Standard TTL Circuits - Electrical Engineering Video Lecture - Electrical Engineering (EE)

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so today we shall continue our
discussion on this 7400 TTL gate and
draw the circuit once again this is the
earliest form of the TTL gate okay and
it is important to understand the
circuit how it works and then once we
understand this we can look at the
modifications you can very easily
understand the modifications in the
later stages in the more advanced
versions of this gate so this TTL always
uses five volt power supply these are
the resistance values okay this is the
TTL NAND gate seven four zero zero okay
so what we shall do first in in today's
class is we shall obtain the input
output characteristics of such a gate
okay this is the output point okay and
we shall take the input at one of these
we shall take one of these inputs as the
input to the circuit all right now
before actually going into the input
output characteristics the better way to
do this is to consider this input here
at the base of t2 alright of the
transistor t2 and obtain the
characteristics of the output voltage
versus
the voltage at the input of our at the
base of t2 that is vo v b2
characteristics once we do that then we
can very easily get the input output
characteristics by obtaining the drop
across the if you know what is the drop
across the t1 transistor all right so
now let us see what is the
characteristics like okay so alright now
let us see when v b2 is say zero volts
okay
all right then what happens to t2 t2 is
off obviously because base voltage is
small and if t2 is cut off t3 will be
also cut off okay because there is no
emitter current for this and if there is
no emitter current t3 cannot have any
base current so t2 and t3 is cut off
okay and what about t4 because there is
a base current 44t fall will be on okay
and so what is going to be the output
voltage now the output voltage would be
the drop across output voltage will be
so when right when v b2 is equal to 0 V
output will be equal to VCC minus I
should call this ib 4 okay this current
which is the base current of t4
ib 4 into 1 point 4 kilo minus vbe 4
minus one diode drop okay all right now
so all we know all this VCC we know 5
volts these we can assume to be around
0.7 volts okay now this voltage drop if
you assume say that the load is depends
on the load current actually how much
current you are drawing okay
is the emitter current of the transistor
t4 so if you assume that this load
current is 1 milliampere okay and if the
beta is 50 of this transistor then the
base current is around 20 micro amperes
all right and 20 micro ampere into 1
point 4 kilo ohm would be a point naught
2 into 1 point 4 that is around it is
less than 0.1 volts okay all right so we
can almost
neglect this term okay all right so it
is almost equal to if you assume 0.7
volts each okay this is almost equal to
3 point 6 4 all right so when BB 2 is
equal to zero output voltage would be
around 3 point 6 volts okay now how long
does this output voltage remain at 3
point 6 volts as long as this situation
prevails okay we can say that this
situation is going to prevail as long as
t2 is cut off all right now if we say
that the t2 the base in input voltage at
which t2 starts conduct the bay I mean
base current starts to flow at t2 is
around say 0.65 volts or something okay
then we can say that when VB to till VB
to is 0.65 volts
okay this output voltage will remain at
3 point 6 volts okay so then we can draw
like this okay so this is point zero
point six five volts
okay alright so the same situation
prevails okay so we can say I just write
it here when vb2 is less than 0.6 five
volts okay C 2 and T 3 is off are off
and T 4 is on okay all right and that is
why you get V output is equal to three
point six all right
now what happens when it exceeds point
six five volts say when this input
exceeds 0.65 volts
bb2 this transistor t2 is going to start
conducting all right and when t2 starts
conducting this part of the circuit okay
which I call it the phase splitter part
okay this behaves as an amplifier okay
with a gain of minus RC by re you know
that okay an amplifier okay so that is
one point four by one that is one point
four so the the voltage at this
collector point okay is going to fall
okay it is going to have a gain of minus
one point four with respect to the base
voltage okay so as the base voltage
keeps on increasing this voltage Falls
by one point four times that okay and
since this point here is basically the
emitter follower point of this
transistor okay this is also going to
follow alright that is the output
voltage here is also going to have a
gain of minus one point four is respect
to V be two because this voltage is you
know point seven point seven they remain
almost constant they do not change okay
so so as we go on increasing
the bass MVB to this output voltage is
going to fall with a gate with the slope
of minus one point four okay and how
long is that going to happen
till the point okay till d3 starts
conducting okay all right now what is
this voltage range in which t2 is on but
