Timing diagram for edge triggered flip flop
While CK is high, Q will take whatever value D is at. The most common type of latch is the D latch. Latches are similar to flip-flops, but instead of being edge triggered, they are level triggered. The timing diagram for the negatively triggered JK flip-flop: The truth table for a negatively triggered JK flip-flop: The JK flip-flop is usually negative edge triggered. If J and K are both 1, the output is inverted. If J and K are both 0, the output stays the same as it was before. J corresponds to a "set" signal, and K corresponds to a "reset" signal. The JK flip-flop has two inputs, labeled J and K. Timing diagram for the positive edge triggered D flip-flop: (↑ indicates a rising edge on the clock pulse X indicates that it has no effect on outcome) The truth table for a positive edge triggered D flip-flop: The D flip-flop is usually positive edge triggered. The D (Data) flip-flop has an input D, and the output Q will take on the value of D at every triggering edge of the clock pulse and hold it until the next triggering pulse. There are several types of flip-flops but the two most important kind are the D and J-K flip-flops. Truth Table for CLR and PRE (active low) PRE The symbols used for clear and preset (the bubble indicates an inverted signal): The CLR and PRE signals can be asserted any time and don't have to be edge triggered they will override any other inputs, including the clock. There are several ways to implement a dual edge-triggered flip-flop. As a result, power consumption is reduced, making DETFFs desirable for low power applications 2-19. These inputs are typically inverted, so they are active when the input signal is low ( Active Low Input). dual edge-triggered flip-flop (DETFF), which can operate at half of the clock frequency while maintaining the same data throughput compared to SETFF 1. Many flip-flops will also have a clear (CLR) and preset (PRE) terminal. The symbols used for positive and negative edge triggering on flip-flops: Flip-flops are edge triggered they either change states when the clock goes from 0 to 1 (positive/rising edge) or when the clock goes from 1 to 0 (negative/falling edge). Let me know if you have any confusion.Because the state of a flip-flop often depends on the previous state of a circuit (for example, the output of one flip flop may be the input to another), and because each flip-flop and logic gate needs a certain amount of time to switch its output, we usually clock the devices, that is, we synchonize all the flip-flops to change states at the same time with a clocked pulse. (b) Assume Q begins at 1, but Clock, J, and K are the same. If the question was supposed to mean that INPUT is given to both J and K simultaneously, then choose the case accordingly. Fundamentals of Logic Design (6th Edition) Edit edition Solutions for Chapter 11 Problem 22P: Fill in the timing diagram for a falling-edge-triggered J-K flip-flop. I have assumed K is not connected to any control input, hence the output values depend on it in certain clock edges.
If output was 0, for INPUT = J = 1, the output becomes 1, for both K=0 or 1. If output was 1, similar case of CLK edge 4 applies. The output in the previous cycle could be 1 or 0, depending on the value of K. The output in the previous cycle was one and INPUT= 1 at edge4.įor J=1, the output will toggle to 0 if K=1, or it will remain at 1 if K=0. The clock input of the remaining flip-flops is triggered by the Q output of the previous flip-flop.
The clock pulse input is given only to the first flip-flop. Since it is a 3-bit counter, 3 negative edge-triggered flip-flops are used. The output in the previous cycle was zero and INPUT =1 at edge3.įor J=1, the output is 1, for both K=0 or 1. The below diagram shows the 3-bit asynchronous down counter. The output in the previous cycle was zero and INPUT= 0 at edge2. So, the output should be zero in this clock cycle. The output was initially zero (or to be precise, high impedance) and at edge1, INPUT = J = 0. I'll consider the following JK-flip flop truth table. This is how I see your question: It seems the INPUT port is your 'J' port, which is being given the signal and you are expected to come out with the output value for the given change in J.īecause it is positive edge triggered, the output value will change only at the positive edge transition with respect to its output value in the previous clock cycle. Firstly, you should not see if it is a 'good' or 'bad' output, it should seem 'correct'.