# Active Low S R Latch and Flip Flop

on 24/2/2012 & Updated on Thursday 17th of May 2018 at 03:44:56 PMThere is one type of latch which is SET when S = 0(LOW), and this latch is known as

Rate the page.**Active Low S R Latch**.This latch is normally designed by using NAND gates. The logical circuit is shown below.In the above logic circuit if S = 0 and R = 1, Q becomes 1. Let us explain how.

- NAND gate always gives output 1 when at least one of the inputs is 0.
- So, when S is applied as 0 the output of gate G1 i.e. Q is 1 irrespective of the condition of second input to the gate.
- Now, Q is input of gate G2 so both the inputs of G2 become 1 as R is already 1. So output of G2 is now or 0.
- So whatever may be the previous condition of Q, it always becomes Q = 1 and = 0 when S = 0 and R = 1. This is called SET condition of the latch.

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In the above logic circuit if S = 1 and R = 0, Q becomes 0. Let us explain how.

- As we already said, a NAND gate always gives output 1 when at least one of the inputs is 0.
- So when R is applied as 0, the output of gate G2 i.e. is 1 irrespective of the condition of second input Q to the gate.
- So whatever may be the previous condition of , it always becomes 1 this 1 is then fed back to input of gate G1. As here S is already 1, both inputs of G1 are 1. Hence output of G1 i.e. Q will be 0. So Q = 0 and = 1 when, S = 1 and R = 0. This is called RESET condition of the latch.

In the above logic circuit if S = 1 and also R = 1, Q remains same as it was. Let us explain how.

- First suppose Q is previously 1.
- Now both inputs of G2 are 1 as R = 1 and Q = 1. So output of G2 i.e. is or 0.
- Now the inputs of G1 are 1 and 0 as R = 1 and = 0. So output of G1 i.e. Q is or 1.
- Now suppose Q is previously 0.
- Now the inputs of G2 are 1 and 0 as R = 1 and Q = 0. So output of G2 i.e. is or 1.
- Now both inputs of G1 are 1 as S = 1 and = 1. So output of G1 i.e. Q is or 0.
- So it is proved that Q remains same as it is when, S = 1 and also R = 1.

In the above logic circuit if S = 0 and also R = 0, the condition of Q is totally unpredictable. Let us explain how.

- First suppose Q is previously 0.
- Now both inputs of G2 are 0 as R = 0 and Q = 0. So output of G2 i.e. is or 1.
- Now the inputs of G1 are 0 and 1 as S=0 and = 1. So output of G1 i.e. Q is or 1. That means Q is changed.
- Now Q is 1. So inputs of G2 are 0 and 1 as R = 0 and Q = 1. So output of G2 i.e. is or 1. That means is unchanged.
- Now the inputs of G1 are 0 and 1 as S=0 and = 1. So output of G1 i.e. Q is or 1. That means Q is unchanged.

**S R latch**normally avoided.Rating = 0 & Total votes = 0