采用Xilinx我已经编码的8051架构VHDL。 在试图提高时钟频率,我曾流水线所有的80C51指令。 该指令能够按预期运行,对如。 正在被处理的第一指令时,所述第二指令获取取出。
不过,我只尽管创造了3流水线深度,从综合报告得到(约+/-赫兹)的稍高的时钟频率。 我想通了,由综合报告指定的瓶颈是由于一个操作,但我不明白的综合报告。
请问什么是“SEQ / decode_3到SEQ / i_ram_addr_7”努力做数据路径? (从我的猜测,我推断,使用的情况下,当语句来检查100+相关操作码,但不知道这是瓶颈,但是我无能)
因此,我只有2个查询是:
首先,是有可能,流水线不增加时钟频率和测试平台是解释在时序减少的唯一途径?
其次,我怎么可能推断出在我的代码路径是从“SEQ / decode_3到SEQ / i_ram_addr_7”的瓶颈。
谢谢你的人谁可以帮助解释我的疑惑!
Timing Summary:
---------------
Speed Grade: -4
Minimum period: 12.542ns (Maximum Frequency: 79.730MHz)
Minimum input arrival time before clock: 10.501ns
Maximum output required time after clock: 5.698ns
Maximum combinational path delay: No path found
Timing Detail:
--------------
All values displayed in nanoseconds (ns)
=========================================================================
Timing constraint: Default period analysis for Clock 'clk'
Clock period: 12.542ns (frequency: 79.730MHz)
Total number of paths / destination ports: 113114 / 2670
-------------------------------------------------------------------------
Delay: 12.542ns (Levels of Logic = 10)
Source: SEQ/decode_3 (FF)
Destination: SEQ/i_ram_addr_7 (FF)
Source Clock: clk rising
Destination Clock: clk rising
Data Path: SEQ/decode_3 to SEQ/i_ram_addr_7
Gate Net
Cell:in->out fanout Delay Delay Logical Name (Net Name)
---------------------------------------- ------------
FDC:C->Q 102 0.591 1.364 SEQ/decode_3 (SEQ/decode_3)
LUT4_D:I1->O 10 0.643 0.885 SEQ/de_state_cmp_eq002111 (N314)
LUT4:I3->O 7 0.648 0.740 SEQ/de_state_cmp_eq00711 (SEQ/de_state_cmp_eq0071)
LUT4:I2->O 3 0.648 0.534 SEQ/i_ram_addr_mux0000<0>11111 (N2301)
LUT4:I3->O 1 0.648 0.000 SEQ/i_ram_addr_mux0000<0>11270_SW0_SW0_F (N1284)
MUXF5:I0->O 1 0.276 0.423 SEQ/i_ram_addr_mux0000<0>11270_SW0_SW0 (N955)
LUT4_D:I3->O 6 0.648 0.701 SEQ/i_ram_addr_mux0000<0>11270 (SEQ/i_ram_addr_mux0000<0>11270)
LUT3_L:I2->LO 1 0.648 0.103 SEQ/i_ram_addr_mux0000<7>221_SW2_SW0 (N1208)
LUT4:I3->O 1 0.648 0.423 SEQ/i_ram_addr_mux0000<7>351_SW1 (N1085)
LUT4:I3->O 1 0.648 0.423 SEQ/i_ram_addr_mux0000<7>2 (SEQ/i_ram_addr_mux0000<7>2)
LUT4:I3->O 1 0.648 0.000 SEQ/i_ram_addr_mux0000<7>167 (SEQ/i_ram_addr_mux0000<7>)
FDE:D 0.252 SEQ/i_ram_addr_7
----------------------------------------
Total 12.542ns (6.946ns logic, 5.596ns route)
(55.4% logic, 44.6% route)
=========================================================================
Timing constraint: Default OFFSET IN BEFORE for Clock 'clk'
Total number of paths / destination ports: 154 / 154
-------------------------------------------------------------------------
Offset: 8.946ns (Levels of Logic = 6)
Source: rst (PAD)
Destination: SEQ/i_ram_diByte_1 (FF)
Destination Clock: clk rising
Data Path: rst to SEQ/i_ram_diByte_1
Gate Net
Cell:in->out fanout Delay Delay Logical Name (Net Name)
---------------------------------------- ------------
IBUF:I->O 444 0.849 1.392 rst_IBUF (REG/ext_int/fd_out1_0__or0000)
BUF:I->O 445 0.648 1.425 rst_IBUF_1 (rst_IBUF_1)
LUT3:I2->O 4 0.648 0.730 ROM/data<1>1 (i_rom_data<1>)
LUT4:I0->O 1 0.648 0.500 SEQ/i_ram_diByte_mux0000<1>17_SW0 (N1262)
LUT4:I1->O 1 0.643 0.563 SEQ/i_ram_diByte_mux0000<1>32 (SEQ/i_ram_diByte_mux0000<1>32)
LUT4:I0->O 1 0.648 0.000 SEQ/i_ram_diByte_mux0000<1>60 (SEQ/i_ram_diByte_mux0000<1>)
FDE:D 0.252 SEQ/i_ram_diByte_1
----------------------------------------
Total 8.946ns (4.336ns logic, 4.610ns route)
(48.5% logic, 51.5% route)
=========================================================================
为了让我更加specfic,我给的例子代码snipplet在1个操作码译码阶段。
解码opdcode,这是一个mov指令时,以下是1这样的情况下。 大约有超过100个的操作码(100+指令),这意味着该情况下陈述有超过100时语句。
案例操作码
--mov A,Rn中
当 “11101000” | “11101001” | “11101010” | “11101011” | “11101100” | “11101101” | “11101110” | “11101111”=>情况下de_state是当E7 =>de_state <= E8; when E8 => de_state <= E9; when E9 => de_state <= E10; when E10 => --Draw PSW i_ram_addr <= xD0; i_ram_rdByte <= '1'; de_state <= E11; when E11 => --Draw from Rn i_ram_addr <= "000" & i_ram_doByte(4 downto 3)& opcode(2 downto 0); i_ram_rdByte <= '1'; de_state <= E12; when E12 => --Place into EDR EDR <= i_ram_doByte; --close rdByte i_ram_rdByte <= '0'; when others => end case;
我希望你能有我的VHDL代码更好的主意。 我将不胜感激任何形式的帮助。 谢谢!