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=Electronics Hardware Questions=


  Q: how to design a divide-by-3 counter with equal duty cycle?


Here is one of the solutions...
Start with the timing diagram. It shows an input clock, an output of a regular divide-by-3 counter and an output of divide-by-3 counter with 50% duty cycle.

It is obvious from the diagram, that we have to use both rising and falling edges of the input clock.
The next drawing is a state diagram.

On this diagram R - is a rising edge of input clock, and F - is a falling edge.
How many FF do we need to implement 6 states? At least 3. In this example I am going to use 4 D-type FF just to simplify the design.
Now, look at the table below. Q0 ... Q3 are the outputs of FFs. Q - is the output of the devider with 50% duty cycle. In the first raw the outputs are in the initial state: 0000. In the second raw - the data after the first rising edge and so on.The status of the FFs' ouputs is changing on every rising or falling edge of the input clock according to the information on D-inputs. So, D-inputs should have the data before the clock edge.


These equations are resulting from the table analysis:
D1 = Q0
D2 = Q1
D3 = Q2
D0 = (Q1+Q2+Q3)'
Q = Q0Q1'Q2'Q3'+Q0Q1Q2'Q3'+Q0'Q1Q2Q3' = Q1Q3'(Q0Q2'+Q0'Q2)+Q0Q1'Q2'Q3'

Next step will be a circuit diagram:

This design is just for illustration and can be optimized further.
Comments from Badri:
Hey Guys

Excellent web site. Thanks.

I had one comment on one of your solutions:

In "Electronics Hardware Questions", the question: how to design a divide-by-3 counter with equal duty cycle ?

I think the suggested solution is unnecessarily complex. My solution is like this (sorry I don't have a schematic tool)

Make a circular shift register with 3 flops, with initial state = 001 (or 100 or 010). (Circular shift register is a snake swallowing its own tail). All the 3 flops in this are clocked on the positive edge of the input clock.

Now take any one of the flop outputs from above (call it A), and feed it to the D input of another flop. This flop is clocked by the inverted version of the input clock. Call this flop's output A_. Then, "A OR A_" is the required divided-by-three clock with 50% duty cycle.

We only need 4 flops, one inverter and one OR gate for this.

Comments from Satish:

Does the solution for divide by 3 counter by Badri work? I tried it and it is not working for me.

I have attached my solution for divide by-3 circuit which i think is simple enough. This is how the circuit works.
We add a gate on the clock to get differential Clock and Clock bar, a flip flop that triggers on the Clock Bar rising edge (Clock Neg.) to shift the output of "B" by 90 degrees and a gate to OR the outputs of FF "B" and FF "C" to produce the 50% output.

I have attached a .jpg file of the schematic

Satish B.


I do have a simpler solution to the question: "how to design a divide-by-3 counter with equal duty cycle ?" under section: "Electronics Hardware Questions". It consists of two D or JK flip-flops (called A & B) with their input's tied high. clocks are falling edge triggered. Input "clk A" is connected to "NEG_Q B", Input "clk B" to a XOR output. One of the XOR inputs is connected to "Q A", witch is also our main output. Main clock signal goes to the other XOR input.

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