In this video we're going to talk about
electronic circuits. This is the modern
topic in electrical engineering, is how to
build circuits that actually provide a
power gain. We're going to talk about
them, and we're going to have to talk
about a new circuit element.
This is called dependent sources, we'll
introduce them. We're going to show how
those are used in model, what are called
operational amplifiers. These are the
workhorses of what are called active
filters. And the word active here has to
do with providing a power gain. It's all
about power. So let's see the difference
between an electrical circuit and an
electronic circuit. So an electrical
circuit like we've been talking about
can't provide any power gain. Now if you
figure out the transfer function between
some output variable and the source you
may well have a case where the gain is
bigger than one. However, when you attach
a low resistor to it, for example, to try
to get that power out, it turns out the
power that's dissipated in the load has to
be smaller than what's produced by the
source. Easy way of seeing that, the only
thing in the circuit that's producing
power, is that source. And we know from
the conservation of power, that all the
other elements have to dissipate, or not.
Dissipate at all but have to dissipate
power, so you pick any one of 'em, that
has to be less than what's the power
that's being produced. So, no power gain.
So it would seem like, when you look at
the conservation of power, that there's no
circuit that could produce a power gain,
and that's true. Well we're going to
introduce a new circuit element, that has
that capability. And so our new circuit
element, it's what's known as the
dependent source. Okay, so here's the
idea. We have a circuit, and I'm drawing
an abstract line. Somewhere inside there's
a circuit element. That has a voltage v
defined for it. Somewhere else in the
circuit is a voltage source whose voltage
depends on what the voltage across that
element was. How this happens is a
mystery. Somehow this voltage source knows
exactly what this voltage over here is.
And produces something that's K times it.
That's why this is a voltage dependent
voltage source. Now this is clearly not
something that you could build just out of
raw elements. how in the world can you
have some sort of thing and this is really
points out that, these are ideal circuit
elements and they're used to describe
something a lot more complicated. Now this
action gets built, I'll show you an
example in a second, and it's not very
simple but it can be well described by our
ideal circuit elements, as we'll see in a
second. I want to also mention that you
can have all four possible configurations
here. We have a voltage-dependent voltage
source. You could have a current-dependent
voltage source. You could have a
voltage-dependent current source, and you
could have a current Dependent current
source. Turns out the last one is
particularly important for modeling
transistors. So they are fundamental to
characterizing transistor circuits, in
particular those circuits that have a
power gain. Let me show you how that's
possible. So, here's a very simple
circuit. It's split in 2 pieces as our
voltage dependent and voltage source and
because it's in two separate parts. they
don't interact with each other, except by
this mystical, dependent source that
mystically somehow figures out what the
voltage is across that resistor. Now,
let's do a little power calculation. I
think it's pretty clear that all the power
dissipated in that resistor is produced by
that source. And all the power dissipated
in that resistor is produced by the
dependent source, okay? Now let's draw a
little box here, okay? So, what's the
power It's produced, by the voltage
source, but that's gotta be what's
dissipating the resistor, so we know
that's Vin squared over R. It has to be
producing that much power. How much power
is being consumed in our resistor over
here? Well that's KVin squared. Over R.
And I really do mean the same resistors
here. So if K is bigger than 1 which is
certainly is allowed to be. this circuit
is producing more power than its, than is
coming into it. There's a power game. So
what's going on? I thought this wasn't
possible and the answer is, it is possible
if inside this circuit is somehow being
provided with power that's not indicated.
And this gets into the idea of what's
called a power supply and these days for
cellphones and things like that. You might
think this has equivalent to a battery
that provides power as part of it's
functioning so that this circuit can
provide the indicated power gain, because
the power supply is an additional source
of power For the output. how that happens,
that's why the dependent source models are
used, because they're very simple ways of
describing somewhat complicated circuits.
But in general, circuits containing
dependent sources can produce a power
gain. Okay.
And here's an example of one. But here's a
practical example, and this is the
operational amplifier I mentioned earlier.
