[Digital IC manual tearing code] Verilog automatic beverage machine | topic | principle | design | simulation

Preface

This series aims to provide 100% Accurate numbers IC Design / Verify the title of the hand tearing code link , principle ,RTL Design ,Testbench And reference simulation waveform , The content of each article is checked by simulation . The quick navigation links are as follows ：

Odd frequency division
Even frequency division
Semi integer batch
decimal / Fractional frequency division
Sequence detector
Mode three detector
Beverage machine
Asynchronous reset , Simultaneous release
Edge detection （ Rising edge , Falling edge , On both sides ）
Full adder , Half adder
Gray code to binary
single bit Cross clock domain （ Two beats , Edge synchronization , Pulse synchronization ）
Sync FIFO

Ought to say , The hand tearing code link is in the interview process Both important and simple A part of , Compared with software jobs , Numbers IC Hand tear code Fixed topic , Limited number , It belongs to a link that must be scored in the whole interview , Outside this series , I also recommend numbers IC Job seekers use “HdlBits” Code Training
Links are as follows
HDLBits — Verilog Practice

Problem of automatic beverage machine

1. Use Verilog Design circuit , Complete the following functions : Each bottle of drink 1.5 element , You can only put in one coin at a time , Available input 0.5 And 1.0 Two kinds of coins , With change function .

Principle of automatic beverage selling machine

First , It is obvious that , This hand tearing code should be used State machine To complete
The distinction between States , We can classify according to the amount remaining in the beverage machine , Because there are only two kinds of coins ,0.5/1 element , therefore The state can be divided as follows

• IDLE： Reset state , Indicates that the balance in the beverage machine is 0 element
• s1: The balance of the beverage machine is 0.5 element
• s2: The balance in the beverage machine is 1 element
• s3: The balance in the beverage machine is 1.5 element （ Export drinks , No change ）
• s4: The balance in the beverage set is 2 element （ Export drinks , Give change )

So for the input and output of this state machine
Input : Two of you input Represents the coin put ,input[1] Pulling up means investing one yuan ,input[0] Higher means investment 0.5 element , By default, only one coin can be inserted at a time .
Output :drink Represents the output of drinks ,coin Represents the output of coins .

Veilog Design

module drink_machine( clk ,rst_n, money,drink,coin);
input clk;
input rst_n;
input [1:0] money;
output drink;
output coin;

parameter IDLE = 3'd0, s1=3'd1 ,s2=3'd2, s3=3'd3,s4=3'd4;

reg [2:0] state,nstate;

[email protected](posedge clk or negedge rst_n)
begin
if(!rst_n)
state<=IDLE;
else
state<=nstate;
end

[email protected](*)
begin
case(state)
IDLE:nstate = money[1] ? s2 : (money[0] ? s1 :IDLE);
s1:  nstate = money[1] ? s3 : (money[0] ? s2 : s1);
s2:  nstate = money[1] ? s4 : (money[0] ? s3 : s2);
s3:  nstate = money[1] ? s2 : (money[0] ? s1 :IDLE);
s4:  nstate = money[1] ? s2 : (money[0] ? s1 :IDLE);
default nstate = IDLE;
endcase
end

assign drink = (state == s3 || state == s4)? 1:0;
assign coin  = (state == s4)? 1:0;


endmodule

Testbench Design

module drink_machine_tb();
reg clk;
reg rst_n;
reg [1:0] money;
wire drink;
wire coin;

drink_machine u1 (.clk(clk),.rst_n(rst_n),.money(money),.drink(drink),.coin(coin));

always #5 clk = !clk;

always #10.001 money = {
    $random} % 3;

initial
begin
clk = 0;
rst_n = 1;
#10
rst_n = 0;
#20
rst_n = 1;
#1000
$stop;
end


endmodule

Simulation results

from 20ns To 80ns In the process of , Cast Five cents a piece , One by one ,drink Output is 1.
Then , Cast a piece , Throw another piece ,drink Output is 1 At the same time coin Also output as 1
Other state jumps also meet expectations , Design establishment .