I'm learning verilog and I think there is something that I must not understand about always #* and always (#posedge clk, ...)
Here is a piece of code supposed to send bits via uart. It fails at synthesization.
The error is
" The logic for does not match a known FF or Latch template. The description style you are using to describe a register or latch is not supported in the current software release."
(and 3 other errors for , and )
If I change the always #(...) by always #*, things fail in the next step ("implement design") because things are not connected.
In the book that I have, they implement an fsmd with an always (posedge clk) for the state, and always #* for the other logic, but I don't understand why this doesn't work.
On another forum, I read that the error could come from too complicated conditions. But I have simplified things too (not code the code here but basically I removed the case(state) and the ifs to have single line assignments with ? : or binary conditions, but it didn't work either)
I have seen this error before in other pieces of code that I wrote but I didn't get to the bottom of it, so if you could help me understand the general problem (with this uart thing as a support for a concrete example), I would be very happy.
Thanks
Thomas
P.S : Im using xilinx spartan 3e starter kit and xilinx ise 14.4
module UART_out #(parameter [3:0] NUM_BITS = 8)
(
input wire baud_clk,
input wire send_tick,
input wire[NUM_BITS-1:0] data_in,
output wire tx,
output wire debug_done
);
localparam
IDLE = 0,
TRANSMIT = 1;
reg[NUM_BITS:0] bits_to_send;
reg state;
reg out_bit;
reg[4:0] cnt;
always #(posedge baud_clk, posedge send_tick)
begin
case (state)
IDLE:
if (send_tick)
begin
bits_to_send <= {data_in, 0};
state <= TRANSMIT;
cnt <= 0;
end
TRANSMIT:
begin
if (cnt < NUM_BITS)
cnt <= cnt + 1;
else
state <= IDLE;
bits_to_send <= {1, bits_to_send[NUM_BITS:1]};
out_bit <= bits_to_send[0];
end
endcase
end
assign tx = (state == IDLE ? 1 : out_bit);
assign debug_done = (state == IDLE);
endmodule
The error:
The logic for does not match a known FF or Latch template. The description style you are using to describe a register or latch is not supported in the current software release.
Is referring to the fact that the synthesis tool does not have any hardware cells to use which match your description.
What hardware do you want from :
always #(posedge baud_clk, posedge send_tick)
This looks like you want a flip-flop with an enable signal. The enable signal (send_tick) should be 1 clock period wide. This is then used to select the path of logic on a clock edge. not as an alternative trigger.
I think that this is all you really need:
always #(posedge baud_clk) begin
case (state)
IDLE:
if (send_tick) begin
//...
end
//...
endcase
end
If send_tick is from another clock domain then you will need to do some clock domain crossing to turn it it to a clock wide pulse on the baud_clk.
You may be getting confused with blocks which have multiple triggers, they are normally a clk and reset. A negedge reset_n or posedge reset are often added for reset (initialisation) conditions.
If adding a reset :
always #(posedge baud_clk or negedge reset_n) begin
if (~reset_n) begin
//reset conditions
state <= IDLE;
//...
end
else begin
// Standard logic
end
end
You will notice that there is a very definite structure here, if reset else ... The synthesis tools recognise this as a flip-flop with an asynchronous reset. The data in the reset condition is also static, typically setting everything to zero.
Related
I'm trying to write a RTL model in which I monitor independent clock sources. These clock sources can have variable frequency (range 5 to 50MHz)
Let us say clk1 and clk2. I'm trying to drive a signal 'toggle' which is set '1' at every posedge of clk1 and is set to '0' at every negedge of clk2. I'm having trouble realizing this model.
I tried using 1 flop triggered at the positive edge of clk1 with inputs of this flop tied to 'high' and another flip flop triggered at the negative edge of clk2 with input tied to 'low'. I sent these outputs to a mux, but I have trouble figuring out how to drive the select signal of this mux
Here is my code snippet :
always_ff #(posedge clk1 or rstb) begin
if(!rstb) begin
flop1_out <= 0;
end else begin
flop1_out <= 1;
end
end
always_ff #(negedge clk2) begin
flop2_out <= 0;
end
assign toggle = sel ? flop1 : flop2;
So, as of now nothing is driving sel and trying to figure this out is where I'm having trouble
If I try to drive the same signal (toggle) from 2 different clock sources, I get an error saying that multiple drivers found for signal toggle, which makes sense.
Please let me know if you have any suggestions
EDIT: fixed a typo and removed rstb from the sensitivity list for flop2
assign rstn = clk2;
always # (posedge clk1 or negedge rstn)
if (~rstn)
toggle = 1'b0;
else
toggle <= 1'b1;
note: depending on the clock frequency and insertion delay relationships this circuit may become metastable. if you can tolerate delay, add a synchronizer on the output. better yet, if you can tolerate distortion, add a reset synchronizer on clk2 to clk1mx, where clk1mx is synchronous to clock1 but x times faster.
