The resistor, capacitor, and inductor are linear circuit
elements in that their
v-i relations are
linear in the mathematical sense. Voltage and current sources
are (technically) nonlinear devices: stated simply, doubling the
current through a voltage source does not double the voltage. A
more blatant, and very useful, nonlinear circuit element is the
diode (
learn more).
Its input-output relation has an exponential form.
it=
I
0
ⅇqkTvt-1
i
t
I
0
q
k
T
v
t
1
(1)
Here, the quantity
q
q
represents the charge of a single electron in coulombs,
k
k
is Boltzmann's constant, and
T
T
is the diode's temperature in
K
K.
At room temperature, the ratio
kTq=25mv
k
T
q
25
mv
.
The constant
I
0
I
0
is the leakage current, and is usually very small. Viewing this
v-i relation in
Figure 1,
the nonlinearity becomes obvious. When the voltage is positive,
current flows easily through the diode. This situation is known
as
forward biasing. When we apply a negative
voltage, the current is quite small, and equals
I
0
I
0
,
known as the
leakage or
reverse-bias
current. A less detailed model for the diode has any positive
current flowing through the diode when it is forward biased, and
no current when negative biased. Note that the diode's schematic
symbol looks like an arrowhead; the direction of current flow
corresponds to the direction the arrowhead points.
Because of the diode's nonlinear nature, we
cannot use impedances nor series/parallel
combination rules analyze circuits containing them. The reliable
node method can always be used; it only relies on KVL for its
application, and KVL is a statement about voltage drops around a
closed path
regardless of whether the
elements are linear or not. Thus, for this simple circuit we
have
v
out
R=
I
0
ⅇqkT
v
in
-
v
out
-1
v
out
R
I
0
q
k
T
v
in
v
out
1
(2)
This equation
cannot be solved in closed
form. We must understand what is going on from basic principles,
using computational and graphical aids. As an approximation,
when
v
in
v
in
is positive, current flows through the diode so long as the
voltage
v
out
v
out
is smaller than
v
in
v
in
(so the diode is forward biased). If the source is negative or
v
out
v
out
"tries" to be bigger than
v
in
v
in
,
the diode is reverse-biased, and the reverse-bias current flows
through the diode. Thus, at this level of analysis, positive
input voltages result in positive output voltages with negative
ones resulting in
v
out
=-R
I
0
v
out
R
I
0
.
We need to detail the exponential nonlinearity to determine how
the circuit distorts the input voltage waveform. We can of
course numerically solve
Figure 2 to
determine the output voltage when the input is a sinusoid. To
learn more, let's express this equation graphically. We plot
each term as a function of
v
out
v
out
for various values of the input voltage
v
in
v
in
; where they intersect gives us the output voltage. The
left side, the current through the output resistor, does not
vary itself with
v
in
v
in
,
and thus we have a fixed straight line. As for the right side,
which expresses the diode's
v-i relation,
the point at which the curve crosses the
v
out
v
out
axis gives us the value of
v
in
v
in
.
Clearly, the two curves will always intersect just once for any
value of
v
in
v
in
,
and for positive
v
in
v
in
the intersection occurs at a value for
v
out
v
out
smaller than
v
in
v
in
.
This reduction is smaller if the straight line has a shallower
slope, which corresponds to using a bigger output resistor. For
negative
v
in
v
in
,
the diode is reverse-biased and the output voltage equals
-R
I
0
R
I
0
.
What utility might this simple circuit have? The diode's
nonlinearity cannot be escaped here, and the clearly evident
distortion must have some practical application if the circuit
were to be useful. This circuit, known as a half-wave
rectifier, is present in virtually every AM radio
twice and each serves very different
functions! We'll learn what functions later.
Here is a circuit involving a diode that is actually simpler to
analyze than the previous one. We know that the current through
the resistor must equal that through the diode. Thus, the
diode's current is proportional to the input voltage. As the
voltage across the diode is related to the logarithm of its
current, we see that the input-output relation is
v
out
=-kTqln
v
in
R
I
0
+1
v
out
k
T
q
v
in
R
I
0
1
(3)
Clearly, the name
logarithmic amplifier is
justified for this circuit.
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