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dieymir
February 12th, 2017, 10:58 AM
look at this card

36095

36097

It's a Hualog Microelectronics HM86314Q (S)VGA card, long time ago somebody gave me it as a gift. It has installed 256KB of DRAM, I know it probably doesn't worth but I want to have 512KB on it. Notice that it has 4 memory chips S514256FP and one HM8694P-304. I could find any datasheet but I suppose those DRAM chips are 256x4 FPM RAM. So, now the question: I cannot understand how memory is arranged on this card, 4 x 256x4 DRAMs = 512KB isn't it? What's wrong with this supposition? Another question, Which DRAM chips I need to have 512KB?

36096

seems that it actually is an 8-bit card since I cannot see that those gold fingers are actually connected to nothing

36098

I have benchmarked it in 8 and 16 bit slot and cannot see any difference in performance which confirms the above observation. In fact it's one of the slowest VGA card I have seen, much in the league of an OAK oti37C, but it seems to be decently compatible and it works for sure on a 8-bit slot.

modem7
February 12th, 2017, 10:20 PM
I could find any datasheet but I suppose those DRAM chips are 256x4 FPM RAM.
It seems a reasonable assumption. Per [here (http://minuszerodegrees.net/temp/3/temp_heay84h893wref8.jpg)], 514256 is commonly used for 256K x 4-bit.


It has installed 256KB of DRAM,
What makes you say that ?

Note that in a 16-bit environment, some software authors use "K" to denote a K's worth of 16-bit data.


seems that it actually is an 8-bit card since I cannot see that those gold fingers are actually connected to nothing
If the PCB is multi-layer, you will not be able to "see" where some of the traces are going to.

alecv
February 13th, 2017, 01:13 AM
It's "uber cheap" VGA. It can use broken memory chips. See a memory chips w/o logo and alot of jumpers near the memory. These jupers are intended for "assemble" 4-bit from the broken memory chips.

dieymir
February 13th, 2017, 02:54 AM
What makes you say that ?

That's what the VGA BIOS shows on POST



If the PCB is multi-layer, you will not be able to "see" where some of the traces are going to.
Hmm, maybe, but after careful inspection I think this is not the case.

dieymir
February 13th, 2017, 02:59 AM
It's "uber cheap" VGA. It can use broken memory chips. See a memory chips w/o logo and alot of jumpers near the memory. These jupers are intended for "assemble" 4-bit from the broken memory chips.

You seem to have got it!! Please, look at this incarnation of the card:

http://www.ebay.com/itm/HMC-HM86314Q-ISA-VGA-card-rare-model-n2/331863200271?_trksid=p2047675.c100011.m1850&_trkparms=aid%3D222007%26algo%3DSIC.MBE%26ao%3D1%2 6asc%3D41451%26meid%3D0ef1ce4872c44d76af88271c4802 5693%26pid%3D100011%26rk%3D1%26rkt%3D5%26sd%3D2319 53984384

512KB and 6 256Kx4 FPM DRAM

but this one has the expected 4 DRAM chips

https://commons.wikimedia.org/wiki/File:HMC_HM86314Q_video_card_EV%2B1.0.jpg

another question What's that HM8694P-304 chip just bellow the DRAM slots?

GiGaBiTe
February 13th, 2017, 06:00 PM
seems that it actually is an 8-bit card since I cannot see that those gold fingers are actually connected to nothing.

There are only two contacts that are obviously not connected to anything on the 16 bit portion of the card (pins 9 and 10), which are DRQ 0 and DACK 5. This just means the card can't use DMA channel 0 or acknowledge DMA transfers on DMA channel 5.

The rest either go to plated VIAs or traces off to somewhere on the top layer of the PCB. The only way to be sure the VIAs didn't go anywhere would be a continuity test to various points on the card.

Some 16 bit cards can run in 8 bit slots, albeit at a reduced speed due to the bus width being cut in half.