16.5 A D D I T I O N A L E X A M P L E S
For review purposes, more examples of both piecewise linear and incremental
analysis are given in the following subsections. No new material is presented,
so readers who do not need additional practice can omit this section without
loss of continuity.
16.5.1 P I E C E W I S E L I N E A R E X A M P L E :
CLIPPING CIRCUIT
The output voltage vo in the diode clipper circuit, shown in Figure 16.21a, will
resemble the input voltage vi , except that the bottom of the waveform will
be clipped off. The circuit has only one diode, so that the Thévenin solution
method discussed in Chapter 3.6.1 can be used, but here we will use the method
of assumed diode states. Assume that the diode in Figure 16.21 is ideal, then
draw the two subcircuits, one with the diode OFF, and the other with the diode
ON, as shown in Figures 16.21b and 16.21c. By inspection, the output voltage
with the diode OFF is constant, because there is a fixed current IO flowing up
through R. Thus
vo1 = −IO R.
(16.35)
The current source and the voltage source are in series, so the voltage source
has no effect on vo1 . Furthermore, when the diode is OFF,
vi = vo1 + vD .
(16.36)
Because the diode is in the OFF state, vD must be negative. It follows that in
the OFF state vi must always be more negative than −IO R.
Next, when the diode is ON, the output is directly connected to the input:
vo2 = vi .
(16.37)
In the ON state, vi must be more positive than −IO R. Hence the valid portions
of the waveforms in the subcircuits are the darkened segments, and the complete
output waveform is as shown in Figure 16.21d. As promised, the circuit has
clipped off the bottom of the input wave.
16.5.2 E X P O N E N T I A T I O N C I R C U I T
The circuit shown in Figure 16.22 produces an output voltage vOUT that is
proportional to the exponential of the input voltage vIN for sufficiently large
vIN . To analyze this circuit, assume that the Op Amp is ideal, the saturation
current of the diode is Is = 10−12 A, and the temperature of the diode is
approximately 29◦ C so that its thermal voltage is VTH = 26 mV. Because
918f
IO
+
+
vi
-
R
vo
vo1
(a)
vi
IO
t
iD = 0
+
+
vi
-
vD
-
+
R
-IOR
vo1
(b)
-
Valid for vi < -IOR
vo2
F I G U R E 16.21 Diode clipper.
IO
+
+
vi
-
vi = vo2
t
-
+
vD = 0
R
Valid for vi > -IOR
-IOR
vo2
(c)
vo
vo
t
-IOR
(d)
vi
918g
R = 100 kΩ
iD
+
F I G U R E 16.22 A diode-based
exponentiation circuit.
vIN
+
vOUT
+
-
-
the Op Amp is ideal, and used in a stable negative-feedback configuration, the
voltage at the inverting terminal of the Op Amp is zero. As a result,
iD = Is evIN /VTH − 1 = 10−12 A evIN /(26 mV) − 1 .
For sufficiently large vIN , for example for vIN ≥ 120 mV, the exponential term
dominates, and this relation simplifies to
iD ≈ 10−12 A evIN /(26 mV) .
Next, because the voltage at the inverting terminal of the Op Amp is
zero, and because the current into that terminal is zero, the output voltage
is given by
vOUT = −RiD ≈ −10−7 V evIN /(26 mV) ,
which exhibits an exponential dependence on vIN .
For example, for vIN = 200 mV, 300 mV, and 400 mV, vOUT =
−0.219 mV, −10.3 mV, and −480 mV, respectively.
