Question: Synchronous 4-bit DOWN counter, for the sequence 12...0, using T flip-flops. Followed as many tutorials as I could and simulated the counter, but it doesn't
Synchronous 4-bit DOWN counter, for the sequence 12...0, using T flip-flops.
Followed as many tutorials as I could and simulated the counter, but it doesn't seem to work - can anyone see a reason why? (In the simulation it just switches between two values 1011 and 0011)
Truth table:

Karnaugh map for each of the toggles: 
Simulation in CircuitVerse:

Present state Toggle state T2 T1 Q3 Q2 Q1 QO Q3-1 Next state Q2:1 Q1.1 0 1 T3 Q 1100 TO Q0.1 1 1 1 0 0 1 0 1 1 1 B 1011 1 0 1 1 1 0 1 0 0 0 0 1 1010 1 0 1 0 1 0 0 1 0 0 1 1 9 1001 1 0 0 1 1 0 0 0 0 0 0 1 8 1000 1 0 0 0 0 1 1 1 1 1 1 1 7 0111 0 1 1 1 0 1 1 0 0 0 0 1 6 LD 0110 0 1 1 0 0 1 0 1 0 0 1 1 5 0101 1 0 1 0 1 0 0 0 0 0 1 0 0 4 1 0 0 0 0 1 1 0 1 1 1 0100 0011 3 0 0 1 1 0 1 0 0 0 0 1 0 0 0 0 0 1 O 0 0 1 0 0 1 1 2 1 0010 0001 0000 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 0 0 Q2,Q3 00 01 11 10 QO,Q1 000 1 1 1 Q2,Q3 00 01 11 10 QO,Q1 od 1 0 0 0 TO 1 01 1 1 Q3 + Q2 + Q1 + QO 1 T2 01 1 0 0 0 Q2'Q3 11 1 - - - 11 1 - 101 1 1 1 101 0 0 0 Q2,Q3 f Q2,Q3 00 01 11 10 00 01 11 10 QO,Q1 000 1 0 0 QO,Q1 od 1 0 0 0 T1 011 1 0 0 Q2'Q3 + Q1Q2' + QOQ2' T3 011 0 0 0 QO'Q2'03 11 1 11 0 - 101 1 0 0 100 0 0 0 o
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