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We need to know a lot more about where you are getting the number(s) that will be displayed. Really simple would a sixteenposition rotary switch. Provide some diodes to select segments at each switch position if you use the rotary switch. Please tell us what you are trying to DO.my teacher ask me to create project using 2 sevent segment to display only 015 decimal. but i think if using 2 sevent segment, the display will show 099 decimal.
It's possible to display only 015?
We need to know a lot more about where you are getting the number(s) that will be displayed. Really simple would a sixteenposition rotary switch. Provide some diodes to select segments at each switch position if you use the rotary switch. Please tell us what you are trying to DO.
im sorry to say this,but before reply, please read my question on top of your reply. yes i know, if i use 4 bit, maximum number is 15. please read my detail question on my reply to hevans1944. thank youhere is a question for you. If your input is 4 bits, and one value is assumed to be 0, what is the maximum number that could be represented?
im sorry to say this,but before reply, please read my question on top of your reply. yes i know, if i use 4 bit, maximum number is 15. please read my detail question on my reply to hevans1944. thank you
Our new job is make the number 1015 to show as decimal using 2 sevent segment. but the display must be end at 15 not more or 99. what components we need to make that project?
what do you mean? i dont get it.Sure, but you also said:
The input to the decoder is binary NOT BCD. That is why the decoder truth table must implement a binarytoBCD function. Since @mod already has the 7447 BCDto7segment decoder/drivers, all that is needed is the lookup table logic to convert from 4bit binary to two binarycodeddecimal (BCD) digits, with the first BCD digit being zero or one. As you pointed out, and as I mentioned in post #7, this truth table could also perform the BCDto7segment conversion, but then you would need to implement the driver function of the 7447.What you are asked to do is build a decoder.
It uses only combinatorial logic (no FFs).
build the truth tables an solve them,if you look carefully you can see some "shortcuts" in the truth tables
Should look like this:
View attachment 26592
The input to the decoder is binary NOT BCD. That is why the decoder truth table must implement a binarytoBCD function. Since @mod already has the 7447 BCDto7segment decoder/drivers, all that is needed is the lookup table logic to convert from 4bit binary to two binarycodeddecimal (BCD) digits, with the first BCD digit being zero or one. As you pointed out, and as I mentioned in post #7, this truth table could also perform the BCDto7segment conversion, but then you would need to implement the driver function of the 7447.
Absolutly,The input to the decoder is binary NOT BCD. That is why the decoder truth table must implement a binarytoBCD function. Since @mod already has the 7447 BCDto7segment decoder/drivers, all that is needed is the lookup table logic to convert from 4bit binary to two binarycodeddecimal (BCD) digits, with the first BCD digit being zero or one. As you pointed out, and as I mentioned in post #7, this truth table could also perform the BCDto7segment conversion, but then you would need to implement the driver function of the 7447.
ignore if my statement "more or 99" was wrong
Nice homework problem.
Hop has suggested the easiest way to proceed. Do you understand it? If so, explain it back to us and we'll see if you've got it right.
i'm sorry sir. how i make the first 7seg to MSB 0/1 .Here is a more detailed block diagram.
You can use a single 7447 ,if you do , you should replace the "MSB 7447" with logic to show 0 and 1.
View attachment 26623
thank you sir, you are very kind.Using @dorke's block diagram in post #16 above you must have a "black box" labeled Decoder that has four inputs that represent the bits on your binary counter. The Decoder has two sets of outputs: 4bits for the MSD (Most Significant Digit) and 4bits for LSD (Least Significant Digit). Each of these 4bit outputs drive a 7447 7segment decoder/driver that will accept the BCD values and display a number on their 7segment display.
Inside of the Decoder are combinatorial logic circuits that you have created from a truth table. For each of the sixteen possible combinations of the 4bit input your combinatorial logic will generate the correct 4bit binary output that the 7447 decoder/driver ICs will convert to segment drive signals for your two 7segment displays. All you need to do is take pencil and paper in hand now and draw up the correct truth table. For each of the sixteen input combinations you will have a line in your truth table that shows the values of each of the eight outputs. There will (potentially) be eight columns associated with those sixteen rows, but as you will see, only five columns are really necessary: one column for the MSD and four columns for the LSD.
Being clever, you notice that the MSD can only be either zero or one and this only requires the lsb (least significant bit) to be sent to its 7447 to display either 0 or 1. The other three bits sent to the 7447 for the MSD are always zero, so you permanently connect those three bits to logic zero level.
