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Home/ Questions/Q 8890583
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T22:32:08+00:00 2026-06-14T22:32:08+00:00

I searched a lot but I didn’t find a good solution. Most answers work

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I searched a lot but I didn’t find a good solution. Most answers work only when the duty cycle is 50% but I am searching for a solution that works for clocks with duty cycles like 40%, etc.

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  1. Editorial Team
    Editorial Team
    2026-06-14T22:32:09+00:00Added an answer on June 14, 2026 at 10:32 pm

    SYNCHRONOUS

    You will find it hard to do this with a synchronous design because the flip-flops can only switch on a clock edge.

    Typical divide by 3 circuits will either:

    1. Use positive clock edges and have a 33% output duty cycle

    2. Use positive and negative edges and have a 50% duty cycle if the input is 50%

    Unfortunately, for a general input duty cycle such as 40%, if you sketch out the location of the clock edges you will find they occur at:

    0,0.4,1,1.4,2,2.4,3
    

    but for a 40% divide by 3 you would require edges at separated by 1.2 (=0.4*3) and 1.8, but such differences are not available to you. (You only have differences such as 0.4,0.6,1.0,1.4,1.6,etc.)

    ASYNCHRONOUS

    The standard way to get a divide by 3 with a different duty cycle is simply to instantiate a delay locked loop or a phase locked loop circuit.

    For example, if you are using an Altera FPGA look at the AltPLL megafunction which has options to set the target duty cycle, and will work across a wide range of input duty cycles.

    However, note that the output duty cycle is set when you configure the PLL and will not track the input duty cycle.

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