What is the output of a logic gate when one input is not connected?

In summary, the output of a logic gate will be set at one of three states, depending on the logic family used.
  • #1
r731
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Hello,

I'm currently studying a course in digital logic. The course is included in my IT program. And I have not taken a course in electronics yet; I will later.

My question basically is: what is the output of a logic gate given that one input is neither connected to a high voltage nor connected to ground? (That is, the input is not connected to anything.) And why?

debounceSR_large.gif


Tomorrow I'm having a lab, where I will connect a mechanical switch to a latch.
Thanks in advance.
 
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  • #2
In general case the output will be set at high state or at low state or somewhere between these two states. It all be depend on the logic family you are actually be using. And this is why we should never leave unused inputs floating. And always use a pull-up or pull-down resistors.
 
  • #3
The power supplies to the logic gates are implicit and so not usually shown. In this case the supplies are assumed to be GND = 0V and +6V. Since four gates, each with two inputs and one output, are all in each package, they can share the two power supply pins. That makes 3*4 + 2 = 14 pins on the IC.

The “>=1” gate is an OR gate, but it has an inverted output, so it is a NOR gate.
The circuit shown is an RS flipflop. If you follow the logic state through the circuit, then flip the switch and watch the new state develop, you will understand why this circuit is called a switch de-bounce circuit.
 
  • #4
r731 said:
My question basically is: what is the output of a logic gate given that one input is neither connected to a high voltage nor connected to ground? (That is, the input is not connected to anything.) And why

It is generally considered a design error to leave logic inputs floating. Especially with CMOS gates, floating inputs can cause problems with increased power consumption and noise generation. Always tie off unused logic inputs. :smile:
 
  • #5
The “green” light emitting diodes in the OP diagram may be a problem because they prevent the input voltage being pulled hard down to zero.
If the LEDs are included then there should be another pull-down resistor in parallel with each (Resistor–LED) pair.
 

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