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Divide by clock
Deepak Floria
deepakfloria@gmail.com
Clock
• Clock refers to any device for measuring and
displaying the time.
• Clock is repetitive in nature after some time
...
• Every modern PC has multiple system clocks.
• Each of these vibrates at a specific frequency,
normally measured in MHz ....
• Some types of work can be done in
one cycle while others require many.
• The ticking of these clocks is what
drives the ...
Clock Period
clock
Off Time= tOn Time = t On Time On TimeOff Time Off Time
2t
• On time = off time
• Total time = T = 2t
D...
`
Positive Level
Negative Level
Positive Edge
Negative Edge
Clock Parameters
Posedge to negedge => posLevel => On time => ...
Divide by Clock
• In SOC some type of job done in one clock and
others in multiple cycle.
• There are many types of Buses ...
Divide by 2N
• Freq divide By 2N
• N=1 => Divide By 2
T = 2t
F = 1/T
T = 2t
F = 1/2T
Reference
Clock
Derived
Clock
• Counter: A counter is a device which works on
each edge of the clock and count the number
of clock pulses.
• Mod 2 Count...
Q
D-FF
d
Q’
Reference
Clock
Reset
Divide by 2
Mod 2 Counter
T = 2t
T = 2T
Reference Clock
Q = Div/2 Clk
Div/2 Clock
Divide by 4
• Freq divide By 2N
• N=2 => Divide By 4
Reference
Clock
Derived
Clock
Clk period T = 4T
Freq F= 1/T ‘=> 1/4T
• Mod 4 Counter: Mod 4
Johnson counter will
count Four clock pulses
of the clock signal.
• Consider the second FF
Q1 outpu...
Divide by 4
D-FF
d0
Q0’
d1
D-FFClock
Reset
Reference
Clock
Reset
Q0
Q1
Clock
Q1’
Divide by 4
• Freq divide By 2N
• N=2 => Divide By 4
Reference
Clock
Q1 Derived
Clock
Q0
T = 2t
F = 1/T
T = 4T F = 1/4T
0
...
D-FF
d0
Q0’
d1
D-FFClock
Reset
Reference
Clock
Reset
Q0
Q1
Clock
Q1’
Alternative way to Div/4
Pass the O/P of the
1st FF t...
Alternative way to Div/4
Q1 = Div/4 Clk
T = 2t
T’ = 4T
Reference Clock
Q0 = Div/2 Clk
Ref clk to 2nd
FF
Divide by 8 counter
• Freq divide By 2N
• N=3 => Divide By 8
• A divide by 8 counter requires
three flip flops
• It has 8 ...
Waveform for Divide by 8
clock
T=2t
T’ = 8T
CLK
Q
Divide by 8 counter Logic Diagram
DFF DFFDFF
CLK
DA
QA DB DC
QB QC
Div/8
Alternative way to Div/8
D-FF
d0
Q0’
d1
D-FFClock
Reset
Reference Clock
Reset
Q0
Q1
Clock
Q1’
D-FF
Clock
Reset
Div/8
Pass ...
Alternative way to Div/8
Q2 = Div/8 Clk
T = 2t
T’ = 8T
Reference Clock
Q0 = Div/2 Clk
Divide by 16 counter
• Freq divide By 2N
• N=4 => Divide By 16
• A divide by 16 counter requires 4 flip
flops
• It has 16 ...
clk Count
Q3 Q2 Q1 Q0
Clock
Cycle
X X X X X 0
1 0 0 0 0 1
1 0 0 0 1 2
1 0 0 1 0 3
1 0 0 1 1 4
1 0 1 0 0 5
1 0 1 0 1 6
1 0 ...
Divide by 16 counter
CLK
Q
T= 2t
8T
T ‘ = 16 T
F = 1/16T
Divide by 16 counter Logic Diagram
DFF DFF DFFDFF
CLK
DA
QA DB DC DD
QB QC QD
QD
Div/16
Divide by 2N
• Freq divide By 2N
• N=N => Divide By N
T = 2t
F = 1/T
T = NT F = 1/NT
Reference
Clock
Derived
Clock
Divide by 2N
Q2
D-FF
d0
Q0’
d1
D-FF
Clock
Reset
Reference
Clock
Reset
Q0
Q1
Clock
Q1’
D-FF
Clock
Reset
Div/N
D-FF
Reset
Cl...