So, the operational amplifier, it's, a
jargon name is op-amp. It's, it's referred
to as an op-amp. And the circuit symbol is
indicated here to the left, has this
triangle that has a plus and minus input
which I've labelled. I'm thinking here of
nodes so if that's node a and node b. And
here's the reference. The output, comes
out the end of the triangle. And it's node
C. And that's the. We're going to plug
that into a circuit in just a second. The
model for it, in terms of ideal circuit
elements, for the operational amplifier is
as follows. There's a resistor Rn between
the two input nodes. And here's our
dependent source, voltage-dependent
voltage source, that is sum gained times
the difference of the two voltages here.
And that's why Ea corresponds to plus And
my e, b here correspondes to the negative
side, because that's how the enter into
this formulea for the voltage produced by
this voltage dependant voltage source.
It's all written in terms of node voltages
that are referenc e to the taken with
respect to that reference. There is a
finally, there's a output resistor And
this is the output. Okay, so, what does an
op-amp look like physically? This is an
op-amp. we'll do a little counting here.
There are one, two, three and four inputs
that have to be available for an op-amp.
Well, this is a chip. That you will see in
all kinds of circuits. And this is a short
name for integrated circuit. Which,
different names for the same thing, chip
is what we commonly refer to it. And
notice, it has four pins over here. And it
turns out there are four pins on the other
side, giving a total of eight. Well that's
more than enough, it turns out it needs
two other inputs to make the chip
functioned correctly and two, the
remaining two are not really used. So
what's inside this chip that can provide
the power gain and is modeled by our
dependent source model? Well I think
you'll be a little surprised by the
complexity. So this is what's inside
what's known as a 741 op-amp, operational
amplifier. It contains 22 transistors, 11
resistors, 1 diode and 1 capacitor. So,
this is the symbol for a transistor and
here is another one, here's a really odd
looking one over here. 22 total
transistors and so it is a very
complicated circuit inside to provide that
game and let's go through what the inputs.
So, so I N plus. That's node A in my
diagram. I N minus is of course node B.
Here is the output which is node C and
these are offsets which are additional
inputs. Which we're not going to talk
about here that have to do with detail
function of the output. But, what I really
point out are these two, Vcc plus and Vcc
minus, standard notation for the power
supply. Turns out for this op-amp, the
power supply has to be about 15 volts. so
these are providing additional power
through the circuit which allow us to use
the deep ended source model. If these are
turned off, you always take the battery
out. It no longer looks like a dependent
source. In fact it probably looks like
very, somethin g very dull and boring. The
output either is flat, which means it's 0,
or it has some distorted waveform. It
doesn't function like it was designed to.
You have to turn on the power. In order to
make the circuit come alive, and be
described by our ideal circuit models.
Now, I want to point out, that this
diagram, the circuit diagram. What's
inside came from a website at Texas
Instruments, which is a well-known,
worldwide producer of chips of all kinds,
analog and digital. And this is a website
they provide for you to download
specification sheets, that's what these
are called. I took this from a
specification sheet for a 741 op-amp. And
you can download and look at all the
different kind of circuits and chips
rather they provide and see how they
operate. And we need to look at more
detail to see what special about an
op-amp, in other words I want to know
something about Rin, Rout and G. I know
the op-amp functions this way, but what
are those parameters? Well, let's dive in
to the specification, for our 741 op-amp.
There it is and what you'll see listed in
general are a bunch of different
parameters that describe how this circuit
behaves. And we'll go through those but I
want to point out something very
important. Notice, that what's specified
in this specification sheet is a minimum
and maximum value, and a typical value,
for every one of these parameters. So.
