Section Number Topic: Counters and Timers A decade down counter is a counter tha

Question

Section Number Topic: Counters and Timers A decade down counter is a counter that counts from 9 down to 0; that is, it counts through the pattern 1001 1000 01 1 1 01 10 0101 0100 0011 0010 0001 0000 1001 etc. The Cout output is a "I' when the count is zero, indicating that the counter is about to roll over" Answer the following questions: 1. What is the period of a 4 MHz square wave? 2. If a decade down counter uses a 4 MHz square wave as a clock, how many microseconds does it take for the counter to count through its pattern once? 3. If you have a 4 MHz clock, how many clock periods are there in a one-second interval of time? (A counter running on a 4 MHz clock would need to count this high to count for one second before rolling over.) 4. How many flip-flops would the one-second counter in (c) need? (Hint: If a counter has to count up to a certain number, it needs enough flip-flops to hold the binary representation of that number. For example, the decade down counter counted as high as 9, and 9 is "1001" in binary, so four flip-flops were needed.) 10

Explanation / Answer

Solution:

1) frequency=4 MHz, period= 1/frequency

So, period= 1/(4*106)=250 nano second=250*10-9 second

2) For each count the clock pulse increases by one, so for the counter to count from 1001 to 1000 an increment of one clock pulse is required. Again for the counter to count from 1000 to 0111 another increment of clock pulse is required. Till now for the counter to count from 1001 to 0111 a total of 2 clock pulse has been incremented. As there are 10 states from 1001 to 0000 so a total of 10 clock pulses are required. At the 10th clock pulse the counter resets to 1001. As the period of the 4MHz clock pulse has been calculated as 250 nano second. So the time required to count through the entire pattern would be:

(period of 4MHz* number of states) = 10*250*10-9=2.5 microsecond

3)The period of 4MHz clock pulse is 250 nano second, which means duration of 1 clock period is 250 nano second.

So in 1 second, number of clock period =1/(250 nano second)= 4*106 clock perods

4) 7 flip flops would be needed for the i second counter, because the counter has to count upto 4000000 to roll over. As there are 7 digits in 4000000 and each flip flop can hold only 1 digit so 7 flip flops would be needed.

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