I’m trying to understand the way PRESET and CLEAR work on a positive edge triggered D flip flop, but I may be missing something that I hope someone can clarify please.

Figure1 below shows the flip flop in question. I am using red for high and blue for low. The positive edge detection device is an AND gate with a NOT gate. The output from the edge detector in this diagram is low so the flip flop cannot have its state changed by a change in D.

Figure2 below is a brief moment when the clock edge is rising. The output from the edge detector is high so Q changes to match D.

And Figure3 below is another moment when the clock edge is rising, so again Q samples D and changes to match.

As long as PRE and CLR are both high, the flip flop behaves exactly as I would expect. A three input NAND gates only outputs a 0 when all three of its inputs are high.

But here’s my query. In Figure4 below, the active low CLR input goes low, while there is a rising edge, so the flip flop is enabled. The inverse of Q is now high but Q is not set to 0 as I would expect. There is still a low input to the top right NAND gate, so Q is still high.

In Figure5 below the active low PRE input has been set to low. This is happening on the rising edge again, while D is low. It will make Q high as it should, but Q’s inverse is also high.

For the majority of the time there is no rising edge, and the PRE and CLEAR behave correctly, as in Figure6 and Figures7. I am concerned because a lot of literature (websites etc.) say that the PRE and CLR inputs are asynchronous and completely independent of the input at D, and of the clock. Can someone please clarify this for me?

enter image description here

  • \$\begingroup\$ They are asynchronous clear and preset. This should be obvious since they interact with the output through a nand gate which is not gated by the clock. \$\endgroup\$
    – jbord39
    Commented Aug 9, 2016 at 22:37

1 Answer 1


They are asynchronous PRESET and CLEAR (active low). In this setup, there is the constraint that PRE and CLR cannot both be active (low) at the same time (or else both Q and Q' are 1). You could give one priority if you wanted by modifying the topology a bit.

Additionally, if either PRE or CLR are active (low) at the rising clock edge, the output states will not necessarily be inverted (as you pointed out). But, because the edge detector pulse is narrow, the PRE or CLR will quickly propagate through Q and Q' after the edge detector pulse ends assuming they are held through the length of the pulse.

In essence, whichever 'holds' longer will win: either the data will pass through if the edge detector pulse stays active longer than the PRE/CLR signals stay active (low), or the PRE/CLR signals will stay active longer than the edge detector pulse and over-write whatever D put in there.

In practice, these constraints would be represented the library characterization files. There would be a setup and hold arc defining the timing between the clock, d, PRE, and CLR to prevent any unwanted states.

Or, if the circuit was used in a more custom way, its designers would need to make sure they understood the operation of the pulse latch (not really a flip flop, imo) and how to properly enable or reset it.

You can tell the PRE and CLR are asynchronous easily by looking at the signal flow. PRESET and CLR pass to the output without any gating by the CLK signal (which would make them synchronous).

To prove this to yourself, assume the clock is not toggling, and both PRESET and CLEAR are '1' (inactive). Also, this circuit cannot have both PRE and CLR low at the same time.

Also assume the initial states for Q and Q':

PRE = 1, CLEAR = 1
Q = 1, Q' = 0

As long as you don't touch anything, everything will stay as it is (latched).

Now, pull CLR down to '0' without toggling the clock or data.


As shown in the image above, this clears Q from '1' to '0'. And, clock has not toggled. This means the circuit is asynchronous. From here the CLR signal can be inactivated (returned high) and the circuit will still hold its state.

An easy well to tell is the gating of the clock relative to the clear/enable/reset signals.

  • \$\begingroup\$ Thanks a million for the clear response. The setup arc you mentioned clarifies this for me - I suppose that either PRE or CLR would be normally be held low for longer than duration of the edge. You said that in your opinion my diagram showed a latch rather than a FF, and I wonder why? I thought that, by definition, a FF was an edge triggered device and that my diagram shows such a device? This begs another question; Why do some D FFs use the master slave configuration for edge detection, rather than the combination of AND and NOT gate in my diagram (which is much simpler)? \$\endgroup\$
    – Drummy
    Commented Aug 10, 2016 at 9:36
  • \$\begingroup\$ I would call it a pulse latch because rather than a true master-slave topology, it is a latch combined with pulse generator. So, the memory element is level sensitive for the duration of the pulse. Unlike a regular master-slave which (if there is no clock skew) is truly edge triggered. They are used in practice a good bit, but have drawbacks. There is built in "time-borrowing" through the duration of the pulse. This could be good if you need a few more % of the cycle (the setup time is lesser and lesser the longer the pulse is held, but this delay shows up in the next cycle). \$\endgroup\$
    – jbord39
    Commented Aug 10, 2016 at 14:47
  • \$\begingroup\$ The most common reason they are not used is because it is not guaranteed with regular CMOS to actually work (they are used commonly in custom blocks, at certain companies). You must margin it for all corners (slow nmos/fast pmos, etc) so that the pulse from the edge detector is wide enough to actually fully capture the input data D. This can be hard to do, especially for low voltage, without resorting to prohibitively large delay chains in the edge detector (1 inverter is usually not enough, would need 3 probably). \$\endgroup\$
    – jbord39
    Commented Aug 10, 2016 at 14:49
  • \$\begingroup\$ @Drummy: Not sure you know about the "Accept Answer" feature. Just in case you thought this was what you were looking for :P \$\endgroup\$
    – jbord39
    Commented Aug 11, 2016 at 1:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.