Hello,

i have a problem designing a 4-bit even number counter.,
the states of the counter is,


0000 = 0
0010 = 2
0100 = 4
1000 = 8

in each number the LSB is '0'
so when i draw the karnaugh Map of J0 is filled with 0s and K0 with 'X' (Don't Care) ?
am i doing something wrong ??

Edi: Problem solved !!

 

"Edi: Problem solved !!"

Care to share your solution for other seekers?

My QnD (Quick & Dirty) solution would be to connect the counter output lines to the next highest input line:

O0 -- I1
O1 -- I2
O2 -- I3
O3 -- Not connected
I0 is grounded.

--Rich

 

Jo is grounded
and its really don't care either Ko is connected to Vcc or grounded because in both situations
the output for the flip flop remains '0'.

Jo=0(ground) , Ko=1 =>> Output = '0' (Reset)
Jo=0(ground) , Ko=0 =>> Output = '0' (No change)

Since Jo is grounded therefore there will be no situation where Output is equal to '1' ,
it always remain 0.