I wanted to create a circuit that would count from 2 to 12.

In order to do so, I created a simple synchronous counter that resets itself when I have logic Highs at the third, second, and zero flip-flops. So that means that it will reset when I try to reach number thirteen. However, it resets when I try to go from seven to eight.

I suspect it has to do with the delay time of elements. I doubt that it's due to the flip-flops, because it's a synchronous counter so the delay due to the flip-flops is the same for every step. So it has to be due to the logic AND gates, but I can't figure it out. This is the schematic:


Any help appreciated!

  • \$\begingroup\$ You are doing an asynchronous set/reset based on the current state. That is inherent unreliable: the reset pulse itself is removed as soon as one of the register output changes. In fact you are making a reset spike. \$\endgroup\$
    – Oldfart
    Commented May 7, 2019 at 13:22
  • \$\begingroup\$ @Oldfart Thank you for your response, if i understood you well, you are saying that reset pulse is removed right after one of the flip flops changes it's state. But if that's the case, how come that reseting to 2 still happens, it just happens earlier than it should? \$\endgroup\$
    – cdummie
    Commented May 7, 2019 at 18:03
  • \$\begingroup\$ @cdummie Just prepare an excitation table for your TFF-wired JKs. Have you done that? \$\endgroup\$
    – jonk
    Commented May 7, 2019 at 18:34
  • \$\begingroup\$ @jonk What is TFF? \$\endgroup\$
    – cdummie
    Commented May 7, 2019 at 18:48
  • \$\begingroup\$ @cdummie It's how you have your JK's wired. Just a "toggle" FF. Have you ever heard of an excitation table? \$\endgroup\$
    – jonk
    Commented May 7, 2019 at 18:51

1 Answer 1


Since you want to start at two, I decided to use the \$\overline{Q_B}\$ as output instead of \$Q_B\$ so that the reset state starts at the right place.

$$\begin{array}{c|c|c} \text{Beginning State} & \text{Ending State} & \text{Excitation}\\\\ {\begin{array}{cccc} Q_D & Q_C & \overline{Q_B} & Q_A\\\\ 0&0&1&0\\ 0&0&1&1\\ 0&1&0&0\\ 0&1&0&1\\ 0&1&1&0\\ 0&1&1&1\\ 1&0&0&0\\ 1&0&0&1\\ 1&0&1&0\\ 1&0&1&1\\ 1&1&0&0\\\\ 0&0&0&0\\ 0&0&0&1\\ 1&1&0&1\\ 1&1&1&0\\ 1&1&1&1\\ \end{array}} & {\begin{array}{cccc} Q_D & Q_C & \overline{Q_B} & Q_A\\\\ 0&0&1&1\\ 0&1&0&0\\ 0&1&0&1\\ 0&1&1&0\\ 0&1&1&1\\ 1&0&0&0\\ 1&0&0&1\\ 1&0&1&0\\ 1&0&1&1\\ 1&1&0&0\\ 0&0&1&0\\\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ \end{array}} & {\begin{array}{cccc} T_D & T_C & T_B & T_A\\\\ 0&0&0&1\\ 0&1&1&1\\ 0&0&0&1\\ 0&0&1&1\\ 0&0&0&1\\ 1&1&1&1\\ 0&0&0&1\\ 0&0&1&1\\ 0&0&0&1\\ 0&1&1&1\\ 1&1&1&0\\\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ x&x&x&x\\ \end{array}} \end{array}$$

You wanted to go from 2 to 12. So you can see the binary codes present for that in the above table. If you take your outputs as I suggested, that table should cover it. (Note that I show \$\overline{Q_B}\$ and not \$Q_B\$. If it's not yet clear, the reason is that the power-on reset state for each \$Q\$ output is 0, not 1. So I'm picking the \$\overline{Q}\$ output of the \$Q_B\$ FF.)

The last column is the Excitation that you need for each of your TFF-wired JK-FFs. (Here, all I just mean is that you've tied J and K together so that they can either both be 0 or both be 1 [toggling occurs with the value 1 used, otherwise the output value remains unchanged.]) This last column represents the value that should be applied to the JK-pair wired together for that FF. (You already are doing something like that, so you are aware of the idea.) You want a 0 presented to the TFF if you want to hold the value and you want a 1 presented to the TFF if you want to change the value (toggle it.) It's pretty simple.

Looking over the table, does all this make sense?

Once you have that much, all you need to do is to lay out four K-map tables.

$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} T_D&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&0&x&x\\ \overline{Q_D}\:Q_C&0&1&0&0\\ Q_D\: Q_C&x&x&x&1\\ Q_D\:\overline{Q_C}&0&0&0&0 \end{array}\end{smallmatrix} & \begin{smallmatrix}\begin{array}{r|cccc} T_C&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&1&x&x\\ \overline{Q_D}\:Q_C&0&1&0&0\\ Q_D\: Q_C&x&x&x&1\\ Q_D\:\overline{Q_C}&0&1&0&0 \end{array}\end{smallmatrix}\\\\ \begin{smallmatrix}\begin{array}{r|cccc} T_B&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&1&x&x\\ \overline{Q_D}\:Q_C&0&1&1&0\\ Q_D\: Q_C&x&x&x&1\\ Q_D\:\overline{Q_C}&0&1&1&0 \end{array}\end{smallmatrix} & \begin{smallmatrix}\begin{array}{r|cccc} T_A&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&1&1&x&x\\ \overline{Q_D}\:Q_C&1&1&1&1\\ Q_D\: Q_C&x&x&x&0\\ Q_D\:\overline{Q_C}&1&1&1&1 \end{array}\end{smallmatrix} \end{array}$$

You can now use those tables (fixed for errors you may catch) to develop the logic required.