You need to now look at the truth table rows that represent all sixteen input bit combinations. The only combinations that require the MSD to be one are 10, 11, 12, 13, 14, and 15 which is represented by binary value 1010, 1011, 1100, 1101, 1110, and 1111. For these six input values, your combinatorial logic must output a one and send it to the lsb of the 7447 connected to the MSD. For any other input value you send a zero to the 7447 connected to the MSD.
But you aren't done yet! For those same six input values, your combinatorial logic must output 0000, 0001, 0010, 0011, 0100, and 0101 and send these to the 7447 connected to the LSD. This completes six rows of your truth table and the combinatorial logic associated with those six rows will display the decimal number 10, 11, 12, 13, 14, and 15.
But you aren't done yet! There are ten remaining rows in your truth table that will represent binary input values of 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, and 1001 which will display as 00, 01, 02, 03, 04, 05, 06, 07, 08, and 09. Hmmm. That looks suspiciously like just passing the 4bit input directly to the 4bit output for the LSD! But you can't just wire it up that way. You have to have logic that also implements the other six combinations of four input bits as described above to display 10, 11, 12, 13, 14, and 15.
So, begin to draw up your truth table. Label the first column INPUT (4 bits), label the second column MSD (1 bit), and label the third column LSD (4 bit). Under the INPUT column enter sixteen rows from 0000 to 1111. Fill in the values on each row under the remaining two columns as discussed above. Once you have a truth table you can think about creating the combinatorial logic that will implement that truth table.
So, draw up your truth table and post it here. Then we can discuss how you can implement that truth table with combinatorial logic gates. Be forewarned: it won't be pretty, but there are theorems and methods you can use to reduce the complexity.
Using @dorke's block diagram in post #16 above you must have a "black box" labeled Decoder that has four inputs that represent the bits on your binary counter. The Decoder has two sets of outputs: 4bits for the MSD (Most Significant Digit) and 4bits for LSD (Least Significant Digit). Each of these 4bit outputs drive a 7447 7segment decoder/driver that will accept the BCD values and display a number on their 7segment display.
Inside of the Decoder are combinatorial logic circuits that you have created from a truth table. For each of the sixteen possible combinations of the 4bit input your combinatorial logic will generate the correct 4bit binary output that the 7447 decoder/driver ICs will convert to segment drive signals for your two 7segment displays. All you need to do is take pencil and paper in hand now and draw up the correct truth table. For each of the sixteen input combinations you will have a line in your truth table that shows the values of each of the eight outputs. There will (potentially) be eight columns associated with those sixteen rows, but as you will see, only five columns are really necessary: one column for the MSD and four columns for the LSD.
Being clever, you notice that the MSD can only be either zero or one and this only requires the lsb (least significant bit) to be sent to its 7447 to display either 0 or 1. The other three bits sent to the 7447 for the MSD are always zero, so you permanently connect those three bits to logic zero level.
You need to now look at the truth table rows that represent all sixteen input bit combinations. The only combinations that require the MSD to be one are 10, 11, 12, 13, 14, and 15 which is represented by binary value 1010, 1011, 1100, 1101, 1110, and 1111. For these six input values, your combinatorial logic must output a one and send it to the lsb of the 7447 connected to the MSD. For any other input value you send a zero to the 7447 connected to the MSD.
But you aren't done yet! For those same six input values, your combinatorial logic must output 0000, 0001, 0010, 0011, 0100, and 0101 and send these to the 7447 connected to the LSD. This completes six rows of your truth table and the combinatorial logic associated with those six rows will display the decimal number 10, 11, 12, 13, 14, and 15.
But you aren't done yet! There are ten remaining rows in your truth table that will represent binary input values of 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, and 1001 which will display as 00, 01, 02, 03, 04, 05, 06, 07, 08, and 09. Hmmm. That looks suspiciously like just passing the 4bit input directly to the 4bit output for the LSD! But you can't just wire it up that way. You have to have logic that also implements the other six combinations of four input bits as described above to display 10, 11, 12, 13, 14, and 15.
So, begin to draw up your truth table. Label the first column INPUT (4 bits), label the second column MSD (1 bit), and label the third column LSD (4 bit). Under the INPUT column enter sixteen rows from 0000 to 1111. Fill in the values on each row under the remaining two columns as discussed above. Once you have a truth table you can think about creating the combinatorial logic that will implement that truth table.
So, draw up your truth table and post it here. Then we can discuss how you can implement that truth table with combinatorial logic gates. Be forewarned: it won't be pretty, but there are theorems and methods you can use to reduce the complexity.