I'm working on one of my projects and I'm having trouble with my flip flops. The project is to design the circuit for a gas pump that has a handle and a sensor. When the handle is down, the pump pumps, but when the sensor is activated, the pump shuts off. It will pump again until a second high pressure is read, at which point the pump shuts down completely until a manual reset is done (not part of the project). I'm working on the Mealey design since it's a little smaller.

Essentially, when my handle is high and my sensor is low, I'm getting an X for my output and I don't know why. I hope someone can provide some input:

Here is everything I have for it:

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Alright, I added the reset. Now I'm getting that my states change every time my clock changes. Any ideas?:

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  • \$\begingroup\$ "It will pump again until a second high pressure is read": do you mean if handle goes to zero and then to 1 again? In other words, how is the pump restarted after first shut-off? \$\endgroup\$ Apr 8, 2013 at 23:07
  • \$\begingroup\$ If I understand your question correctly, then yes. Once a high pressure is read while the handle is pulled, the pump goes to 0. The handle must be released and pressed again in order to start pumping until a second high pressure is read. At which point, you cannot pump again. \$\endgroup\$
    – rphello101
    Apr 8, 2013 at 23:27
  • 1
    \$\begingroup\$ If I understand your drawing, your "Handle" input connects only to the AND gate at lower right. And the output of that AND gate goes nowhere. So I don't see any way for the Handle input to affect the state machine. \$\endgroup\$
    – The Photon
    Apr 9, 2013 at 0:46
  • \$\begingroup\$ @The Photon, because this is a Mealey design where the the states are as follows: S0 = no high pressure warning, S1 = 1 warning, and S2 = 2 warnings, the handle does not effect state transition, only whether or not the machine is pumping. The problem rested in the Set/Reset as Tim mentioned below. \$\endgroup\$
    – rphello101
    Apr 9, 2013 at 17:25

1 Answer 1


Seems to me like your flip flops need to have a reset value.

When power is applied to a DFF at "time=0", the output state is X, or unknown. A reset is needed to bring the DFF to a known state.

In your case, check out the FF DB. It's input is:

(DB && !DA) || (Sensor && !DB && DA). 

When I substitute your time zero values into this I get:

= (X && !X) || (0 && X && X)
= ( X || 0 )
= X

So until something happens that actually forces the inputs to a non-X value, the circuit will have an X state forever. A 'reset' would typically be used to clear the Xs.

  • \$\begingroup\$ I think I understand what you mean. Any suggestions on where best to put a reset? \$\endgroup\$
    – rphello101
    Apr 8, 2013 at 23:12
  • 2
    \$\begingroup\$ Ideally you could replace your DFF with resettable DFFs which have a dedicated reset pin, but if that's not supported to your tool you can add an extra logic gate at the D pin of the flop. If I assume that you want to reset the flops to 0, and have an active low reset reset_, then I would just add an AND gate to the D-pin of each flop (D = original input && reset_). Then just pull reset low for a clock cycle and the flops will fall to zero. @rphello101 \$\endgroup\$
    – Tim
    Apr 8, 2013 at 23:20
  • \$\begingroup\$ This did turn out to be my problem. My second problem that I had was in my wiring. My SOP's were correct, but if you look at my wiring, there are a couple wrong connections. Thanks for the advice. \$\endgroup\$
    – rphello101
    Apr 9, 2013 at 17:27

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