Lecture 32: Edge-Triggered Flip-Flop

Hello everybody, in today’s class we shall
discuss Edge-Triggered Flip F lop. So, in the last classwe introduced ourselves to sequential
logic circuitthe fundamental primary elements of it we discussed SR latch and SR clockedah
clocked SR latch. And we found thathow alevel triggeredSR flip
flop is made ok. Now,why we require aah S triggered flip flop
ok. We require it because the when it is level triggered during when it is theclock level
the clock output is high so the output can change. And if there is any feedback okfrom
output input or some inputis changingI mean there is some transients and other things
that is coming into the picture from different other part of the circuit. So, that might
cause more than 1 change in the state during the period the clock remains enabled is it
clear. So, instead of 1 state change we can have
more than 1 state change taking place. And that willthat is not desired and that might
put the differentyou know calculation out of gearin the design process or in the implementation
process. So,we need to ensure that only 1 state change takes place ok and that place
synchronous with the clock. Andfor that we are looking for something called edge triggering.
So, edge occurs only you know very short period I mean when the clock is changing. So, either
positive edge triggered or negative edge triggered if that is the way it works then we can ensure
that only 1 state change will takes place per clock cycle is not it. So,to do thatpeople
have thought aboutvarious means. So, we shall discuss them ah. So, 1 thing that immediately
comes to mind is to make theclockpulse width during which it remains enabled very narrow
very narrow ok. So, narrow that it is of the order of the propagation delay and by the
which the feedback or some changes that is coming fromother part of the circuit because
ofclocking in that particular cycle. So, by that time it it is it comes it has
become disabled ok see if the clock the pulse high has become low if it is a positive high
level triggeredflip flop ok. This is the idea that affect of thisclock clocking rightah
that will be not getting reflected back in this particular cycle itself that is the idea.
So, for that people have thought about the clock is the normal clock here and a pulse
forming a circuit ok. And this is giving a positiveyou knowah thisedgetriggering ok.
So, this pulse forming a circuit so 1 way of you know getting it something like this
we had seen in the discussion related to hazard before if you remember ok. So, this is effectively
a not gate. So, the clock is goinghigh here from say low to high at that time. So, this
a not gate output after the propagation delay it will go from high to low right. And when
you and these two then for a short period after this propagation delay right you can
see a small positive going pulse. So, this clock will be available only for
this circuit ok. So,because directly getting a clock with a very narrow pulse it is difficult
I mean getting very stable clock. So, a pulse forming circuit like this ensure a small a
narrow pulse width ok. Similarly if you want anegative goingyou know negative going edge
a negative a pulse a small pulse width narrow pulse width ok for negativetriggering ok.
So, if it is low level triggered basic circuit basicflip flop ok.
Then at that time you can have not andor combination ok. And you can see that we can get a narrow
pulse width which is negative going ok. So, for this particular circuit though the basic
part of the circuit islevel triggered, but effectively you are getting a edge triggering
ok this is not actual edge triggering, but effectively it is edge triggering ok. So,
this is thethis we canconsider andgo ahead this ensures 1this thing.
So, again what we shall see what we shall try to do is to againmake a compact representation
and by which for edgetriggeredcircuitis edge triggered SR flip flop ok we shall have a
representation like this ok where this you can see this triangle like shape here thatindicates
that it is a edge triggered ok. So, absence ofanybubble here inverter means it is positive
edge triggered and presence of a bubble like this that is a inverter ok. So that means,
it is a negative edge triggered ok is it clear and for the truth table ah.
We can indicate thisby putting athisarrow mark going upward that it is positiveedge
triggered and when this edge comes. So, 0 0 thenow we are indicating that only 1 state
change is taking place per clock cycle it is we are ensuring. So,then the previous value
is written ok 0 1 it is 0 1 0 it is 1 and 1 1 as was discussed before it is forbidden
ok andso the correspondingtiming diagram ok. So, the way we had seen before now we cansee
that wheneverin the earlier case if there were when it is high more changes in S and
R occurs I mean then it would have been reflected. But here it will not be reflected only in
this edges this positive going edges ok in S triggered flip flop thecircuit isstate is
allowed to change ok. So, at that time it will see what is the value of S and R. So,
here it was allowed to change Q as 0. So, that time S and R are 0 so no change ok. So,
here again it is at T 1 it is allowed to change at that time it finds that S is equal to 1
and R is equal to 0 ok. So, then the output becomes 1. So, after that here the clock is
high this has become 0 this has become 1 you can see that thing if I draw the diagram.
