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Time multiplexing and refresh rate calculation by counter

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shikharmakkar

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I was going through a verilog code for time multiplexing with LED patterns. I am attaching the screenshot of the problem. I could not understand why they took the 18 bit counter because that would divide the frequency by 2^18 and not 2^16.
 

srizbf

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the counter msbits[17:16] changes for every 2^16 counts.

that selects your in0 ... in3.
 

FvM

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I could not understand why they took the 18 bit counter because that would divide the frequency by 2^18 and not 2^16.

I agree with your calculation. The refresh rate with an 18 Bit counter is 200 Hz, not 800 Hz.
 

std_match

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The text in the screenshot is confusing. One digit at a time is on and the change rate is 800 Hz. Each digit is only on for one of every four 800 clock cycles. This means that the update rate for the complete display is 200 Hz, which corresponds to an 18-bit counter. The enable signal for each digit will be a 200 Hz signal with 25% duty cycle.

The human eye can only detect flashing up to about 50-70 Hz, so 200 Hz will be a steady light.

- - - Updated - - -

Edit: It should be "one of every four 800 Hz clock cycles".
 

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