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Now let us consider nonlinear circuits; a
circuit with a two terminal nonlinear element
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such as diode, and see how to analyze it.
And let me take this circuit with a voltage
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source, a resistor – R, and a diode. And
of course this could be any other nonlinear
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element as well. I will call the voltage across
this V D, and the current through it as I
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D. Let us see how to analyze this. Now a linear
circuit with this same structure is to have
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R and R 1. Now, of course this one after you
have done your basic electrical circuits or
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even maybe before, it does not even require
a moment thoughts right, you know what the
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solution is, it is just the resistive divider.
Now we will see how to solve this and you
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will understand that the analysis of nonlinear
circuits is considerably harder than linear
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circuits.
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So, now really this is an extremely simple
circuit. So, this is an independent voltage
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source and all way have to do is write k c
l at this node that will give us the solution.
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So k c l at that node says that the current
flowing through R in this direction equals
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I D which is current flowing through a diode
in the downward direction. Now, the diode
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itself tells you that I D is f of V D, this
is the general case for nonlinear element;
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and in case of the diode, it is I s exponential
V D by the thermal voltage minus 1. And the
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current in the resistor R, this equals V S
minus V D which is the voltage across the
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resistor divided by the resistance value.
So, our k c l equation simply says that V
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S minus V D by R equals f of V D in case of
general nonlinearity or in the specific case
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of the diode, it is I s exponential V D by
V t minus 1.
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Now there is only one variable here; V s is
the source, we know that. The only one variable
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is V D; so all others are parameters of the
components. R is the parameter of the resistor,
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and I s is the parameter of the diode, and
V t is the fundamental constant. So, you may
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be familiar with solving nonlinear equations,
you may be familiar with methods like bisection
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method or Newton-Rhapson iteration and so
on. So, if you do then you know how to solve
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this numerically and you know that unlike
linear equations, there is general inverse
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in formula or general solution formula for
nonlinear equations. So, if you have a set
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of linear equations by a matrix inversion
you can find the solutions; but for nonlinear
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equations, there is no such general method
and depending on the type of nonlinearity,
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you may have to choose the different methods.
Usually the solution is numerical, and there
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may be multiple solutions, there may be no
guarantee that you will reach the solution
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and so on. But usually the ways to solve nonlinear
equations are numerical and iterative.
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Of course, in this course we would not go
through that; we just assume that the solution
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can be found somehow. What we will do is to
bypass explicitly solving the nonlinear equation,
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at least for hand calculations. So, this can
be solved numerically. Now what is also useful
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for engineers is to find the solution pictorially
or graphically, this is for the sake of visualization.
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Now usually you do not do graphical solutions
when you want accuracy in the solution, but
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you do it when you want to visualize the behavior.
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So let us do that now for this particular
circuit. Now that technique is also limited,
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so when you have too many variables, you cannot
draw multi-dimensional graphs. In this particular
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case, it turns out you can do it, because
you have only two variables – one voltage
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and one current. So the equation to be solved
is V S minus V D by R which equal to I s exponential
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V D by V t minus 1. So, what I can do is I
can essentially plot the right hand side and
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left hand side separately. So, there is some
function on the left hand side, there is some
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other function on the right hand side, both
are functions of V D which my independent
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variable. And I can find the point of equality
or the point of intersection that will be
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the solution. Now the right side is nothing
but the diode current, so that is why I marked
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it as I D versus V D, so plotting the right
side, it simply means plotting the diode characteristic,
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which are like that; it is the exponential.
And plotting the left side that is what I
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want to plot is V S minus V D by R versus
V D.
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So, this is just the straight line you can
see that this is the linear function of V
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D, so when V D equals zero, this function
will be V S by R; and when V D equals V S,
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this function will be zero. And it obvious
that it also has a negative slope and the
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slope is minus 1 by R. Remember V D is the
independent variable here, and the coefficient
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of V D in this function is minus 1 by R. So,
very easy to plot also; so when V D is zero,
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it is V S by R; when V D equals V S, it will
be zero. And it will be a line – straight
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line connecting these two points, and the
slope of this is obviously minus 1 by R. Now,
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clearly in this case, there is only one point
of intersection and that is the solution.
