Today, we will continue discuss our A to D
converter test, because we have discussed
so far about dual-slope A to D and then successive
approximation, A to D converter and then flash
A D C. In flash A D C, we also discussed about
pipelined architecture. Now, we discuss the
other A to D converter that is sigma delta
A to D converter.
So, sigma delta A D C, because in the other
converter, that is 
in dual-slope we had the dual-slope noise
rejection is good, but very slow noise rejection
is the advantage, but slow in the case of
successive approximation A D C. It is the
question of about 2 microsecond per bit conversion
or 1 microsecond per bit conversion level
of time, but then it needs sample and so on,
but to get 8 bit 16 bit accuracy is possible,
but then going for 24 bit again it takes longer
time and then we have to also make; for example,
24 bit converter means 24 bit accurate D to
A converter need to be made and that is not
easy.
So, successive approximation, if you take
getting 24 bit accuracy, 24 bit converter
converter is difficult because we need difficult,
because we need 24 bit accuracy difficult
because we need because we need 24 bit accuracy
DAC. So, to get high accuracy, but not high
accuracy then there is a possibility because
that kind of application is there in audio
video products. So, if you are looking for
say 24 bit resolution, but I am not looking
for 24 bit accuracy then sigma delta A D C
is the kind of converter that we should look
for.
So, when we need say 24 bit or 32 bit resolution,
but satisfied with 16 bit accuracy then we
can look for sigma delta A D C.
So, let us see how the sigma delta A D C works.
Now, in these converters what is done is that
you have the input voltage. Now, in the input
voltage, I will actually give; for example
to summing amplifier where the difference
between these two voltages are coming out
here and this I will connect to in a integrator.
This integrator output is actually, I put
here A D C probably 16 bit A D C then this
16 bit is converted to a DAC so 16 bit DAC,
so you have this. It has it own reference
and then this output of the DAC, which is
analog output is actually given to this and
this DAC output is actually processed further,
this is A D C output, A D C this A D C 16
bit 16 bit A D C. Now, if you see how the
circuit works see initially, what is happening
that when the voltage is say, if you have
some voltage then if this is for example zero,
then the entire voltage appears as it is.
Suppose, we have V input voltage V then the
integrator actually slowly charges up to minus
and of course it can be converted into plus
having inverter here. For example; I can have
a inverting stage here and I convert into
plus. So, where as the voltage is applied
here the input voltage will be continuously
raising with time, so if I look at the output
of that it will be continuously raising with
time.
Now, the 16 bit A D C converts this in to
a digital value and gives it back to this
analog input, so the to the DAC. So, essentially
what is happening is that whatever voltage
is there at the output of the A D C that it
will return back at this point. So, this you
will get back the original voltage get back
the original original voltage at this point,
we call this is V zero original V zero. Now,
this V zero is actually again subtracted and
then you get the difference between these
two, so you get V it is the input voltage
V minus V naught what it is coming here.
Now, the crux of this usually this, suppose
if I mucking with the 1 volt input voltage
assume that 1 volt is a full scale voltage
assuming A D C and DAC reference is 1 volt
that means the full scale value of A D C and
DAC are 1 volt.
Now, if the input voltage is exactly 0.5.
For example; if the input is exactly 0.5,
assume that V input is exactly 0.5 volt. That
means, if I look at the A D C output then
if V zero is also equal to 0.5 at sometime,
then A D C output would be, you will have
one if a 16 bit converter there all other
bits are zeros. So, you have only the top
alone is 1 and all rest of the 15 bits are
zero then DAC output would be equal to 0.5
volt, assume this is also accurate 16 bit,
the DAC output will be equal to 0.5 volt.
So, you had given 0.5 volt at this point and
you got back 0.5 volt here and the input also
0.5 and this is 0.5. So this V naught, this
is we called V x that is equal to zero. Since,
V x is zero, then the input voltage the input
to the integrator is zero, so input integrator
will not change state, so this makes adder
output to zero, adder output equal to zero,
so integrator will stop charging. So, the
integrator output remains constant, so integrator
remains V out remains constant 
and A D C and DAC, everything DAC are also
give fixed output.
Now, that this is exactly 0.5, suppose if
it is a 16 bit converter we said the input
is 0.5 assume the input is not 0.5, but it
is little more than 0.5, if V in is not equal
to 0.5, if V in is equal to 0.5 plus some
10 microvolt. For 16 bit converter 1 L S B
will be equal to 1 by 2 power 16 that is 1
by 64000 that actually is volt that will come
as 1 by 64 millivolt that will be equal to
1000 by 64 microvolt that is nearly equal
to 16 microvolt. So, 1 L S B is 16 microvolt
for a 16 bit converter. So, if I give the
input voltage V in is equal to 0.5 then point
V zero, actually V zero is 0.5 that is if
I give for A to D here at this point 0.5 volt,
then I will get exactly the output as it must
be 1 all the other bits will be zero. So,
if that is if that is A D C input is 0.5 then
it is output will be 
same all the other 15 bits are zero.