• A divide by 3 clock
requires A mod 3 Counter.
• It can be constructed
using 2 FF.
• It has 4 possible states
and it need...
Div/
3 clk
D-FFD-FF
d0
Q0’
d1
D-FF
Clock
Reset
Reference
Clock
Reset
Q0 Q1
Clock Q1’
d Q
Q’
Divide by 3
• Pass the
second ...
Waveform for Divide by 3
• Freq divide By3
Reference
Clock
Q1
Q0
T = 2t
F = 1/T
T = 3T
0
0
1
0
0
1
0
0
1
0
0
1
0
1
1
1
Q
D...
• A divide by 5 counter requires can be developed using Mod 5
Counter in similar method.
• To get 50% duty cycle output on...
2T
Waveform for Divide by 5
T =2t
t’= 2+1/2 T
T’=5T
CLK
QB
QD
QB + QD
QA
QA
QB
QB
QC
QC
QDDA DB
DC DD
. .
CLK
Y
.
Divide by 5 Clock Logic Diagram
Divide by 6 counter
• Div/6 can be constructed by johnson counter.
• A Div/6 Johnson counter requires 3 bit FF.
Clk Count
...
Wavaeform for Divide by 6
counter
T=2t
3T
T’ = 6T
Clk
Q
Divide by 7 counter
• A divide by 7 counter
requires Mod 7
counter.
• It has 8 possible
states and it needs
only 7 states....
Divide by 7 counter
2t
6T
7T
T’=7T
CLK
QA
QD
QA + QD
1 2 3 4 5 6 7 1 2 3 4 5 6
Divide by 7 counter Logic Diagram
.
QA
QA
QB
QB
QC
QC
QDDA DB
DC DD
.
.
CLK
Y
Divide by 9 counter
• A divide by 9 counter
requires Mod 9 counter.
• It has 16 possible states
and it needs only 9 states...
Divide by 9 counter
2T
4T
4+1/2T
T’=9T
CLK
QB
QE
QB + QE
1 2 3 4 5 6 7 8 9 1 2 3 4 5 6
Div/9 counter Logic Diagram
DA DB DC DD DEQA
QA
QB
QB QC
QC QD
QD
QE
CLK
Y
.
.
.
. .
.
ThankYou
Deepak floria
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Divide by N clock

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this presentation is based to construct different frequency divide by clock with reference to the system clock.

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Divide by N clock

  1. 1. Divide by clock Deepak Floria deepakfloria@gmail.com
  2. 2. Clock • Clock refers to any device for measuring and displaying the time. • Clock is repetitive in nature after some time period.
  3. 3. • Every modern PC has multiple system clocks. • Each of these vibrates at a specific frequency, normally measured in MHz . • A clock "tick" is the smallest unit of time in which processing happens, and is sometimes called a cycle. System Clock clock
  4. 4. • Some types of work can be done in one cycle while others require many. • The ticking of these clocks is what drives the various circuits in the PC, and the faster they tick, the more performance you get from your machine. System Clock
  5. 5. Clock Period clock Off Time= tOn Time = t On Time On TimeOff Time Off Time 2t • On time = off time • Total time = T = 2t Duty Cycle = [(On Time/Total time) * 100 ]% Duty Cycle = t /T * 100 % = t/2t *100 % = 50%
  6. 6. ` Positive Level Negative Level Positive Edge Negative Edge Clock Parameters Posedge to negedge => posLevel => On time => High Level Negedge to posedge => neg level => Off time => Low Level Clock Period => Posedge To Posedge or Negedge to Negedge
  7. 7. Divide by Clock • In SOC some type of job done in one clock and others in multiple cycle. • There are many types of Buses inside SOC system. • These buses works at different clock signal but take reference from the main system clock. • Taking Reference as the main System clock we perform the Divide by Clock operations. Reference clock
  8. 8. Divide by 2N • Freq divide By 2N • N=1 => Divide By 2 T = 2t F = 1/T T = 2t F = 1/2T Reference Clock Derived Clock
  9. 9. • Counter: A counter is a device which works on each edge of the clock and count the number of clock pulses. • Mod 2 Counter: Mod 2 counter will count two clock pulses of the clock signal. • A mod 2 counter is exactly working for two clock cycle. Divide by 2 Clk Count Clock pulses X X 0 0 0 1 1 1 2