16.5.3 P I E C E W I S E L I N E A R E X A M P L E : L I M I T E R
The circuit in Figure 16.23 is useful for making square waves out of sine waves,
and for limiting the amplitude of an output waveform when the input waveform
amplitude varies over a wide range. To analyze the circuit, we note that the
Thévenin approach is not helpful, and a graphical solution might be messy
because of the two diodes (not so, in fact, but that is not obvious yet). So resort
to analysis by assumed diode states. The subcircuits for the four states, assuming
an ideal-diode model, are shown in Figures 16.23b, 16.23c, 16.23d, and 16.23e
918h
R
D1
+
V
-
+
vi
-
D2 +
V vo
+
-
(a) Limiter circuit
vo1
vi
R
+
+
+v
OFF v- D1 OFF D2
vo1
+
V
V+ -
+
vi
-
+V
t
0
-V
(b) Both diodes OFF
vo2
R
+
vi
-
+
V -
+
OFF
vo2
V
+ -
+V
t
0
(c) D1 ON
F I G U R E 16.23 Diode limiter.
vo3
R
+
OFF
+
V-
+
vi
-
V
+
vo3
-
(d) D2 ON
-V
t
0
vo
vi
R
+
vi
-
ON
+
V
-
+
ON
- vo4
V
+
-
(e) Both diodes ON
vo
0
t
(f) Complete waveform
918i
along with the appropriate subcircuit output voltages, obtainable by inspection.
From Figure 16.23b,
vo1 = vi
(16.38)
because there is no current through R. When either diode is ON, the output
voltage is independent of the source voltage vi . For D1 , ON, for example,
vo2 = V. The fourth diode state, Figure 16.23e, cannot be reached with this
topology, (assuming V is a positive quantity) because there is no value of vi that
will force both diodes ON at the same time. Now we must identify the valid
segments of these waveforms. In Figure 16.23b, both diodes are assumed OFF,
so vd1 and vd2 must both be less than zero. Hence, using KVL:
vi − V = vD1 < 0
(16.39)
vi < V
(16.40)
−vi − V = vD2 < 0
(16.41)
vi > −V.
(16.42)
Thus vi must be between −V and +V. Likewise vo1 , from Equation 16.38.
This range of validity is indicated by the darkened segments of the waveform
in Figure 16.23b. It follows that the complete output wave must be as shown
in Figure 16.23f. If the peak amplitude of vi is ten or twenty times V, then vo is
a reasonable approximation of a square wave.
16.5.4 E X A M P L E : F U L L - W A V E D I O D E B R I D G E
Figure 16.24 shows one of the most common rectifier circuits found in electronic equipment, the full-wave diode bridge. We assume therefore that vi is a
60-Hertz sinusoid with 10-volt peak amplitude, and we wish to find the waveform vo across the output resistor. A full-blown assault using assumed diode
states would yield 16 subcircuits, but it will turn out that only two of these
are possible, suggesting a more insightful approach. Suppose that vi is a small
positive voltage. Then current must flow down through the bridge. The only
available path is D1 , RL , and D4 , because of the orientation of D3 and D2 .
Similarly, for vi negative, current must flow up, and thus must follow the path
D3 , RL , D2 . The two corresponding subcircuits, assuming ideal diodes, are
shown in Figures 16.24b and 16.24c. Now, by inspection, for vi positive,
vo = vi
(16.43)
vo = −vi .
(16.44)
and for vi negative,
918j
D1
+
vi
-
RL
(a)
+
vi
-
D2
+ vo -
D3
+
voA
D3 OFF
(b) Subcircuit
for vi positive
D1 OFF
-
D4
+
vi
-
D1 OFF v
+ oB D4 OFF
(c) Subcircuit
for vi negative
vi
+
vi
-
+
voC
-
D3 OFF
D4 OFF
F I G U R E 16.24 Full-wave diode
bridge.
(d) D1 and D2 ON
vo
t
t
(e)
All other subcircuits are degenerate. Consider, for example, the subcircuit for
both D1 and D2 ON, as in Figure 16.24d. Clearly no current can flow in RL .
Also, current can’t flow down through D3 , or up through D4 , so all diode currents must be zero in this state. A similar argument holds for all adjacent diode
pairs, whether ON or OFF. Hence the two states depicted in Figures 16.24b
and 16.24c are the only ones we need to consider.
