0
\$\begingroup\$

The literature states that a SR flip flop is a sequential device and that sequential devices are those whose output depends on its current inputs and prior state. However, that doesn't make sense to me: flip flops are the basic building blocks of some memories and I wonder how useful a memory can be if its next state depends on its prior state. That would be a very bizarre kind of storage! You have to have the ability to write anything you need to and, in fact, that is exactly what a flip flop allows you to do. So, my question is: am I missing something (conceptually speaking) or in fact a flip flop does not comply with the definition of sequential logic? Thank you very much in advance!

\$\endgroup\$
  • \$\begingroup\$ Virtually all functional devices are initialized to a known state on power up (or stay electrically alive to maintain a known state. For memory, being able to set or reset a specific bit is what it is all about. What are you not understanding? \$\endgroup\$ – R Drast Mar 2 '17 at 13:20
  • \$\begingroup\$ i actually understood your question.it is an interesting one. i will take some time when i get home to research and answer. i also recommend adding more relevant tags to the question. \$\endgroup\$ – yoyo_fun Mar 2 '17 at 13:28
  • 1
    \$\begingroup\$ First, sorry if my question sounded a bit confusing, English is not my first language!! The way a sequential device is defined ("unlike combinational devices, sequential devices are those whose current output depends on the current inputs and prior state") doesn't seem to properly describe a flip flop because its output is uniquely defined by its current inputs. So, I ask why on Earth flip flops are considered sequential once they doen't meet the definition. \$\endgroup\$ – Humberto Fioravante Ferro Mar 2 '17 at 13:29
  • 2
    \$\begingroup\$ "... its output is uniquely defined by its current inputs." No! See my answer below. \$\endgroup\$ – Dave Tweed Mar 2 '17 at 13:35
4
\$\begingroup\$

The output of a FF depends on not only the present values of its inputs, but also on the previous values of those inputs. This is the definition of a sequential circuit.

In other words, there are combinations of input values for which the output can be in either of two states — which one is determined by the past history of the input values. Internal feedback gives the circuit "memory".

\$\endgroup\$
  • \$\begingroup\$ Hum, I think I've got what I missed so far: indeed you can write anything you want to the Q output by properly setting the S or R inputs, regardless the prior state. HOWEVER, once you did that, the FF will store the value you wrote even if you unassert the inputs; i.e., it will "remember" what was wrote and that's why it is sequential (and not combinational). Is that the idea? \$\endgroup\$ – Humberto Fioravante Ferro Mar 2 '17 at 13:47
  • \$\begingroup\$ Yes, exactly right. \$\endgroup\$ – Dave Tweed Mar 2 '17 at 13:48
5
\$\begingroup\$

What is the current output state of this FlipFlop?

   _______
0-|S     Q|-?
0-|R      |
  |_______|

It depends on the previous state!

\$\endgroup\$
  • \$\begingroup\$ Christian, that case woundn't be the one in which the output is “undefined” since the output it is inconsistent (both Set and Reset inputs asserted)? \$\endgroup\$ – Humberto Fioravante Ferro Mar 2 '17 at 13:39
  • \$\begingroup\$ No, he's showing them as both negated. \$\endgroup\$ – Dave Tweed Mar 2 '17 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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