t3 is off okay all right now as you go
on increasing this voltage what is going
to happen is this voltage is falling but
this voltage is increasing okay that is
the base voltage of t3 or the emitter
voltage of t2 that voltage is increasing
because that is the emitter follower
point okay so so as we if you say that
point six five volt is the cutting
voltage so when this voltage becomes
0.65 that is the at the base of t3 this
t3 is going to turn on okay and then
when if you say that at that point t to
the base voltage of t2 is 0.7 volts okay
because it is conducting okay so this is
0.7 volts VB - is point vbe 2 is 0.7
volts then we can say that when VB 2 is
1 point 3 5 volts because this is 0.7
and at the base of t3 is 0.65 so when I
can write here when vb 2 lies between
0.65 and 1 point 3 5 volts okay
t2 is t2 + x-l 64 on t3 okay
all right so at this range what is the
output voltage the output voltage Falls
with a gain of minus one point four okay
so this one in this range you know when
db2 becomes one point three five volts
here okay this is going to fall okay
so this is a 0.7 volts difference okay
if the slope is minus one point four
okay this would be around one volt okay
so this is 2 point 6 volts okay alright
okay
now what happens this has become 0.65
volts and now if you go on increasing VB
to even further
alright this output voltage will fall
with a the slope is going to increase
even further why because if you look at
this amplifier now okay this transistor
okay input of this transistor is coming
in parallel with this one kilo ohm
alright and so what happens is
effectively the emitter resistance is
going to be reduced
alright and so the slope is going to be
increased alright okay so the the output
voltage is going to fall more sharply
okay
and so both t2 and t3 Arkin is
conducting so what is going to happen is
ultimately okay because t3 gets is base
current from the emitter current of t2
alright mostly a part of it is of course
bypass but most of it flows in as base
current of t3 t3 is going to go to
saturation okay before t2
all right because if the base current is
much larger in the case of t3 because it
is derived from the emitter current of
t2 which is okay I mean all right you
understand so it's a large base current
so t3 in fact goes to saturation before
t2 okay all right so at around
vb2 is equal to 1 point 4 volts ok 1.45
I should say if you say that it is the
saturation at saturation vbe is 0.75
volts okay okay it's 0.75 and 0.7 so at
around 1.45 volts okay t2 t3 goes to
saturation and when t3 goes to
saturation what is the output voltage
it's 0.2 volts okay so that is a very
sharp fall you can see it actually goes
down like this okay
at one point four five volts okay this
is one point four five volts the output
voltage has reached 0.2 volts okay this
is point two volts alright so okay and
now of course if you go on increasing VB
to even further okay
this at 1.5 volts if you assume that
0.75 volts is the base emitter voltage
at which the transistor goes to
saturation okay
at 1.5 volts t2 also goes to saturation
okay so it's just you know it it keeps
on like this but one you must be you
must have understand the fact that okay
BB - you are not applying VB - this is
not there is no external voltage source
and vb2 cannot exceed something like 1.5
volts okay because it would mean that
the base emitter drop of a forward bias
Junction okay is very high which is not
possible again as we said okay so this
if you plot the V output VB two
characteristics okay you must have to
truncate it here alright because this
voltage is
not cannot go more than that alright
okay is it clear so what happens the V
output VB two characteristics okay
so this is this characteristics now what
we are going to do is we will try to
translate this this VB two axis okay
you basically like to draw the
corresponding via input axis okay and
see what is this value okay what are
these values on the V input axis okay
so what I mean is that when VB to is
0.65 what is VI when VB to is 1.35 what
is the value of VI alright and then we
will have the input output
characteristics okay so basically we
have to understand the state of this
transistor t1 that is the multi emitter
transistor here okay all right okay now
when VI is quite large okay when VI is
quite large I mean where the input
voltage is quite high I should say okay
what is the state of this transistor
this transistor is acting in the active
inverse mode you know that that we have
already discussed and this base current
flows into t2 of course and the this t2
this VB 2 will be around 1.5 volts okay
and this output voltage is going to be
0.2 volts okay so when V input is quite
large okay you have output voltage is
0.2 volts
okay now suppose you are reducing V
input from a very high value okay what
is what we want to know is at what value
of VI okay does this say t2 start coming
out of saturation okay all right because
t2 will come out of saturation before t3
because we have already said that t2
goes into saturation later than t3 when
you are increasing the voltage okay so
