Electronic – Why a potential divider at high speed LVDS clock outputs

I'd say that the rule of thumb is: set either input port of the top module, or Q pin of an internal flip-flop as the source of generated clock.

Example Verilog code:

module top (

input clk,
input rst,

...

);

...

always @(posedge clk or negedge rst)
begin
if (rst == 1'b0) 
    div_2_clk = 1'b0;
else
    div_2_clk = ~div_2_clk;           
end

...

endmodule

Example SDC code:

create_clock -name clk -period 5 [get_port clk]
...
create_generated_clock -name slow_clk -source [get_port clk] -divide_by 2 [get_pins div_2_clk_reg/Q]

I did no test the above syntax's. Also note the extension _reg added to the RTL name of the signal - this is the extension added by synthesis tool when it detects that the signal must be represented by a flip-flop. This extension may vary between tools (I don't know for sure).

If you use any RTL wrapper around flip-flops - set the source of the generated clocks to be the internal Q pin of the flip-flop, not the output pin of the wrapper.

If you follow these simple rules you need not worry about any buffers added by the synthesis or P&R tools.

You can't properly describe an LVDS receiver or transmitter in VHDL. LVDS interfaces have electrical requirements that typically cannot be met by just using plain digital input and output pins. Your FPGA needs to have dedicated input and output macrocells for those functions and you would just instantiate them. Exactly how that is done depends on the exact FPGA and toolset you are using. Within the FPGA, there is just one signal that would come from the LVDS receiver and one signal that would go to the LVDS transmitter.