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basic digital circuit
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claudiu_tatar2000
Hi all,
I have 2 digital circuits that perform a given task. Both have the same input (4 bits) and contains the logical gates (AND, OR, XOR, NAND, etc). i1---- i2---- i3---- i4---- In the case that i1 is 0 I want to have the answer from the first circuit and in the case i1 equals to 1 I need the answer from the second circuit. How can I mixed them? I don't know how to check if the input is 0 or 1 and then to process the corresponding circuit. Please help, Clau |
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Stefan Trethan
On Wed, 07 Apr 2004 08:32:02 -0000, claudiu_tatar2000 <claudiu_tatar2000@...> wrote:
Hi, i do not understand your question. you have 2 logic circuits, and one input determines which one determines the output? it is simple, you AND one output with i1 and the other one with NOT i1. then you or these signals. i think that would result in what you want.. (make the truth table to see it it is correct. A1 A2 I1 | O1 000|0 001|0 010|0 011|1 100|1 101|0 110|1 111|1 for 1 you get (I1 AND A2) OR ((NOT I1) AND A1) i think everyone can see this from the truth table. you could also use a multiplexer (or integrated analog switch) if you use ics. I hope i understood your question correct. Ths solution will most likely not be the easiest one. normally you write a truth table for te whole circuit (both logic circuits) or make a "kv diagram". This way you find the most simple logic for the whole circuit.. good luck.. ST |
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If I understand this, you want to have three inputs to a circuit
(i2,i3 and i4) the output is the result of logic gates as you describe, but one output will only be active if i1 is a 1 while the other will be active if i1 is zero. Is that right? If so, take output 1 through an AND with i1 ad take output 2 through a NAND with i1. Effectively, the circuits are both processing the data all the time, but the output is blocked until i1 is in the correct state. I suspect I am missing something here. If so, tell us some more about the application and you will get some more detailed answers. Peter --- In Electronics_101@..., "claudiu_tatar2000" <claudiu_tatar2000@y...> wrote: or 1 and then to process the corresponding circuit. |
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claudiu_tatar2000
thank you,
I have another question in this case. How can I use only the circuit that I want to use? I mean if the i1=1 then I want to use only the first circuit without using the second... Could I design such behavior using only some basic gates? Regards, Clau --- In Electronics_101@..., Stefan Trethan <stefan_trethan@g...> wrote: On Wed, 07 Apr 2004 08:32:02 -0000, claudiu_tatar2000or determines1 and then to process the corresponding circuit.Hi, i do not understand your question. the output?i1. then you or these signals. i think that would result in what youwant.. (make the truth tablesee this from the truth table.you use ics.write a truth table forway you find the most |
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Stefan Trethan
On Wed, 07 Apr 2004 08:56:44 -0000, claudiu_tatar2000 <claudiu_tatar2000@...> wrote:
thank you,i thought that was what i described. both circuits will be physically there, but if I1=0 the output will be determined by the first circuit, if it is 1 it will be the second. you can't "rewire" the circuits dependent of I1 (not so easy and it depends on the circuits). but you can let both circuits produce a result and then take the result from the circuit you want (which is determined by I1). You could "switch off" the second circuit which is not in use, but it does not have any advantage. ST |
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claudiu_tatar2000
Hi Peter,
I have 2 circuits with 4 inputs and 4 outputs each. i1---- ----o1 i2---- [I have some gates here] ----o2 i3---- ----o3 i4---- ----o4 Note that both circuits have the same input (2^4 = 16 cases). If i1 = 1 I want (as the output of the global circuit) the result provided by the first circuit. If i1==0 I want the result provided by the second circuit. So, I have to build another bigger circuit (that contains the other 2 described above) that will have only 4 inputs and only 4 outputs. Hope that you can help me. Thanks, Clau --- In Electronics_101@..., "peterhawken" <peterhawken@s...> wrote: If I understand this, you want to have three inputs to a circuitthrough a NAND with i1. Effectively, the circuits are both processing the |
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Stefan Trethan
I thought you have 1 output only....