Same thing happens, if V zero is equal to
0.5 plus 10 microvolt then A D C will be again
it will be all the other 15 bits will be 0
0, but same the DAC also will give only 0.5
volt. But now, the problem is that if the
summing amplifier because you will have only
0.5 volt at the input. If you look at the
figure again that you know we have the input
voltage, we are summing it here that plus
minus summing it and that is given to the
integrator. The integrator output is given
to A to D. Now, if this 0.5 input is 0.5 volt
plus 10 microvolt this is 0.5 volt, because
the DAC output DAC gives you from the A D
C output what you are getting is again only
0.5 volt. So, it is sitting at 0.5 volt. Since
this is 0.5 and this is 0.5 you get the error
voltage of 10 microvolt here that will be
charging, because we said there is a inverter
here, so we have a the analog inverter at
this point which will actually, if this is
10 microvolt then the integrator will not
stop charging, now the integrator will be
charging and then the V zero what we have
here will be increasing.
So, if I redraw this it will look like this.
So, you have the input voltage 0.5 plus 10
microvolt and you have the summing amplifier,
so this was giving 0.5 volt, you got 10 microvolt
signal coming out here that is actually integrated
here and this is of course inverted if you
required, because that makes the A D C work
in the positive side. So, with time that your
voltage will be increasing. Now, you got 10
microvolt so which was stopped earlier at
0.5, so 16 bit A D C the input to this and
then the DAC receive the output of 0.5 volt
16 bit DAC, so you got the input voltage coming
in.
Now, what happens this output will start increasing
with time, because you have here 0.5 volt
here 0.5 volt plus 10 microvolt, which makes
the output increase with time and that makes
after sometime if the voltage goes to at this
point V zero, so V zero would be if we look
at V zero, so V zero at that time versus voltage
that actually would be start raising up. If
the V zero goes when V zero becomes 0.5 plus
16 microvolt then A D C output becomes 1 that
you have the and this becomes 1. Because now
you got 16 microvolt, 16 will be recognized
as a 1 L S B and then the output will go high
and that will make DAC output that DAC will
give you 0.5 plus 16 microvolt where 1 L S
B extra is the DAC is getting that will give
this.
Now, that means the input to the summer becomes
16 microvolt. That means, now the new condition
for the A D C looks like this that you have
a summing amplifier here and then you have
a input here for 0.5 volt and here it gives
you 0.5 plus 16 microvolt, here is 0.5 plus
10 microvolt, so that actually becomes minus
6 microvolt and now the output of this would
be minus 6 microvolt. That actually when you
give it to the integrator 
and then the inverter here then I give it
to A D 
C and that is given to the DAC here you get
this actually becomes 16 bit DAC. Now, here
you get the in here the output is 0.5 plus
10 microvolt, so you got minus 6 microvolt
that will make the integrator discharge and
this voltage started decreasing.
So, the V zero since it is the integrator
input 
is negative, so integrator started discharging,
so 
V zero started decreasing. Then V zero becomes,
when V zero becomes 0.5 volt then again A
D C becomes 1, again it comes zero that DAC
also gives you exactly 0.5 volt. This actually
what happened then because of this minus 6
microvolt this started decreasing and then
this again become 0.5 volt. Once A D C become
DAC becomes 0.5 volt then summing amplifier
output you know we have here 0.5 plus 10 micro
here actually 0.5 then output becomes plus
10 microvolt.
That means that is, it will make it will make
integrator to charge again to charge. So,
V zero also will increase will start increasing.
That is it will make integrator to charge
again, so V zero will start increasing, so
this will make again A D C output as, A D
C output will come as again the last 1 L S
B will change to 1 and DAC again will give
you 0.5 plus 16 microvolt. So, if you look
like this, this 1 bit of A D C will be continuously
changing 1 and zero, that is A D C output
if you look at in time that you will get continuously
changing the last one bit and next time it
becomes 0, 0, 0 then 0, 0 1 and so on it will
be keep changing it. Now, if I am seeing the
output in time domain you know A D C output
in time domain, what is the A D C output in
time domain what is the A D C output A D C
output A D C output in time domain. Actually
the A D C output, if we look it take 100 samples,
for 100 samples, how many L S B's will be
1 and how many L S B's will be zero. This
is the question that is asked, because how
many will be 1 out of 100, how many will be,
1 how many will be 100.
