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Switch debouncer using SR Latch

We use switch in our day to day life to switch ON/OFF a bulb or a fan or  any electrical devices. But when we use switches in digital circuits, we observe a phenomenon called bouncing. This occurs because, when we turn the switch, the mechanical parts vibrate. i.e It toggles between ON and OFF state for some time until the mechanical contact attain equilibrium. this vibrations are minute and are not at all noticeable in electrical circuit. where as in digital circuits, these vibrations create pulses. which are detected by circuits which results in an error.
              In the above circuit, there is a switch connected to VCC. You can toggle it between terminal 'a' and terminal 'b'. which as a voltage drop of 'Va' and 'Vb' respectively. In digital circuits if you can observe, it takes a finite amount of time to toggle between terminal 'a' and 'b'. which might approximately take 15ns. Now let us consider the terminal is at 'a'. we will switch it to 'b'. what we can observe are...
  • It takes finite time (approximately 15ns) to make contact with terminal 'b'.
  • Once the metal contact touches the terminal 'b'. It bounces several times before it reaches equilibrium.

So for the above circuit, the timing diagram is as follows...


From the timing diagram it is evident that there are several unwanted pulses which are present due to bouncing. This can be eliminated by using SR latch using NOR gates.
Switch debouncer


TRUTH TABLE for SR Latch using NOR gates

S           R          Q             Q'
1            0           1              0
0            1           0              1
0            0           No change
1            1          monostable state

The timing for the switch debouncer is shown below.

Here we can see that during switching time, 'Va' is high. this is because, during switching time Va and Vb are zero. According to the truth table, at S=0 and R=0, the output is the previous state. Hence 'Va' continues to be high.when the metal contact touches the B terminal. S=0 and R=1. Hence 'Vb' becomes high, because of the bounce effect, Vb and Va becomes 0. but from truth table we know there is no change in output when S=0 and R=0. hence Vb continues to be in high state irrespective of number of bounces in the metalic contact.


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