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So far we have seen that, when you have a
nonlinear circuit and you have a certain operating
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point already computed; when the values of
the input change or when there is some change
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in the stimulus to the circuit, you can represent
all quantities as increment over the operating
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point. And if you do this and neglect the
higher order terms in the Taylor series of
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the nonlinearities, the relationship between
the incremental quantities alone, this does
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not apply to the operating point, but only
for the incremental quantities, the relationship
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will be linear. So, for instance for this
particular circuit, assuming that the diode
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current is a nonlinearity f of the diode voltage
then the incremental voltage is V S and V
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D have this relationship. Now, this V D is
the variable, the incremental diode voltage
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can be obtained as V D equals V s 1 by f prime
at V D 0 divided by R plus 1 by f prime at
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V D 0 that is the reason I write it in this
form.
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This form looks very similar to that of a
voltage divider. The voltage across the diode
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is the source voltage times some number divided
by resistance plus some other number. And
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it is very clear that this number has dimensions
of resistance also, because f of V D is the
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diode current, now f prime is nothing but
the derivate of I D with respect to V D which
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has dimensions of conductance, and the reciprocal
of this is a resistance. So, essentially it
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appears that the incremental quantities can
be found from this circuit, I have not derived
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this, I am just putting down this circuit.
So for instance, if I had a linear circuit,
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where the input source is V S, and there is
a resistive divider, this resistance is R
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exactly the same as here, and this resistance
is one by f prime of the diode characteristic
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calculated at the operating point. You will
get the same solution right, because clearly
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the voltage that you see between these two
terminals is equal to that this you know from
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the resistive divider formula.
So what I want to show is that this is not
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a coincidence, it is not that for particular
simple circuit this happen to be the case.
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Now it turns out that in every circuit the
nonlinear element can be replaced by some
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linear equivalent. Now this is not very surprising,
we replace the nonlinear characteristic by
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a straight line around the operating point.
So the nonlinear element can be itself replaced
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by a linear element, and this is easier. So
while this business of expanding the nonlinearity
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in Taylor series and neglecting higher order
terms, this is easy, which still do not want
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to setup the equation every time and then
go about this business of neglecting higher
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order terms. We want even most straight forward
way of arriving at the solution for the incremental
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quantities and that we do by considering the
incremental equivalent for every element.
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So, the reason I showed the equation is to
show that it is a general method. Now, what
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I will show is that every nonlinear element
can be replaced by a linear equivalent and
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the resulting equations will of course be
the same as what you would get if you expanded
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every nonlinearity in Taylor series and neglected
higher order terms.
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So instead of considering the whole circuit,
what we can do is to just consider a single
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element. So we have a voltage V, and a current
I and we know that I is nonlinearly related
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to V, by the way we could also express the
voltage as some function of the current. So,
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I just use this formulation most of the time,
but this is perfectly valid. And these are
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like equivalent statement show ohms law, you
could write I equals V times the conductance
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or V equals I times the resistance, so this
is for linear element and this is for a nonlinear
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element. Now, let us say the nonlinear element
is in some operating point condition. We have
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V 0, and a current I 0 flowing through it,
obviously I 0 equals f of V 0, these two quantities
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satisfies this equation and let us call this
the operating point.
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Now, any other value of V can be thought of
as the operating point plus an increment,
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and any other value of I can be thought of
as the operating point plus an increment.
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So, now this is V naught plus lower case v,
this is I naught plus lower case i. And these
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quantities also obviously have to satisfy
the constraint imposed by the element, which
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is it nonlinear characteristic. Now I expand
this in a Taylor series about the operating
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point, so I get f of V naught plus f prime
calculated at V naught times v plus higher
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order terms. Again if this deviation v, the
increment v is small enough, I can neglect
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all the higher order terms and also I already
know that I naught is f of V naught, so this
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goes with that. So the incremental current
i is nothing but the derivative of the characteristic
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at the operating point times the incremental
voltage v. So, as far as the increments are
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concern the current is proportional to the
voltage which means that it behaves like a
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resistor. So this is only for increments,
the incremental voltage v and the incremental
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current i are related by this expression.
Now, there is something dependent on the operating
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point here that is the value of the resistance
itself. So, let me denote the resistance by
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lower case r, this is just to indicate that
it is the incremental resistance and that
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will be equal to the reciprocal of that slope
at v naught. Now, it is the reciprocal of
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the slope, because the function gives you
current as the function of voltage or the
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incremental conductance g is f prime at the
operating point.
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So, every two-terminal nonlinear element
is a resistor for incremental quantities provided
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the increment is small. So, it is extremely
important to remember that this equivalence
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is only for the increments. And the value
of the resistor depends on the operating point.
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And this is called the incremental resistance,
and because it is valid only for small increments,
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it is called the small signal incremental
resistance or similarly the quantity g, which
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is f prime this is known as the incremental
conductance.
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If the nonlinearity were represented like
this; voltage as a function of a current then
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the incremental resistance would be the slope
of h calculated at the operating point, this
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is whatever the operating point is and the
incremental conductance would be the reciprocal
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of this one. So, every nonlinear element can
be approximated by a linear element. Now,
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you see why this method is easier, we do not
even need to write the nonlinear equation
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of the original circuits. You have to do it
for calculating the operating point, but that
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point onwards for increments all you have
to do is look at each element and replace
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it by its incremental equivalent.
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So, what do I do my original problem was to
calculate the solution to this, where this
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V naught come from the operating point that
is I have already calculated the solution
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for V naught. I have solved for this and of
course, I can also include the currents here,
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so this is I naught plus i and this is I naught.
So, one thing I could do of course is go through
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the process of solving the nonlinear equation
again, but what I do instead is make a new
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circuit, which consist only of incremental
quantities. So, I have the incremental V S
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here, and I have R, I will come to the incremental
equivalent of a linear element, the diode
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I replace by its incremental equivalent at
the operating point. So, it is extremely important
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to remember that the incremental resistance
or incremental equivalent in general depends
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on the operating point. So, if you today calculated
the incremental equivalent of a diode at an
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operating point current of 1 milli amp and
tomorrow you have a different operating point
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current let say 10 milli amps, the incremental
equivalent will be different, so that you
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have to keep in mind.
So, this will be the incremental diode voltage,
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and this will be the incremental current in
the circuit it is very easy. So, if you want
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the incremental diode voltage, it will come
from the voltage divider formula; and if you
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want the incremental current, it will come
from ohmâ€™s law and so on. This is of course
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the very simple circuit, but regardless of
number of nonlinear elements you have, you
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can do this for any circuit. So, every nonlinear
element has to be replaced by its incremental
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equivalent at its operating point. So, you
could have two diodes in the circuit with
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different operating points, that is one could
be having operating point current of 1 milli
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amp, the other one could have 10 milli amp,
they will have different equivalent resistance
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in general. And this circuit, which relates
only the incremental quantities and it is
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linear. This is known as the
small signal incremental equivalent circuit.
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So, sometimes it calls the small signal equivalent,
sometimes the incremental equivalent and so
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on.
You have to remember that this is valid only
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for increments about the operating point,
and also it is valid only when the increments
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are sufficiently small, so that you can neglect
the higher order terms. Now, what is sufficiently
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small is very dependent on the nonlinear you
have and so on. The only definition I can
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give is that it should be small enough so
that second order and higher terms should
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be negligible. And also when you say negligible,
you have to specify some accuracy, so depending
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on the accuracy of the solution you want,
you can restrict the size of the increment.