The manufacturing process guarantees that
the circuit will be within these minimum
and maximum values, whichever parameter
you're talking about. But what it is
exactly will vary depending which
operational amplifier you pick out of the
batch. So this raises an issue in circuit
design. Suppose I want to, and circuit
with a very well-defined gain although we
can't get it by just using an op amp as it
is. We're going to have to put circuitry
around it to make it function and behave
exactly like we want, not like the
manufacturer can sort of approximately
tell us it's going to behave. We're going
to see that in just a second. Let's g o
through and see how the various parameters
of our model for 741. So, I'm going to
point out here ri, what the spec sheet
calls ri, turns out, corresponds to Rn and
the whole, and the specification sheet
says that the typical value is two
mega-ohms. And it could be as small as .3,
and the maximum, it could be bigger. But
the point is, is that this is big, that
resistor is big and it's in the megaohm
range, more or less, and that turns out to
be a good thing, as we'll see. Okay? Let's
go on to the output resistor, which is
labeled Ro, output resistance. It is
typically 75 ohms, so it's small, and
that's good, okay? Now here comes the
goody, and the large signal, differential
voltage amplification constant, that's G.
The typical value is 200, but look at the
units, volts per millivolt. So what this
is saying is that it's 200 times 10 to the
3 typically, which means that the gain.
And be about 2 times 10 to the 5th,
200,000 but it could be as small as
50,000. well, this to me says that the
gain is big, which has important
consequences. So big gain, small incoming,
big input resistors, small output
resistor. There's something else about the
operation of the amplifier, has to do with
the power supply and this is worth
pointing out. So, the vertical parameter
here turns out its maximum output voltage
swing. How big can this voltage at node C
be? Okay. And it turns out it depends
somewhat on the value of the load resistor
which you're going to attach out here but
notice it's about equal to the power
supply voltage, plus and minus 14. Okay,
let's take the big number. So suppose the
output voltage, C so we get G times some
voltage, equal to 14. And what's that
voltage? Well if you divide this out and
you use the V number, you get 70
microvolts. So, the largest that the input
could be, if you stuck it in node a, is 70
microvolts. If you exceed that input,
what's going to happen? If you put in a
sign wave because we want the circuit to
work in a linear way. Our output is going
to be sign wav e, but if its amplitude is
bigger than 14, what's going to happen is
that the output's going to get what we
call clipped. Is the amplitude will be
chopped off. So, that's not good. That
means sign wave in something looking more
like a square laid out and it's a clear
sign of distortion. So, how do we handle
this? The point is the threshold voltage
for the clipping phenomenon is pretty
small and that's where the external
circuitry comes in. We have to learn how
to use the op amp in the context of
surrounding circuitry to control it. But
the big thing that's going to be very
important to understanding that is that
the gain is big.
We'll see that, how that works in just a
second. Okay, so here is the classic
op-amp circuit, the reason op-amps were
developed. Here's our circuit model and
I've now stuck that in the circuit. And
it's, I have a source and I have my own
resistor that I'm using, here's the load.
And here is our F. What does the F mean?
What does the F signify? What do you
think? Well F, is for feedback. It's
where, it means it's regarding the input
to the output. Now one thing to notice
here, is that, the op-amp is put in upside
down. So, you'll notice this is plus and
minus up top, notice here is minus plus
and that's intentional, you'll see that in
a second. , also note that the plus side,
node a has been tied down to the
reference, okay?
And node b is where the input voltage is
coming in. Okay.
So, maybe it becomes a little bit clearer
when we use the circuit model for the
op-amp, and stick it into our circuit,
okay.
So, the way that works here is Rn, and
we'll define a voltage V across it, and
because the op-amp, if you will, is
plugged in upside down. The output is
minus Gb where G is this big, positive
gain, that we, I had showed you in the
previous slide. Okay.
Now that I have the model for my op-amp
that's been turned on, I need to find how
the input and output voltages are related
to each other. We need to solve the
circuit. What we're going to use is the
node metho d because it can solve
anything. The dependent source means, the
presence of the dependent source means we
can't use the serial and parallel rule
This voltage is either not current while
they're gone. So we have to use node
method, that's why we introduced it so we
could talk about this circuit. Okay.
How many nodes are there? Okay. What I
get. there's 1, 2, 3, and 4 and of course,
here's our reference node, pretty obvious.
