Thank for your clear explanation, I learned a lot. Will the code design by finite state machine better? Because your code mixing sequential circuits and combination circuits in one block. Looking forward to your reply🥹
I would recommend an improvement to this code: set the shift register BACK to zero after the conversion is done in the last 'if' block shift_register [19:8] = 0; This will avoid the shift register continually continuing up to a larger and larger number every time the 8-bit number changes. This, I think, will save data/time as the number of times the <number> changes will continually increase Great video. The verbosity in your code made it easier to follow. Very clear!
I used this algorithm successfully in verilog but when I tried adapting it to vhdl I haven't been successful, anyone know why?
In the first if statement: "if (i==0 & (OLD_eight_bit_value != eight_bit_value)) begin" why do you assign the temp_* values to the upper bits of the shift_register? This is confusing. Why not assign to all 0's which is what they should be at the beginning of the binary to BCD conversion? Thank you for the tutorial!
If you could post videos about PISP, PIPO, and SIPO that would appreciated.
Plz make a vedio about digital alarm clock.
hello! could you tell me why addition by 3 is necessary for when one's, ten's or hundred's place is greater than or equal to 5?
Hope you enjoyed this tutorial! Let me know what was something you learned through this video!
What is the shift register value to be given for 12 bit integer?
if 2's compliment 8 bit binary number is there than what changes should I do in the above code? I want to make 2's compliment 8 bit binary to BCD converter in verilog. In 2's Complement 8 bit number, MSB [bit 7] is sign bit.
do we need a testbench to run our code ?
it's REPresents. not REEPresents. lol 1:04 1:13
I used this algorithm successfully in verilog but when I tried adapting it to vhdl I haven't been successful, anyone know why?
@SK17470