![]() Using this method requires seven clock cycles after initialization for an eight digit result. The remainder from the last subtraction is the one's digit. Repeat this procedure down to input7 minus 90, 80, 70. Take the remainder from largest factor where the result is not negative (test the top bit of the result), record the factor as the top digit and use the remainder as the input to the next stage, which is again 9 subtractions in parallel. How? By doing long division in parallel.įor an eight digit result from a 27-bit input, start by doing 9 subtractions in parallel, input1 minus: 9000_0000, 8000_0000, 7000_0000. A reachable goal for an N-digit result is N-1 clock cycles. For an 8-digit result from a 27-bit input Double-Dabble requires 35 clock cycles. While Double-Dabble works, the example given here uses 11 clock cycles after initialization to arrive at a 3-digit result from an 8-bit input. FPGA's and ASIC's are parallel in nature: we need to use their strengths to our advantage. The problem with Double-Dabble is that it was written for software programs, so the solution is inherently serial. Here is an example, stolen from this Wikipedia Page:Ġ000 0000 1010 10011000 Add 3 to ONES, since it was 7Ġ000 0001 1000 00110000 Add 3 to ONES, since it was 5Ġ000 1001 0000 11000000 Add 3 to TENS, since it was 6īuilding this process in Verilog is relatively straight forward. This approach basically is a way to overflow any values greater or equal to 10 in a digit into the next one without too much hardware. Essentially the approach is to scan through the binary representation, then any time you see a number which is 5 or higher in each group of 4 bits, you add 3 to it. The simplest way of converting from binary to BCD is an algorithm called "Shift-Add-3" or "Double-Dabble" (both names for the same thing). ![]() This is an ideal format for outputting to 7-segment displays, screens, in fact anything that needs a decimal number to be displayed. In BCD each digit is 4 bits, but those 4 bits are only used to represent the numbers 0-9 (hence the decimal bit). The problem you are having is quite a common one - how to convert a binary number to something called "Binary Coded Decimal" (BCD). I want it so that if I pass in binary 10, ( 8'b00001010), then HEX1 HEX0 should be 1 0, not 0 A (as hex works) The only problem is that after 9, the values will become in hex letters. ![]() To output on HEX0 and HEX1, I can do something like this: hex_decoder h0(.hex_digit(value), HEX0) Suppose I have the module as follows: module hex_decoder(hex_digit, segments) Suppose I have an 8 bit value, say 8'b00000001 (1)
0 Comments
Leave a Reply. |