226. A certain JK flip-flop has tpd = 12 n sec. What is the largest MOD counter, that can be constructed from these FFs and still operate upto 10 MHz?
     

1. 1. 16
      

   
   
   

   2. 64
      

   
   
   

   3. 128
      

   
   
   

   4. 256

   
   
   

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1. View Hint View Answer Discuss in Forum

   f ≤	1
   	ntpd

   
   

   or MHz ≤	1
   	n × 12 × 10-9

   
   

   or 10×106 ≤	1
   	n × 12 × 10-9

   
   

   or n ≥	1000
   	120

   
   or n ≥ 8.33 or n = 8
   The largest MOD counter = 2⁸ = 256.

   Correct Option: D

   f ≤	1
   	ntpd

   
   

   or MHz ≤	1
   	n × 12 × 10-9

   
   

   or 10×106 ≤	1
   	n × 12 × 10-9

   
   

   or n ≥	1000
   	120

   
   or n ≥ 8.33 or n = 8
   The largest MOD counter = 2⁸ = 256.

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227. The counter starts off in the '0000' state, and then clock pulses are applied. Some time later the clock pulses are removed and the counter FFs read '0011'. How many clock pulses have occurred?
     

1. 1. 3
      

   
   
   

   2. 35
      

   
   
   

   3. 51
      

   
   
   

   4. Any of them

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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228. Assume that counter is holding the count 0101. What will be the count after 27 clock pulses?
     

1. 1. 0101
      

   
   
   

   2. 0001
      

   
   
   

   3. 1111
      

   
   
   

   4. 0000

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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229. How many flip-flops are needed for MOD-16 ring counter and MOD-16 Johnson counter?
     

1. 1. 16, 16
      

   
   
   

   2. 16, 8
      

   
   
   

   3. 4, 8
      

   
   
   

   4. 8, 16

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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230. For the ring oscillator shown the propagation delay of each inverter is 125 pico sec. What is the fundamental frequency of oscillator output?
     
     
     
     

1. 1. 10 MHz
      

   
   
   

   2. 100 MHz
      

   
   
   

   3. 1 GHz
      

   
   
   

   4. 2 GHz

   
   
   

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1. View Hint View Answer Discuss in Forum

   Total propagation delay of the ring oscillator = 4 × 125 ns = 500 ps
   Fundamental frequency,
   

   f =	1	=	1	= 2 GHz
   	tpd		200 PS

   
   

   Correct Option: D

   Total propagation delay of the ring oscillator = 4 × 125 ns = 500 ps
   Fundamental frequency,
   

   f =	1	=	1	= 2 GHz
   	tpd		200 PS

   
   

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