As far as I can understand that the hardware required to implement the code below is not supported in Xilinx ISE Web Pack. I'm trying to implement only the functionality of the 8-bit adder using an always block. Here's the code:
module Addr_8bit(Clk, Rst, En, LEDOut
);
input Clk;
input Rst;
input En;
output reg [7:0] LEDOut;
always #(posedge Clk or posedge Rst) begin
if(Rst)
LEDOut <= 8'b00000000;
if(En)
LEDOut <= LEDOut + 8'b00000001;
end
endmodule
The error is on the line where the non-blocking assignment: LEDOut <= LEDOut + 8'b00000001; is located.
Particularly it says that:
ERROR:Xst:899 - "Addr_8bit.v" line 33: The logic for <LEDOut> does not match a known FF or Latch template. The description style you are using to describe a register or latch is not supported in the current software release.
I am trying to make the LEDOut's 8-bit output to correspond to the each single one of 8 LEDs on the BASYS2 FPGA Board(Spartan-3E).
Thank You.
Change your behavioral description (the code inside the always block) as follows:
always#(posedge CLK or negedge RST) begin
if(!RST) begin // Reset condition goes here
LEDOut <= 0;
end
else begin // Everything else goes here
if(En)
LEDOut <= LEDOut + 1'b1;
end
end
The reason your code won't synthesize is because you generally can't assign to the same register under the same edge of two different signals. (You can't have your always block trigger on the low to high transition of CLK and RST if you're assigning to a variable in both cases.) So you can't trigger the reset condition on the positive edge of RST, but you can do it on the negative edge. This is due to the way the physical register elements (called flip-flops) are designed.
I am trying to write a program in Verilog that should "move" a light LED on an array of LEDs. With a button the light should move to the left, with another one it should move to the right. This is my code:
module led_shift(UP, DOWN, RES, CLK, LED);
input UP, DOWN, RES, CLK;
output reg [7:0] LED;
reg [7:0] STATE;
always#(negedge DOWN or negedge UP or negedge RES)
begin
if(!RES)
begin
STATE <= 8'b00010000;
end
else
begin
STATE <= UP ? STATE>>1 : STATE<<1;
end
end
always # (posedge CLK)
begin
LED <= STATE;
end
endmodule
The problem is in STATE <= UP ? STATE>>1 : STATE<<1; and the error the following:
Error (10200): Verilog HDL Conditional Statement error at led_shift.v(34): cannot match operand(s) in the condition to the corresponding edges in the enclosing event control of the always construct
I tried to modify the code without using that kind of if:
always#(negedge DOWN or negedge UP or negedge RES)
begin
if(!RES)
STATE <= 8'b00010000;
else
begin
if(!DOWN)
STATE <= STATE<<1;
else
begin
if(!UP)
STATE <= STATE>>1;
else
STATE <= STATE;
end
end
end
It compiles, but does not work: the LED "moves" only to the left, when I press the other button all LEDs shut down. Probably there is a problem in my code, but I cannot understand why my first code does not compile at all.
Thank you for any help!
harrym
It is not clear for the synthesizer to know how to control STATE with the 3 asynchronous control signals.
Most likely your your synthesizer is trying to map STATE to a D flip flop with asynchronous active low set and reset. For example it might be trying to synthesize to something like:
dff state_0_(.Q(STATE[0], .CLK(DOWN), .SET_N(UP), .RST_N(RES(, .D(/*...*/));
In a real flop with asynchronous set and reset, the default should be consent and would explain the error in your first code. In your second attempt, UP becomes part of the combination logic cloud along with DOWN. DOWN is also being used as a clock. Since UP is not a clock, shifting continuously happens while UP is low, completely shifting the on bit out instantly. Another error for the second case would actually be more appropriate.
For the synthesizer to do a better job, you first need to synchronize your asynchronous control signals. Use the same technique as CDC (clock domain crossing; A paper by Cliff Cummings goes into detains here). A basic example:
always #(posedge clk) begin
pre_sync_DOWN <= DOWN;
sync_DOWN <= pre_sync_DOWN;
end
Now that the controls signals are synchronized, make STATE the output of your combination logic. Example:
always #* begin
if(!sync_RES)
STATE = 8'b00010000;
else
case({sync_UP,sync_DOWN})
2'b01 : STATE = LED>>1;
2'b10 : STATE = LED<<1;
default: STATE = LED;
endcase
end
With everything running on one clock domain and explicitly defined combination logic, the synthesizer can construct equivalent logic using flops and basic gates.
FYI:
To shift only on a negedge event you need to keep the last sync value and check for the high to low transition. Remember to swap sync_ with do_ in the combination logic that drives STATE.
always #(posedge clk)
keep_DOWN <= sync_DOWN;
always #*
do_DOWN = (keep_DOWN && !sync_DOWN);
Recently, I had seen some D flip-flop RTL code in verilog like this:
module d_ff(
input d,
input clk,
input reset,
input we,
output q
);
always #(posedge clk) begin
if (~reset) begin
q <= 1'b0;
end
else if (we) begin
q <= d;
end
else begin
q <= q;
end
end
endmodule
Does the statement q <= q; necessary?
