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weeds
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Can anyone please help with the design of a BCD 2 out of 5 encoder (74210 code) using basic logic gates
Okay i follow you now.. The ABCD is a 4 bit binary-coded-decimal, and your P7,P4,P2,P1,P0 is the 5 bit (74210 code)weeds said:Thanks for the reply.My input is a 4 bit number i.e 0001,0010,0011 etc till count ten.The 74210 code goes something like this,11000,00011,00101,00110,01001,01010,01100,10001,10010 and 10100(these correspond to P7,P4,P2,P1,P0).
Are you also incorporating the parity check into your digital logic??It has got two 1's for parity checking or error detection.
It would be better, if you can show the expressions you have figured out so far, and I'll try to show you where you are going astray.ALJ said:Ouabache,
What were the expressions that you produced from your Kmaps. I am also struggling with the same problem as weeds.
I have produced the truth table but am having a bit of difficulty with the Kmpas and logic diagrams for the BCD to 74210 encoder.
Any assistance would be appreciated.
Kenneth Mann said:The difficulty seems to have less to do with the problem itself than with working with Karnaugh Maps. Maybe one of us should give an introductory tutorial in using these (we would assume that basic logic is understood). I could do so (estimate four or five or so sections), if there is interest in it. The problems I see are two. First, the space allowed in this forum is only about 25K bytes for an attached file, and the maps I have take about 30 to 90K. Second, by the time it is done it would likely be too late for ALJ's assignment. In any case, if anyone knows how to input larger attachments, and if anyone is interested, I can show how "Reflective Karnaugh Maps work.
KM
Ouabache said:It would be better, if you can show the expressions you have figured out so far, and I'll try to show you where you are going astray.
(that will also give me a chance to go dig up where i put my scratch paper for that problem... you're lucky i didn't throw that away)
this is what I got for my output.can this be simplified any further.No use of exclusive gates.Do not tell the answer just hints
Thanks
P0=A'BC'D'+A'B'C'D+A'BCD+A'B'CD'
P1=AB'C'D'+A'B'D+A'C'D
P2=AB'C'D+A'B'C+A'CD'
P4=A'C'D'+A'BC'+A'BD'
P7=A'BCD+B'C'D'+AB'C'
MirrorM said:Hi
Im having trouble drawing the circuit in micro cap. Would someone please happy me with this or drop a few hints. Would it be correct in saying that I have to have 2 leds turn on for each of the outputs?
Ive also been told that this circuit can be build with 8 or 9 ics only, do you believe this to be correct.
Thanks
weeds said:this is what I got for my output. Can this be simplified any further.
P0=A'BC'D'+A'B'C'D+A'BCD+A'B'CD'
P1=AB'C'D'+A'B'D+A'C'D
P2=AB'C'D+A'B'C+A'CD'
P4=A'C'D'+A'BC'+A'BD'
P7=A'BCD+B'C'D'+AB'C'
If you look at the earlier posts, you will see that Weeds successfully verified his equations with micro cap.MirrorM said:Ive tested these in micro cap and I not believe they are correct, have you tested them?
Weeds said:"Using micro cap I drew the circuit (logic) fed my inputs A,B,C and D and what I got as outputs (P7...) matched my Truth table.
Were you successful deriving your equations from your Kmaps?MirrorM said:Ive just completed a working design using 8 AND gates and 1 OR gate plus 4 inverters. Do the Inverters count as ic's?
Kenneth Mann said:This makes it easier to put it into an ordered/reflected K-Map.
Ouabache said:If you look at the earlier posts, you will see that Weeds successfully verified his equations with micro cap.
Were you successful deriving your equations from your Kmaps?
How will you invert a signal without using an IC ??
Kenneth Mann said:Your answers are correct, however, I'd like to note that the usual procedure is to put the A, B, C and D terms in the truth table with "A" in the low order position (to the right). This makes it easier to put it into an ordered/reflected K-Map. I have attached two truth tables (Example01) and the K-Maps and logic equations (Example02and Example03) for term "P7".
Note that the logic generated is different for the two cases, but both cases are correct (showing that there is more than one right answer – actually 24). They all give the same configuration of gates. It is easier, however with the standard arrangement of A, B, C and D, because in this order, the terms go easily into the reflected K-Map (You can probably see why).
Sorry for the ugly maps, but they they had to be drawn in under 25K each, which is not the most versatile. I also have better for these maps in Word (which looks better) and Pagemaker (which looks even better), but these result in large files. If anyone wants copies I'd be glad to send them. I can also show how the reflected K-Map is used. (The number of variables a map can handle is limited only by the size of the map that can be drawn.)
Note, also that the terms of the equations derived can be combined by grouping, however a quick examination showed no apparent advantages to that. (I'm not saying that something simpler cannot be found, just that I didn't readily see anything.)
KM
P.S: The arrangement you have now will take 10 ICs counting the inverter. If the inverted signals are already provided, the inverter can be omitted, leaving 9.
Kenneth Mann said:This approach will let you set-up a 10-input K-Map if you want to.
KM
Rev Prez said:How can you include more than four inputs in a K Map? I thought you had to use some other method?
Kenneth Mann said:When you have a logic equation like:
ABC' + D'E'F' + G'HI
Here, we use three 3-input ANDs (one IC), and feed their outputs to a 3-input OR (1/3 IC).
MirrorM said:So if you had a question on the number of IC's used, would the answer be 1 1/3 (three 3-input ANDs (one IC), and feed their outputs to a 3-input OR (1/3 IC)) or 2 IC's because is can't have 1/3 of an IC?( can you?)
A BCD 2-out-5 encoder is a digital circuit that takes in a binary-coded decimal (BCD) input and converts it into a 5-bit output. It is used to encode decimal numbers into a binary format for digital systems to process and manipulate.
A BCD 2-out-5 encoder works by using logic gates, such as AND and OR gates, to compare the input BCD code with the 74210 code. The output of the encoder is determined by the input code and the 74210 code, which is a predetermined sequence of binary numbers.
The logic gates used in a BCD 2-out-5 encoder are AND, OR, and NOT gates. These gates are used to compare the input BCD code with the 74210 code and generate the 5-bit output.
The purpose of a BCD 2-out-5 encoder is to convert decimal numbers into a binary format for digital systems to process and manipulate. It is commonly used in applications such as digital displays, calculators, and data storage systems.
To design a BCD 2-out-5 encoder using logic gates, you need to first determine the truth table for the encoder based on the input BCD code and the 74210 code. Then, use the truth table to create a logical expression using AND, OR, and NOT gates. Finally, draw a circuit diagram using the logical expression and implement it using logic gates.