But no change in the output has taken place ok had it been level triggered immediately
there would been a change. So, that is not what is happening and it will
happen again I mean it will depend only again in thethis edge whenever it comes anddepending
on the value of S and R at that time so S is equal to 0 and R is equal to 1 ok. So,
it gets reset it becomes value become a 0. So, this is the way it progresses this is
a difference between edge triggered and level triggeredflip flop and their timing diagram
ok . So, now we look atother varieties offlip flopthat
isused in thesequential logic design ok so one is called D flip flop. So, what ishow
D flip flop isdesigned how does it look like. So,the D flip flop the we know this is basic
the SR flip floppart of it ok So, this was S input and this was R input and this is the
clock that is already known to us. So, you put a inverter between S and R by which thisthere
becomes only 1 in 1 input to it because S and R are connected in this manner ok.
So, this is a single input of course, the clock trigger is thereI mean other than thatclock
and the Q is the output and Q bar of course, a complementary output is there. So, this
is the way the flip flop is arranged ok. So, for D flip flopah whenever the clock whenever
the clock is you know 0 no or 1 I mean the no edge has come ok this is normal say 0 or
1 ok. So, irrespective of the previous value will be written. So, whenever it is 0 ok so
S is 0 inverted here. So, R is 1 and edge comes what will happen it will get reset.
So, if there is a 0, 0 will be there in the output and instead of these if this was 1
then this is S is 1 and this will be 0 because of the inverter right. So, the output will
be when the h comes output will be 1. So, whatever is the input that gets transferred
to the out and remains stored there remains stored there ok. Andthis is how the D flip
flop works and more of it it is application we shall see later. Now you will see that
all these circuit when we employ inum complex sequential logicah you know implementation.
So, we needin certain casesclearing the circuit or presetting the circuit clearing the state
orpresetting the state okat a given time 1 importantrequirement is during initialisation
ok. So, you have connected many flip flops and you have made a complex circuit out of
it. So, how would you start where from which state do you start usually you may want to
start it from say 0 0 0 or say 1 1 1 whatever it is or it could be from 0 1 0 something
which you know is predefined ok. So, in that casewe need something by which it can be initialised
in that manner ok. Orah um some state can beintroduced without the application of the
clock. So, that is called asynchronous presetting
or synchronous clearing so for that what we can do. So, this is the clock part of the
circuit right. So, we can just put aanother such or gate right right and whenever we put
preset is equal to 1 and a clear is equal to 0. So, this is 1 and this is 0 ok which
will make irrespective of the clock. Now clocking is not required thats why it is asynchronous
whenever it is made like this then this is yourthis will be set to 1 and this will be
0 and only when both of them are 0 this or gate will be acting on this input when both
of them are 0 right this is a nonforcing input for the or gate.
So, whatever is coming from here this and gate and this and gate through clocking and
all through whatever is present in the D flip flop ok so that will be deciding. What is
the output? So, similarly for clearing will make it over 1 this is 1 and this is 0. So,
that will make it0 ok and this will become 1 right this is clear. So, this is shown for
a D flip flop and then in the case of symbol again to make a compact representation. So,
in we putting here pr andclear that is preset and clear.
So, if it was a this because it is active high preset active high clear. So, if active
low then we would have put an put a bubble here and the put a bubble here. So, instead
of you know this high if there was a not gate put before this ok righ. Then then we would
have put a bubble over there is it clear. So, this has been shown for D flip flop it
could be for SR flip flop also it could be possible for any flip flop this kind of you
know asynchronous preset clear can bedesigned .
