Electronic – The TI quad-XOR gate acts as it is short circuit

Possible reasons:

1. Load Balancing
   • The driver of A has an unknown number of fan-out to drive. Fan-out within the circuit and the parasitic it induces can be calculated for the specific circuits, but we do not know the other circuits that are connected the driver. Essentially the inverters are being used as buffer equivalent. and help manage the parasitic.
2. Timing and total current
   • To reduce the transition glitch, the second state inverters can be sized for a faster transition switch. Doing so makes the NAND gates input update near the same time. With the inputs changing less periodically, power can be saved and transition glitches can be reduced.
3. Signal boosting and power
   • Lets say VDD = 1.2V but the input is 0.9V. The input is still a logical 1, but considered weak which causes slower switching and burns more power. The first inverters can be sized to handle transitions better, making the voltage more predictable for the rest of the design.
   • There is also a possibility of the change in the voltage domain. In this case the inverters in the first state can act as a step down, e.g. a 5V input domain to 2V domain.
4. Any combination of the above

Don't listen to caveman.

^{simulate this circuit – Schematic created using CircuitLab}

Flip-flops can be synthesized from gates, although you may well decide to take Spehro's advice.

In the schematic, Switch 5 is a reset, and pressing it causes all 4 outputs (at the LED) to go high, turning off the LEDs. Pushing any of Switches 1 - 4 will cause the corresponding LED to turn on, and disable any other switch activations by way of the AND gates.

If you want to use discrete logic, such as 7400/74LS/74HC, etc, the total package count is 5 ICs - 1 7400, 2 7410 and 2 7411.

ETA - You'll notice that the LEDs are driven by being pulled down, with current coming from VCC. This is important if you decide to use 7400 or 74LS - those technologies do not source current well, but they sink current just fine. If you decide to go with a CMOS family such as 74HC, this will still work. Or, if you like, you can use the other output from each latch, the 2-input NAND gate, and drive the LED to ground (being careful to get the LED polarity right). Just a thought.