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I'm working on designing a simple "retro computer" based on the NXP 68SEC000. The "SEC" variant (by NXP) is essentially a Motorola 68000 with a few differences. Such as the inclusion of A0 and it's a static/CMOS version.

I'm also reading through the excellent Microprocessor System Design book by Alan Clements. This book does not directly cover the 68SEC000 as it covers the "generic" 68000 from Motorola.

In the book, it references a minimal computer that has 74LS244 bus drivers between the address pins and other devices. It also has 74LS245 bus transceivers between the data bus and other devices.

My understanding is that back then, many of these chips could not drive other components very well so these bus drivers were used to boost the outputs (forgive me, I'm a newbie when it comes to electrical engineering).

However, I have designed another computer in the past based on the 65C02 which doesn't need any drivers because it's modern, static and I assume just "stronger" in its ability to drive other components. This computer proved to be very reliable and I have not seen the need for any drivers on it (yet).

I've tried to find reference designs that use the newer 68SEC000 but I cannot find anything. My searches always leads me back to the generic Motorola 68000 (just like the book).

So my question is, is there any real benefit of using these drivers and transceivers on a new design that utilizes the 68SEC000? Or, does the very nature of the CMOS/static design negate the need for them?

Some points of interest is that my design uses all new and modern parts (except for the 68SEC000). My SRAM, UART's, etc. are all modern devices.

Thanks for any help.

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  • \$\begingroup\$ When I looked at a 68000 design from Convergent Technologies (CT), one of the greatest Silicon Valley computer companies to die in the 80's from Burroughs buyout, they improved clock speed and bus performance by adding active bus terminator resistors to match impedance to threshold voltage to double the speed, that Motorola specified. It was also my 1st personal computer that was amazing with CTOS and 5 programming languages of choice and a great word Processor "Write One". So learn how to measure and match impedances. e.g. Zol=Vol/Iol \$\endgroup\$ Jan 22 '20 at 17:18
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The solution to your answer lies in the datasheet for the chip you want to use. Look for \$Voh, Vol, Ioh, Iol\$. These characteristics tell you how the chip will drive the busses.

Next look at the chips you will be hanging on those busses and look at their complementary attributes \$Vih, Vil, Iih, Iil\$. Especially the \$Iil\$ and \$Iih\$.

Then add up all the \$Iil\$ for each chip on the bus and see if it is less than or equal to \$Iol\$ for the CPU. Then do the same for \$Iih\$ and compare the total to \$Ioh\$.

If the total input current for all the chips on the bus is less than what the CPU can provide, then you do not need drivers. If the total current is more than the CPU can provide then you do need drivers.

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  • \$\begingroup\$ Thanks! I don't know what all of those values mean but you've given me a clear-cut formula to research. That's the kind of information I was looking for! Although, my question may not have been worded as such. :-) \$\endgroup\$
    – cbmeeks
    Jan 22 '20 at 21:14
  • \$\begingroup\$ Looking in the datasheet...if running VCC at 5V, it says VOH is VCC - 0.75V. So does that mean as long as my SRAM (for example) has a IOH of anything LOWER than 4.25V, then I do NOT need any drivers? Is it that simple? \$\endgroup\$
    – cbmeeks
    Jan 22 '20 at 21:28
  • \$\begingroup\$ You're on the right track, but you need to compare I to I and V to V. You cannot compare VOH and IOH. \$\endgroup\$
    – Aaron
    Jan 23 '20 at 1:22

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