For example; if the input is 0.5 volt then
all 100 L S B's will be zero. If the input
is 0.5 volt plus 8 microvolt that is half
L S B and input is 0.5 8 microvolt then you
will see then 50 L S, 50 will be L S B's 50,
L S B's will be 1 and 50 L S B's will be
0. There is out of 100 samples you will find
50 L S 50 times will be, 1 L S B will be 1
and 50 times L S B will read as zero. So,
this is because we had given input as 8 microvolt.
So, when the input is 8 microvolt then if
you look at the summing amplifier and the
integrator that it looks like this, because
we have input here, so we have input here
0.5 plus 8 microvolt. Here actually the input
is given, which is coming if it is 0.5 volt
then output is actually plus 8 microvolt.
Now in the other case, if the comparator we
have the input 0.5 plus 8 microvolt and then
at the A D C, from A D C this is from DAC,
so this is also from DAC if this comes 0.5
plus 16 microvolt, because 1 L S B is 16 microvolt
then L S B is 1, for this for L S B zero,
this is the case for L S B 1, L S B equal
to 1 this is the case, so now you will get
minus 8 microvolt. So, if you look at the
output at this point it will be changing plus
8 microvolt next it goes to minus 8 microvolt,
so charging and discharging will be equal,
so that means the integrator sees equal amount
of charging and discharging.
That is integrator output, if you see integrator
output you will see that it will have, if
I look at the integrator here this is V input,
so you get plus 8 microvolt of that time and
minus 8 microvolt of that time. So, you have
a basically a square wave equal going plus
8 microvolt and this is going minus 8 microvolt.
That will make the integrated output look
sitting at some point going up and down, the
charging equal time going up and again it
will discharge back to the actual value, so
it will have equal time going up and down
because when it is plus if I look at inverter
output that is we have. If you look at the
output V zero at with respect at time then
you will see that it is going equal amount
of time going lower and higher. So, this will
be 1 L S B up L S B is equal to 1 this will
correspond to L S B is equal to zero. If you
Since the slopes are equal so you will get
any digitize you will get equal number of
time 100 times if I sample the A D C output,
if the A D C this integrated output is given
to A to D converter, so you have A D C and
this is given to the input. So, if I look
at the A D C output I will get whatever may
be sample half the time it will be high and
half the time it will be low, that is if I
look at L S B, so L S B out of 100 times 50
times will be higher and 50 times will be
lower.
But if the for example; the input is not 8
microvolt that is if I take the summing amplifier,
if the input is 0.5 volt plus 1 microvolt
then the input to this will be 0.5 volt then
at that time I will get here 1 microvolt plus
1 microvolt. Whereas, if I take the other
option have the other option that if I have
0.5 plus 1 microvolt here and then when L
S B is on I will get 0.5 plus 16 microvolt
when the L S B is 1 for L S B is equal to
1 for L S B is equal to zero.
So, this is the condition that you get for
L S B is equal to zero and L S B is equal
to 1 and here the output will be minus 15
microvolt. That means now the discharging
will be faster than charging. So, for this
condition the discharge will be faster than
for this integrator discharge is faster than
charging. So, you will get the integrator
output like this. So, integrator output V
with a time versus V zero by plot that you
will get the charging is very slow, so you
will have the normal voltage here the charging
actually takes time to goes to 16 microvolt
up then the discharge is very rapidly equal
to this point, so this is L S B 1, L S B is
equal to 1 and the L S B will be zero, so
you will get discharge very rapid charging
is very slow.
So, discharging charging is slowly rising
up and discharge is very rapid. So, this is
actually charging and this is actually discharging.
So, discharging is very rapid, charging is
very slow, that means you will find that if
the transition is taking place half way point
then you will find L S B 1 will be for less
number of time and L S B zero will be obtained
for more number of times. This is because
we will find that when the discharging is
very rapid which gets down below the threshold
value require for the convertor quickly and
then it slowly rises back to 1 volt slowly,
so you will find L S B equal to 1 is for only
very less amount of time and then L S B is
equal to zero will be there for more amount
of time.
Now, that means if the input voltage is because
it is just 1 microvolt, whereas if I take
with the input voltage equal to 0.5 plus 15
microvolt for V in is equal to 0.5 plus 15
microvolt then the condition would be that
you have 0.5 volt plus 15 microvolt and the
input coming here, if it is zero then you
will have 15 microvolt plus with microvolt.