when you reduce the voltage this step 2
will come out of saturation first okay
and then t3 so at what value of input
okay does this t2 come out of saturation
that is well somewhere around this point
okay this is one point five is where the
transistor goes to saturation okay so
let us look at that now let us assume
again that when t2 comes which is just
at the edge of saturation okay let VCE
of this transistor is almost around say
0.3 volts okay is very close to point 2
volts 8 it is point 3 volts okay all
right and so and what is this voltage
this transistor is in saturation so here
it is 0.75 volts okay all right so what
we want to know is what is the collector
current of t2 or what is the current
which is flowing here okay so we can
write so if you want to calculate the
collector current of t2 IC of T to add
that can under that condition
okay then IC of t2 we can write is equal
to 5 minus 0.75 minus 0.3 okay point
seven five is the voltage at the base of
t2 and point three is the drop across d2
so that is the drop across the one point
four kilo ohm resistance so this is the
this is the collector current of t2 this
is around you can say 1 volt so it is 4
by 1 point 4 kilo ohm okay so this works
out to be around how much 4 volts by 1.4
kilo ohm okay so it is around
I think 2.8 million bit because 1 by
this is okay
Oh point 7 into 1 point 4 is 1 so it's
okay so it is around 2.8 million bits
okay so what is the base current of t2
which is actually equal to the collector
current of t1 okay so this is going to
be this divided by beta alright so if
you assume a beta of 50 say okay so this
is going to be 2 point 8 milliampere by
250 okay which works out to be 56 micro
amperes
all right so what is happening is at
that point of times a current of 56
micro amperes is flowing into the base
of t2 alright okay and what is the base
current of t1 that is if you assume this
is 1.5 volts okay around 1.5 volts and
this this Junction is also forward
biased so this voltage can at the most
be two point two volts say okay so you
have a point seven million pair of base
current flowing okay so in this
transistor a base current t1 transistor
a base current of 0.7 milli ampere is
flowing and a collector current okay
that is a reverse collector current
basically okay negative collector
current okay normally in a normal
transistor a collector current would
flow into the collector okay in an NPN
transistor so there's very small
negative collector current is flowing
all right off 56 micro amperes very much
smaller compared to the base current
alright and then what happens if you go
on reducing the input voltage even
further what is happening is slowly that
even that 56 micro ampere is going to
reduce and finally it will be a zero
current that current goes to zero all
right
the collector current of t1 reduces and
finally goes to zero and this t2 is cut
off alright so if you look at this
region of the characteristics okay at
this end when that t2 is just at the
edge of saturation if if this very small
collector current a negative collector
current is flowing all right and as you
can reduce the input voltage okay or all
right what is happening is that negative
current goes to zero okay so at what
region of the characteristics is this
transistor operating
okay you have to recall our discussion
on the bipolar transistor
characteristics okay and I had referred
to one particular region of the
characteristics okay instead we are
going to discuss it or we are going to
come back to it when we discuss TTL okay
all right so if you look at the
transistor characteristics okay
you draw usually draw it like this you
know but I said if you magnify this
region all right it is actually there is
it is something like this the
characteristics actually start off
something like this and this region is
called the offset region okay so this is
VC this is output characteristics okay
all right so in this region IC is
negative okay so negative and then it
slowly goes to zero so basically I see
goes to zero at a small positive VC
which is called the offset voltage of
the transistor okay and we have also
seen the reason for that when we
discussed the bipolar transistor okay so
and so basically what is happening is it
is operating in this region so a small
negative current of my 56 micro ampere
which goes to zero so it is almost
confined to a small region here okay
throughout the entire operation of this
region
okay 41 it is confined in that region
okay so approximately we can assume that
the drop across the VCE of t1 okay is
around 0.1 volts okay along for the
entire region here
okay approximately it's so it's a very
small voltage you know around 0.1 volts
say okay now if we see is 0.1 volts 41
in this entire range okay so what is
going to basically we have to translate
or shift this axis by 0.1 volts to get
the input output characteristics okay of
course it is going to change by a small
amount it is not constant but it is
almost a constant okay it's changes by
very small amount because this entire
range is very small okay so it's around
point one volts and it goes like this
okay so if you look at it like that okay
