Hi, Hope this is the right group for this. I've made a big circuit in digital works that makes a 7 Segment digital display count from 0 to 9. I now have to make the Boolean equations from the circuit. One for each segment. Here is the first segment (8kb): http://www.magic-carpet.freeserve.co.uk/thb/digi_circuit.GIF Would it be: _ _ _ _ _ _ _ _ _ _ SEG A = A B C D + A B C D + A B C D Or is that total crap?! TIA, Jon
Logic gate into boolean equation
Started by ●November 17, 2003
Reply by ●November 17, 20032003-11-17
"Jibbed" <jons1@SPAMMAGEmagic-carpet.freeserve.co.uk> wrote in message news:bpajkl$o8o$1@newsg3.svr.pol.co.uk...> Hi, > Hope this is the right group for this. > > I've made a big circuit in digital works that makes a 7 Segment digital > display count from 0 to 9. I now have to make the Boolean equations fromthe> circuit. One for each segment. > > Here is the first segment (8kb): > http://www.magic-carpet.freeserve.co.uk/thb/digi_circuit.GIF > > Would it be: > _ _ _ _ _ _ _ _ _ _ > SEG A = A B C D + A B C D + A B C D > > Or is that total crap?! > > TIA, > Jon >Well, the alignment of the bars didn't work. You need to use a fixed font... Assuming that the first vertical "connection" doesn't connect input A to input D, I get: _____ _ _ _ _ _ _ _ _ SEG A = A B C D + A B C D + A B C D which, if I didn't make a mistake, readily simplifies to: _____ _ _ _ _ _ SEG A = A B D + A B C D Here's how I approached it: Assign E,G,H to the inputs to the last NOR Simply trace back to find what the inputs to each AND are... _ _ _ E = A B C D and so forth..... _____ SEG A = E + G + H so, _ _ _ SEG A = E G H Fred
Reply by ●November 17, 20032003-11-17
Jibbed wrote:> Hi, > Hope this is the right group for this. > > I've made a big circuit in digital works that makes a 7 Segment digital > display count from 0 to 9. I now have to make the Boolean equations from the > circuit. One for each segment. > > Here is the first segment (8kb): > http://www.magic-carpet.freeserve.co.uk/thb/digi_circuit.GIF > > Would it be: > _ _ _ _ _ _ _ _ _ _ > SEG A = A B C D + A B C D + A B C D > > Or is that total crap?! > > TIA, > JonI don't suppose that this is the right group, but many of us here know what you want to. Posting a link to your circuit was the right thing to do. Unfortunately, I see only the logic for one of the segments. (I can imagine the rest easily enough.) You need to provide a map before anyone can help you. Which segment is Segment A? (I know there is a standard mapping, but I don't remember it with certainty, and I won't trouble myself to look it up. Besides, it may not be what you have in mind.) Using the same symbols to represent the segments and the outputs is likely to be confusing. I suggest computerese for the inputs: "INPUT A" becomes "D0", "INPUT B" becomes "D1, "INPUT C" becomes "D2", and "INPUT D" becomes "D3". They are, in fact, data lines. There are two sets of logic equations you might want. One is a description of what your decoder ought to accomplish, another is a report of how it does that. Start with the first, and pictures of which segments are to be lit with each input code. The equations will practically fall out of a hat. Keep in mind that real 7-segment displays are often active low. That is, the anodes are common, and a segment turns on when its cathode is grounded. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 17, 20032003-11-17
