Foreword:
This column aims to record high-frequency written interview hand-torn code questions for the digital front-end autumn tricks. All articles in this column provide principle analysis, codes and waveforms, and all codes have been verified by myself.

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topic description

Verilog RTL implementation of a divider. 16bitA, 8bitB. C=A/B

Solutions

In hardware, they are all binary numbers. Binary division is similar to decimal division. It is a process of shifting and comparing sizes.

calculation steps:

① Compare the high-order data of the dividend with the divisor. If the former >= the latter, the quotient of the corresponding bit can be obtained as 1, and the difference between the two can be used to obtain the remainder of the first step; otherwise, the corresponding quotient is 0, and the former is directly used as remainder.

② Splicing the remainder in the previous step and the remaining 1-bit data of the dividend into new data, and then comparing it with the divisor, a new quotient and remainder can be obtained.

③ Repeat process ② until the lowest digit of the dividend also participates in the calculation.

For example 375/23 = 16'b0000_0001_0111_0111 / 8'b0001_0111

Take the first digit of the dividend, compared with the divisor, 8'b0000_0000<8'b0001_0111, the quotient is 0, and the former is used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0000< 8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0000< 8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0000< 8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0000< 8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0000< 8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0000< 8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0001< 8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0010< 8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0101< 8'b0001_0111, the quotient is 0, and the former continues to be the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_1011< 8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue the comparison, 8'b0001_0111<=8'b0001_0111, the quotient is 1, and the difference between the two is used as the remainder.

Then take the number shifted one bit to the right and continue the comparison, 8'b0000_0000<8'b0001_0111, the quotient is 0, and the former is used as the remainder.

Then take the number shifted one bit to the right and continue the comparison, 8'b0000_0001<8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue to compare, 8'b0000_0011<8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder.

Then take the number shifted one bit to the right and continue the comparison. 8'b0000_0111<8'b0001_0111, the quotient is 0, and the former continues to be used as the remainder, which has been compared to the last digit. The output result: the quotient is 1<<4=16, and the remainder is: 8'b0000_0111=7.

Therefore, the calculation is 375/23=16...7.

the code

module divisor(
    input     [15:0]    A       ,
    input     [7:0]     B       ,
    output    [15:0]    result  ,
    output    [7:0]     remain
    );
    reg [15:0]  a_reg   ;
    reg [7:0]   b_reg   ; 
    reg [31:0]  temp_a  ;  # 其实这里取16+8+1bit就够了，取32位是为了好看。
    reg [31:0]  temp_b  ;

    integer     i;
    always@(*)begin
        a_reg = A;
        b_reg = B;
    end

    always@(*)begin
        temp_a = {16'h0,a_reg};
        temp_b = {b_reg,16'h0};
        
        for(i=0;i<16;i=i+1)begin
            temp_a = temp_a <<1;
            if(temp_a >= temp_b)begin
                temp_a = temp_a-temp_b+1;
            end
            else begin
                temp_a = temp_a;
            end
        end
    end

    assign remain = temp_a[31:16];
    assign result = temp_a[15:0];

endmodule

testbench

module divisor_tb();

reg   [15:0]    A       ;
reg   [7:0]     B       ;
wire  [15:0]    result  ;
wire  [7:0]     remain  ;

initial begin
  #10
  A <=  16'd375;
  B <=  8'd23;
  #10
  A <=  16'd557;
  B <=  8'd57;
end

divisor u_divisor(
  .A      (A)       ,
  .B      (B)       ,
  .result (result)  ,
  .remain (remain)
);

endmodule

Waveform

Formula 1: 375/23=16...7

Formula 2: 557/57=9 ...... 44

The results are consistent with what we considered.

Digital IC tearing code--Xiaomi Technology (divider design)

topic description

Solutions

the code

testbench

Waveform

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