Now, we come to anothertype of flip flop which is called JK flip flop. So, JK SRah was set
reset there were two inputs. So, JK flip flop you can see thattwo inputs are there. So,
this JK flip flopif you see how the connection is made. So, this is thenormal SR flip flop
these SR latch and the SR flip flop with clock so this third input was not there ok. So,
without this this is a standard SR flip flop please understand this part.
So, this feedback is eliminate this a feedback then this is your S and this is your R isnt
it is a standard SR flip flop. So, what what has been done to get the JK flip flop is Q
is connected to this R and we are giving a different name of the input K to distinguish
to differentiate it from SR flip flop. And Q bar is coming to where the S was there as
a third input to the and gate ok. And we are giving the name J is it fine. So, this is
how the JK flip flop is made right. So, how the truth table now becomes different
from the truth table we had for SR flip flop ok. So, if we investigate we see that whenever
J and K are 0 and 0 this is and gate 0 is the forcing input. So, this input will be
0 and 0 irrespective of what is feedback or whatever you know the other things ok I mean
whenever the clock appropriate clocking is there alright. So, what does it mean? The
last value will be there clear first part right .
Then if this is 0 and this is 1 again 0 is a forcing input . So, this will be 0 right
and this input this input if Q was 0 and this is 1 please understand this part; if Q was
0, 0 will be feedback here. So, this will be 0. So, 0 0 means previous value will be
written. So, if 0 was there it will it will at 0 1 will remain at 1 ok. Instead if previous
value was 1 ok andthen this was 0. So, this 0 comes over here this 1 comes over here ok
and clock is enabling it so this will become 1.
So, 0 and 1 what does it mean the output will become 0 it is reset. So, in either case irrespective
of the pulse value was 0 or 1 the output becomes 0. So, 0 1 output becomes 0 is it clear . So,
from symmetry you can say for 1 0 the output will become 1. And then comes the last case
for which actually this feedback is given and SR flip flop is different from JK flip
flop otherwise thee remain the other threerows of this truth table is same.
So, when it is 1 and 1 and say this is 0 and this is 1 the previous value and you have
made it to 1 1 and the clock is is enabled now. So, this 0 comes over here ok, this clockthis
1this 1 goes over there and we are just in a you know activated the clock at that time.
So, what what has happened to the output? So, 1 1 1 this output is 1 and get and this
is 0 so this is 0. So, 1 and 0 S and R the output will become
1 ok and this output will become 0 is it clear so if this was 0 it becomes 1. Again from
symmetry you can figure out because it is identical right. So, if the previous outputwas
1 and this is 0 and it would have been 0 and this would have been 1. And so basically it
toggles at that time is it clear. So, this is no more forbidden which wasthe case for
SR flip flop the 1 1 input. Now if 1 1 is given the previous state is
just inverted ok . So, this is what is the JK flip flop truth table and the symbol will
be like this. And if you are having a what is that called active low preset and clear
so this is the way the symbol will be presented. So, this isbasically mayreferring to negative
edge triggered JK flip flop ok . Now, we look at something called master slave
flip flop ok. So, when you say master and slave so there will be something which is
drivingthe circuit and slave means the one which is just following it just carrying it
to forward ok. So,the JK flip JK flip flopexample we can take. in that in this case you see
thatthis is one SR flip flop so the slave is always SR flip flop ok. And the master
is depending on the flip flop type here we are talking about JK flip flop right.
So, this is theah JK flip flop with the feedback and all and this feedback is coming over here
which will be carried to the next level ok. So, there is an intermediate output at this
two points . And what is notable here is that that there is aclock which is connected I
mean what about the clock goes here the inverted theclock if it is active forthis is logic
high ok so when it goes to logic low right, so when it goes to logic low .
