Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!tut.cis.ohio-state.edu!ucbvax!husc6!m2c!umvlsi!killoran From: killoran@umvlsi.ECS.UMASS.EDU (Mik I Killoran) Newsgroups: comp.sys.ibm.pc Subject: IBM PC memory problems Message-ID: <101@umvlsi.ECS.UMASS.EDU> Date: 3 Apr 89 19:48:37 GMT Reply-To: killoran@umvlsi.ECS.UMASS.EDU (Mik I Killoran) Organization: University of Massachusetts, Amherst Lines: 37 I have a very old IBM PC that has 256K of on-mother-board memory in the form of rows of nine 64K chips. There are four of these rows on the mother board with the first being permanent and the last three rows socketed in place. I have installed a expansion memory board that allows up to 384K additional memory bringing the total up to a sum of 640K. Well, the memory that I bought for the expansion board had a mess of BAD chips in the batch. I set the system board (those two banks of switches) to let the system know that it has 640K and I set the expansion board to let it know there is 256K on the mother board. Problems used to occur upon bootup with a message of #### 201 being displayed on the screen. I was told that the 201 meant that there was a memory problem and the #### told where the bad chip was. Using #### I substituted new chips for defective chips and have gotten to the point where the system will bootup. BUT, Things only work if a version of DOS <=2.0 is used Sometimes big programs lock up There is occasionaly a error of PARITY CHECK # where # might be 1 or 2 (when this happens the screen goes into a width of fourty characters) A rather large spreadsheet of mine won't run giving the error internal error. Things will work ok if I use DOS 2.0 and little programs. But I would really like to use what I have. I have used Norton Utilities' SYSINFO to check how much memory DOS reports and Active memory found, and both say 640K. Is there a way to find out which memory chips are bad? If it is a bad memory chip wouldn't it show up as a #### 201 error upon booting up? Any thoughts on my problem would really help me as I am quite stumped. Thanks in advance, Michael Killoran