  10. 10. Q D-FF d Q’ Reference Clock Reset Divide by 2 Mod 2 Counter T = 2t T = 2T Reference Clock Q = Div/2 Clk Div/2 Clock
  11. 11. Divide by 4 • Freq divide By 2N • N=2 => Divide By 4 Reference Clock Derived Clock Clk period T = 4T Freq F= 1/T ‘=> 1/4T
  12. 12. • Mod 4 Counter: Mod 4 Johnson counter will count Four clock pulses of the clock signal. • Consider the second FF Q1 output which is high for two Clock & low For Two Clock Cycle Divide by 4 Clk Count Q1 Q0 Clock pulses X X X 0 1 0 0 1 1 0 1 2 1 1 1 3 1 1 0 4
  13. 13. Divide by 4 D-FF d0 Q0’ d1 D-FFClock Reset Reference Clock Reset Q0 Q1 Clock Q1’
  14. 14. Divide by 4 • Freq divide By 2N • N=2 => Divide By 4 Reference Clock Q1 Derived Clock Q0 T = 2t F = 1/T T = 4T F = 1/4T 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1
  15. 15. D-FF d0 Q0’ d1 D-FFClock Reset Reference Clock Reset Q0 Q1 Clock Q1’ Alternative way to Div/4 Pass the O/P of the 1st FF to the next FF as Clk signal
  16. 16. Alternative way to Div/4 Q1 = Div/4 Clk T = 2t T’ = 4T Reference Clock Q0 = Div/2 Clk Ref clk to 2nd FF
  17. 17. Divide by 8 counter • Freq divide By 2N • N=3 => Divide By 8 • A divide by 8 counter requires three flip flops • It has 8 possible states • The Q’ output of the third FF is given as an input to the first flip flop clk Count Q2 Q1 Q0 Clock Cycle X X X X 0 1 0 0 0 1 1 0 0 1 2 1 0 1 0 3 1 0 1 1 4 1 1 0 0 5 1 1 0 1 6 1 1 1 0 7 1 1 1 1 8 O/P of the 3rd FF is high for 3 clk cycle & low for 3 clk Cycle. This is the required Div/8 Clk signal
  18. 18. Waveform for Divide by 8 clock T=2t T’ = 8T CLK Q
  19. 19. Divide by 8 counter Logic Diagram DFF DFFDFF CLK DA QA DB DC QB QC Div/8
  20. 20. Alternative way to Div/8 D-FF d0 Q0’ d1 D-FFClock Reset Reference Clock Reset Q0 Q1 Clock Q1’ D-FF Clock Reset Div/8 Pass the O/P of the one FF to the next FF as Clk signal
  21. 21. Alternative way to Div/8 Q2 = Div/8 Clk T = 2t T’ = 8T Reference Clock Q0 = Div/2 Clk
  22. 22. Divide by 16 counter • Freq divide By 2N • N=4 => Divide By 16 • A divide by 16 counter requires 4 flip flops • It has 16 possible states. • The Q’ output of the last flip flop is connected to input to the first flip flop
  23. 23. clk Count Q3 Q2 Q1 Q0 Clock Cycle X X X X X 0 1 0 0 0 0 1 1 0 0 0 1 2 1 0 0 1 0 3 1 0 0 1 1 4 1 0 1 0 0 5 1 0 1 0 1 6 1 0 1 1 0 7 1 0 1 1 1 8 1 1 0 0 0 9 1 1 0 0 1 10 1 1 0 1 0 11 1 1 0 1 1 12 1 1 1 0 0 13 1 1 1 0 1 14 1 1 1 1 0 15 1 1 1 1 1 16 The last FF O/P value is low for 8 clk cycle & high for 8 clk cycle. This O/P is the required Div/16 clk signal.
  24. 24. Divide by 16 counter CLK Q T= 2t 8T T ‘ = 16 T F = 1/16T
  25. 25. Divide by 16 counter Logic Diagram DFF DFF DFFDFF CLK DA QA DB DC DD QB QC QD QD Div/16
  26. 26. Divide by 2N • Freq divide By 2N • N=N => Divide By N T = 2t F = 1/T T = NT F = 1/NT Reference Clock Derived Clock
  27. 27. Divide by 2N Q2 D-FF d0 Q0’ d1 D-FF Clock Reset Reference Clock Reset Q0 Q1 Clock Q1’ D-FF Clock Reset Div/N D-FF Reset Clock Q2’ QN’ d2 dN
  28. 28. • A divide by 3 clock requires A mod 3 Counter. • It can be constructed using 2 FF. • It has 4 possible states and it needs only 3 states Divide by 3 Clk Count Q1 Q0 Clock pulses X X X 0 1 0 0 1 1 0 1 2 1 1 0 3 Observe the OP of 2nd FF
  29. 29. Div/ 3 clk D-FFD-FF d0 Q0’ d1 D-FF Clock Reset Reference Clock Reset Q0 Q1 Clock Q1’ d Q Q’ Divide by 3 • Pass the second FF O/P to one more FF which is triggered as negedge of clk. • Make ORing of Q1 & Q. • This is the require Div/3 50 % duty cycle Clk circuit.
