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I'm building a simple ring counter with two jk flip-flop. I'm using 74HC76 integrated that contains two jk.

This is the integrated pinout:

enter image description here

This is my schematic (very basic):

schematic

simulate this circuit – Schematic created using CircuitLab

To build a working circuit it's necessary that you connect pin 5 to Vcc (5V) and pin 13 to GND. But...

The problem is that in the circuit that I'm observing on my desk the ring counter only works when pin 5 (Vcc) is unwired. It means that if I wire pin 5 to Vcc I got always 10 output. Is that possible? If no what's the matter of my circuit?

I've tried all:

  • I've substituted the integrated -> same problem
  • I've checked another time pin connection -> same problem
  • I've changed wires -> same problem

EDIT: I initialize the first jk with 1 using a button and preset and clear pin.

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    \$\begingroup\$ First: Do not, ever, disconnect the VCC pin when debugging, because that breaks chips. Second: what are you doing with the Preset and Clear pins? (i.e. J and K propagate 1 to 1, if you don't preset and clear one on initialisation, you will not know what happens, but them both starting with 1/0 or both with 0/1 is quite likely) \$\endgroup\$ – Asmyldof Dec 9 '15 at 22:54
  • \$\begingroup\$ JK1 and JK2 need to initialize in opposite states, don't they? How do you ensure this? Also, you should use resistors in series with the LEDs \$\endgroup\$ – user28910 Dec 9 '15 at 23:04
  • \$\begingroup\$ @Asmyldof: Thanks for your advice. About Preset and Clear pin I'm managing them well: I've a button that sets the starting 1 in the first jk, so I think that the problem isn't about Preset or Clear pin. \$\endgroup\$ – xdola Dec 9 '15 at 23:04
  • \$\begingroup\$ @user28910 : I've just edited my question \$\endgroup\$ – xdola Dec 9 '15 at 23:07
  • \$\begingroup\$ On CMOS parts, all unused inputs must be connected, directly or through a resistor, to either Vcc or ground, whichever is required to make the part work as required. For the Preset pin where you have a switch to pull it to ground, you must also have a resistor (5K - 10K) to Vcc to ensure the preset pin is high when the switch is not taking it low. \$\endgroup\$ – Peter Bennett Dec 9 '15 at 23:27
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Without a resistor in series with each LED is is likely that the corresponding Q output is being dragged down to a voltage that is not high enough to be recognized as a legitimate high level by the following J input.

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On CMOS parts, all unused inputs must be connected, directly or through a resistor, to either Vcc or ground, whichever is required to make the part work as required. For the Preset pin where you have a switch to pull it to ground, you must also have a resistor (5K - 10K) to Vcc to ensure the preset pin is high when the switch is not taking it low.

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