Hi. "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:gfSdnTzurtFukSSiRVn-tg@centurytel.net...> > "Jibbed" <jons1@SPAMMAGEmagic-carpet.freeserve.co.uk> wrote in message > news:bpajkl$o8o$1@newsg3.svr.pol.co.uk... > > Hi, > > Hope this is the right group for this. > > > > I've made a big circuit in digital works that makes a 7 Segment digital > > display count from 0 to 9. I now have to make the Boolean equations from > the > > circuit. One for each segment. > > > > Here is the first segment (8kb): > > http://www.magic-carpet.freeserve.co.uk/thb/digi_circuit.GIF > > > > Would it be: > > _ _ _ _ _ _ _ _ _ _ > > SEG A = A B C D + A B C D + A B C D > > > > Or is that total crap?! > > > > TIA, > > Jon > > > > Well, the alignment of the bars didn't work. You need to use a fixed > font...Yeah sorry was a bit stupid to have typed it out.> > Assuming that the first vertical "connection" doesn't connect input A to > input D, I get: > > _____ _ _ _ _ _ _ _ _ > SEG A = A B C D + A B C D + A B C D > > which, if I didn't make a mistake, readily simplifies to: > > _____ _ _ _ _ _ > SEG A = A B D + A B C D > > Here's how I approached it: > > Assign E,G,H to the inputs to the last NOR > Simply trace back to find what the inputs to each AND are... > _ _ _ > E = A B C D and so forth..... > > _____ > SEG A = E + G + H > > so, > _ _ _ > SEG A = E G H > > Fred > >Thanks for your help. This is for my uni course and I've never been taught it that simplified way but it makes perfect sense actually. BTW is the first equation you wrote before you simplified the same as: http://www.magic-carpet.freeserve.co.uk/thb/equation.GIF ?? Thanks for your help! Jon
Reply by ●November 17, 20032003-11-17
Sorry. I shouldn't have assumed you knew what segments where which (SEG A is the top one by the way) Anyway using your help and some from others I get it now. Thanks a lot! Jon "Jerry Avins" <jya@ieee.org> wrote in message news:bpb4hl$7ua$1@bob.news.rcn.net...> Jibbed wrote: > > > Hi, > > Hope this is the right group for this. > > > > I've made a big circuit in digital works that makes a 7 Segment digital > > display count from 0 to 9. I now have to make the Boolean equations fromthe> > circuit. One for each segment. > > > > Here is the first segment (8kb): > > http://www.magic-carpet.freeserve.co.uk/thb/digi_circuit.GIF > > > > Would it be: > > _ _ _ _ _ _ _ _ _ _ > > SEG A = A B C D + A B C D + A B C D > > > > Or is that total crap?! > > > > TIA, > > Jon > > I don't suppose that this is the right group, but many of us here know > what you want to. > > Posting a link to your circuit was the right thing to do. Unfortunately, > I see only the logic for one of the segments. (I can imagine the rest > easily enough.) You need to provide a map before anyone can help you. > Which segment is Segment A? (I know there is a standard mapping, but I > don't remember it with certainty, and I won't trouble myself to look it > up. Besides, it may not be what you have in mind.) Using the same > symbols to represent the segments and the outputs is likely to be > confusing. I suggest computerese for the inputs: "INPUT A" becomes "D0", > "INPUT B" becomes "D1, "INPUT C" becomes "D2", and "INPUT D" becomes > "D3". They are, in fact, data lines. > > There are two sets of logic equations you might want. One is a > description of what your decoder ought to accomplish, another is a > report of how it does that. Start with the first, and pictures of which > segments are to be lit with each input code. The equations will > practically fall out of a hat. Keep in mind that real 7-segment displays > are often active low. That is, the anodes are common, and a segment > turns on when its cathode is grounded. > > Jerry > -- > Engineering is the art of making what you want from things you can get. > ����������������������������������������������������������������������� >
Reply by ●November 17, 20032003-11-17
On Tue, 18 Nov 2003 00:48:51 -0000, "Jibbed" <jons1@SPAMMAGEmagic-carpet.freeserve.co.uk> wrote:>Sorry. I shouldn't have assumed you knew what segments where which (SEG A is >the top one by the way)Hmmm. Since there is only one standard way to name the segements, I think that was a reasonable assumption. +-A-+ F B +-G-+ E C +-D-+ Regards, Allan.
