VHDL project : 5 bit shift reg

Michael wrote:
> Code is attached too .
Please attach the 5 files unzipped. Then I'm able to read it on a phone 
also...

Sorry , just registered to be able to upload them all .

Thanks

Do you have to use all of that components for that shift register?
Or can you do it simply like the rest of the world?

I have to use these components so i can have all the "options" that 
final product needs to have .

I have a schematic but unfortunately it is still a draft , and not 
completed .

Here is a non complete diagram that may help .

The schematic is wrong .

Black lines going upwards , were meant to connected on the inputs .

I will upload the right schematic later today , and now i started to 
understand how to make it happen .

> Reset is load with all zero?

Yes

> Right,left shift ok but where is input and output?

You get a 5 bit input an a standard output, and every time you select 
what you want to do with input .

> Circular shift, ok, input = output?

yes

> Try to describe each "function" in HDL and then pack it in a case
> structure.

I have to do it using different modules

- - - - - -

BTW , i think i made it working !

You give a 5 bit input and the number of the selection (what to do) and 
it gives output based on them of course .

I have uploaded the file . Of course this one doesn't use the prescaler 
, it's just for simulation .

Wow, lots of Errors, you did not even try to simulate.

But i fixed them ... and still it does not work as intended. Am i right 
that with SEL = "010" it should shift right with every clock?

Sorry for the late reply .

I don't get any errors , only 2 warnings if i remember corectly.

Sel='010' shifts left every clock , circular .

Here are the final's programms options (sel) , sorry .


000 : Parallel loading /
001 : Right shift /
010 : Left shift /
011 : Circular right shift /
100 : Circular left shift /
101 : 1s' complement /
110: 2s' complement

Can you help me fix the errors you found ?


Thamnks a lot

Michael K. wrote:
> I don't get any errors , only 2 warnings if i remember corectly.

I get errors for

I(5) =>not(INP(0)),

in port map.
and for

clear =>not(res),

and for

Sout =>sum2,
because Sout is not defined in half_adder.

But if it works for you every think seems to be fine.

I changed S to Sout in half-adder , to avoid problems .

I uploaded the files again (ignore prescaler that it is inside) , i 
would really appreciate it if you test it and help me a little with the 
test bench file (i can only check 1 input and 1 option (sel) per time) .

Here the Testbench and Project folder.

Thank you very much for your help .

It works OK from what i see in the simulation .

I will have to get the fpga (spartan 3) this coming week to test it , so 
i'll probably need to figure out how to upload it to the fpga too (will 
update here too) .

Thanks a lot Gustl and everyone else that helped with their answer !

Thank you! And yes, it is a hard long way :-)

So ... i think it is not working. Here pass many clockcycles while 
nothing changes, no shifts or so.

HDLs like VHDL can be used even on a more abstract level. Ne need to 
describe singe FlipFlops. A shift register with 5 bits can be written 
as:

signal register: std_logic_vector(4 downto 0); --5 bits total
...
begin
...
register <= register(0) & register(4 downto 1); --circular right shift
register <= register(3 downto 0) & register(4); --circular left shift

And the MUX can be written as

if .... then
elsif ... then
endif;

or with the case statement.

Thanks . I'm new to all this so i'll have to learn a lot .

Check the file , for me the project you sent me works OK . EKS is the 
output , and everytime it is as requested .

Really? I understand shift registers a bit different:

When it is in the state "shift right" the contend is shifted right one 
step with every clock cycle. So in the waveform sel is 1 ("001") and 
there are many clock cycles but is does not shift with each cycle.

It should look like 
http://www.cse.psu.edu/~kxc104/class/cmpen297B/08f/hw/hw7/serialShift2.png
or the contend:

sel = "001"
it gehts loaded with "11000" and then it should shift:
"11000"
"01100"
"00110"
"00011"
...

But maybe we have different understanding and it is all right.

To be honest i don't know if it is requested to be in that way . But i 
understand now what you mean .

Is it difficult to modify current project to work as you suggest ?

Thanks

Hm, the way you started is very educating, but not easy. Yes, you can 
describe every single FlipFlop and MUX and half-adder, but you could use 
the language constructs from VHDL instead. I don't know your constraints 
if you must describe each element on its own.

I attached a waveform which shows what i guess it should show :-)

The register contains "11000"
Then res goes high for one clock
The register contains "00000"
Then sel goes to "001" and the registers loads the input "11000" again.

