Providing delay using counter - muneeb-mbytes/FPGABoard_edgeSpartan6 GitHub Wiki

Find the number of bits required for a counter with a delay of 10ms of a push button using a 50MHz clock, we can use the following formula:

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N = log2(Tdelay/Tclk)

Where:-

  • N is the number of bits required for the counter
  • Tclk is the clock period (in seconds)
  • Tdelay is the delay of the push button (in seconds)
  • First, we need to convert the clock frequency from 50MHz to the clock period Tclk:
  • Tclk = 1/fclk
  • Tclk = 1/50MHz
  • Tclk = 20ns

Next, we need to convert the delay of the push button to seconds:

  • Tdelay = 10ms
  • Tdelay = 0.01s Now we can substitute these values into the formula to find the number of bits required:
  • N = log2(Tdelay/Tclk)
  • N = log2(10ms/20ns)
  • N = log2(500000)
  • N = 18.87
  • Since we cannot have a fraction of a bit in a counter, we need to round up to the nearest integer. Therefore, the number of bits required for the counter would be 19 bits.

If we have a push button delay of 20ms and a clock frequency of 50MHz :

  • N = log2(Tdelay/Tclk)

  • where:

  • N is the number of bits required for the counter

  • Tdelay is the delay of the push button (in seconds)

  • Tclk is the clock period (in seconds)

  • First, we need to calculate Tclk:

  • Tclk = 1/fclk = 1/50MHz = 20ns

  • Next, we can calculate the number of bits required:

  • N = log2(Tdelay/Tclk)

  • N = log2(20ms/20ns)

  • N = log2(1000000)

  • N = 19.93

  • Since we cannot have a fraction of a bit in a counter, we need to round up to the nearest integer. Therefore, the number of bits required for the counter would be 20 bit

  • So for a push button delay of 20ms with a clock frequency of 50MHz, a counter with at least 20 bits is required to accurately count the delay.