how to design circuit count even numbers 0-2-4-6-8-0-2 ...

[MrAnhKieu]

use flip flop.
Please help me , thanks

 

[Electroenthusiast]

i'm not sure, 7 segment decoder should be different...(so that the successive odd number count should be displaying the previous even number - All possible if the logic inside decoder is changed)... it can be done by this way.
But the actual way used might be different...

 

[MikeMl]

use flip flop.
Please help me , thanks

Sounds like homework. One approach: build a state machine with five states (takes only three flip flops). Allocate the states so that they create the bit patterns for 0 to 4, but wire the outputs left shifted by one bit, creating the correct BCD patterns for 0,2,4,6, and 8

 

[mmaters]

You'll need to be more specific: count the output from another circuit or generate the numbers? And what format BCD, Hex, binary.
If simply digital, then just trigger on the least significant digit, if that's a 0 then you have an even number...

 

[Electroenthusiast]

Sounds like homework. One approach: build a state machine with five states (takes only three flip flops). Allocate the states so that they create the bit patterns for 0 to 4, but wire the outputs left shifted by one bit, creating the correct BCD patterns for 0,2,4,6, and 8

You'll need to be more specific: count the output from another circuit or generate the numbers? And what format BCD, Hex, binary.
If simply digital, then just trigger on the least significant digit, if that's a 0 then you have an even number...

these would give him what he/she's searching for...
what would be the answer if its PRIME Number?

 

[Hero999]

Use a divide by four counter.

Connect the A output to the 7 segment decoder's B input, the B output to the C input etc. and leave the D output and A input unused (connect the input to 0V and leave the output unconnected).

You'll need an and gate to reset the counter when the A and C outputs are high, you can use a diode AND or a real gate. I'd only bother with the gate if I was using a quad NAND IC to make an oscillator.

 

[MikeMl]

Use a divide by four counter.
....

Cant be done with a divide by four counter; OP needs a minimum of five states.

 

[Hero999]

Sorry, I meant a divide by eight counter, I'd use a divide by 16 counder because it's a common IC.

All I was saying is that there's no need for a shift register or anything fancy like that, just shift the connections to the decoder by one bit.

 

[MikeMl]

Sorry, I meant a divide by eight counter, I'd use a divide by 16 counder because it's a common IC...

OP stated the solution must start with flip-flops, so I suspect this is a homework assignment.

 

[mmaters]

you'll only need to built a 2-bit counter, the last bit is always a low (0)......

 

[mmaters]

There it is with JK-flipflops, as said the least significant digit needs to be generated seperate and is always low
Counter - Wikipedia, the free encyclopedia

Right?

 

[Hero999]

No, you need at least three bits because there are five states, 0, 2, 4, 6 and 8.

 

[mmaters]

My Hero,

Ur right, you'll have to add another bit [just another JK_flipflop] and figure out how to reset all after the cycle when the most significant bit goes high.
Should be quite a number of examples out there on the internet I'd think.
[I'm not an expert here]

 

[MrAnhKieu]

Thanks all!
The need to do is count the number of transfer switches 0-2-4-6-8-0-2 ... when it counted 1-3-5-7-9-1-3 ...

Only a single giant's all. But I do not know where to start.

 

[Hero999]

Sory I don't know what you mean.
Do you want it to count in even or odd numbers?

What's a transfer switch?

To count 1, 3, 5, 7, 9 it's the same as 0, 2, 3, 4, 8 but you make the first bit permanently high.