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Depending on your exact application, you might also use two counters at 200MHz, one with inverted clock, the other without. Then you can add the values of both counters. (You should ensure that the tsu/th from the pin to the counters is about the same for both counters. If the enable-input is directly a device-pin, you should add some synchronizer-stages, regardless if you use one or two counters.)
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Not quite: because both counters will be enabled by the same input pin you have a redundancy: at any moment in time the count in the two counters will only differ by maximum 1. One counter will have count 'n', the other will then either be 1 less, equal or 1 more. The sum will then either be 2n , 2n - 1 , or 2n + 1. The second case 2n + 1 equals 2(n +1) - 1. So you can obtain the same effect with a single counter and a smaller prescaler (of 1 bit size) circuit, which could be operated with both edges of the clock or double the frequency. You would need a equivalent circuit to drive the proposed 2-counter solution as well.
Regards,
someone :)
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