Hi, there are three JK-FF, with Q1 as MSB, Q3 as LSB. J1,K1 ist input to Q1, etc. . There are 8^6 possibilties (6 inputs to the Qx or QxNOT or to HIGH or to LOW) of which 2069 generate a cycle length of 5.
The following wiring will generate the cycle 1 3 5 2 4 : J1=Q2 K1=Q1 J2=Q3 K2=Q2 J3=Q2NOT K3=Q1 Q3 is output with a 3/5 duty cycle. The unused state 0 will migrate to 1,6->0,7->0. So the machine does not get stuck. Cheers Detlef _______________________________________________ time-nuts mailing list -- time-nuts@lists.febo.com To unsubscribe, go to http://lists.febo.com/mailman/listinfo/time-nuts_lists.febo.com and follow the instructions there.
