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ECEN 2250: Simple Models for Diodes
The linear passive circuit elements (resistors, capacitors and inductors)
that we have studied so far are bilateral: that is, they behave the same way
for negative voltages and currents as they do for positive ones. A diode is the
simplest example of a unilateral element, one whose behavior for negative
voltages is very different from that for positive voltages. The following is
a summary of the essential concepts needed to analyze circuits containing
diodes.
Diode Models
The three most common models for a diode are shown in the figure below.
anode
+
cathode
forward bias
region
reverse bias
region
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1. A diode is a nonlinear, unilateral circuit element whose element law is
given most accurately by the Shockley diode equation:
h
i
i = IS ev/(nVT ) − 1
where IS is the reverse saturation current (typically very small, on
the order of 10−16 A), and n is the ideality factor (typically between
1 and 2, and often approximately equal to 1), both of which depend
on properties of the particular diode. The voltage VT is called the
thermal voltage and depends on the ambient temperature (at room
temperature, it is about 26 mV). The diode equation is valid so long
as the magnitude of the diode voltage is not large enough to cause
breakdown.
2. A simplified model that is often used to approximate the behavior of
a diode is the ideal diode, for which the current is zero when v < 0
(reverse-bias condition), while the voltage is zero when i > 0 (forwardbias condition). In this model, the diode functions as an idealized
controlled switch, allowing current to flow in one direction only: the
switch is open if v < 0 and closed if i > 0.
3. The piecewise-linear diode model is an attempt to more accurately
model the diode under forward-bias conditions. The current is equal
to zero when v < Vt , the so-called turn-on or threshold voltage of the
diode. When v > Vt , the i-v curve is a straight line with a finite
slope 1/Rs , where Rs is the forward-biased resistance of the diode.
This model cannot accurately match the actual diode behavior at all
voltages, and the values of Vt and Rs will depend on the range of v for
which the model is to be used. This model can be thought of as an
equivalent circuit consisting of an ideal diode in series with an ideal
voltage source Vt and resistor Rs as shown in the figure.
For our purposes, the ideal model will be used in all problems and examples.
Power Supply Circuits
Household and industrial AC electrical power is supplied as sinusoidal alternating current (AC) form. However, semiconductor-based consumer electronics universally require direct-current (DC) power. Although AC voltages
can be increased or decreased by the use of transformer circuits, conversion
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of AC voltage into DC voltage in power supplies is commonly done using
diodes. The simplest such circuit is the half-wave rectifier circuit shown in
the figure below.
vin(t)
V
+
Assuming the diode to be ideal for simplicity, a sinusoidal input voltage
vin (t) = VA sin ωt (say) is converted to the rectified waveform vout (t) as
shown. With an ideal diode, the peak value of the input voltage waveform
is the same as that of the output. Although the output voltage is not DC,
it can be expressed as the sum of its average value VDC and a fluctuating
part ∆V whose time-average is zero:
vout (t) = VDC + ∆V
where
VDC
1
=
T
Z
T
0
VA
vout (t) dt =
T
Z
T /2
0
or, since ωT = 2π,
VA
cos ωt T /2
sin ωt dt =
−
T
ω
0
VA
π
By connecting a large capacitor C in parallel with the load resistor R, the
fluctuating part of the output voltage can be eliminated, and the output
waveform smoothed out to be close to VDC as shown below.
VDC =
3
+
vin(t)
+
VDC
C
R
Although this circuit “works”, it is inefficient in that VDC ≪ VA , largely
because the negative half of the original AC waveform is “wasted”.
A more efficient way to make use of the power available from the AC
input is to use a full-wave rectifier circuit using four diodes as shown below.
+
vin(t)
+
VDC
R
vout(t)
vout(t)
t
Careful analysis of the behavior of the diodes during the positive and negative half-cycles of vin (t) shows that the negative half cycle has its polarity
reversed and inserted in between the positive half cycles as shown. Thus,
the entire AC waveform is used to create an output voltage whose average
value is twice that of the half-wave rectifier circuit:
2VA
VDC =
= 0.6366VA
π
which is reasonably close to the RMS value of the original AC voltage:
VA
Vrms = √ = 0.7071VA
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Once again, the fluctuating part of the output voltage can be substantially
eliminated by placing a large capacitor in parallel with the load resistor R.
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