How JK flip flop works

digital-logicflipflop

Consider below JK flip flop circuit and truth table:

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I was guessing how Qn+1 column in truth table is calculated.

Interpretation 1

One text book says:

Consider the case J=1, K=0, Qn=0, Qn'=1, line 5 in truth table.

  1. K=0 forces output of G4=1.
  2. All inputs (J, CLK, Qn') of G3 are 1 which makes output of G3 to 0.
  3. G3 output is fed to G1 input. This forces output of G1, Q(n+1) to 1.
  4. This (output of G1) is fed to input of G2 along with output of G4, both of which are 1 (from point 1 and 3). Hence output of G2, Q'(n+1) will be 0.

So in this case output is correctly Q(n+1)=1, Q'(n+1)=0, which is correctly a Set behavior.

Interpretation 2

I was guessing why it doesnt behave like this:

  1. K=0 forces output of G4=1.
  2. Qn=0 which is fed to G2. This forces output of G2, Q'(n+1) to 1.
  3. Q'(n+1)=1 which is fed to G3.
  4. All inputs (J, CLK, Qn') of G3 are 1 which makes output of G3 to 0.
  5. This output of G3 is fed to G1, which forces output of G1, Q(n+1) to 1.

So in this case both Q(n+1) and Q'(n+1) becomes 1, which is invalid.

Why it doesnt behave like interpretation 2 and give invalid state?

Best Answer

You are not done yet.

After Q changes to 1 at step 5 you need to go back to revisit step 2. Q' now becomes 0. This feeds back to G1 and keeps Q at 1.

There are not necessarily just \$n\$ and \$n+1\$ values of the signals. You need to keep propagating the changes through the circuit until no more gate outputs change state. At that point the circuit is stable.

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