數(shù)字設(shè)計(jì)筆試Verilog手撕代碼—累加器
已知一個(gè)加法器IP，其功能是計(jì)算兩個(gè)數(shù)的和，但這個(gè)和延遲兩個(gè)周期才會(huì)輸出?，F(xiàn)在有一串連續(xù)的數(shù)據(jù)輸入，每個(gè)周期都不間斷，試問最少需要例化幾個(gè)上述的加法器IP，才可以實(shí)現(xiàn)累加的功能。
設(shè)計(jì)分析
實(shí)現(xiàn)累加器的加法器例化的個(gè)數(shù)。按照原文大佬的設(shè)計(jì)方法，因?yàn)閿?shù)據(jù)連續(xù)且加法器的延遲周期是2，使用使用一個(gè)實(shí)現(xiàn)累加，會(huì)有一半的數(shù)據(jù)丟失。那這樣設(shè)計(jì)他就將奇數(shù)偶數(shù)的數(shù)據(jù)進(jìn)行了分開做一級(jí)累加，然后第二級(jí)將奇數(shù)偶數(shù)的累加結(jié)果再累加。這樣做共需消耗3個(gè)加法器。
這樣設(shè)計(jì)當(dāng)然沒問題，但是這樣設(shè)計(jì)是否是最少呢？我先拋出我的思考，我認(rèn)為在允許少量邏輯設(shè)計(jì)的情況下，最少需要例化兩個(gè)上述的加法器IP可以實(shí)現(xiàn)累加。
如果比較極限的情況下，一個(gè)都可以，先把一串?dāng)?shù)據(jù)使用寄存器緩存，然后一個(gè)一個(gè)取出來慢慢算即可，但這樣是不太可取的，首先，數(shù)據(jù)是連續(xù)的并沒有給出數(shù)據(jù)的極限長度，也就是說不論用任何涉及存儲(chǔ)結(jié)構(gòu)進(jìn)行緩存，都沒法確保該次數(shù)據(jù)能完全被存儲(chǔ)。如果題目改成一串連續(xù)數(shù)據(jù)輸入，長度最大為10，那我認(rèn)為用寄存器緩存這樣的設(shè)計(jì)是合理的。
設(shè)計(jì)架構(gòu)
回到設(shè)計(jì)思路：用兩個(gè)加法器的結(jié)構(gòu)如圖示。
設(shè)計(jì)實(shí)現(xiàn)
加法器設(shè)計(jì)
假設(shè)兩個(gè)時(shí)鐘周期延時(shí)加法器代碼如下，通過例化加法器進(jìn)行構(gòu)建累加器。
//加法器IP
module adder
#(parameter DATA_WIDTH = 8)(
input clk,
input rst_n,
input [DATA_WIDTH-1:0] a_in,
input [DATA_WIDTH-1:0] b_in,
output reg [DATA_WIDTH-1:0] out
);
reg [DATA_WIDTH-1:0] sum;
always @(posedge clk or negedge rst_n)begin
if(rst_n == 'd0)begin
sum <= 'd0;
out <= 'd0;
end
else begin
sum <= a_in + b_in;
out <= sum;
end
end
endmodule
累加器設(shè)計(jì)
//累加器實(shí)現(xiàn)
module adder_for_acc
#(parameter DATA_WIDTH = 8)
(
input clk,
input rst_n,
input [DATA_WIDTH-1:0] din,
input din_valid,
output reg dout_valid,
output reg [DATA_WIDTH-1:0] dout
);
reg [DATA_WIDTH-1:0]din_r0;
//打一拍
always @(posedge clk or negedge rst_n)begin
if(rst_n == 'd0)begin
din_r0 <= 'd0;
end
else if(din_valid==1'B1)begin
din_r0<= din;
end
else begin
din_r0<='d0;
end
end
//adder0_valid信號(hào)
reg adder0_valid;
always @(posedge clk or negedge rst_n)begin
if(rst_n == 'd0)begin
adder0_valid <= 'd0;
end
else if(din_valid==1'B1)begin
adder0_valid<=!adder0_valid;
end
else begin
adder0_valid<='d0;
end
end
wire[DATA_WIDTH-1:0] a_in = (adder0_valid && din_valid)?din:0;
wire[DATA_WIDTH-1:0] b_in = (adder0_valid)?din_r0:0;
wire[DATA_WIDTH-1:0] ab_sum;
adder adder0_dut (
.clk (clk ),
.rst_n(rst_n ),
.a_in (a_in ),
.b_in (b_in ),
.out (ab_sum)
);
//第一級(jí)加法器輸出有效信號(hào)
reg [1:0]adder0_valid_dly;
wire ab_sum_valid = adder0_valid_dly[1];
always @(posedge clk ) begin
adder0_valid_dly<={adder0_valid_dly[0],adder0_valid};
end
wire [DATA_WIDTH-1:0] sum_in;
wire [DATA_WIDTH-1:0] ab_sum_in = (ab_sum_valid)?ab_sum:0;
wire [DATA_WIDTH-1:0] accsum_in = (ab_sum_valid)?sum_in:dout;
adder adder1_dut (
.clk (clk ),
.rst_n(rst_n ),
.a_in (ab_sum_in),
.b_in (accsum_in),
.out (sum_in )
);
//第二級(jí)加法器輸出有效信號(hào)
reg [3:0]din_valid_r0;
reg [1:0]adder1_valid_dly;
wire adder1_outvld = adder1_valid_dly[1];
always @(posedge clk ) begin
adder1_valid_dly<={adder1_valid_dly[0],ab_sum_valid};
end
//輸出
always @(posedge clk ) begin
din_valid_r0<={din_valid_r0[2:0],(din_valid || adder0_valid)};
end
always @(posedge clk or negedge rst_n) begin
if(rst_n == 'd0)begin
dout <= 'd0;
dout_valid <= 'd0;
end
else if(adder1_outvld == 1 && (din_valid_r0[3]==1 && din_valid_r0[2]==0))begin
dout <= sum_in ;
dout_valid <= 'd1;
end
else begin
dout <= dout ;
dout_valid <= 'd0;
end
end
endmodule