  30. 30. Waveform for Divide by 3 • Freq divide By3 Reference Clock Q1 Q0 T = 2t F = 1/T T = 3T 0 0 1 0 0 1 0 0 1 0 0 1 0 1 1 1 Q Div/3 clk
  31. 31. • A divide by 5 counter requires can be developed using Mod 5 Counter in similar method. • To get 50% duty cycle output one more flip flop is added and it is negative edge triggered. • Pass the output of the second Divide by 5 clock Clk Count Q2 Q1 Q0 cycle X X X X 0 1 0 0 0 1 1 0 0 1 2 1 0 1 0 3 1 0 1 1 4 1 1 0 0 5 Observe the output of second FF. It is High for 2 cycle & low for 3 cycle. • Pass the output of the second FF to one more FF which is triggered with negedge of clk then make ORing of these two.
  32. 32. 2T Waveform for Divide by 5 T =2t t’= 2+1/2 T T’=5T CLK QB QD QB + QD
  33. 33. QA QA QB QB QC QC QDDA DB DC DD . . CLK Y . Divide by 5 Clock Logic Diagram
  34. 34. Divide by 6 counter • Div/6 can be constructed by johnson counter. • A Div/6 Johnson counter requires 3 bit FF. Clk Count Q2 Q1 Q0 cycle X X X X 0 1 0 0 0 1 1 1 0 0 2 1 1 1 0 3 1 1 1 1 4 1 0 1 1 5 1 0 0 1 6 The O/P of the 1st FF is high for 3 clk cycle & low for 3 clk cycle. This is the Required Div/6 clk signal.
  35. 35. Wavaeform for Divide by 6 counter T=2t 3T T’ = 6T Clk Q
  36. 36. Divide by 7 counter • A divide by 7 counter requires Mod 7 counter. • It has 8 possible states and it needs only 7 states. Clk Count Q2 Q1 Q0 cycle X X X X 0 1 0 0 0 1 1 0 0 1 2 1 0 1 0 3 1 0 1 1 4 1 1 0 0 5 1 1 0 1 6 1 1 1 0 7 The O/P of the 3rd FF is High for 3 clk cycle & low for 3 clk cycle. Pass this O/P to one more FF which will work negedge of Clk then make ORing of these two O/P.
  37. 37. Divide by 7 counter 2t 6T 7T T’=7T CLK QA QD QA + QD 1 2 3 4 5 6 7 1 2 3 4 5 6
  38. 38. Divide by 7 counter Logic Diagram . QA QA QB QB QC QC QDDA DB DC DD . . CLK Y
  39. 39. Divide by 9 counter • A divide by 9 counter requires Mod 9 counter. • It has 16 possible states and it needs only 9 states clk Count Q3 Q2 Q1 Q0 Clock Cycle X X X X X 0 1 0 0 0 0 1 1 0 0 0 1 2 1 0 0 1 0 3 1 0 0 1 1 4 1 0 1 0 0 5 1 0 1 0 1 6 1 0 1 1 0 7 1 0 1 1 1 8 1 1 0 0 0 9 1 1 0 0 1 10 The 3rd FF O/P value is low for 5 clk cycle & high for 4 clk cycle. This O/P is the required Div/9 clk signal but not 50% duty cycle. Pass this O/P to one more FF triggered with negedge clk and then make Oring fo these two signal to 50% duty cycle.
  40. 40. Divide by 9 counter 2T 4T 4+1/2T T’=9T CLK QB QE QB + QE 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6
  41. 41. Div/9 counter Logic Diagram DA DB DC DD DEQA QA QB QB QC QC QD QD QE CLK Y . . . . . .
  42. 42. ThankYou Deepak floria

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