Then 3 clocks right shift:
"0110"
"00110"
"00011"

Then 3 clocks left shift:
"00110"
"01100"
"11000"

Then 6 clocks Circular right shift
"0110"
"00110"
"00011"
"10001"
"11000"
"01100"


Then 6 clocks Circular left shift
"11000"
"10001"
"00011"
"00110"
"01100"
"11000"

But i don't know what is meant by
1s' complement and 2s' complement. They are representation of numbers. 
Should the shift register count or somethig like that? Or should the 
contend be converted?

Edit:
I attatched the schematic view from xilinx ise (the software you use) 
and from the newer vivado (which does not support spartan FPGAs). Looks 
like your hand-drawn schematic is very close to the schematic ise made 
from my hdl code which is working. but on more intensive look not sooo 
close. i recommend to write it in a more abstract way, not describing 
instantiating single FFs and MUXes. Let the toolchain do that for you.

Just asked and you're totally right . It has to change every cycle.

Thing is i can't really figure out how to do it .

Btw , 1s complement is the opposite . for example , if you give 11001 it 
should give 00110 and 2s complement is : 1s complement plus 1 (previous 
example gives 00111).

OK, so ... i attatched the Testbench and a startingpoint for the 
fivebit.vhd description.

The only data which is stored in this description are the bits in the 
register:

1

signal data: std_logic_vector(4 downto 0):=(others => '0');

The output is directly connected to the content of the register:

1

EKS <= data;

then we use a clocked process, so everything is triggered with the 
rising edge of clock:

1

process begin

2

  wait until rising_edge(clock);

3

.

4

.

5

.

6

end process;

The reset "res" is synchron with the clock, read as "if res = '1' and a 
rising edge of clk happens" then the reset is executed. asynchron reset 
should not be used in modern FPGAs.

After the reset there is the MUX which can be written as many

1

if SEL =  "001" then data <= whatever_you_like;

2

elsif SEL =  "010" then data <= whatever_you_like;

3

else data <= whatever_you_like;

4

end if;

it can also be written like

1

case SEL is

2

  when "001" => data <= whatever_you_like;

3

  when "010" => data <= whatever_you_like;

4

  when others => data <= whatever_you_like;

5

end case;

The "when others" case is mandatory. in the fivebit.vhd i used if .. 
else description.

I wrote the first 4 entrys:

1

    if    SEL = "000" then data <= (others => '0'); --Reset

2

    elsif SEL = "001" then data <= INP; --Parallel loading

3

    elsif SEL = "010" then data <= '0' & data(4 downto 1); --Right shift

4

    elsif SEL = "011" then data <= data(3 downto 0) & '0'; --Left shift

the rest is left for you. circular right/left shift is like normal 
right/left shift but with the bit falling out at one end shifted in the 
other end again.
I still don't really understand what should happen to the bits in the 
register with 1s or 2s complement selected. And with normal right/left 
shift this

1

elsif SEL = "010" then data <= '0' & data(4 downto 1);

shifts right, and the leftmost bit is filled with '0'. Is this how it 
shold be or should the leftmost bit be for example a new bit from the 
input?
like data <= INP(0) & data(4 downto 1) or so ...

Gustl B. wrote:
> OK, so ... i attatched the Testbench and a startingpoint for the
> fivebit.vhd description.
Nice job, but I think it must be cleared whether those components from 
the initial post MUST or CAN be used.
When they MUST be used then it is a fairly stupid exercise for 
structural description from an old fashioned teacher (because in real 
life nobody will instantiate a flipflop or a adder by using components 
explicitly).
If they CAN be used, then i would kick then beside and do it like your 
simple examples in less then 30 lines VHDL code.

Up to now for a pity it looks more like they MUST be used... :-/

Lothar is totally right (and i also wrote it):

Gustl B. wrote:
> I don't know your constraints
> if you must describe each element on its own.

Please ask your Teacher.

Unfortunately they must be used :/

Because of cluelessness and lazyness i still would go the easy way. 
Describe the logic in simple vhdl, look at the RTL view of your 
toolchain (i posted a picture above somewhere) and then implement this 
picture (= a working schematic) with the components you have to use.

I will take a look tomorrow and see if i can actually do it this way .

Thanks again

Hm, the schematic from ISE uses a different type of MUX, the schematic 
from VIVADO a different type of FF (with CE Pin).

Wohoooo! It works (in simulation)! But ... it was tricky. Use the VIVADO 
schematic because VIVADO uses the right MUXes. And then you have to dive 
deep in vivado and get the details for all connections.

000 : Reset /
001 : Parallel loading /
002 : Right shift /
003 : Left shift /
004 : Circular right shift /
005 : Circular left shift /

can be implemented this way with 5 MUXes and 5 FFs and one inverter (for 
res) because reset in your FlipFlop HDL has an active low clear.