Reply by ●November 18, 20032003-11-18
Allan Herriman wrote:> On Tue, 18 Nov 2003 00:48:51 -0000, "Jibbed" > <jons1@SPAMMAGEmagic-carpet.freeserve.co.uk> wrote: > > >>Sorry. I shouldn't have assumed you knew what segments where which (SEG A is >>the top one by the way) > > > Hmmm. Since there is only one standard way to name the segements, I > think that was a reasonable assumption. > > +-A-+ > F B > +-G-+ > E C > +-D-+ > > Regards, > Allan.It's long enough since I wired up a display that I forget. I bet you don't remember the basing diagram for a 6SN7. That's standard too! (Is it plate, cathode, grid, heater, heater, plate, cathode, grid? I forget too.) A minimum implementation of a ten-digit decoder will have don't-care states. That can save a few gates. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 18, 20032003-11-18
Jerry Avins wrote:> Allan Herriman wrote:>>> Sorry. I shouldn't have assumed you knew what segments where which >>> (SEG A is >>> the top one by the way)(snip)>> +-A-+ >> F B >> +-G-+ >> E C >> +-D-+> It's long enough since I wired up a display that I forget. I bet you > don't remember the basing diagram for a 6SN7. That's standard too! (Is > it plate, cathode, grid, heater, heater, plate, cathode, grid? I forget > too.) > > A minimum implementation of a ten-digit decoder will have don't-care > states. That can save a few gates.The 7447 was always one of my favorite in the 7400 series TTL, though possibly the 74247. The TI databook gives pretty good descriptions for the logic. The 7447 gives 16 unique outputs, while reasonably minimizing the logic. There is another, I believe 9368, that gives A-F for the higher values, though. -- glen
Reply by ●November 18, 20032003-11-18
"Jibbed" <jons1@SPAMMAGEmagic-carpet.freeserve.co.uk> wrote in message news:bpbq13$9q5$1@newsg1.svr.pol.co.uk...> Hi. > > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message > news:gfSdnTzurtFukSSiRVn-tg@centurytel.net... > > > > "Jibbed" <jons1@SPAMMAGEmagic-carpet.freeserve.co.uk> wrote in message > > news:bpajkl$o8o$1@newsg3.svr.pol.co.uk... > > > Hi, > > > Hope this is the right group for this. > > > > > > I've made a big circuit in digital works that makes a 7 Segmentdigital> > > display count from 0 to 9. I now have to make the Boolean equationsfrom> > the > > > circuit. One for each segment. > > > > > > Here is the first segment (8kb): > > > http://www.magic-carpet.freeserve.co.uk/thb/digi_circuit.GIF > > > > > > Would it be: > > > _ _ _ _ _ _ _ _ _ _ > > > SEG A = A B C D + A B C D + A B C D > > > > > > Or is that total crap?! > > > > > > TIA, > > > Jon > > > > > > > Well, the alignment of the bars didn't work. You need to use a fixed > > font... > > Yeah sorry was a bit stupid to have typed it out. > > > > > Assuming that the first vertical "connection" doesn't connect input A to > > input D, I get: > > > > _____ _ _ _ _ _ _ _ _ > > SEG A = A B C D + A B C D + A B C D > > > > which, if I didn't make a mistake, readily simplifies to: > > > > _____ _ _ _ _ _ > > SEG A = A B D + A B C D > > > > Here's how I approached it: > > > > Assign E,G,H to the inputs to the last NOR > > Simply trace back to find what the inputs to each AND are... > > _ _ _ > > E = A B C D and so forth..... > > > > _____ > > SEG A = E + G + H > > > > so, > > _ _ _ > > SEG A = E G H > > > > Fred > > > > > > Thanks for your help. This is for my uni course and I've never been taught > it that simplified way but it makes perfect sense actually. BTW is thefirst> equation you wrote before you simplified the same as: > > http://www.magic-carpet.freeserve.co.uk/thb/equation.GIF >Well, yes except that the whole equation is negated. Fred
Reply by ●November 18, 20032003-11-18
On Tue, 18 Nov 2003 18:54:31 GMT, glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:>Jerry Avins wrote: > >> Allan Herriman wrote: > >>>> Sorry. I shouldn't have assumed you knew what segments where which >>>> (SEG A is >>>> the top one by the way) >(snip) >>> +-A-+ >>> F B >>> +-G-+ >>> E C >>> +-D-+ > >> It's long enough since I wired up a display that I forget. I bet you >> don't remember the basing diagram for a 6SN7. That's standard too! (Is >> it plate, cathode, grid, heater, heater, plate, cathode, grid? I forget >> too.)I never used one of those (they were a little before my time). I did use some tubes / valves in transmitters though: 4cx250 and 3cx3000 come to mind. The latter only had four terminals (the cathode was directly heated) so there was no no basing diagram to forget!>> A minimum implementation of a ten-digit decoder will have don't-care >> states. That can save a few gates. > >The 7447 was always one of my favorite in the 7400 series TTL, though >possibly the 74247. The TI databook gives pretty good descriptions for >the logic. The 7447 gives 16 unique outputs, while reasonably >minimizing the logic. There is another, I believe 9368, that gives A-F >for the higher values, though.Data sheets here: http://www-s.ti.com/sc/ds/sn7447a.pdf The '247 is like the '47 but with tails on the 6 and 9. Regards, Allan.
