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HIMEMV2 or equivillent PROM

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    Originally posted by Chuck(G) View Post
    Nah, it's not big--it'll fit easily into a small flat-rate Priority Mail box.
    Well, sounds like ibmapc is determined to send you big bucks, so I'll stand by in case his wife says no.


      I will say that I was amazed at how slow stuff shipped first class from Canada takes to get across the border. I picked up a bunch of EEPROMs--good price, but they took three weeks to get here. In contrast, I typically get stuff from Hong Kong in about 5 days or less.

      I thought the US and Canada were part of NAFTA, so the hold-ups in shipping are really strange. Fast, cheap shipping is good for business, or hasn't Canada Post/Postes Canada figured that out yet?
      Reach me: vcfblackhole _at_ protonmail dot com.


        Tell me about it.

        Somebody sent me some diskettes from N Dakota a while back; took 4 days and cost $4.00. I copied them and sent them on in the same box to a pal in Winnipeg (just across the border from N Dakota, so the same distance); cost me $10 and took 3 1/2 weeks.


          Hey Chuck(G),
          Please check your PM's.


            Not sure if I should put this here or start a new thread

            Hey Chuck,
            I've been stairing at the equations in U44V2M.EQN(That you created for programming the GAL to replace U44), trying to figure out what they mean. I've also been studying the document by Cy Atkinson, trying to reconcile the two, but I just can't get my head around it. One thing that confuses me is the equation:
            would that be the same as:
            From info on the net, I've learned that" /"=NOT, "*"=AND,"+"=OR, but I just haven't had that "A-HA" moment. Maybe you could shed some light on this to get me beyond my mental blockage.

            /Q2 =
            /Q1 =
                       A7*A3*A2*A1*/A0 +
                       /A3*/A2 + /A2*/A1 + /A7*/A2
            /Q0 =
                      /A7*A5*/A4*A3*A2*/A1 +
                      /A7*A5*A4*A3*A2*/A0 +
                      A4*A3*A2*/A1*/A0 +
                      A7*A5*A3*/A2*A1 +
                      A7*A3*A1*A0 +


              The rules for EQN2JED are a little strange. If a pin (input or output) is declared as low-active in the pin assignment list, then if the pin is used as a negated term in the equations, it becomes high-active. Confusing? You bet.

              Fortunately, in the EQN file for U44, all pins are declared high-active, so Q2 is a straightforward assignment of A2. You could just as easily have said Q2 = A3, but since the GAL uses active-low outputs, it makes sense to keep them all that way. For that matter, any De Morgan equivalent assignment could be used on any of the pins, but since they're talked about as SOP terms, it makes sense to keep the equations looking like that.
              Reach me: vcfblackhole _at_ protonmail dot com.


                Originally posted by ibmapc View Post
                One thing that confuses me is the equation:
                would that be the same as:
                /Q2 = /A3
                As Chuck says the answer is yes. The reason to have an equation where the output simply follows the input is to 'buffer' the input signal. The GAL output has good drive capability and can drive a lot of loads.
                Last edited by dave_m; January 17, 2012, 09:51 AM. Reason: typo


                  Originally posted by dave_m View Post
                  As Chuck says the answer is yes. The reason to have an equation where it output simply follows the input is to 'buffer' the input signal. The GAL output has good drive capability and can drive a lot of loads.
                  Well, actually, the idea here was to replace a bipolar PROM with a GAL with a minimum of fuss. In this case, all one has to do is extend the ground pin to handle the move from a 16-pin PROM to a 20-pin GAL. All of the other lines match up. I really like programmable logic!

                  Although it probably doesn't matter in this case, the other reason would be to match up propagation times of the signals. But it can matter in other cases.
                  Reach me: vcfblackhole _at_ protonmail dot com.


                    So, if we look at the first part of the next equation:

                    /Q1 = A7*A3*A2*A1*/A0

                    If A7=1 and A3=1 and A2=1 and A1=1 and A0=0 then /Q1=1 so Q1=0 ?

                    Have I got that right?
                    Or am I all backwards?


                      No, that's right.

                      If you're handy with C or even BASIC, you can take the equations and construct your own truth tables to verify the equation.
                      Reach me: vcfblackhole _at_ protonmail dot com.