Well, this node has a voltage source
attached to it so we don't' need to define
a node voltage there. It turns out the
same thing applies to the dependent source
because there's a voltage source between
that node and the reference. we don't need
it to find the node voltages, so the only
voltages we need to define are. v, at node
voltage, and that node voltage which turns
out to be v out. So we need to write the
node equations. So, here they are. And
let's go over one of them, I've writ, the
first one I've written applies to this
node right there, so let's see. The
current going south, is V out over RL. And
the node voltage headed west is V out
minus, the minus GV, which is the voltage
over there, divided by R out. And then
finally, V out minus V divided by RF, so
that's the node equation. At this node,
and, you can check to make sure I've got
the right node equation for that node.
Okay.
So, it's pretty easy to write. It's
always, the node voltage has to come in
with a plus sign for that particular node
equation, and the algorithms have to come
in with a minus sign. So that all checks,
so we're ready to go. We want to find the
relationship to VN and VL. Which means we
have to eliminate the voltage V by
combining these equations and simplifying.
And the results you get unfortunately, is
not something simple but that's okay. So,
after some manipulation what you get is
the following result. So, as you might
expect, the result is going to depend on
every circuit element, and the gain of the
op-amp.
This is where the fun comes. What do we
know about the parameters of the op-amp?
The circuit model. We know that R out is
small, and we know that R in is big. Now,
the critical thing is right over here, so
R out is small, and G is big. So GRf is
certainly bigger than RL so we can forget
about the RL. Well that simplifies things.
That means the RL cancels so this term in
turn is Rout over G times these, this
complicated expression. Raw.
Because G is big, and all resistors have
values basically bigger than one, all
these things are small, and further more
dividing them by G is going to make them
smaller cause G is huge. Which means we
can forget about everything here. All that
is Negligible compared to this, which
leaves us with a very simple answer. So,
we have, we have found the input output
relationship, under the assumption that G
is big. So, the output is equal to the
input voltage times a negative number. It
is given by RF over R the ratio of these
two resistor values. This is a important
result because now we control the gain of
our amplifier by specifying the feedback
and input resistor values, we control the
gain. So we're not, sensitive anymore to
the variations in G of Rn or RF. They
don't matter, as long as we have a big
input a big gain. This result applies so
we control the gain. Do note that it
inverts. Remember, a negative sign
corresponds to inversion in electrode
engineering. So, this is a inverting
amplifier. That's the name for it. I do
want to point out that if you had put the
op-amp in the other way, if you had put it
in plus to minus like this, this answer
would not have changed. If you note how
you can see that mathematically, if I put
in minus G in place of G which corresponds
of putting it in right side up, if you
will. This term is still small. And you
still would have forgotten about it.
However, the result would not have been
consistent. The input is going in this
side. And, if you had plus to minus like
this, and built it, what would have
happened, is that you would, it would not
have inverted, instead, would it happen ,
the output would hit the power supply
values, positive or negative and it would
not have looked like a sign wave. So it is
important that we, you put the op-amp in
right away. In mathematics, when you
figure it out tells you which way it had
to be put in. It has to be put in. If you
will, upside down because of that minus
sign. So, that minus sign is very
important. It tells us that the input had
to go in the minus side. Okay.
So, there's another thing to point out.
That this result does not depend on RL. So
again, it's only these two resistors that
determine the gain. And those two values
control the negative gain of the inverted
amplifier. Now you may not like the fact
that it inverts, that it negates. And I'll
show you a little bit later how you can
fix that. Turns out to be pretty easy.
Alright so, we now because of impedances
that everything follows. If I replace my
resistors in my circuit, IRLC circuits.
Okay.
I haven't shown an RL here or a ZL, if you
will, because the answer as we just found
for the transfer function doesn't depend
on that. And we know right away, I don't
have to do any more analysis because in
the analogy with resistor circuits that
the transfer function for this is minus ZF
over Z. The word active filter comes in
because we now can provide a gain that is
active. It has to have the power supply.
That's where the word active comes from.