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Now if you solve this equation numerically,
you would find this value of V D, and if you
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find the value of the current corresponding
to that that is V S minus V D by R or this
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side, it will be this value of current. So
this is the solution. So, by plotting different
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parts of the equation and finding the point
of equality that is finding the point of intersection,
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you can find the solution. So, say somehow
you find the solution either numerically or
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graphically. Now I will just give you an illustration;
if you want, you can solve this numerically
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and find the solution. So, let me take a case,
where V S is 5.7 volts, R is 5 kilo ohms and
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the diode has a saturation current of 10 to
the minus 15 amperes. It turns out that if
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you solve this numerically, you will find
that V D is approximately 0.72 volts, and
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the diode current is approximately 1 milli
amp that is the solution; let say we have
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solved it for a particular value of V S, which
is 5.7 volts.
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And now the value of V S changes, so let say
it becomes 6.7 volts. So, what you have to
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do, you have to solve this all over again
that is pretty clear right. So, with let say
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this particular curve, whatever I have shown
here, this is the part the straight line is
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what is dependent on V S right in this particular
case. This exponential the right hand side
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of it it is independent of V S, it happens
to be so in this particular case. And the
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left hand side is dependent on V S, it is
this line. So, let say this particular one
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which I have already drawn corresponds to
V S equals 5.7 volts. If you change the value
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of V S, what you have to do, you have to redo
this. So, let say V S changes to 6.7 volts,
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you will get a line like that and you will
get a different solution. So let say V S is
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6.7 volts or alternatively if V S is 4.7 volts,
you would get a solution like that and you
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will get yet another solution. So this corresponds
to V S equals 4.7 volts. So, the point is
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that any time the input changes, you have
to solve the nonlinear equation all over again.
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Now this is the big difference from linear
circuits.
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So, if I go back to the linear circuit, I
showed earlier again this is extremely simple
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and if you want to solve it, you can do it
from scratch also right. So, in this case,
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let say you knew the solution for V S equals
5.7 volts then if V S changes to 6.7, you
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know that the new solution is simply proportional
to the new value of V S, so linearity holds.
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Even when you have multiple sources, the solution
will be linear combination of all the sources.
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So, if you find solutions, if you find the
solution to the circuit for one particular
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value of V S, you can find it for any other
value simply by scaling. Now in case of nonlinear
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circuits, first of all finding the solution
is the very complicated; and top of it, if
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you change the value of V S, you have to go
through the nonlinear calculations all over
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again.
Now this case where the input is changing
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is a very relevant case for us, because we
want to amplify some signals in our amplifier.
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And signal usually is a time varying signal
that is when I am speaking into the microphone,
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it is giving a time varying voltage to the
circuit. So we have to be able to analyze
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our circuit varying inputs.
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So, I will show the same
example with let say two sources in series,
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so this could be let say 5.7 volt source,
and this could be a sinusoidal source, this
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is what I call a signal. This could be let
say some 100 milli volt sin omega t or maybe
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one volt sin omega t. Now if you want to find
V D in this particular case, what will you
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have to do. So, the way we set it up, the
circuit is exactly the same, so the voltage
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source that is applied to the circuit, it
is now time varying. So this is the V S right,
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this is the value of V S, and V S in this
case is 5.7 volts plus 1 volt sin omega t.
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And if I plot V S versus time, it will look
like that. The maximum value will be 6.7 during
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positive peak of the sin wave and minimum
value will be 4.7 volts. So, how would you
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find the solution or the response V D to an
input like this, you would have to find it
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for 5.7, there is some value of V D that you
get that is see from the previous graph. So
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this is the value for 5.7. You have to do
it for 6.7 and 4.7 and so on, and also many
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points in between.
So, essentially this is V S, you have to keep
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varying V S, and keep drawing these straight
lines, find the solution and plot the solution
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V D from that one. So, there will be some
solution here, something else for 5.7, something
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else for 4.7, some other solution for intermediate
value and so on. It is extremely cumbersome.
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And you can feel that to get a reasonable
plot, you have to take many points in this
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cycle. So, first of all, finding the solution
for each point is very tedious and doing it
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for so many points this is just next to impossible.
So, you can find
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V D like this in principle , we know that
value of V D, when the input is 5.7 that is
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about point seven volts. So, the 0.72 volts,
so that is what I said and when it is 6.7,
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you can calculated numerically, it turns out
to be something slightly higher. And for 4.7,
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it will be something slightly lower. And you
can calculate the intermediate points and
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find that the solution looks somewhat like
that.
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Now, this while ay in principle, it is simply
not practical. So, what we will do is to find
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an alternative way of analyzing nonlinear
circuits which is approximate but at least
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something that we can carry out by hand and
that we need to able to do in order to get
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any insight into our circuits.