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A solid state diode consists of a junction of either dissimilar semiconductors (pn junction diode)
or a metal and a semiconductor (Schottky barrier diode). Regardless of the type, the circuit
symbol for a diode is as shown in Fig. 8.1a, and the corresponding device in Fig 8.1b
Fig. 8.1a
Fig 8.1b
If V is positive, the diode is forward-biased. Then, the diode can conduct a significant
positive current I even though V is a small voltage of typically 0.7 V for the most common diode
(silicon diode). If V is negative, the diode is reverse-biased. This negative current is so small
that it is often considered to be zero. Thus, the usual function of a diode is to allow current to
flow in the direction of the arrow (the forward direction) for positive V’s, but not allow any
current to flow in the reverse direction for negative V’s.
Only a small forward bias (positive V) is required to cause a diode to conduct a
significant current I, and the less this voltage, the better. Ideally, this voltage would be zero
volts. Also, ideally, a diode can conduct any value of current I in the forward direction, with this
value being determined not by the diode, but by other components in the circuit in which the
diode is connected. Also, ideally, a diode conducts zero amperes for a negative V, regardless of
the voltage magnitude.
Put another way, an ideal diode is a short circuit for a voltage V that tends to be positive
(but it cannot be more than 0 V). Also, an ideal diode is an open circuit for a negative V. Thus,
an ideal diode acts like a switch that is closed for current flow in the direction of the arrow in the
diode circuit symbol, and open otherwise. Essentially, it is an electronically operated switch.
This ideal approximation is satisfactory for analyzing many circuits that contain diodes, provided
that the voltage levels are much greater than 0.7 V. Fig. 8.2 shows the I-V characteristic for an
ideal diode.
Fig. 8.2
Fig. 8.3 shows the I-V characteristic of an actual, physical diode. The part of the curve in
the first quadrant is the forward characteristic, and the part in the third quadrant is the reverse
characteristic. The current Is is called the reverse saturation current. For a reverse voltage VB,
the diode “breaks down” and draws a large reverse current.
Fig. 8.3
The diode forward characteristic is shown on an expanded scale in Fig. 8.4. Observe the
“turn-on” voltage VT. For forward voltages less than VT, a diode conducts very little current.
Also, in the normal forward operating range, the diode voltage is approximately VT, almost
irrespective of the current value. For the common silicon diode, VT is approximately 0.7 V.
Fig. 8.4
Except for the reverse-breakdown region, the I-V characteristic of a diode may be
expressed analytically as
I = Is(e40V - 1)
at room temperature (20C). For a silicon diode, the saturation current Is is of the order of 1 nA.
Although the forward characteristic is exponential, because of the large factor 40 in the
exponential exponent, the characteristic appears to be almost vertical for a forward voltage
slightly greater than VT, as can be seen in Fig. 8.4.
Usually, a stripe on the diode casing designates the cathode (-) end. If there is any doubt,
a DMM ohmmeter can be used to measure the diode resistance in both directions. A diode has a
small resistance in the forward direction, which is for current flow from anode to cathode. It has
almost an infinite resistance in the reverse direction, for current flow from cathode to anode.
Measurement of Diode I-V Characteristics
An oscilloscope display of the diode I-V characteristic can be obtained using the circuit of
Fig. 8.5 where the device is a diode. However, this circuit configuration can be used to displace
the I-V characteristic of any two-terminal device. It is important to note this method but in this
lab, we will use one of the built-in panels available in VI ELVIS.
Fig. 8.5
The Two-Wire Current-Voltage Analyzer panel is a stand-alone instrument that is a basic
two-wire I-V curve tracer. It is capable of measuring four quadrant IV signals within ±10V and
±40mA. The procedure for obtaining an I-V graph is as follows:
Select a diode and place it on the DMM readout of the ELVIS
protoboard. Connect one end of the device to DUT + and the other end to
DUT - as shown in Fig. 8.6.
Set the current limits to ±10 mA and the voltage sweep range from 0V
(start) to1.2V (stop) with increments of 0.10V.
Run the tool.
Comment on how the characteristic obtained in Step 1 compares to that of
an ideal diode.
DUT Fig. 8.6
Diode Circuits
Diodes are used in many types of circuits. For example, they are used in rectifier circuits to
convert AC to DC. Also, they are used in clipper circuits to select for transmission that part of a
waveform that is either greater or less than some reference value.
The Half-Wave Rectifier
Consider the circuit of Figure 8.7. Assume that vi represents a DC voltage source Vi in series
with the diode and the resistor. We monitor the voltage V0 across the resistor. If we consider the
diode to be ideal (VT = 0) then current will flow in the diode only if Vs is positive, the diode will
act as a short circuit and V0 = Vi. For Vi < 0, the diode will be reverse biased, allowing no
current flow and V0 will be zero. If the diode is not ideal, then for the current to flow, Vi must be
equal to or greater than VT. When the current flows in the diode, there is a voltage drop VT
across it and the voltage across the resistor is given by V0 = Vi  VT. Again, for Vi < 0, V0 = 0.
Now reconsider the circuit of Figure 8.7, where we have an AC voltage source vi(t) of angular
frequency  (Period T = 2/) and amplitude Vp.
vi(t) = Vp sin(t)
During the time 0  t  T/2, vi(t) is positive and the diode, if ideal, will act as a short circuit,
resulting in v0(t) = vi(t). However, in the second half of the period, vi(t) is negative and the diode
will remain open during this time, resulting in v0(t) = 0. If the diode is not ideal, then the diode
current will remain zero during the time vi remains less than VT. However, when vi > VT, v0(t) =
vi(t)  VT.
Consider the circuit of Figure 8.7. Let vi(t) = 5.0sin(377t) V and VT = 0.7 V. Sketch the
waveform v0(t) and show the times when v0 is zero. Does your answer depend on the
value of the resistor?
Consider the circuit shown in Figure 8.8. Let vi(t) = 5.0sin(377t), VT = 0,
Vr=2.0V. Sketch the output waveform v0(t) and show the maximum and minimum
values of the output voltage.
Consider the circuit shown in Figure 8.9. Let vi(t) = 5.0sin(377t), VT = 0 and Vr1 = 3.0 V
and Vr2 = 2.0 V. Sketch the output waveform v0(t) and show the maximum and minimum
values of the output voltage.
1. Construct the half-wave rectifier circuit shown in Fig. 8.7. Use a 10V peak to
peak, 1-kHz sinusoidal voltage for Vi.
2. Observe the output voltage Vo on an oscilloscope and sketch it. Be sure that
the AC-GND-DC switch (input coupling) for CH-0 and CH-1 are both set to
DC to avoid shifting of the waveforms in a vertical direction. Can the diode
turn-on voltage be determined by looking at the output voltage waveform? If
so, what is it?
Fig. 8.7
The Diode Clipper Circuits
A clipping circuit is shown in Fig. 8.8. Let VR = 2 V and the input voltage
Vi be a 5-peak, 1 kHz sinusoidal voltage. Observe and sketch the output
voltage Vo. This circuit transmits that part of the vi waveform that is more
negative than VR + VT.
Fig. 8.8
Reverse the direction of the diode in the circuit of Fig. 8.8. