The only fakes I've seen are those soldered on the motherboard. Looks like your board doesn't have them (upper right corner, next to SIMM).

The module looks genuine to me.

If you have it you can easily confirm it by running cachechk, speedsys or a similar tool.

The fake cache boards were soldered cache. Being that is a COAST it should be real. One thing: It says 3v so you must use a 3.3v 486, not a 5v as the cache is powered off the cpu VRMS and at 5v you would burn up the cache chips over time --- Unless they are 5v tolerant.
Here is a topic here at Vogons about those coast modules: Re: Lets make new M919 Cache sticks?

edit: I type to slow 🤣

Thanks. Glad to know that the modules didn't have fake chips on it. Will probably be able to test tomorrow.

I think this board is actually an Amptron DX-9700, which would explain the sticker on the back of the module.

The box also has a logo that's a big "A."

Horun wrote on 2020-05-07, 23:25:

The fake cache boards were soldered cache. Being that is a COAST it should be real. One thing: It says 3v so you must use a 3.3v 486, not a 5v as the cache is powered off the cpu VRMS and at 5v you would burn up the cache chips over time --- Unless they are 5v tolerant.
Here is a topic here at Vogons about those coast modules: Re: Lets make new M919 Cache sticks?

Good to know! Glad I haven't fired it up yet.

how much did you pay for it, that cache module alone must be worth like 100 bucks now

SETBLASTER wrote on 2020-05-08, 01:40:

how much did you pay for it, that cache module alone must be worth like 100 bucks now

I was thinking something like that myself earlier as a real working coast for a DX9700 in USA is as much as a bare board without one...if you can find just the coast.

SETBLASTER wrote on 2020-05-08, 01:40:

how much did you pay for it, that cache module alone must be worth like 100 bucks now

I paid about $75 shipped for everything.

This is so strange..

We can see clearly the onboard cache running in circles going nowhere... but this board as them and, a cost module... so. i'm thinking. This version as the cost modlue and not the soldered chips. but, there where others, with the fake cache chips and the cost module not soldered !!!

really confusing this board

those traces do have a certain beauty though, I'll give them that ^^

There are many revisions of M919 board. This one is V3.4B/F and it's the last one. It never had soldered cache chips just the COAST module. So you either have cache via COST module or you don't have it at all. Performance penalty with 16KB L1 486 chips is not that big, but it is still noticeable.

Since you probably got the board as-is, the price is fair.

About ready to test this thing, but I need some help with memory. I have a bag of mismatched 72-pin SIMMs and have always been bad at deciphering the markings on the chip to determine their size.

I sorted them by type (FPM or EDO), speed, and whether there were chips on both sides. I ended up with two sets, one FPM and one EDO (I think). Can anyone tell me the size of these SIMMs (how many megabytes)?

FPM

EDO

I'm leaning toward trying the FPM set first. I assume that if the sizes of the sticks are different, it won't be a problem. I am going to use an Intel 486DX4-100 and have set the jumpers for a 3.3V CPU.

no clue on how to read memory specs and count that specification of the chip and multiply by #of chips on the memory to get the total amount of memory on 1 stick. if anyone can explain in detail i would really appreciate in order to learn.

The km44c1000cj-6 chips in top pic should be 1Mx4, 60nS FPM. 8 chips = 4Mb SIMMS (8x1Mx4 /8), if double sided would be 8Mb SIMMS.
The gm71c17403cj6 chips in bottom pic should be 4M x 4bit 60nS EDO. 8 chips = 16Mb SIMMS or 32MB if double sided

SETBLASTER wrote on 2020-05-09, 03:52:

no clue on how to read memory specs and count that specification of the chip and multiply by #of chips on the memory to get the total amount of memory on 1 stick. if anyone can explain in detail i would really appreciate in order to learn.

Added: All ram is measured in Bytes (8bits=1Byte) no matter if 16bit (30pin), 32bit (72pin) or 64bit (DDR, etc) so in above the km44c1000cj chips are 1M x 4Bit chips. So two would equal 1M x 8bit (or 1M Byte) so if 8 then 4Mbyte. The real quirk is finding datasheets to tell you what each chip is..

Horun wrote on 2020-05-09, 04:53:

The km44c1000cj-6 chips in top pic should be 1Mx4, 60nS FPM. 8 chips = 4Mb SIMMS (8x1Mx4 /8), if double sided would be 8Mb SIMMS.
The gm71c17403cj6 chips in bottom pic should be 4M x 4bit 60nS EDO. 8 chips = 16Mb SIMMS or 32MB if double sided

Thank you! I look at the datasheets but then get confused. I think I understand now. These are double sided.

I also found these, which thankfully, are marked as 8MB EDO SIMMS (double sided). These chips are 1Mx16, so four chips per SIMM would be 64 Mb or 8 MB.

Yep ! Some older boards do not like the high density chips in your example, is why you see a lot of 8 chips per side versus 2 chip per side 72pin SIMMS for 486. Most Pentium boards are ok with the high density but many 486 do not like them. You can try them but if you get a mem error (beeps on the speaker) that is usually why,

I'm going to stick this heatsink/fan combo onto the 486DX4. The fan has a two-pin power connector. Ideally, I'd like to use a 2-pin to Molex adapter, but I don't have one of those.

The M919 does have a fan connector. I'm sorry if this is an obvious question, but connecting the fan is as simple as plugging in the fan header where the red wire is "12V" and the black wire is "G," correct? Just don't want to blow anything up at this point.

It should be that simple. You can always verify voltage at the pins with a multimeter.

chrismeyer6 wrote on 2020-05-09, 15:11:

It should be that simple. You can always verify voltage at the pins with a multimeter.

Thank you for the quick reply. I figured why risk it when someone else might confirm soon.

You can power on for a minute even without the fan on the cpu.