Amiga memory bandwidth

Question

Looking at a timing diagram for the various kinds of memory access occurring on the Amiga http://amigadev.elowar.com/read/ADCD_2.1/Hardware_Manual_guide/node02D4.html it seems to be saying that video data is being fetched at the rate of one word per color clock = 280 ns.

However, my understanding is that early eighties RAM chips typically took 375 ns for a full access cycle. The amortized access time can be reduced by using page mode, but the timing diagram doesn't seem to suggest this is happening.

What am I missing?

Answer 1

The Amiga A500 used 41256-15 DRAM chips for its onboard memory. According to the datasheet these had a cycle time of 260ns, so could easily match the 280ns timing constraint with 20ns to spare (although the board had two banks of them, and switching between the banks likely used up most of that 20ns).

The 375ns limit really only applied to the 4116 and 4164 type chips that were manufactured in huge bulk and were extremely cheap. Faster chips were available even back to the early 80s, but cost more money so designers of low-end micros tended to avoid using them where possible (cf the 4816A chips in the BBC micro we discussed previously, which were necessary in order to let a 6502 run at 2MHz with capacity left for video access between the CPU's accesses), especially when working with either 1MHz 6502 or 4MHz Z80 designs where faster RAM would have been useless.

Answer 2

To understand how Amiga memory bandwith should be calculated and how data fetching to Amiga Chipram works.

The timingslot schematic that rwallace refering to does not tell anything about how individual chipram ic' operates or are supposed to operate; it just explains how the Amiga timing system is layed out.

The Amiga is a synchronized multiprocessor platform/architecture. It share same memory to some extension. That shared memory is called chipram. The shared memory require some rules to be followed to make it work and to avoid something called bus contention. Bus contention occurs when more than one processor want access to the common address-, data-, or registerbus.

To solve this Amiga has a nested (interleaved) access to the buses (and some other solutions too). It means that the bus access is divided into odd and even timeslots. The whole timeslot concept is based on duration of one video raster-scanline, and that one timslot has a duration of approximatly 280ns. Why 280ns? Because it's the speed of the color clock driving the amiga chipset (Agnus, Paula, and Denise. Agnus contains the DMA controller responsible for memory access to chipram).

1000 000 000 / 3546895hz (colorclock) = 281.9367ns (PAL)

Due to the interleave approach to solve the bus contention problem the individual processors does not fetch data from chip memory with a bandwidth of 7mb/s but instead half of that amount - 3.5mb/s. Now this it not the whole truth. In graphic intensive tasks the amiga chipset can take full control of the buses and run at full bandwidth (7mb/s) at the expense of the cpu not having any access at all (no timeslots to its disposal). The DMA sytem can also be turned off and let the cpu have full access and do all the work (every other timeslot that is). Note: horisontal and vertical blanking period is not included in these situations.

To calculate the memory bandwidth for the cpu (mc68000):

Following calculations applies to OCS/ECS (16-bit) chipset:

1. Master clock (MC): 28.37516Mhz (PAL), 28.63636Mhz (NTSC)
2. Color ClocK (CCK): MC/8 = 3.546895Mhz (PAL), 3.579545Mhz (NTSC)
3. CPU speed is (MC/4): 7.093790Mhz (PAL), 7.159090Mhz (NTSC)
4. It takes 4 cpu cycles for main cpu (Motorola 68000) to fetch data from external memory (Fast or Chip mem) -->

7.093790Mhz / 4 = 1.773448Mhz (PAL),

7.159090Mhz / 4 = 1.789773Mhz (NTSC)

5. Bandwidth:

1.773448Mhz * 2byte = 3.546895mb/s (PAL)

1.789773Mhz * 2byte = 3.579545mb/s (NTSC)

It actually takes 560ns - two time slots (one time slot being 280ns) for cpu to fetch data from memory (Fast OR Chip mem).

An example:

(PAL)

• 1000 000 000 / 7093790Mhz = 140.96837ns
• 140.96837ns * 4 cpu clockcycles = 563.87347ns

How much is the different clock-speeds in terms of nanoseconds (ns)?

Always divide the (Mhz) by 1000 000 000 (1000 000 000 / Mhz) and you will get ns. When you read the inscription on memory ic's the access speed is written in nanoseconds e.g. 41256-15 means 150ns (access time).

To increase bandwidth to chip memory three 'rules' must be fulfilled:

1. Cpu clock need to be synchronized with masterclock.
2. Chip memory need to be fast enough to handle the new clockspeed.
3. The cpu clock need to be at least four times as fast, i.e. at the same speed as masterclock to double the bandwidth. The synchronization require cpu clock to be a multiple of masterclock.

Why does cpu need to be four times as fast to increase the bandwith? Because of the interleaved approach to chipram between cpu and amiga chipset; this together with the fact that - as mentioned above - it takes 4 cpu clock cycles to fetch data from chipram (and fastram). In other words; cpu access chipram (and chipregisters) every other timeslot and therefore it needs to be quadrupled in clockspeed.

NOTE: This will only increase bandwidth between cpu <-> chip-ram and not the speed between Amiga Chipset <-> chip-ram.

Answer 3

A500 accesses its chip 16bit memory at 3.5MHz rate, therefore the bandwidth is 3.5*2=7Mbytes/second.

In A1200 there is a page mode optionally used to access 2 words instead of one, and the memory itself is 32bit, while the access rate is the same. Therefore the bandwidth is 28Mbytes/second.