![]() ![]() Top Cognos Interview Questions And Answers.Top Microstrategy Interview Questions And Answers.Top Data Warehouse Interview Questions and Answers.Top 65+ Tableau Interview Questions and Answers in 2023.QlikView Interview Questions and Answers.Top MSBI Interview Questions and Answers.Top Pentaho Interview Questions And Answers.Top Obiee Interview Questions And Answers.Top SAS Interview Questions and Answers.Top Sqoop Interview Questions – Most Asked.Top Hive Interview Questions – Most Asked.Top PIG Interview Questions - Most Asked.Top Couchbase Interview Questions - Most Asked.Top HDFS Interview Questions And Answers.Top Mapreduce Interview Questions And Answers.Top 40 Apache Spark Interview Questions and Answers in 2023.Top Apache Storm Interview Questions And Answers.Top Apache Solr Interview Questions And Answers.Top 70 Hadoop Interview Questions In 2023.Top Splunk Interview Questions and Answers.Top 50 Automation Anywhere Interview Questions and Answers in 2023.Top 50 UiPath Interview Questions and Answers in 2023.Top RPA Interview Questions and Answers.This problem is avoid by SR = 00 and SR = 1 conditions. Hence S = R = 0 or S = R = 1, these input condition will never appear. S and R will be the complements of each other due to NAND inverter. Due to this data delay between i/p and o/p, it is called delay flip flop. The input data is appearing at the output after some time. The master slave flip flop will avoid the race around condition.ĭelay Flip Flop or D Flip Flop is the simple gated S-R latch with a NAND inverter connected between S and R inputs. This avoids the multiple toggling which leads to the race around condition. So it does not respond to these changed outputs. But since clock = 0, the master is still inactive. These changed output are returned back to the master inputs. ![]() So S and R also will be inverted.Ĭlock = 0 − Slave active, master inactive. Therefore outputs of the slave become Q = 1 and Q bar = 0.Īgain clock = 1 − then it can be shown that the outputs of the slave are stabilized to Q = 1 and Q bar = 0.Ĭlock = 1 − Master active, slave inactive. That means S = 1 and R =0.Ĭlock = 0 − Slave active, master inactive. Therefore outputs of the master become Q 1 = 1 and Q 1 bar = 0. Thus we get a stable output from the Master slave.Ĭlock = 1 − Master active, slave inactive. Hence with clock = 0 and slave becoming active the outputs of slave will remain Q = 0 and Q bar = 1. Therefore even with the changed outputs Q = 0 and Q bar = 1 fed back to master, its output will be Q1 = 0 and Q1 bar = 1. Therefore outputs of the slave become Q = 0 and Q bar = 1.Īgain clock = 1 − Master active, slave inactive. That means S = 0 and R =1.Ĭlock = 0 − Slave active, master inactive. ![]() Therefore outputs of the master become Q 1 = 0 and Q 1 bar = 1. Therefore outputs will not change if J = K =0.Ĭlock = 1 − Master active, slave inactive. But since the S and R inputs have not changed, the slave outputs will also remain unchanged. ![]() When clock = 0, the slave becomes active and master is inactive. Whereas when clock = 0 (low level) the slave is active and master is inactive. Hence when the clock = 1 (positive level) the master is active and the slave is inactive. But due to the presence of the inverter in the clock line, the slave will respond to the negative level. Master slave JK FF is a cascade of two S-R FF with feedback from the output of second to input of first. Hence the Race condition will occur in the basic NAND latch. This is the reset condition.Īs S = 1, R = 1 and E = 1, the output of NAND gates 3 and 4 both are 0 i.e. Hence output of S-R NAND latch is Q n+1 = 1 and Q n+1 bar = 0. R' = 1 and E = 1 the output of NAND-4 i.e. Since S' and R' are the input of the basic S-R latch using NAND gates, there will be no change in the state of outputs. If S = R = 0 then output of NAND gates 3 and 4 are forced to become 1. ![]()
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