it's just a very con it's a very small
region so what you have here is this
0.65 will become on five five
okay one point three five would become
one point two five one point four five
would become one point three sides so on
okay and of course when you draw the
input output characteristics you do not
have we are not constrained here you can
go to high voltages all right is it
clear so basically if I draw the input
output characteristics okay what I am
going to do is this remains the same I
am just going to make this VI now okay I
just make this 0.55 this is one point
two five this is one point three five
and this can go on because as you
increase the input voltage okay
this transistor this remains in
saturation okay so alright and the
output voltage is 0.2 volts okay so this
is the input output characteristics of
this TTL gate all right okay fine
now well yeah
okay the next thing we shall see is
about some further specifications of
this gate okay now if you look at this
manual of the seven four zero zero C
gate okay and look at the different
specs of this gate okay you will come
across the values okay these terms we
have already defined VI L vol VIH voh
okay you know what they are in terms ok
VI l the values which are given is vol
is 0.4 this is 0.8 V output to point 4
volts and this is VI H is 2 volts so
these are the specifications provided by
the manufacturer all right which means
that V output low
okay the maximum value the output can
have when it is output is low is 0.4
volts okay and V input low that is the
maximum value of the input which will be
considered a logic low is 0.8 volts okay
it is somewhere here all right
okay Oh point 8 volts and but the thing
is they define voh as 2 point 4 volts
not 3 point 6 volts because if you
define voh as 3 point 6 volts okay you
would have to define v IL as 0.55 volts
or even less okay so now so you know
actually this if you look at these
specifications you will find that they
are very conservative in the sense that
one could if you look at the
characteristics no give a better
description you know in the sense that
you could say you could put via voh at a
much higher value and so on and so forth
but why they do it is that
a manufacturer a large number of chips
slam in a batch processing and if they
give very stringent conditions they will
have to reject a lot of their output so
these values are more conservative okay
now so they say that vo h is sorry VI l
is 0.8 volts and vo H is 2 point 4 volts
it means that if the input voltage is
less than 0.8 volts they guarantee
output voltage greater than 2 point 4
volts and if the out the input voltage
is greater than 2 volts the output
voltage will be less than 0.4 volts okay
and we see that that is a is going to be
satisfied by this characteristics okay
because at 0.8 volts okay output voltage
obviously is going to be greater than 2
point 4 and at input voltage greater
than 2 volts output voltage obviously is
going to be greater than point 4 volts
and what does this mean in terms of
noise margin the noise margin
definitions I hope you remember okay so
the noise margin high noise margin is
given by 2 point 4 minus 2 okay
this means that it's noise margin high
this point 4 volts and noise margin low
is also point 4 volt that is another
reason for giving this type of
specifications because there's no point
making 1 noise margin high and the other
very low okay so these are the 1 part of
the specifications you will also find
another set of specifications which is
terms of current okay you will this is a
il iih I oh L IOH okay III L is the
input current when the input is low okay
for this it is 1 point 6 milliampere iih
is 40 micro amperes okay IOL is 16 milli
amperes
and IOH is 0.4 million okay so these are
another set of specifications which is
given now iil is the input current when
the input is low okay it is the maximum
value okay it is the maximum value of
the input current when the input is low
okay so the input current can be at the
maximum one point 6 milliampere okay iih
is 40 microampere is the maximum value
okay of the input current when the input
is high so obviously you see that if you
look at go back at this circuit here
okay when input is low say 0.2 volts
this would be 0.9 volts okay which would
mean actually a current of 1 1
milliampere which the which is the same
as the since this transistor collector
current is almost zero so base current
is almost equal to the emitter current
so a current would flow in this way
equal to slightly greater than 1
milliampere
but of course they give a somewhat
higher current because you know this in
the IC technology the resistances cannot
be fabricated with such definite values
okay there is always a tolerance okay so
if the resistance is less than much less
than 40 ohm
okay the current is going to go up okay
so they say that the maximum input
current okay when the input is low is 1
point 6 milliampere on the other hand
when the input is high alright this
transistor is operating in the inverse
mode okay and the collector current
which this behaves is the collector in
