Normally, for any counter, we can use both PET (Positive edge triggering) or NET (Negative edge triggering). But, most of the time we use NET. The reason is that in case of PET, considering a standard asynchronous counter ( n-bit), the output of nth flip-flop will change it's state, as it's preceding flip-flop changes it's state from 0 to 1 or low to high. In general, we take the output of flip-flop from Q input, and, the counter will seem to run in reverse order, if PET is used.
This can be understood by an example. Let us take a 2-bit counter with PET J-K Flip-Flops. Let the clock input is applied to the first Flip-Flop. Now taking a look at it's states. Assume that, initially all the flip-flops are at logic low. At the first rising edge of the clock pulse, FF0 (First flip-flop) will change it's state from 0 to 1. This transition from 0 to 1 by FF0 is the clock input to FF1, which sees a rising edge at it's clock input, will also change it's state from 0 to 1. So, at the first clock pulse, the counter will go from it's initial state of 00 to 11. On, second clock pulse, FF0 will transit from 1 to 0, but FF1 will remain at 1 because it is seeing a falling edge. So, the state will shift from 11 to 10. At third pulse, FF0 will shift from 0 to 1, and FF1 will shift from 1 to 0. so, the state will shift from 10 to 01. Finally, at fourth clock pulse, FF0 will shift it's state from 1 to 0 and, FF1 will remain at it's previous state. so, the new state will be 00. Now, taking a look at complete cycle we will have, 00,11,10,01....... so on. So, it is clear that counter is behaving as down counter or reverse counter.
It is a general practice to take output from Q output and not from Q-bar (~Q). So, a better method is to use NET. Since, the output of FF1 in that case will shift it's state when FF0 will shift from 1 to 0. So, the state cycle in that case will be, 00,01,10,11..... so on. It is clear that output of FF1 is changing only when FF0 is changing from 1 to 0 ie. falling edge.
So, to conclude, we can say that in general, we use negative edge triggering in case of counters.
This flip-flop toggles (Q changes state) on the negative going edge of the clock pulse. T acts as an ENABLE / INHIBIT control. Q will only toggle on the negative edge of the clock pulse, when T is high. Below is shown a D type flip-flop connected as a toggle type. On each clock pulse positive going edge, Q will go to the state bar Q was before the clock pulse arrived. Remember that bar Q is the opposite level to Q. Therefore Q will toggle.
Level Trigger:1) The input signal is sampled when the clock signal is either HIGH or LOW.2) It is sensitive to Glitches.Example: Latch.Edge Trigger:1) The input signal is sampled at the RISING EDGE or FALLING EDGE of the clock signal.2) It is not-sensitive to Glitches.Example: Flipflop.
counters are used to store, display or sometimes count the pulses in a circuit.
negative indent
A: it does not apply to only flip-flop but to all kinds of logic circuits where a [+] transition from a less negative to a more positive level occurs [-] and the other way around meaning a +/- transition must occurs to transfer states.
Yes, there is a difference between edge triggering and pulse triggering. Edge triggering occurs when a circuit changes state based on the transition of an input signal (e.g., from low to high or high to low). Pulse triggering, on the other hand, involves triggering a circuit based on the detection of a specific pulse width within the input signal.
An edge-triggered flip-flop changes states either at the positive edge (rising edge) or at the negative edge (falling edge) of the clock pulse on the control input.
flipflop is edge triggering and latch is level triggering
No, negative counters do not have the ability to eliminate creatures with indestructible.
Zero.
negative triggering relay.
The characteristics of a UJT are: stable triggering voltage, negative resistance, high pulse current capability and low value of triggering current.
No, Dijkstra's algorithm does not work for graphs with negative edge weights because it assumes all edge weights are non-negative.
Dijkstra's algorithm does not work with negative edge weights in a graph because it assumes all edge weights are non-negative. Negative edge weights can cause the algorithm to give incorrect results or get stuck in an infinite loop. To handle negative edge weights, a different algorithm like Bellman-Ford should be used.
Not all are: JK flip-flops use a master-slave triggering for example.
This flip-flop toggles (Q changes state) on the negative going edge of the clock pulse. T acts as an ENABLE / INHIBIT control. Q will only toggle on the negative edge of the clock pulse, when T is high. Below is shown a D type flip-flop connected as a toggle type. On each clock pulse positive going edge, Q will go to the state bar Q was before the clock pulse arrived. Remember that bar Q is the opposite level to Q. Therefore Q will toggle.
$15,000 and up more for a negative edge pool