfers on the same bus half (like 8088 did internally), but that would increase the cost. Even if these compatibility problems were not too difficult to solve, most preferred to avoid them, in the quest for minimum cost.

gary_0 3y ago

It's mentioned at the bottom of this article[0] that Intel had an obsession with using the smallest packaging and pin count possible, while competitors did not see any need to be that conservative. I wonder if that has something to do with it. [0] https://www.righto.com/2023/07/8086-pins.html https://www.righto.com/2023/07/8086-pins.html

phire 3y ago
The 8088 and 8086 actually have the same number of pins and almost identical pinouts, despite the the difference in data bus width. This is because the data and address are multiplexed onto the same pins, because of intel's obsession with lowest pincount (each pin does add quite a bit of cost to packaging). On the 8086, the bottom 16 address pins (of 20) are multiplexed with data, on the 8088, only the bottom 8 address pins are multiplexed. So you had to demultiplex the bus no matter which chip you chose. I think the cost savings mostly come from being able to configure systems with only 8 DRAM chips per bank (instead of 16 DRAM chips per bank with the 8086). A 16 bit bus also requires that your ROM chips are in pairs, and it probably increases the motherboard routing complexity. And a small bit of extra decoding logic. A 16 bit bus would have also required IBM to skip straight over the 8 bit ISA standard with it's smaller sockets and forced up the complexity of all PC expansion cards (which would also require double the ROM chips, double the RAM chips and extra logic) Edit: And now that I think about it, the fact that the upper bit of the address bus aren't multiplexed might actually allow you to simplify the DRAM row/column addressing logic... But only if you put the column into the upper bits of the address.

kens 3y ago

According to Dave Bradley, one of the creators of the IBM PC, "We chose the 8088 [over the 8086] because of its 8-bit data bus. The smaller bus saved money in the areas of RAM, ROM, and logic for the simple system." Another thing is that even if almost nobody bought the minimal RAM configuration, having a low-cost configuration can be very important from a marketing standpoint. (By the way, it's kind of amazing that the base RAM for an IBM PC was just 16 kilobytes, and now 16 gigabytes is a base RAM configuration.) https://www.tech-insider.org/personal-computers/research/1990/09.html https://www.tech-insider.org/personal-computers/research/199...

adrian_b 3y ago

The cost saving was more a marketing trick. Most of the savings were not from memory but from the reuse of the existing 8-bit peripherals without additional hardware and without software changes. On a 16-bit bus, you could connect an 8-bit peripheral to one of its halves, but then the internal registers that previously were at consecutive addresses now were spread at multiples of 2. When doing only 8-bit transfers, you could rewrite the software drivers to use the modified addresses. However you could not do 16-bit transfers, because the bytes were no longer in the same word. This could be fixed with additional buffers, decoders and registers, to convert 16-bit transfers into pairs of 8-bit transfers on the same bus half (like 8088 did internally), but that would increase the cost. Even if these compatibility problems were not too difficult to solve, most preferred to avoid them, in the quest for minimum cost.