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So, what are we really doing when we approximate
the curve based straight line, so let us see.
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So, in general terms, the equation for this
circuit were V S minus V D by R equals f of
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V D, and for the particular case of the diode,
this f of V D was I s exponential V D by V
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t minus 1. Now, let me say that the first
solution or the operating point that I calculate
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is for V S equals V S 0. So, that means that
this blue stuff here calculated when V S was
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equal to some value V S 0, it does not matter
what it is, it is basically the first thing
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00:00:57,990 --> 00:01:04,659
I calculated. So this is V S 0 and this is
V S 0 by R, and this is the operating point
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solution or the original solution. Let me
call those values as V D 0 and I D 0. So,
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V D 0 and I D 0 are the operating point of
the diode and this has to be got by nonlinear
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analysis
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and this is of course usually numerical.
And let me say that at the operating point
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the diode voltage is V D 0, I denoted like
this; and the diode current is I D 0. We have
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taken an example earlier where if I s is ten
to the minus 15 amperes, and V s is 5.7 volts
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and R is 5 kilo ohms, so V D 0 turns out to
be about 0.716 volts, and I D 0 is approximately
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1 milli amp. So, this I got from numerical
analysis and that is the operating point.
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Now, let say that we have a different value
of V S, and I will represent the new value
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of V S as V S 0 plus some lower case v s so
that means that new V S represented as old
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value plus an increment lower case v s. I
can always do this, I mean my original value
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of V S is 5.7 volts, if V S changes to 6.7
volts, I will say that it is one volt increment
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over 5.7 volts. I can do this for any V S
right. And similarly I represent every quantity
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in this circuit, that is the actual diode
voltage as the original diode voltage, the
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operating point diode voltage plus an increment
in the diode voltage and so on.
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So
all quantities are represented as the original
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value which of course, it is called the operating
point plus increments over the operating point.
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So now I can write my equations in terms of
this definition. So I had V S minus V D by
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R equals I S exponential V D by V t minus
1. The operating point of course satisfies
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the equation, so V S 0 minus V D 0 by R equals
I S exponential V D 0 by V t minus 1. And
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the new value also satisfies this that is
V S 0 plus v s, I am writing this V S new
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value as V S 0 plus an increment minus V D
0 minus v d by R equals I s exponential V
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D 0 plus v d by V t minus 1.
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Now let me write it in general terms as well
not just for the diode. So this would be in
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general some function of V D, and this will
be function of V D 0 – the operating point.
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So this is the general equation. And in the
third case, I represent the general case as
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the operating point plus an increment, so
I have f of V D 0 plus v d. Now what I do
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is expand this in a Taylor series about the
operating point, this is the significance
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of the operating point. I choose some operating
point which is the original case, for which
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I calculate the nonlinear solution exactly.
And above that operating point, I expand the
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new case in a Taylor series. And what do I
get, I will get the function at the operating
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point plus the first derivative calculated
at the operating point times this increment
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00:07:07,960 --> 00:07:15,360
V D plus higher order terms, because I am
going to neglect this, I am not including
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them.
Now what is this mean, we have already seen
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this in the graphical solution, so let say
this is the operating point V D 0. So the
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00:07:40,750 --> 00:07:51,690
very first term here, f of V D 0 is nothing
but this part, this is f of V D 0. And the
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00:07:51,690 --> 00:08:00,370
second term means a linear dependents on V
D or increment over V D 0, so that corresponds
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00:08:00,370 --> 00:08:08,310
to this straight line that is the sum of these
terms corresponds to this straight line. And
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then you can add higher order terms, the next
one will be parabolic and the next one will
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be cubic and so on. And if you add all of
that you will get the exact nonlinearity,
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but we will stop here and use the straight
line. So whatever I said earlier if you are
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not move too far from the operating point,
you can use the straight line approximation,
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and the straight line is the tangent to the
curve at the operating point, this is what
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I meant. I expand the nonlinearity in a Taylor
series above the operating point, and neglect
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all of the higher order terms that is second
order and higher than second order terms.
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The Taylor series expansion goes like this
right f of V D which is f of V D 0 plus the
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00:08:50,810 --> 00:09:02,800
increment will be f of V D 0 plus f prime
calculated V D 0 times V D plus the second
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00:09:02,800 --> 00:09:20,700
derivative by two factorial times V D square,
of course everything is calculated V D 0 and
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00:09:20,700 --> 00:09:34,570
so on. Now, you can see that these terms the
higher order terms reduce more rapidly with
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00:09:34,570 --> 00:09:40,330
decreasing value of V D compared to the first
one. So, you can always find a sufficiently
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00:09:40,330 --> 00:09:47,750
small value of lower case v d – the increment
where these terms are negligible, as long
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00:09:47,750 --> 00:10:13,750
as you have a smooth continuous function.
So, this approximation is always valid, the
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only thing is you have to restrict the range
of V D over which it is valid, that is all.
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00:10:22,000 --> 00:10:36,170
Now, what does it mean if I neglect these,
I will be left with this f of V D 0 and the
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first order dependence on this lower case
v d or the incremental v d. Then, what I do
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00:10:43,680 --> 00:10:54,310
is
I subtract the operating point from this case
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00:10:54,310 --> 00:10:57,520
that is I simply take this equation and that
equation and subtract.
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What do I get
in easily verify that we will get V S minus
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00:11:10,800 --> 00:11:28,050
V D by R to be equal to f prime calculated
at the operating point times V D. Now what
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00:11:28,050 --> 00:11:45,529
is the significance of this, this is linear
in the increment V D. Now there is this constant
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that dependents on the operating point, but
as far as the equation is concern that is
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just the constant; V D is the variable and
this equation is linear in V D, so that means
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00:11:59,470 --> 00:12:05,880
that it is much easier to solve than the nonlinear
equations. So for a changed value of V S,
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we have to solve the nonlinear equation all
over again. Now we do not do that, we find
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the approximate solution using a linear equation,
now that is a great simplification. Also because
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there are lots of nice techniques for handling
linear equations and we can use all of those
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00:12:24,270 --> 00:12:51,130
things as far as the incremental quantities
are concerned.
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00:12:51,130 --> 00:12:58,260
So this is the great simplification; so in
summary, we first calculate the exact solution
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00:12:58,260 --> 00:13:02,520
that is the solution from the nonlinear equation
for the operating point, and that we will
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do numerically or graphically later we will
see how to do it by hand. Then for any other
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value of the input, there can be more than
one input it does not matter. You express
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every quantity in the circuit both inputs
and internal variables as the original the
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00:13:17,960 --> 00:13:26,700
operating point cases plus increment. And
then you find that wherever you have a nonlinearity
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00:13:26,700 --> 00:13:30,830
you can expanded in a Taylor series about
the operating point and neglect higher order
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00:13:30,830 --> 00:13:35,130
terms that is second and higher order terms
then you will be left with a linear equation
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00:13:35,130 --> 00:13:39,890
relating the increments. So now you can use
everything that you know about linear circuit
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00:13:39,890 --> 00:13:45,370
analysis whatever you learned in basic electrical
circuits or some equivalent course to analyze
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00:13:45,370 --> 00:13:48,920
nonlinear circuits as well in a approximate
way.