So, let's suppose I wanted to design a
lowpass filter. And for the simple kind of
filters we've been talking about, this is
the trans, let's say that this is the
transfer function we want. Okay.
So let's go over it a little bit. So, the
negative sign comes in because we know
this is going to invert, so we have to
live with that. K. What's K? Well if you
look at the denominator when f is equal to
0 then the, the gain of our of our lowpass
filter at the origin, is now K in
magnitude, so our transfer function starts
out at K. You look at the rest of the
formula, when F is equal to FC that makes
this imaginaary number here component one,
which mean that corresponds to the cut-off
frequency, so. We have some FC out here
and so in between we get our nice low pass
filter where as 1 over the square root of
two times K, So this formula is a little
bit different for the RC filter because
now I'm going to have to pick element
values to fit this transfer function that
hasn't gained but has a cut off frequency
of f c. So how would we do that? Well one
way is just to mimic the formula that we
have. We could set ZF equal to our gain,
because it's upstairs. And we'll set Z
equal to that, because it's downstairs.
what's the impedance 1 plus j times f,
constant times f. And that looks like a
resistor in series with an inductor. Now
for all kinds of reasons, inductors are
not used very frequently and filter
circuits can be larger than capacitors.
And we tend to want to shy away from using
the inductor unless we really have to. So
this may satisfy the if you built this
with a, inductor and resistor in series in
the feedback part I'm sorry, in the input
part, you certainly would come up with a
low-pass, but it's not the best way to do
it. Let's think about this a little. Okay.
How about this suggestion? Suppose that
feedback had this for impedance. And
suppose the input of impedance was 1 over
K, well that means that it's a resistor
with a value of 1 over K. And the question
is, how do you interpret this thing? And
what that's telling me is that I have two
elements in parallel because 1 over, you
have a parallel combination of two
elements, two impedances. That's 1 over 1
over Z1 plus 1 over Z2. Which will give us
the old parallel form of the product where
the sum. Okay? So, what this looks like is
that Z1 is a 1, so it's a resistor. And
Z2, at 1 over Z2 looks like that which
means it's a capacitor. So, this is what I
would like to put in the feedback circuit,
is a resistor in parallel with a
capacitor. Right now we don't have very
good values here, if we want to gain
bigger than 1, that means the input here
is a really s mall resistor. And I
certainly don't like this 1 ohm resistor
sitting here. But I know the structure I
want to head for, I'm going to put that in
the feedback of the resistor in input and
let's just reanalyze it with those, so we
can make our choices. So here's the
circuit I want to consider. There's the
parallel combination of the resistor and
capacitor on the feedback part. And here's
the input resistor. And we know that the
transfer function using s is got RF in
parallel with a capacitor Okay, you see
it. That's the shorthand notation. And
it's going to be divided by R, and there's
our minus sign. And we start plugging
things in, like, particularly, what's the
impedance of the capacitor, and simplify,
you'll finally get this. Okay, so now
let's see in terms of these three circuit
element values the transfer function
changes. So clearly at 0 frequency, the
gain k is r f over r. So those two element
values control the gain at 0 frequency.
The cutoff frequency is 1 over 2 pi RF CF
so these two control the cutoff frequency
and you should not be surprised by that
this happened for the RC low pass 4 over 2
there was a product of R and C that gave
us what we want. So, we can pick element
values. And we have three elements, but
we're only constrained by two values, so
we have some flexibility to pick and get
the gain we want, and to pick the cutoff
frequency we want for our low pass. And
this is a very commonly used circuit. For
low-pass filters. Okay.
So.
Electronics.
What are electronic circuits? They're the
ones that provide power gains. They have
to have a power supply, or a battery,
that's in the circuit to provide that
power gain. Otherwise you cannot have a
power gain with what's called electrical
circuits. And that's only possible to
power supply. And most circuits today are
designed based on transistors. They could
be integrated circuits or chips, but
inside they're based on transistor theory.
You have to learn about transistors in
another course. We're not going to talk
about them in any great detail at all.
With other things we want to do, and that
is to understand signals and systems based
on designing and using active filters.