What is the most optimum way of making a mod-counter that is not a power of 2? The counter increment is a variable

My approach on a mod-13 counter for example:

wire [3:0] counter_d ;
reg  [3:0] counter_q ;
wire [4:0] counter_plus_increment ;
wire overflow ;

assign counter_plus_increment = counter_q + increment ; //small increments
assign overflow = counter_plus_increment[4] ;
assign counter_d = overflow ? 4'd3 : counter_plus_increment ;

always @ (posedge clk or negedge resetn) begin
  if (!resetn)
     counter_q <= 4'd3 ;
     counter_q <= counter_d ;

do you think this is correct? Any better, more efficient suggestions?


2 Answers 2


I have written thousands of counters which are not a multiple of two. The procedure is always the same:

if (counter==MAX-1)
   counter <= 0;
   counter <= counter + 1;

Plus a reset of course.

Your counter does not start at 0 which is not usable in 99.9% of the cases. Also you solution for mod-13 is not generic usable so in my option there is no 'most efficient' way. (Also in what aspect 'most efficient'? Area? Speed? Power? )

In the industry you can't spend your time on optimizations like that unless you really, really need them. Modulo 13 running at 40GHz: Build a shift register with 13 registers, loop the output to the input. Reset with 13'b0000000000001

A different way to count is us count-down.

if (counter==0)
   counter <= MAX-1;
   counter <= counter-1;

This makes that the compare logic becomes a big AND gate instead of AND/OR. But the gain in today's logic is neglectable.

  • 1
    \$\begingroup\$ I know it will count its way out of an illegal state of counter >= MAX eventually, but if it's a 32 bit counter implementing MAX=7, that could take some time. However I guess counter==MAX-1 generates much less logic than counter>=MAX-1, is that right? Or are you assuming the reset will always start it in the legal range? \$\endgroup\$
    – Neil_UK
    Jun 17, 2018 at 11:40
  • \$\begingroup\$ @Neil_UK I always reset to zero. Left it out here so they can choose for a sync. or a-scync reset. I also added more text dealing with "the most optimum". I would never use 32 bit for a MAX=7. I sometimes use a count-down: load with 'MAX-1, count down to zero to reduce the compare logic to AND gate. I'll add the latter to the answer'. \$\endgroup\$
    – Oldfart
    Jun 17, 2018 at 11:47
  • \$\begingroup\$ Both of you are assuming counter increments of 1. However, the increment could be any number...and thats what makes the modulo problem interersting \$\endgroup\$
    – frank_010
    Jun 17, 2018 at 12:37
  • \$\begingroup\$ @frank_010 That's a critical detail that you failed to put in your question. If that's important to you, edit the question accordingly. \$\endgroup\$ Jun 17, 2018 at 12:48
  • \$\begingroup\$ @elliot-alderson In the code I gave for my attempt, I had used a variable to denote increments. But anyways, you are correct it should've been explicitly mentioned and I have made the edit. \$\endgroup\$
    – frank_010
    Jun 17, 2018 at 13:17

While Oldfart's counter is straightforward, this is better in some instances

if (counter==0)
   counter <= MAX-1;
   counter <= counter - 1;

Which instances?

Testing == 0 will generally generate less logic than equal to an arbitrary N, and is likely to be faster.

When N is variable, especially on a cycle by cycle basis, the propagation delay from clock to == 0 is independent of the N. This may seem quite subtle, but is vital if the divider is part of a fractional N synthesiser. If the propagation delay is N-dependent, then noise will be downconverted to in-band.

A reset may or may not be necessary. If the counter powers up in the illegal range of counter > MAX-1, then it will eventually count its way down to the legal range. If this time is unacceptable in your application, then a reset is necessary. However, if you're going to test this, then a reset is highly advisable anyway.

  • \$\begingroup\$ That's a bit curious. I would think detecting zero requires one gate input per bit, whereas detecting a specific count on a count-up counter requires only one gate input per '1', so guaranteed to be better. Unless the carry logic can be put to double duty(?). \$\endgroup\$ Jun 17, 2018 at 13:02
  • \$\begingroup\$ @SpehroPefhany, essentially you're suggesting testing cnt and val == val instead of cnt == val? \$\endgroup\$
    – ilkkachu
    Jun 17, 2018 at 13:27
  • \$\begingroup\$ Yes, that's correct. Maybe that's an optimization that has to be done manually. \$\endgroup\$ Jun 17, 2018 at 13:30
  • 1
    \$\begingroup\$ Your answer is wrong. Testing against zero creates the most hardware resources, because you have to test all bit of a counter. It requires a N-bit NOR-gate. The method of @SpehroPefhany reduces the number of tested bits to only the ONEs in the test value. The best implementation regarding testing is to count towards -1 and test the sign-bit. That's only one test bit. It needs an init value of MAX-3 compared to the usual MAX-1. \$\endgroup\$
    – Paebbels
    Jun 17, 2018 at 15:54

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