the inverse mode and the collector
current which is flowing is actually
beta reverse times the base current
which is going to be pretty low okay
because as I said the beta reverse of
this transistor is deliberately made
very low so you have a very low input
current of around 40 micro amperes all
right okay what are these IOL and IOH
the other two specifications
IOL refers to the maximum load current
okay that the output can sink
as so that the output voltage is at
logic low all right okay
again I'd repeat the definition I owe l
is the maximum current that the output
can sink okay so that the output voltage
remains at logic low okay the important
thing is it remains at logic low what is
going to happen is if you sink too much
current okay if you look at again I go
back to this circuit here okay
now if it sinks too much current okay
basically you if you connect another
gate here you know okay you have say
this in this NAND gate okay is connected
to another NAND gates here okay you have
more than one NAND gates so this is
logic low say all right now this is the
current I I L okay input low current
which is flowing okay where is it
flowing into this flowing into into this
gate so this current is sinking
okay the output is thinking that current
so if you have more than one fan-out all
of them is flowing in here all right
okay now if you if you have too much
current flowing in you see this
transistor is in saturation okay and the
condition for saturation is that the
collector current should be less than
beta times IB isn't it now if you have
too much collector current flowing in
this transistor you know you can come
out of saturation okay all right and so
if it does that then the output voltage
is going to rise okay so in order to
ensure that this transistor is in
saturation okay you have made this very
high effective load okay but if you have
too much current flowing in basically it
means that the you are reducing the load
the collector resistance effectively
okay which means that the transistor can
come out of saturation all right so
there is a limit on the collector
current which it can sink okay in you
know and the voltage specifications are
retained that is the output voltage
remains less than vol okay
similarly IOH okay if you the definition
of ìoh is that it is the maximum current
that the output can source okay it is
the source
okay current flowing out okay I mean
when the output is a logic high so that
the output voltage is maintained above
voh okay so that it alright that is the
definition of IOH okay that is the
maximum current the output can source
when it is at logic high okay so that
the output voltage is maintained above
voh okay because that is the definition
of output high that is the output
voltage greater than voh okay so the
manufacturer guarantees that if you do
not draw more than 0.4 milli amperes
when the output is high okay the output
voltage will be retained above voh but
if you draw more than that it does not
guarantee that okay that output is going
to be above vh okay similarly for AOL if
you sink less than 16 milli amperes
output is going to be less than vol okay
above if you above that it does not
guarantee so there is no guarantee of
retaining the proper logic okay so this
is the input and output current
specifications and they are very
important from 2n 2 to understand or the
fan-out criteria okay how much how many
gates can you connect at the output of a
particular gate okay
so how much how many so you have to look
at both the logic low and the logic high
okay so you have to see for example IOL
by I il okay so you have to take what
are the maximum minimum sorry off
so the fan-out I shall write it here
separately
we'll be minimum of i/o l biii l and
whichever is minimum is going to be the
maximum is the maximum fan-out okay so
in this case of course both are ten so
there is no problem okay but in many
cases especially this is very important
when you are connecting to different
logic families okay for example you are
having in a circuit you have both CMOS
as well as you are using CMOS NAND gate
as well as seven four zero zero series
for it perhaps okay now if you are
connecting all right you must be very
careful okay whether a CMOS gate which
you are having can drive a TTL gate okay
you have to see look at these current
specifications you cannot just connect
them because they are NAND gates okay
and you know that the circuit you know
you just connect them like that get it
it will work but it doesn't okay you
must be very careful of the fan-out
criteria so you have to look at say if
you are connecting to NAND if you are
connecting NAND gates like this okay
suppose this is CMOS and this is TTS all
right so you must see what is the
specifications the output specifications
of the CMOS gate all right and the input
specifications of the TTL gate okay and
you must see how many you can drive all
right so this is very important gives
you an idea okay this type of
specification so when you are connecting
any circuit logic circuits using these
chips okay of different logic families
here of course if you buy TTL they will
