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![Shift Register Parallel In Serial Out Verilog Code For Digital Clock >>> http://shurll.com/77rkt
The shift path is shown above when SHIFT/LD is logic highmodule nand3(input one, input two, input three, output out); assign out = !(one&two&three); endmodule module nand2(input one, input two, output out); assign out = !(one&two); endmodule module DFlipFlop(input clk, input reset, input D, input set, output Q); reg Q; initial Q 8 are set, bits 7 -> 0 of bus are reset assign VCC = 1; nand3 control(!CP, !CE, PL, bus[16]); nand2 set1(!PL, setbits[7], bus[15]); nand2 set2(!PL, setbits[6], bus[14]); nand2 set3(!PL, setbits[5], bus[13]); nand2 set4(!PL, setbits[4], bus[12]); nand2 set5(!PL, setbits[3], bus[11]); nand2 set6(!PL, setbits[2], bus[10]); nand2 set7(!PL, setbits[1], bus[9]); nand2 set8(!PL, setbits[0], bus[8]); nand2 reset1(!PL, bus[15], bus[7]); nand2 reset2(!PL, bus[14], bus[6]); nand2 reset3(!PL, bus[13], bus[5]); nand2 reset4(!PL, bus[12], bus[4]); nand2 reset5(!PL, bus[11], bus[3]); nand2 reset6(!PL, bus[10], bus[2]); nand2 reset7(!PL, bus[9], bus[1]); nand2 reset8(!PL, bus[8], bus[0]); endmodule module testbench; reg [7:0]setbits; reg [7:0]setbits2; reg CE, PL, CP, D; wire [16:0]bus; wire [7:0]Q; always #5 CP = !CP; initial begin CP = 0; D = 0; setbits = 8'b11001011; PL = 0; CE = 1; #10 PL = 1; CE = 0; end initial #500 $finish; initial begin $display("CP, CE, PL, b16 b15 b14 b13 b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b1 b0 Q7 Q6 Q5 Q4 Q3 Q2 Q1 Q0"); $monitor("%b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b", CP, CE, PL, bus[16], bus[15], bus[14], bus[13], bus[12], bus[11], bus[10], bus[9], bus[8], bus[7], bus[6], bus[5], bus[4], bus[3], bus[2], bus[1], bus[0], Q[7], Q[6],Q[5],Q[4],Q[3],Q[2],Q[1],Q[0]); end controlmux CM(setbits, CP, CE, PL, bus); DFlipFlop FF(bus[16], bus[7], D, bus[15], Q[7]); DFlipFlop FF2(bus[16], bus[6], Q[7], bus[14], Q[6]); DFlipFlop FF3(bus[16], bus[5], Q[6], bus[13], Q[5]); DFlipFlop FF4(bus[16], bus[4], Q[5], bus[12], Q[4]); DFlipFlop FF5(bus[16], bus[3], Q[4], bus[11], Q[3]); DFlipFlop FF6(bus[16], bus[2], Q[3], bus[10], Q[2]); DFlipFlop FF7(bus[16], bus[1], Q[2], bus[9], Q[1]); DFlipFlop FF8(bus[16], bus[0], Q[1], bus[8], Q[0]); endmodule verilog shareimprove this question asked Nov 2 '15 at 13:16 HighAndroid 11 add a comment 1 Answer 1 active oldest votes up vote 1 down vote You have misinterpreted the schematicIt is important to note that with this type of data register a clock pulse is not required to parallel load the register as it is already present, but four clock pulses are required to unload the dataReply Eugene Lisovy (+8),4 years ago Score:1 Example of Universal Shift Register [4]AND Operation Logical AND OperationMagnetic FluxDigital Logic Gates
We only show threeasked 1 year, 8 months ago viewed 1,831 times active 1 year, 8 months ago Related -1Verilog interfacing with C#0I tried to wrote verilog code of 4-bit register with parallel loadThe output from each flip-Flop is connected to the D input of the flip-flop at its rightIt is abbreviated from our previous terminology, but works the same: parallel load if low, shift if highTest Bench of Shift Register In Verilog Published in : 2015-12-31 by VHDL Language Test Bench of Shift Register In Verilog by manohar mohantaAfter t3, QA QB QC = 001J K Flip FlopAfter the slash, the 3 (arrow) indicates shifting
All Rights ReservedIf CLKINH is high, the clock is inhibite](http://cdn07.dayviews.com/500/_u3/_u8/_u7/_u6/_u6/_u0/u3876609/62944_1501189070/Shift_Register_Parallel_In_Serial_Out_Verilog_Code_For_Digital_Clock_http_shurll_com_77rkt_The.jpg)
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Shift Register Parallel In Serial Out Verilog Code For Digital Clock >>> http://shurll.com/77rkt