Does that also make sense?

Let's start with \$T_A\$, since it's pretty easy. All of the \$x\$ values on the \$Q_D\: Q_C\$ row can be set to 0 (since it doesn't matter.) And the remaining \$x\$ values can be set to 1. This makes it very easy to work out that a NAND gate suffices: \$T_A=\overline{Q_C\: Q_D}\$:

$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} T_A&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&1&1&1&1\\ \overline{Q_D}\:Q_C&1&1&1&1\\ Q_D\: Q_C&0&0&0&0\\ Q_D\:\overline{Q_C}&1&1&1&1 \end{array}\end{smallmatrix} \end{array}$$

Next up is \$T_B\$. I think you can just spot the changes I made to the \$x\$ values for this table, by inspection. Now I think you can see that \$T_B=Q_A+Q_C\: Q_D\$:

$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} T_B&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&1&1&0\\ \overline{Q_D}\:Q_C&0&1&1&0\\ Q_D\: Q_C&1&1&1&1\\ Q_D\:\overline{Q_C}&0&1&1&0 \end{array}\end{smallmatrix} \end{array}$$

Now for \$T_C\$. Again, spot the changes by inspection and you'll see why \$T_C=Q_A\:\overline{Q_B}+Q_C\:Q_D\$:

$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} T_C&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&1&0&0\\ \overline{Q_D}\:Q_C&0&1&0&0\\ Q_D\: Q_C&1&1&1&1\\ Q_D\:\overline{Q_C}&0&1&0&0 \end{array}\end{smallmatrix} \end{array}$$

Finally, \$T_D\$. And again, inspect the following chart to see that \$T_D=Q_A\:\overline{Q_B}\:Q_C+Q_C\:Q_D\$:

$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} T_D&\overline{Q_B}\:\overline{Q_A}&\overline{Q_B}\: Q_A&Q_B \:Q_A&Q_B \:\overline{Q_A}\\ \hline \overline{Q_D}\:\overline{Q_C}&0&0&0&0\\ \overline{Q_D}\:Q_C&0&1&0&0\\ Q_D\: Q_C&1&1&1&1\\ Q_D\:\overline{Q_C}&0&0&0&0 \end{array}\end{smallmatrix} \end{array}$$

So the equation summary from the above work is:

$$\begin{align*} T_A&=\overline{Q_C\: Q_D}\\ T_B&=Q_A+Q_C\: Q_D\\ T_C&=Q_A\:\overline{Q_B}+Q_C\:Q_D\\ T_D&=Q_A\:\overline{Q_B}\:Q_C+Q_C\:Q_D \end{align*}$$

Let's set up some temporary outputs and modify the above equations as we go:

Step 1: $$\begin{align*} T_0&=\overline{Q_C\: Q_D}\\ T_A&=T_0\\ T_B&=Q_A+\overline{T_0}=\overline{\overline{Q_A}\: T_0} \end{align*}$$

Already, you can see that with just two NAND gates we've got both \$T_A\$ and \$T_B\$ covered. (This is because your flip-flops have both \$Q\$ and \$\overline{Q}\$ outputs. So we don't even need to add an inverter.) Not bad, so far.

Step 2: $$\begin{align*} T_1&=\overline{Q_A\: \overline{Q_B}}\\ T_C&=\overline{T_1}+\overline{T_0}=\overline{T_0\:T_1}\\ T_D&=Q_C\left(\overline{T_1}+Q_D\right)=Q_C\:\overline{T_1\:\overline{Q_D}} \end{align*}$$

And here we find we need just three more NAND gates plus an AND.

So the total required will be five NAND gates and an AND gate.

The resulting schematic is:


simulate this circuit – Schematic created using CircuitLab

Using Neemann's "Digital" program I created some test vectors. Here's the resulting conclusion from his program:

enter image description here

  • \$\begingroup\$ This is more than good, i just simulated this circuit in simulation software and it works just fine. However, i am wondering what was the problem with my circuit, what was the reason for reseting after 7 instead reseting after 12? \$\endgroup\$
    – cdummie
    Commented May 8, 2019 at 7:46
  • \$\begingroup\$ @cdummie All you have to do is apply your logic gates to the states, just as I did in the very first table I generated. Try it by hand and apply your logic gates. See what states it takes you to. You will immediately see the problem. By the way, it isn't enough to show you how to succeed? But I must also show you how you failed, too? \$\endgroup\$
    – jonk
    Commented May 8, 2019 at 8:13
  • \$\begingroup\$ @cdummie Mostly, the problem will be found in your approach to the solution. If you follow the above approach religiously, you'll always get the correct result. It simply works every time. \$\endgroup\$
    – jonk
    Commented May 8, 2019 at 8:25
  • \$\begingroup\$ I see, i tried to find whats the deal in some digital electronics books, because i cannot see wheres the logical mistake in my circuit, however i will keep in mind your approach since it seems like it will work for every possible combination \$\endgroup\$
    – cdummie
    Commented May 8, 2019 at 9:11

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