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simplest is to take a multiplexer. this is only one ic for all the functionality. if you can't use a multiplexer use 1 and, one not and, and a or on each of the 4 (8) output lines. ST On Wed, 07 Apr 2004 10:12:31 -0000, claudiu_tatar2000 <claudiu_tatar2000@...> wrote:
Hi Peter, |
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claudiu_tatar2000
How do I use the multiplexer?
How many gates has? I'm not very familiar with multiplexer... What I know is that a multiplexer decodes some addresses... Clau --- In Electronics_101@..., Stefan Trethan <stefan_trethan@g...> wrote: I thought you have 1 output only....each of the 4 (8)provided by other 2the second circuit. thedescribed above) that will have only 4 inputs and only 4 outputs. answers.other will be active if i1 is zero. Is that right?through gates first(AND, OR, XOR, NAND, etc). thecircuit and in the case i1 equals to 1 I need the answer from is 0second circuit. Service.or1 and then to process the corresponding circuit. |
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claudiu_tatar2000
The problem is that I want a short circuit.
I don't know what complexity a multiplexer has. Regards, Clau --- In Electronics_101@..., Stefan Trethan <stefan_trethan@g...> wrote: I thought you have 1 output only....each of the 4 (8)provided by other 2the second circuit. thedescribed above) that will have only 4 inputs and only 4 outputs. answers.other will be active if i1 is zero. Is that right?through gates first(AND, OR, XOR, NAND, etc). thecircuit and in the case i1 equals to 1 I need the answer from is 0second circuit. Service.or1 and then to process the corresponding circuit. |
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Stefan Trethan
On Wed, 07 Apr 2004 10:35:43 -0000, claudiu_tatar2000 <claudiu_tatar2000@...> wrote:
if you don't build it with ICs then a multiplexer is no help. As said you can simply make it with gates. the first circuit makes A1 B1 C1 D1 as output. the second A2 B2 C2 D2. you simply make (A1 AND I1) OR (A2 AND NOT I2) = output A (B1 AND I1) OR (B2 AND NOT I2) = output B (C1 AND I1) OR (C2 AND NOT I2) = output C (D1 AND I1) OR (D2 AND NOT I2) = output D Make a truth table and you will see it works. ST |
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Roy J. Tellason
On Wednesday 07 April 2004 04:49 am, peterhawken wrote:
If so, take output 1 through an AND with i1 ad take output 2 throughI've seen two folks suggest this, and it's wrong. The difference between AND gates and NAND gates for any given set of inputs is only the sense of the output. What's wanted instead is to invert the controlling input to one of two AND gates, and OR the outputs of them together. |
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claudiu_tatar2000
Or, is there some switch to turn on/off one of the circuits?
I'm not very familiar with the hardware implementation of the circuits so maybe you have some basic tricks to enable/disable the use of one of the circuits. Thanks, Clau --- In Electronics_101@..., "Roy J. Tellason" <rtellason@b...> wrote: On Wednesday 07 April 2004 04:49 am, peterhawken wrote:throughIf so, take output 1 through an AND with i1 ad take output 2 between ANDa NAND with i1. Effectively, the circuits are both processing theI've seen two folks suggest this, and it's wrong. The difference gates and NAND gates for any given set of inputs is only the senseof the output. What's wanted instead is to invert the controlling inputto one of two AND gates, and OR the outputs of them together. |
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Stefan Trethan
On Wed, 07 Apr 2004 14:14:23 -0000, claudiu_tatar2000 <claudiu_tatar2000@...> wrote:
so you are making a hardware implementation? if you use standard logic ics you can of course again use the multiplexer. this is a single ic instead of the 8 AND 4 inverter and 4 OR i suggested.... some ics have a input pin called CE (chip enable) or OE (output enable) which switches on/off the outputs, this is present with tri-state chips. (standard gates don't have this). It is very hard to help you if you keep us guessing, tell us what you want do and then we can help. if you can not use a multiplexer the and not and and or solution i provided will work, but be more complicated than a multiplexer (more ICs). Also keep in mind that if you use more than ? three? seperate logic ics it micht already be cheaper and more efficient (space on board etc) to use some sort of programmable logic. (pal, gal, cpld, eprom). you gain the added benefit of "software changeable wiring". good luck ST |