The shift path is shown above when SHIFT/LD is logic highmodule nand3(input one, input two, input three, output out); assign out = !(one&two&three); endmodule module nand2(input one, input two, output out); assign out = !(one&two); endmodule module DFlipFlop(input clk, input reset, input D, input set, output Q); reg Q; initial Q 8 are set, bits 7 -> 0 of bus are reset assign VCC = 1; nand3 control(!CP, !CE, PL, bus[16]); nand2 set1(!PL, setbits[7], bus[15]); nand2 set2(!PL, setbits[6], bus[14]); nand2 set3(!PL, setbits[5], bus[13]); nand2 set4(!PL, setbits[4], bus[12]); nand2 set5(!PL, setbits[3], bus[11]); nand2 set6(!PL, setbits[2], bus[10]); nand2 set7(!PL, setbits[1], bus[9]); nand2 set8(!PL, setbits[0], bus[8]); nand2 reset1(!PL, bus[15], bus[7]); nand2 reset2(!PL, bus[14], bus[6]); nand2 reset3(!PL, bus[13], bus[5]); nand2 reset4(!PL, bus[12], bus[4]); nand2 reset5(!PL, bus[11], bus[3]); nand2 reset6(!PL, bus[10], bus[2]); nand2 reset7(!PL, bus[9], bus[1]); nand2 reset8(!PL, bus[8], bus[0]); endmodule module testbench; reg [7:0]setbits; reg [7:0]setbits2; reg CE, PL, CP, D; wire [16:0]bus; wire [7:0]Q; always #5 CP = !CP; initial begin CP = 0; D = 0; setbits = 8'b11001011; PL = 0; CE = 1; #10 PL = 1; CE = 0; end initial #500 $finish; initial begin $display("CP, CE, PL, b16 b15 b14 b13 b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b1 b0 Q7 Q6 Q5 Q4 Q3 Q2 Q1 Q0"); $monitor("%b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b %b", CP, CE, PL, bus[16], bus[15], bus[14], bus[13], bus[12], bus[11], bus[10], bus[9], bus[8], bus[7], bus[6], bus[5], bus[4], bus[3], bus[2], bus[1], bus[0], Q[7], Q[6],Q[5],Q[4],Q[3],Q[2],Q[1],Q[0]); end controlmux CM(setbits, CP, CE, PL, bus); DFlipFlop FF(bus[16], bus[7], D, bus[15], Q[7]); DFlipFlop FF2(bus[16], bus[6], Q[7], bus[14], Q[6]); DFlipFlop FF3(bus[16], bus[5], Q[6], bus[13], Q[5]); DFlipFlop FF4(bus[16], bus[4], Q[5], bus[12], Q[4]); DFlipFlop FF5(bus[16], bus[3], Q[4], bus[11], Q[3]); DFlipFlop FF6(bus[16], bus[2], Q[3], bus[10], Q[2]); DFlipFlop FF7(bus[16], bus[1], Q[2], bus[9], Q[1]); DFlipFlop FF8(bus[16], bus[0], Q[1], bus[8], Q[0]); endmodule verilog shareimprove this question asked Nov 2 '15 at 13:16 HighAndroid 11 add a comment 1 Answer 1 active oldest votes up vote 1 down vote You have misinterpreted the schematicIt is important to note that with this type of data register a clock pulse is not required to parallel load the register as it is already present, but four clock pulses are required to unload the dataReply Eugene Lisovy (+8),4 years ago Score:1 Example of Universal Shift Register [4]AND Operation Logical AND OperationMagnetic FluxDigital Logic Gates
We only show threeasked 1 year, 8 months ago viewed 1,831 times active 1 year, 8 months ago Related -1Verilog interfacing with C#0I tried to wrote verilog code of 4-bit register with parallel loadThe output from each flip-Flop is connected to the D input of the flip-flop at its rightIt is abbreviated from our previous terminology, but works the same: parallel load if low, shift if highTest Bench of Shift Register In Verilog Published in : 2015-12-31 by VHDL Language Test Bench of Shift Register In Verilog by manohar mohantaAfter t3, QA QB QC = 001J K Flip FlopAfter the slash, the 3 (arrow) indicates shifting
All Rights ReservedIf CLKINH is high, the clock is inhibited, or disabledDigital ElectronicsFigure 1 shows a PISO shift register which has a control-line (SH/) and combinational circuit (AND and OR gates) in addition to the basic register components (flip-flops) fed with clock and clear pinsThe alarm reads the remote keypad every few tens of milliseconds by sending shift clocks to the keypad which returns serial data showing the status of the keys via a parallel-in/ serial-out shift register555 Timer and 555 Timer WorkingAfter t2, QA QB QC = 010The parallel data inputs of the shift register are pulled up to +5V with a resistor on each input b84ad54a27
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