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Question about DRAM IC's

I'm in the process of redesigning a RAM expansion board for an old Apple IIe series computer to help me better understand how electronics work. Some of you may remember the Apple IIe from back in the 80's. I've found it's a great platform to learn on. Anyway, the RAM expansion board uses 1 Meg x 1 bit DRAM's and I want to modify the board to use 1 Meg x 4 bit DRAM's. Just a little homebrew project to further my knowledge of DRAM IC's and the circuits that drive them.

Here's the datasheet for the 1 Meg x 1 bit DRAM's:

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Here's the datasheet for the 1 Meg x 4 bit DRAM's:

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A friend of mine has been helping me a bit along the way too. After we studied the two data sheets we noticed that the 4 bit DRAM has an Output Enable pin which the 1 bit DRAM does not. It would seen to me that if I invert the Write signal with a 74LS05 then I could use it to drive the Output Enable pin. So I wired the 74LS05 in to the circuit . I then wired up the rest of the signals accordingly. I took IO1 and connected it to where the first 1 bit DRAM I/O would be, IO2 and connected it to where the second 1 bit DRAM I/O would be, and so on. On my board the 1 bit DRAM's have pin 1 and 17 wired together. I then have the four remaining sockets in the bank of eight DRAM's filled with 1 bit DRAM IC's.

Next I insert my board in to the computer, run my RAM test and... it fails. The program tests each DRAM separately, so the 1 bit DRAM's pass the test but the 4 bit DRAM fails. I've even tried several different brands of 4 bit DRAM's too. I can't seem to see a reason why it would fail, at least not according to the data sheets.

Does anyone have any experience with what I'm doing or maybe can see something I'm missing from the data sheets? As you can probably guess I'm a little stumped and have been trying all sort of things for the past couple weeks. Any help would or guidance would be appreciated.

Henry

Reply to
Henry
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"Henry" schreef in bericht news:-62dnUtYgZofIznZnZ2dnUVZ snipped-for-privacy@comcast.com...

Hmm... Can't say too much without schematic but inverting /WR to achieve /OE may be too simple. Print out and compare the waveforms of the datasheets. There are a lot of possibilities and you need to know which ones are actually used. So what kind of read, write and refresh methods are implemented on the RAM-board? As for the refresh make sure the board provides a nine bits refresh address.

petrus bitbyter

Reply to
petrus bitbyter

Hey guys. Thanks for the reply.

petrus bitbyter: I did compare the waveforms of the datasheets and they do match up, or I should say they match to my eye. The board I'm working with is actually an expansion board to another card. I guess I'll have to draw the schematic to the main card to really understand how and what the main card is doing and what it expects from the daughter board. Thanks for the advice!

Helmut Sennewald: Concerning the refresh cycles - If I'm looking in the right place on the data sheets I see the 1bit DRAM requires 512 cycles every

8 ms and the 4bit DRAM requires 1024 cycles every 16 ms. Isn't this the same thing or am I confusing something? Thanks.

Henry

Reply to
Henry

The problem is that two lots of 512 cycles, do not activate the extra address line, that a '1024' cycle refresh implies...

Best Wishes

Reply to
Roger Hamlett

Hello Roger.

I'm not sure I understand what you mean about 'The problem is that two lots of 512 cycles, do not activate the extra address line, that a '1024' cycle refresh implies...'.

Well, from what I believe is going on my card is set to refresh 512 address lines every 8 milliseconds - or at least that is what the data sheet for the

1bit DRAM says it needs. Since the 1bit DRAM's work I will assume those are the numbers then. The DRAM I believe has a built in counter to advance the CAS address lines automatically from the LAST line it had accessed in refresh mode - so you don't need to apply all the possible address to the DRAM and eat up CPU time. Now if the DRAM is just idle and not being accessed then the current scheme will work just fine - i.e. - 512 refreshes in 8ms, 2 cycles of 512 = 1024 in 16ms. So that's good, or so I would think.

Now what happens when the DRAM is at line 0 and the refresh kicks in? Of course it receive 512 cycles and the counter will now be on line 511. Now I access the DRAM in between the first 512 refresh and the second refresh and read or write data to line 0. The refresh then kicks in again after my read or write and advances the counter from the LAST line I accessed which was 0, to 511, thus missing the other 512 lines that still need to be refreshed. Now what if I access line 0 again just after that refresh and then the refresh kicks in again?

I think this is what both of you have been talking about. Where if the board was designed to refresh 1024 times per refresh cycle then it's guarantied that every line of the array would be refreshed no matter what line in the array was last accessed. Am I following correctly?

On another side test I tried a RAM board that used 256k x 1bit DRAM's. I replaced 4 of them with a 256k x 4bit DRAM. Same type of issue as with the

1Meg DRAM's. Also the same refresh numbers issue as the 1 Meg DRAM's - that the 4bit DRAM refresh cycles are double that of the 1bit DRAM. It's the only thing that I can see is really that different between the two datasheets. In every case that I can see the 4bit DRAM specs are well within operating in a 1bit design - all but the refresh cycle numbers. As for the technical stuff, the 2 timing values that are relevant, that I see, are tCSR and tCHR. On the 1Mx1 chips, their min values are 10 and 15, respectively. With the 1Mx4 chips, I see 5 and 15. Looks to me that the 1Mx4 chips are just as compatible (note: there's no value listed as in the max column for either chip). Even the tRPC value, which I'm not sure is a factor, is a min of 10 on the x1 and a 0 on the x4. The only real difference seems to be the refresh cycles which I now assume correspond directly to the memory array in the DRAM IC.

I also believe my board to be doing /CAS-before-/RAS refresh but I can't say for sure. Any good 'test' to see if my assumption is correct? Wish I had a scope.

Any clarification to any of my thoughts/statements is always appreciated.

Henry

Reply to
Henry

Your first problem is that you've cross-posted to a slew of groups. I have very little desire to reply to some fool who feels they need to do that.

But not only do you overly cross-post, but you lack comprehension.

Inside the dynamic ram is an array of rows and columns. The refresh is to activate each location in turn, because dynamic memory is not static. It needs to be access in order to keep it's content. Since it's an array, you don't need to go through the full address space, just whatever is specified, the row or column.

To use a simple example, say it's a really small RAM, 8 columns by 7 rows, ie 56 address locations. You have to refresh all those columns each time for a full refresh to happen. If you only had a counter with 7 bits output, one of those columns would never see a refresh, so a significant amount of the memory space would never be refreshed.

So you get one of those new DRAMs with twice as much space, having 9 columns by 7 rows, all of 63 bits. Your old refresh circuit will no longer be good enough, because it has 8 outputs, but you now need 9 bits.

That's what they are all trying to tell you as you spew this thread over a bunch of newsgroups.

The details are specific to the DRAM. Some may require column refresh, some row refresh, and the arrangement of the array will vary. You have to track down the details and then match that.

Michael

Reply to
Michael Black

Pardon ?

Cross-posting to just *3 relevant groups* is the recommended practice. Far better than mutli-posting.

Graham

Reply to
Eeyore

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The problem may not relate to refresh...

Think about where each of the Apple's raw address bits are going. Remember that some address bits are used as Row bits and some are used as Column bits.

The host system (Apple IIe?) controls how the bits are multiplexed... you can't just put chips with a different number of Row & Column bits than the system was designed for and expect it to work. (hint: Lost Address Bits)

Dennis

Reply to
Barbarian

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