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--- In Electronics_101@..., "Roy J. Tellason"
<rtellason@b...> wrote: On Wednesday 07 April 2004 04:49 am, peterhawken wrote:throughIf so, take output 1 through an AND with i1 ad take output 2 between ANDa NAND with i1. Effectively, the circuits are both processing theI've seen two folks suggest this, and it's wrong. The difference gates and NAND gates for any given set of inputs is only the senseof the output. What's wanted instead is to invert the controlling inputto one of two AND gates, and OR the outputs of them together.You are quite right. The NAND is wrong. An inverted i1 into a second AND will do the job. |
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Or use a microcontroller. a PIC12F629 costs $1.08/1 and has 6 pins
for input or output. I hardly bother with random logic anymore as its easier to make the PIC circuit and you can change it easily. Plus its an 8 pin package. I know the cost of development tools makes this cost prohibitive (30- 50 USD) but once you have them, its cheap-n-easy. --- In Electronics_101@..., Stefan Trethan <stefan_trethan@g...> wrote: On Wed, 07 Apr 2004 14:14:23 -0000, claudiu_tatar2000multiplexer. this is a single ic instead of the 8 AND 4 inverter and 4 OR ienable) which switches on/off the outputs, this is present with tri-statechips. (standard gates don't have this).you want do and then we can help. if you can not use a multiplexer the andnot and and or solution(more ICs).ics it micht alreadyof programmable logic. |
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Stefan Trethan
On Wed, 07 Apr 2004 17:15:33 -0000, Phil <phil1960us@...> wrote:
Or use a microcontroller. a PIC12F629 costs $1.08/1 and has 6 pins6 I/O isn't going to be enough for 8 in and 4 out.... Who sold you a 30USD development tool? you can make a programmer for a few cent and the software is free. A pic is not the holy grale, a processor is not the holy grale. The problem to solve is obviously simple straightforward logic. no point in making it unneccesary complex with a sequential working processor. there are so many programmable logic devices out there which are not at all more expensive than a pic (even cheaper, especially if you need many pins). It is also much more straightforward to program them than programming a pic. i do refuse to encourage to use processors where there is absolutely no need for. you add unneccesary complexity, and room for error. ST |
¿ªÔÆÌåÓýIn a message dated 4/7/2004 5:53:08 AM Central Standard Time, claudiu_tatar2000@... writes:
I want a short circuit ?
You mean a "SIMPLE circuit", don't you?? A "short circuit" is an "unwanted" connection, often resulting in blown fuzes, etc.! |
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I see no need to be so snide.
who said a processor was the holy grail? just another way to do it that, frankly, I think is far superior to a bunch packages (1 vs 1). Especially if you dont fully use all the gates. way lower pincount and errors are easy to fix. but each to his own. you can build programmers for almost nothing or pay well over $100. 30 will get a reasonable pre-built one. You can build microprocessors from sand, that's really cheap. maybe you confused this group with how-cheap-can-you-build-it? actually the way he drew it, it looks like only one line has logic on it. if one needs 4 in, 4 out, try a 14 pin pic for about $1.50. and you are stating opinion on the programming aspect, not fact. have a nice day --- In Electronics_101@..., Stefan Trethan <stefan_trethan@g...> wrote: On Wed, 07 Apr 2004 17:15:33 -0000, Phil <phil1960us@y...> wrote:PlusOr use a microcontroller. a PIC12F629 costs $1.08/1 and has 6 pins (30-its an 8 pin package. for a few50 USD) but once you have them, its cheap-n-easy.6 I/O isn't going to be enough for 8 in and 4 out.... cent and thenot at all more expensiveprogramming a pic.absolutely no need for. |
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