Now, in the other case you know if this for
L S B zero so this will be 0.5 volt, this
is for L S B is equal to zero and L S B is
equal to 1, so you get here the same 0.5 volt
plus 15 microvolt and you get 0.5 volt plus
16 microvolt and minus will be minus 1 microvolt.
So now, the integrator will be there, integrator
will reach quickly, charging will be very
fast, discharging will be slow. Now charging
is fast and 
discharging is slow that will make 
more number of times you will see more numbers
of times L S B is 1 and less number of times
L S B equal to zero. So, obviously if I sample
over sample the A D C output then if I find
how many number of times L S B 1 the fraction,
which was there between zero and 16 microvolt
can be decoded.
So, if you find the ratio between 
is equal to 1 and L S B is equal to zero is
same as is same as ratio between V input and
16 microvolt. That is the fraction ratio as
the same as ratio between fractional V in
fractional V in at 16 microvolt. For example;
in this case if it is 0.5 and 1 microvolt
for example; for V in is equal to 0.5 plus
1 microvolt then the fraction is the fraction
is 1 by 16, for V input is equal to 0.5 plus
5 microvolt the fraction 
is 5 by 16. So, the ratio between 1 and zero
L S B's 1 and zero's will also be the
same fractional ratio, which will be the A
D C L S B 1 and L S B equal to 1 and L S B
is equal to zero ratio also would be also
will be same as 1 by 16 and 5 by 16.
So, by over sampling, that means if I take
100 samples and find out what is the average
between these two that will tell me the average
of the input that will give me the same the
input signal ratio. That is all that I do
is I oversample the A D C, so oversample the
the A 
D C and find out N and N plus 1 at the output.
So, that N by N plus 1 gives you the same
ratio as that of the fraction ratio. For example;
it will be 1 by 16 in one case that is for
5 plus 1 microvolt, N by N plus 1 is equal
to 5 by 16 for 0.5 plus 5 microvolt, for N
by N plus 1 would be equal to 8 by 16, for
input of 0.5 plus 8 microvolt; that will be
exactly half the time will be this.
So, one can get higher resolution A D C by
oversampling. So, A D C resolution can be
increased by oversampling by this way. So,
the oversampling and decimation and the averaging
all that is done by normally D S P, so using
D S P oversampling and the calculations are
carried out oversampling and 
the required calculations are carried out.
So, one can get more than get resolution of
resolution more than 16 bit using 16 bit A
D C and DAC, but accuracy is limited to 16
bit only. This kind of A D C 
can be used in signal processing. For example;
audio video processors where accuracy is not
a prime concern the resolution is all that
matters. In that application higher bit resolution
can be easily obtained using sigma delta A
D C. So, this is how the sigma delta A D C
works. There are different types of sigma
delta A D C are there, which I am not discussing
here, like there are dual integrated sigma
delta A D C are there dual integrated single-slope
A D C are there, which we will not discuss
at this point, but the principal remains the
same.
So, we had seen in this course from the beginning,
how the transistors are working and how the
operational amplifiers are working, how can
we use the operational amplifier signal conditioning
and then error budgeting and so on, but various
things we have discussed. Now, let us list
what are the things we have discussed in this
course and then we complete the course with
this.
So, if you look at the look back what we had
discussed so far then the following items
will emerge. So, we have started our discussion
with transistor applications. Here, we have
discussed about how to use a transistor as
a switch how to use. We had given a example
and then shown how to use the switch then
transistor as a latch, shown how to use transistor
latch, then how to use a transistor to boost
the current 
and then the transistor for increase the transistor
for basically to increase the power. So, in
this we had discussed these applications in
transistor and then I had also shown you transistor
switch in two types that is transistor as
a switch keeping the beta more than one and
then beta less than one. So, the transistor
switch we have seen two different items for
beta greater than one and then beta less than
one. That is in this case it is collector
bias collector is reverse bias in this case
bias collector is forward biased. So, this
is one application we had discussed in the
beginning.
And here we had also shown an example how
to use a transistor to get a voltage regulator.
So, one can also recollect that if you want
to make a voltage regulator using a transistor
we had shown you one circuit like this that
we have a error amplifier for transistor,
so the classical transistor circuit, so we
had the input here and then bias this and
then we had shown the output given a feedback,
so that is V zero so what is this, these applications
in transistor.
Then after discussing about the transistor,
we had discussed about operational amplifier.
