# Timing Diagrams for D Flip-Flops

• STEMucator
In summary, the circuit has a timing diagram with three clocks, CLK, RCLK, and FS. The circuit has two D flip-flops, one that is triggered by CLK and one that is triggered by RCLK. The purpose of the circuit is to allow two inputs to be compared. The first D flip-flop is triggered by the rising edge of CLK, the second D flip-flop is triggered by the rising edge of RCLK. The first D flip-flop is also triggered by the first FS clock, the second D flip-flop is also triggered by the second FS clock.
STEMucator
Homework Helper

## Homework Statement

Complete the timing diagram for the following circuit:

## The Attempt at a Solution

So I was doing some pre-midterm studying, and I was slightly confused with the operation of this circuit.

I think part of the solution provided is incorrect, but I'm not positive so I thought I would ask. Here is my attempt so far at filling in the waveforms:

I believe the value of ##B## provided at that clock edge is incorrect (the place where I have not connected the lines).

At that rising clock edge, the rising edge triggered D flip-flops will start by copying the value of ##\bar x## into ##A##. Hence why ##A = 1## at that clock edge.

Then looking at the NOR gate before the second D flip-flop, the inputs should be ##A = 1## and ##x = 0##. This results in ##0##. Should this not imply that ##B = 0## and not ##B = 1## as they have provided?

EDIT: I also forgot to ask, does synchronous reset have any impact on a rising edge triggered D flip-flop for a falling clock edge? I have ignored the reset in the early portions of the waveform because it was a falling clock edge.

Last edited:
Wait wait, I think I see how I was reading the diagram incorrectly before. Would this be the answer:

These are positive edge-triggered?

To me it looks like B should be low at all times, since either x or A is high on CLK. C should then be high after first CLK until reset, since x AND B is never true.

milesyoung said:
These are positive edge-triggered?

To me it looks like B should be low at all times, since either x or A is high on CLK. C should then be high after first CLK until reset, since x AND B is never true.

Yes, these are rising edge triggered (probably master-slave d-latch) D flip-flops. So when the CLK is high, the master is transparent and the slave is latched.

I also thought ##B## should be low at all times at first, but they provide part of the solution, and ##B## goes high at one point. This changed the way I interpreted the diagram.

If ##B## really does go high at the point they have indicated, then the only way that could happen is if they are sampling the values of the input variables ##x## and ##A## just before the rising clock edge.

So for the point where ##B## goes high in the solution they provided, I read it like so:

##x = 0## before the rising edge.
##A = 0## before the rising edge.

##0## NOR ##0## is ##1##, therefore ##B = 1## just after the rising edge.

I think this is because of the operation of the actual D flip-flop itself. They way it sends the values through just after the clock goes high, but it samples them a very small instant just before the clock goes high.

Therefore, I believe the image in post #2 is correct.

Last edited:
I agree with the wave form for B in post #2.

The data on D is sampled on the rising edge of the clock only. The data must be valid for a short time (Tsetup) before the rising edge and held for a short time after the rising edge (Thold). Apart from that changes to D at any other time do not effect Q.

So looking at the second latch...

Just prior to the first rising edge of clk... X = 0 and A = 0. So the data input to the latch will be 1. That means after the rising edge B = 1.
Just prior to the next rising edge of clk .. X = 0 but A = 1. So the data input to the latch will be 0. That means after the rising edge B = 0.
etc

PS Normally the output of a latch will take time to change after the clock edge. This delay plus the NOR gate delay helps ensure that Thold is met for the next latch in the chain.

STEMucator
Zondrina said:
I think this is because of the operation of the actual D flip-flop itself. They way it sends the values through just after the clock goes high, but it samples them a very small instant just before the clock goes high.

Therefore, I believe the image in post #2 is correct.
I took the view that the flip-flops were transparent on positive-edge CLK, but that doesn't make any sense, since they then wouldn't function as delay elements. Apologies.

An example would be the 7474.

## 1. What is a timing diagram for a D flip-flop?

A timing diagram for a D flip-flop is a graphical representation of the inputs, outputs, and internal signals of the flip-flop over time. It shows the timing relationships between the clock signal, input D, and output Q.

## 2. How do I read a timing diagram for a D flip-flop?

To read a timing diagram for a D flip-flop, you need to understand the timing relationships between the clock signal, input D, and output Q. The horizontal axis represents time, while the vertical axis represents the signal values. The rising edge of the clock signal indicates when the flip-flop will capture the input D and update the output Q.

## 3. What is the purpose of a timing diagram for a D flip-flop?

The purpose of a timing diagram for a D flip-flop is to illustrate the timing relationships between the clock signal, input D, and output Q. It allows you to visualize and analyze the behavior of the flip-flop over time, which is crucial for designing and troubleshooting digital circuits.

## 4. How can I use a timing diagram to troubleshoot a D flip-flop circuit?

A timing diagram can be used to troubleshoot a D flip-flop circuit by comparing the expected timing relationships with the actual behavior of the circuit. If the timing diagram shows unexpected results, you can use it to identify potential issues and make adjustments to the circuit design.

## 5. Are there any limitations to using a timing diagram for a D flip-flop?

One limitation of using a timing diagram for a D flip-flop is that it only shows the behavior of the flip-flop on a high-level. It does not provide information about the internal workings of the flip-flop, such as the propagation delays and setup/hold times. Additionally, timing diagrams can become complex and difficult to interpret for circuits with multiple flip-flops and inputs.

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