Hello, I am stuck with the following problem:
Generate a module 20 Johnson counter using 74198 ICs. Make sure that when the circuit is energized it starts in a code state.

I need the counter to be modulo 20, therefore, modulo = 2 * n then n = 10, my output would have 10 bits, right?
The 74198 is an 8 bit shift register, therefore I would need to use two ic's to implement the modulo 20 counter.

The first thing that occurred to me (I'm not sure if it's okay) is to propose the following table, where my initial state would be 0000000000 and I would go right shifting until completing the sequence of 1111111111, then I would make another right shift but this time entering a 0, which would be 0111111111 and as the zero runs, it would reach 0000000001 and finally it would return to the initial sequence of 0000000000.

I have a doubt, because according to the table I made, for the Q8 output, I would have for the same combination, two different values of Q8, that is why I think what I did is wrong, would someone be so kind to help me with the design ?

Thanks!!

 

You need three 8-bit shift registers to make a 20 bit shift register. 4 of the bits will be unused. You may have other problems with your design, but this is the big one.

 

Generate a module 20 Johnson counter using 74198 ICs. Make sure that when the circuit is energized it starts in a code state.

Please post the entire text of the problem. I wouldn't consider a shift register to be the same as a Johnson counter.

 

Please post the entire text of the problem. I wouldn't consider a shift register to be the same as a Johnson counter.

With feedback it is exactly how you make a Johnson (aka twisted-ring) counter.

 

With feedback it is exactly how you make a Johnson (aka twisted-ring) counter.

But a decade Johnson counter doesn't take 10 flip flops.

 

But a decade Johnson counter doesn't take 10 flip flops.

I never said it did. We are talking about a modulo 20 counter.