I also attatched the mux.png it shows how the mux is connected in the 
schematic_vivado.png it can be implemented as 5 independed muxes with 
the mux HDL description you provided.

what i could not implement was the thing with 1s and 2s complement. i 
understand what 2s complement is, but not how it should behave in the 
shift register. the question is:

contend of the shift register is data(4 downto 0)
now SEL = "110" (6).
what should happen with every rising clock?
data <= ?
And what should happen if SEL = "111" (7)?

Gustl B. wrote:
> contend of the shift register is data(4 downto 0)
> now SEL = "110" (6).
> what should happen with every rising clock?
> data <= ?

i asked the same and i'm waiting for reply from my teacher .

> And what should happen if SEL = "111" (7)?

nothing .

-------

Can i do the process you mentioned in ISE ? Because i have to

Michael K. wrote:
>> And what should happen if SEL = "111" (7)?
>
> nothing .

Souldn't this be the 2s complement thing?
1s complement is easy but for 2s complement you need an adder or more 
complex logic.
My guess is that with SEL = 110 with every clock the content in the 
register should be converted like 1s complement. and so with SEL = 111, 
with every clock it should be converted in 2s complement.

Michael K. wrote:
> Can i do the process you mentioned in ISE ? Because i have to

I don't know. i might be hard because ISE does not use 8-input MUXes in 
the RTL schematic. the HDL is independent from ISE or VIVADO, so if you 
once wrote the HDL files in vhdl you can use an simulate them with ISE 
or VIVADO.
Here i only needed VIVADO to get the information about what is connected 
to what.

Edit:
I just added the SEL = 110 and SEL = 111 cases as:
[vhdl]    when "110" => data <= not data; --1s' complement
    when others => data <= std_logic_vector(unsigned(not data) +1); 
--2s' complement (SEL = "111")[vhdl]

and ... now ISE uses 8-input MUXes in the RTL schematic. Just like 
VIVADO. In the waveform you can see that with SEL = 110 all bits in the 
register are inverted with every clock and with SEL = 111 all bits are 
inverted and 1 is added with every clock. But i don't know if this is 
intended.
If this is ok, then you may use ISE, it looks good.

Gustl B. wrote:
> Michael K. wrote:
>>> And what should happen if SEL = "111" (7)?
>>
>> nothing .
>
> Souldn't this be the 2s complement thing?
> 1s complement is easy but for 2s complement you need an adder or more
> complex logic.
> My guess is that with SEL = 110 with every clock the content in the
> register should be converted like 1s complement. and so with SEL = 111,
> with every clock it should be converted in 2s complement.

I was confused because in my project i had 7 options , and reset was 
connected straight to tue flip flops . Yes if you use 000 option as 
reset , 111 will be 2s complement .

I'll ask my teacher and/or an older student tomorrow if they have any 
idea if it's possible to do it from my current project . I'll also have 
to check the the last testbench file you uploaded earlier today .

Thanks again

Edit:
i edited my post above and added images. i missed an / in the closing 
[/vhdl] tag ... so here it is again:

1

when "110" => data <= not data; --1s' complement

2

when others => data <= std_logic_vector(unsigned(not data) +1); --2s' complement (SEL = "111")

I really appreciate your help !!

It seems and it really is easy that way , let's hope i can do it like 
this (not really i guess :( ) .

More tomorrow (it's night here).

Thanks !

it's night here too (germany). your handwriting on the schematic photo 
looks as if you were from greece?

Gustl B. wrote:
> it's night here too (germany). your handwriting on the schematic photo
> looks as if you were from greece?

You're right . Thanks Gustl

No Problem, i learn something too, good Night!

Edit:
And it works. Schematics attatched. But it was not easy to dig through 
the very strange signal names in ISE. And it's very sag that neither 
vivado nor ise show signal names in the schematic, you have to klick on 
the lines. you also can't show busses as individual wires.

I think i finally made it !

My Diagram in post #6 has the solution ... In 000 option (parallel 
transfer) you transfer the input to the output . Then EVERY time , you 
give as input to the mux's from the output of option 000 (parallel 
transfer) , and so every time it gets the output without the user doing 
something . Basically to change input you have to give a new 5 bit input 
and re-run 000 option .

Ok, very good! Respect! In Simulation are too few clock for circular 
shifts, so you don't see if it shifts really circular. But this is easy 
to fix in the Testbench.

It was a simple test .

Here is full cycle , with different input

Perfect! Now even with 1s and 2s complement! Well done :-)