So, this will go from logic low to logic high. So, when this is active at 1 this is a level
triggered each 1 of them is level trigger ok you can see that there is no pulseyou know
forming circuit and other things are there. So, when this is active this is inactive and
when this is inactive this is active. So, how does it actually it thenhelp. So, when
it is active so because of the input whatever the input is present the output comes up to
the master output; it cannot go to the final output because this part of the slave is now
inactive ok. Now slave when this becomes inactive 0 now
slave becomes active right. So, the output of this master whatever was there this inactive
means the previous state will be written this these are all 0 is means 1 1. So, previous
state will be written so master output is written. So, that master output now goes to
the final output ok. So, now, even if the feedback is there right and it is asking for
to toggle and all. But the clock I meanif the both the inputs are 1.
So, the value will be coming and resting I mean value will be coming here, but it will
not be able to change the output. Because the master is now disabled because clock is
low and when clock becomes high slave becomes disabled. So, because of which this is a toggles,
but the output will come over here ok, but it will not be going to the next stage slave
stage. So, only 1 state change isyou know guaranteed ok within this. But if something
is coming from different other places when this clock is high it need to be stable ok
so that is another part of the story. And effectively then we are getting that whenever
it goes it was high master has changed whenever it has gone low this slave is changing ok.
And after that there is effectively no change can occur slave can change here, but no way
the output can alternate at time because the master has gone low. So, effectively you are
getting a negative edge triggering over here is not it. So, that is another way of gettingthe
edge triggering in a roundabout way you can say ok, where; individually they are level
triggered and we do not have restriction on narrow pulse width andthings like that ok.
But there areconditions like the input need to be a bit stableinput need to be stable
and all ok. And instead of JK flip flop you want SR master slave or D master slave then
we need to havethis initial stage to be D or SR and the slave will always be SR flip
flop because there are two inputs coming from Q and Q bar of the master ok. And this is
one waypeople have I mean this is thecompact way people havemake made a presentation ofrepresentation
of master slave flip flop. So, the clock is level triggered, but the output is changing
at the following edge ok. So, this is your flip flop indicator and with there is a preset
and clear available ok. thenasynchronous so it will be indicated in this manner ok .
Now, we shall look atedge triggering by input lockout this is something which is perhapsyou
would beah finding very close to whatan edge triggering circuitshould look like. So, no
down therefore, to a no narrow pulse width and the pulse forming circuit or master slave
andbasically stable input at the masters. So, all those things apart can beyou knowconditions
can be eliminated and let us see how it works and we take a D flip flopas an example ok.
So, the basic circuit you can see that there is 1 kind of you know latch here this is another
latch and this is another latch ok. So, 1 2 3 4 and 5 6 right and one point to be noted
thathere there is a this is a this 3 is 3 input rest all are 2 input ok . Now we see
how this circuit works how this circuit works ok we shall go through examples ok . So, you
consider that D is initially 0 ok D is initially 0 and the output could be 0 or 1 that part
we shall we it can be any 1 of them ok. These output is only linked here ok it is not feed
back to any other stage right. So, whenever GD is 0 ok this output is because
it is a forcing input for the nand gate so this output is 1. And we are talking about
the clock remaining this is a positive edge triggering we are talking about. So, clock
is at 0 means it is a it is inactive if the output is not changing the value. So, let
us see how it is happening. So, clock is 0 means this is 1 and this is also 1 right.
So, this 1 is feedback here this 1 is there coming from this feedback and clock is also
1 right. So, this 1 is again going back to this place this 1 is over here so 1 1 is making
it 0 right. So, 1 1 the previous value will be written
if it is 0 it will be 0 it will be it 1 it will remain 1 and vice versa I mean the Q
n will Q n will be just the opposite of it is it clear . Now what happens changes from
0 to 1 ok? So, what are the gates that will get affected first? The gates that are connected
directly to the clock ok, so clock becomingso sorry this is 0 and this is 0 and this 0 is
feedback here ok. So, clock becoming 1 means this is becoming 1 ok so this 3 ones this
3 ones make it make it 0. Now clock has become 1 here, but this particular
gate this 0 is over there. So, this 0 is holding it to 1 ok is there any other change? No,
because whenthis 0 is it has become here. So, 0 feedback here this 0 is holding back
it to 1 there is no such change. So, this 1 and this 1 together it is 0. So, what we
see that 1 and 0 will be occurring ok. So, that will make this 0 go to 1 because for
the nand gate this is forcing input and this 1 1 feedback it is 0. So, whatever be the
previous value 0 or 1 this will be 0 and this is 1 ok. So, if D I is D is 0 Q becomes 0
that is what the flip flop how it should look like.
