Working with numbers and simple arithmetic in an up/down counter. • modularizing code by instantiating VHDL modules in other VHDL modules. • driving your circuit with a. VHDL Module” to create a module called “counter” with the following inputs and outputs entity counter is. Two Speed Counter. Add another input to. Rivatuner statistics server msi. I need to design an 8 bit up counter in VHDL using T flip flop and generate statement. Worms 2 no cd in drive errors download. I know how the counter works, but I am not able to design it in VHDL. The main problem is 'using generate statement'. As I can see, in an up counter (ripple or synchronous; either is OK; I decided to use ripple ), the previous output is fed as a clock. So I can map Q(i -1) for clock in the i'th generated flip flop. But what about the first flip flop? It has a clock explicitly applied. So if I use generate statement, I cannot simply map the clock to previous output; the first flip flop will always create problem. Another thought which came was to initialise first counter explicitly and then use generate statement for remaining 7 flip flops. But here as well, I think, the first (or rather, second) flip flop using generate statement will pose a similar problem ( ie mapping clock). Am I incorrect or missing something important here? Here is the code ( planned code, not actual code): component tff is Port ( t: IN BIT; clk: IN BIT; q: OUT BIT ); end component --Other irrelevant stuff For i IN 0 TO 7 GENERATE tffx: tff PORT MAP ( tIn, q(i-1), q(i) ); end GENERATE; --More irrelevant stuff Thank you. Here is a simple generate statement, which generates 8 TFFs and connects the clock input of the tff to the q output from previous FF. Because you are using indices calculations (i+1 or i-1), you need to wider range for the tff_clocks range or you must shorten the generate loop. I'm using a loop from 0 to 7 so I extended tff_clocks by 1. Index 0 is connected to the original system clock. Discover releases, reviews, credits, songs, and more about Six Ft Ditch - Unlicensed Cemetary at Discogs. Complete your Six Ft Ditch collection. View credits, reviews, tracks and shop for the 2005 CD release of Unlicensed Cemetary on Discogs. Six ft ditch unlicensed cemetery music. Signal tff_clocks: std_logic_vector(8 downto 0); begin tff_clocks(0) tff_clocks(i), t => '1', q => tff_clocks(i + 1) ); end generate; async_counter_result. I'm doing a college's task that asks for implementing in VHDL an up/down asynchronous counter. My implementation consistis of using a control variable ctrl so when it's 0, the counter counts in ascendant order, else in descendent one. The code I've implemented (In the discipline, we use Quartus 13 and FPGA Cyclone IVE EP4CE129C7 for simulation) is followed in this. The resulting, however, exhibits only '0' for outputs q0 and q1. So, where can be a possible bottleneck in the code so that this is occuring? I'm not a quartus user so have no familiarity with the limits of your waveform viewer, but this looks like it's caused from migrating from type BIT to type Std_Logic. Where the default value (unless otherwise specified would be 'U' for Std_Logic (Std_ULogic) while it would be '0' for type BIT. That results in you assigning NOT 'U' for the FFT q's to the input of the MUX. The result is your flip flops never get assigned a value other than 'U'. From the std_logic_1164 package: -- truth table for 'not' function CONSTANT not_table: stdlogic_1d:= -- ------------------------------------------------- --| U X 0 1 Z W L H -| -- ------------------------------------------------- ( 'U', 'X', '1', '0', 'X', 'X', '1', '0', 'X' ); NOT of a 'U' is a 'U'. With a waveform viewer that shows 'U' values: This was done with ghdl and gtkwave, and an added testbench. The only change to architecture arc of Taraefa09 was to add sq0_n and assign it in a concurrent signal: INV: sq0_n.
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