Note that the current always flows in the same direction through RL ,
regardless of the polarity of vi . For sinusoidal input the waveforms appear
as in Figure 16.24e. The circuit is called a full-wave rectifier because current
flows through RL on both halves of the input wave. Neglecting diode voltage drops, the average value of the output voltage, that is the DC voltage, is
0.637 times the peak of the input sinusoid. Further, by symmetry there is no
frequency component in the output waveform at the input frequency, 60 Hertz
in our example, or odd multiples thereof. Hence the circuit has a much higher
percentage of DC relative to harmonics compared to the half-wave rectifier
discussed in Section 4.3.
918k
R
i
+
i
F I G U R E 16.25 Zener-diode
regulator.
+
-
v
v
+
-
50 mV AC
vo
20 V DC
(a)
(b)
16.5.5 I N C R E M E N T A L E X A M P L E : Z E N E R - D I O D E
REGULATOR
All semiconductor diodes will break down and conduct appreciable current
under reverse bias conditions if the reverse voltage across the diode is large
enough. This breakdown is non-destructive if the current is not excessive; the
diode returns to normal reverse-bias behavior if the voltage is reduced. A Zener
diode is a semiconductor diode in which this breakdown under reverse bias is
carefully controlled by the manufacturing process so that the breakdown occurs
at a specified voltage, the so-called Zener voltage of the diode. A typical v i
curve is shown in Figure 16.25a.
Because the breakdown voltage of a Zener diode can be carefully controlled by the manufacturing process, and the incremental resistance in the
breakdown region is quite small (around 10 to 50 ), Zener diodes are
quite useful as voltage regulators. A simple example is shown in Figure 16.25b.
Equation 4.74 is clearly inappropriate for finding the incremental resistance in
breakdown, because this part of the characteristic is not an exponential. Hence
the value must be obtained from the data sheet for the Zener diode in question.
Integrated-circuit regulators, with transistors, Zener diodes, and resistors all
on a single chip, will certainly outperform either of the crude regulator circuits
discussed here.
16.5.6 I N C R E M E N T A L E X A M P L E :
DIODE ATTENUATOR
It should be clear from Section 4.5, and particularly from Equations 4.63
and 4.74, that for small increments of voltage or current, the semiconductor diode looks like a linear resistor whose value depends on the DC current
flowing through it. Thus it should be fairly easy to build an attenuator with
an attenuation constant that can be changed by means of an external voltage
or current. Figure 16.26 shows a simple example. Here a diode is used in the
shunt branch of a voltage divider on a small-signal source vi . The DC current through the diode is controlled by the DC voltage VC through the large
918l
RC
Rt = R1||RC
R1
+
VC
-
R1
- +
VOC = VC ------------------R1 + RC -
+
vi
-
(a)
(c)
F I G U R E 16.26 Diode
attenuator.
RC
RC
+
VC
-
+
- 0.6 V
Rd
R1
+
R1
0
ID0
(b)
+
vi
-
(d)
vo
rd
-
resistor RC . If we assume that vi produces less than a 5 mV change in the
diode voltage, then the incremental analysis approach discussed in Section 4.5
can be applied.
First, draw the circuit for the calculation of the DC current ID0 is to form the
Thévenin equivalent of the linear part of the circuit, as shown in Figure 16.26b.
An easy way to solve for ID0 is to form the Thévenin equivalent of the linear
part of the circuit, as shown in Figure 16.26c. At the same time we replace the
diode by its piecewise linear model. Then
ID0 =
(VOC − 0.6)
Rt + R d
.
(16.45)
The incremental resistance rd of the diode will thus be a function of VC :
rd =
kT
qID0


kT  (R1 RC ) + Rd 
.
=
q
VC R1 − 0.6
(16.46)
(16.47)
R1 +RC
Now draw the subcircuit that relates the incremental variables. Replace the
diode in Figure 16.26a by the incremental resistance rd , and set all DC sources,
in this case VC , to zero, as indicated in Figure 16.26d. By inspection,
vo = vi
RC rd
R1 + (RC rd )
.
(16.48)
The attenuation is clearly dependent on the DC voltage VC , as desired.
918m