Now whether it isedge triggered or not let us see . Now the clock is at 1 right we have
seen that when clock becomes high it is changing like this clock has become 1 right. So, what
is the what are the outputs at that time this was 0 D was 0 ok. So, D was 0 means this isthis
was 0 andthis is 1 clock is at 1 ok this is 0 means this was 1 ok this was 0 right andwhat
else? So, this is 1 andthis 0 is coming over here right I am talking about this stage I
am not have not talked about you know changes ok. And this 1 is coming over here. So, this
0 is here clock is 1 ok. So, now, all the inputs have been understood
and we are over here at this place right before changing the D. So, clock is 1 now we have
changed from 0 to 1. What is what happens when clock changes from 0 to 1? So, we have
to see where it gets first affect where it first affects. So, it first affects the this
particularnand gate because D is connected only to that right. So, 0 to 1 when it becomes
so this becomes 1. So, what is the corresponding effect? So, these gate already the other input
of it is 0 ok. So, 0 and 1 it will remain at 1 only isn’t
it. So, and this 1 is fed back here and it remains 1 only so it remains 0. So, there
is no nothing no other gate is getting affected. So, at this change you see the 1 that are
important so all of them are remaining at the previous value. So, even if input has
changed the input is locked out ok . So, this is what is happening even if the clock is
high change over here is not making any difference right. Only 1 clock from goes from 0 to 1
there is a differencethere is a change occurring .
And you can see the same thing occurs if the situation wasfor D is equal to 1 and clock
was 0 ok. You canyou knowsee thatthe same way you canlook atif youthe changes that are
taking place ok. Thatthis output will become 1 and this output will become 0 ok atah when
the clock goes from 0 to 1 and when it is clock is at 1 and ifinput changes from 1 to
0 ok. This output will change from 0 to 1 this output will change from 0 to 1.
So, this 1 gets feedback here, but the other input of it is 0 so 0 will hold back it to
1 and no further change that gets transferred to the final stage ok. So, similar analysis
like what we had done before so if you do we shall see that andthe immediate outputs
are shown. That 0 to 1 there is a change the output becomes 1 if this is 1 and when clock
is at high then at 1 when it is1 to 0 ok the out the changes are taking place only here.
But after that no further changesthere. So, basically though there is a change in
input the input is getting locked out ok. So, this is so what we note andthis is way
you can get edge triggering forD flip flop by locking out the input. So, similar thing
we can do for SR JKother type of flip flop ok .
And finally,for a D typeedge triggered flip flop. Ok we can put asynchronous preset and
clear andah this is a standardcircuitthatI show you IC 744 ok 7474 ok. um it is a dual;
that means, two such say positive edge triggered flip flop is there with asynchronous active
low preset and clear. So, this preset and clear are there you can see this preset input
is going directly over here ok. So, if it is 0 and this is 1 so this will
become 0 and this will become 1 right. Andthen this output we become 1 and this output will
become 0 right ah. So, that is active low so when it is low it is becoming high right.
So, if you look at theah cases when the the clock is 1. So, basically the clock will be
1 at that timesorry clock is inactive clock. So, this is 0 ok andthis is 1 and so this
many things you can see is it fine ok. So, with thisah we conclude today’sdiscussion.
What we have seen? In a level triggered flip flop there can be more than 1 or unintended
state change in 1 clock cycle if clock is passed through a pulse forming circuit that
generates a very narrow pulse width effectively. An edge trigger circuit can be obtained master
slave arrangement uses two level triggered flip flopwhich work in two different phases
of the clock. And edge triggering using input lockout does
not require separate pulse forming circuit or master slave arrangement. And D and JK
flip flops we have seen; how they worktheir truth table. And we also noted that asynchronous
preset clear inputs are useful in setting resetting a flip flop without clock trigger
ok . Thank you.

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