Does the statement q <= q; necessary?
No it isn't, and in the case of an ASIC it may actually increase area and power consumption. I'm not sure how modern FPGAs handle this. During synthesis the tool will see that statement and require that q be updated on every positive clock edge. Without that final else clause the tool is free to only update q whenever the given conditions are met.
On an ASIC this means the synthesis tool may insert a clock gate(provided the library has one) instead of mux. For a single DFF this may actually be worse since a clock gate typically is much larger than a mux but if q is 32 bits then the savings can be very significant. Modern tools can automatically detect if the number of DFFs using a shared enable meets a certain threshold and then choose a clock gate or mux appropriately.
In this case the tool needs 3 muxes plus extra routing
always #(posedge CLK or negedge RESET)
if(~RESET)
COUNT <= 0;
else if(INC)
COUNT <= COUNT + 1;
else
COUNT <= COUNT;
Here the tool uses a single clock gate for all DFFs
always #(posedge CLK or negedge RESET)
if(~RESET)
COUNT <= 0;
else if(INC)
COUNT <= COUNT + 1;
Images from here
As far as simulation is concerned, removing that statement should not change anything, since q should be of type reg (or logic in SystemVerilog), and should hold its value.
Also, most synthesis tools should generate the same circuit in both cases since q is updated using a non-blocking assignment. Perhaps a better code would be to use always_ff instead of always (if your tool supports it). This way the compiler will check that q is always updated using a non-blocking assignment and sequential logic is generated.
I am very new to HDL language. I have a question about how to program a shift register. (i know i shift to the other direction). Why does the book use wire[N-1:0] r_next? what's drawback of my implementation?
thanks
my first try is as following
module lesson04#(parameter N=8)(
input wire clk, reset,
input wire data,
output wire out
);
reg [N-1: 0] r_reg;
always #(posedge clk or negedge reset)
begin
if(!reset)
r_reg =0;
else
r_reg[0]=data;
r_reg = r_reg<<1;
end
assign out =r_reg[N-1];
endmodule
but the book gives:
module lesson04#(parameter N=8)(
input wire clk, reset,
input wire data,
output wire out
);
reg [N-1: 0] r_reg;
wire[N-1:0] r_next;
always #(posedge clk or negedge reset)
begin
if(!reset)
r_reg =0;
else
r_reg <= r_next;
end
assign r_next= {data, r_reg[N-1:1]};
assign out =r_reg[N-1];
endmodule
First of all, don't forget your begin-ends around sections of code:
else begin
r_reg[0]=data;
r_reg = r_reg<<1;
end
Without this, only r_reg[0]=data will be in the else clause of the if statement. This will work, but is considered bad style due to the blocking statements in a sequential logic description...
Second, for modeling sequential blocks, use nonblocking assignments (<=) or your calculations may 'fall through' (google nonblocking vs blocking for more info). Your example may very well work (did you try it in a simulator?) but if things get more complicated and more variables are added things can break.
always #(posedge clk or negedge reset)
begin
if(!reset)
r_reg <= 0;
else begin // This is horrible! Don't write code like this!
r_reg[0] = data; // blocking
r_reg <= r_reg<<1; // non-blocking
end
end
For the above reason, it is sometimes recommended that combo logic is separated from sequential logic so that you can write nonblocking assignments to registers in sequential blocks, and blocking in combo blocks and never have to worry about the scheduling.
To code in this way, you need to calculate what the next output should be using the current state, hence the r_next bus in the answer. I think it tends to help the synthesis tool out too if all the flip-flops are separated from surrounding combo logic in this way.
Also, if your reset is active low (ie LOW resets ) it should be named as such, eg resetb or reset_n.
Your implementation produces quite a different output from the book's. You should prove this to yourself by constructing a simple testbench to drive your inputs and run a simulation. You will see that the book's output shifts the input data by a single clock cycle, whereas your output shifts the input data by eight clock cycles.
By the way you have indented your always block, I am led to believe that it is not what you wanted. This is how your block really behaves:
always #(posedge clk or negedge reset)
begin
if(!reset) begin
r_reg =0;
end else begin
r_reg[0]=data;
end
r_reg = r_reg<<1;
end
I always explicitly use the begin/end keywords in if/else statements to avoid this confusion.
The way it simulates, r_reg is always 0 because you clobber the 1st assignment (r_reg[0]=data;) with the 2nd (r_reg = r_reg<<1;). Another difference is that the book assigns data to the MSB of the shift register, but you assign it to the LSB.
If you are using decent linting and synthesis tools, you would probably get a bunch of warnings for your code. This would alert you to make some changes.