Q= Design a three-bit up/down counter using D flip-flops. It should include a control input
called Up/Down. If Up/Down = 0, then the circuit should behave as an up-counter. If Up/Down
= 1, then the circuit should behave as a down-counter.
Answer=> FOLLOWING DIAGRAM; can please someone explain me how do we draw it with
all the details!
Solution
The digram give 3 bit up counter,and its implemented using MUX and exor gate and D flip flop
Truth table for
from truth table:
D0=Q0
D1=[(Up/Down\' .Q0)+(Up/Down .Q0\' )]xor Q1
D2=[(Up/Down\' .Q0.Q1)+(Up/Down .Q0\' .Q1\')]xor Q2inputPresent stateNext stateExcitation
input for
DflipflopUp/DownQ2Q1Q0Q2Q1Q0D2D1D00000001001000101001000100110110011100100
01001011010101110110011011111101110000001000111111100100000010100010011011010
0101100011011110110010011101011011111110110.
Q= Design a three-bit updown counter using D flip-flops. It should .pdf
1. Q= Design a three-bit up/down counter using D flip-flops. It should include a control input
called Up/Down. If Up/Down = 0, then the circuit should behave as an up-counter. If Up/Down
= 1, then the circuit should behave as a down-counter.
Answer=> FOLLOWING DIAGRAM; can please someone explain me how do we draw it with
all the details!
Solution
The digram give 3 bit up counter,and its implemented using MUX and exor gate and D flip flop
Truth table for
from truth table:
D0=Q0
D1=[(Up/Down' .Q0)+(Up/Down .Q0' )]xor Q1
D2=[(Up/Down' .Q0.Q1)+(Up/Down .Q0' .Q1')]xor Q2inputPresent stateNext stateExcitation
input for
DflipflopUp/DownQ2Q1Q0Q2Q1Q0D2D1D00000001001000101001000100110110011100100
01001011010101110110011011111101110000001000111111100100000010100010011011010
0101100011011110110010011101011011111110110
![Q= Design a three-bit up/down counter using D flip-flops. It should include a control input
called Up/Down. If Up/Down = 0, then the circuit should behave as an up-counter. If Up/Down
= 1, then the circuit should behave as a down-counter.
Answer=> FOLLOWING DIAGRAM; can please someone explain me how do we draw it with
all the details!
Solution
The digram give 3 bit up counter,and its implemented using MUX and exor gate and D flip flop
Truth table for
from truth table:
D0=Q0
D1=[(Up/Down' .Q0)+(Up/Down .Q0' )]xor Q1
D2=[(Up/Down' .Q0.Q1)+(Up/Down .Q0' .Q1')]xor Q2inputPresent stateNext stateExcitation
input for
DflipflopUp/DownQ2Q1Q0Q2Q1Q0D2D1D00000001001000101001000100110110011100100
01001011010101110110011011111101110000001000111111100100000010100010011011010
0101100011011110110010011101011011111110110](https://image.slidesharecdn.com/qdesignathree-bitupdowncounterusingdflip-flops-230326121804-b84295d8/85/Q-Design-a-three-bit-updown-counter-using-D-flip-flops-It-should-pdf-1-320.jpg)
