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.