We had also shown how the operational amplifiers
are evolved from transistor. The operational
amplifier basically had come mainly to remove
the problem that encountered in the transistor
amplifiers, because initially when the transistor
amplifiers were used this is the kind of circuit
they were using for amplify the signal and
you have A C signal where amplified using
this, so you have a plus voltage and then
amplified output was appearing here. Nevertheless;
when they want to amplify A D C voltage then
they had encountered problem, because drift
was very high because the bias emitter voltage
drift was there, so V B E drift created problem
for D C amplification.
So, from there only we had come out with the
operational amplifier circuit, where we can
use the three transistor op-amps were used,
so we had shown you one example with the three
transistor op-amp, so where in you get the
output voltage. So, we have the input here
V input that become V output, but this output
voltage is drift free, this V zero is not
drifting due to temperature. This was one
of the main achievement and this is how the
operational amplifier was born. So, this is
basically a mini operational amplifier and
then we explained you this, from there the
operational amplifier were come in and then
we had discussed about how to use the operational
amplifier and then how to use them for the
various circuit application we had discussed.
So, in that we had shown you that the operational
amplifier can be used for various purpose
like we had shown you the from a simple amplifier
like this to how to use this amplifier in
a different configurations. So, one of the
application that we had shown in this was
that how to make a voltage regulator using
op-amp then we also discussed about constant
current sources 
how to make with op-amp. In this concern,
we have also shown you 4 to 20 milliampere
current transmitters we had discussed.
Then we also have shown you different types
of 4 to 20 current transmitters. We also discussed
during the process about the transformer,
how the transformer works and how to use the
transformer for the voltage regulator and
so on then we also discussed about how to
make L V D T. For example; L V D T signal
conditioning was discussed then we also discussed
in elaborate manner capacitive transducers
signal conditioning for capacitive transducers.
So, in this process we also discussed about
ratio transformer bridge applications. So,
we had explained you how to make a capacitance
measurement that to very low capacitance measurement
without involving the stray capacitance, because
the basic concept involved in the capacitance
measurement is that, if I have two methods
we had discussed, one is that if I have an
amplifier with input capacitance here then
the output is connected to the output. For
example; if I have a signal generator here
then the stray capacitance one and stray capacitance
two has no effect here, this is stray capacitance
one and then this is the stray capacitance
two, these two capacitance has no affect on
the output signal. So, this was the main concept
used in capacitor circuit design and this
was actually compared with ratio transformer
bridge, how the ratio transformer bridge works
and then the same concept was brought out
in capacitance transducer and then how to
make the capacitance measurement using op-amp.
So, this was elaborately discussed in this
lecture.
Then we also discussed various other applications
like solar based battery charger we have discussed.
We had shown a workout working worked out
example how to charge the battery using a
solar panel and then we also discussed about
how to use the MOSFET as a switch that also
was discussed. During that process, we also
discussed how to provide a short circuit production
for the MOSFET, then we also discussed elaborately
about errors involving op-amp usage. So, error
budgeting concept was brought in error budgeting,
because error budgeting is an important aspect
in analog circuit. So, we had shown you in
the op-amp errors involved in the op-amp like
offset voltage drift offset voltage drift
then input resistance affect then three output
resistance 
then we also explained bias current errors
then C M R R related errors. So, these errors
are discussed in detail and I had shown you
how to make error budget.
So, error making error budget is an important
task for analog circuit designer making an
error budget for the circuit is an important
task for the analog circuit designer, then
we had discussed about different types of
A to D and D to A converters, so then we had
discussed about A D C and D to A converters.
In this we have discussed about dual-slope
A to D converter, dual-slope A D C dual-slope
A D C with processor also we have discussed,
A D C with micro controller, then errors involved
in this also discussed, errors involved in
this converters also discussed also discussed.
Then we have discussed about other three converters
that is successive approximation A to D converter
then flash A D C and then we have also discussed
sigma delta A D C today. So, if you see this
one, we have started our journey long back
and then we have come this 40 lectures various
items where discussed. Particularly, we concentrated
more on the error budgeting and then the circuit
design skill. So, as the analog circuit designer
we have to acquire minimum three skills major
skills that one have to acquire would be that
one is error budgeting 
then second one is circuit design, you know
how to use a correct component or how to select
the correct circuit for given operation circuit
design skill. So that, you know how to connect
various components in a correct working manner
and third one is various techniques involved
like for example; we have discussed how to
measure a capacitance, how to get rid of stray
capacitance, so various techniques involved,
for example; how to measure current various
techniques involved. So, these are the three
skills one have to acquire in analog circuit
design. I introduced you all three skills
with examples. So, one have to take the thread
form here and learn in this direction that
will help you to go good to go as a good circuit
analog circuit designer to the industry, so
with this I will complete this course.