give if you have completely one logic
family maybe they will specify on this
spec sheet you know that maximum fan-out
ten you are very sure but when you have
to connect different logic families okay
you must be very careful okay so you
must look at the specification sheet
okay
the currents okay and then only connect
otherwise
you may run into problems clear okay so
this is the this gives you the complete
specifications of the TTL 7400 series
seven four zero zero NAND gate all right
so I think we have discussed this gate
in great detail and we shall move on now
to the next gate in this next gate in
this series of the or shall I say the
next series in this TTL gate family okay
that is the Schottky TTL so this seven
four zero zero came out as I said in the
mid-sixties okay
and then the next improved version of
TTL which came out was the Schottky TTL
series which was the seventh for s0 0
alright Schottky TTL series I shall draw
the circuit of this gate okay and then
we shall list out the differences okay
between this these two circuits and then
we shall see the reasons for this these
changes
you
this is quite obvious that by the name
you know s short key TTS so it is this
uses Schottky transistors all right so
and we have already studied what is the
short key transistor okay
so I am just this is the symbol of the
short key transistor is you well know so
you will see that this this looks like
an S so there are lots of modifications
which went into the next version okay
you can see for yourself now I hope I
have incorporated all of them shall see
so well these are the values I think two
point eight kilo ohm this is 900 ohms
this is a 250 ohms 500 ohms so this is
the circuit of the Schottky TTL 7 4 s 0
0 gate NAND gate ok all right now if you
look at this circuit ok we can now list
out if you we have actually we have both
the circuits on the blackboard ok the
seven four zero zero here and by the
side you have seven for s0 0
now looking at both of them ok alright
you can now list out the differences
okay all right I can give you a hint
there are actually six major differences
ok so it's like playing a game you know
having two figures and finding out the
six differences okay all right so can
you just point out what are the six
differences yes
yeah there are two diodes at the input
okay two extra diodes at the input all
right so - they are Schottky diodes okay
as you write Schottky barrier diodes at
input okay then number two yeah there is
the output transistor is okay the one
major difference okay I should point out
is all the transistors okay okay this is
also short key transistor so all the
transistors are Schottky barrier
transistors our Schottky transistors
okay except of course one okay this is
not this need not be a Schottky
transistor because of the fact that
across the collector base Junction you
have this transistor and since VC of
this transistor is also always going to
be positive which which ensures that
this collector emitter of this
transistor is always collector to base
junction of this transistor is always
going to be positive so it is always
going to be in the active region okay so
it's not necessary
so Schottky transistors so that is the
next modification number three the
values of the resistances if you look at
them okay they are all lower value
resistances okay the red resistance
values have been reduced slower
resistance all the resistances have been
effectively slashed obviously I shall
tell you that this Schottky transistor
the Schottky TTL the focus of the
Schottky TTL is to achieve higher speed
okay and lower resistances give you the
higher speed we have already seen that
okay now when we talk to our inverter
transistor inverter okay one way to
improve speed is lower resistance higher
currents which means higher currents
yeah what are we left with this part of
the circuit okay so instead of
resistance here at the base of the
output transistor you have a act
if this is called active pull down okay
all right you had an active pull up so
this part okay instead of a resistance
you have a transistor circuit here so
this is called an active pull down okay
so you introduce what is called an
active pull down number five this one
this is what is this configuration
called Darlington pair and then what is
the remaining one the diode is missing
one diode is missing okay missing diode
okay between the two I should say so
these are the six major differences
which you will look which you see okay
between the two between the two gates
okay now what we shall do in the next
class we shall take each of take up each
of them alright
see the reasons for that and what
happened how does it improve our the
performance alright so we shall take up
this seven for l00 then we shall go on
with the newer series seven for LS which
came up later okay the further
modifications okay and then the even the
more recent ones the advanced series of
TTL gates okay which have very good
characteristics okay so we shall finally
we'll go through the entire range of TTL
series gates okay
and with that we shall conclude our
discussion on TTL in the next class
Oh
you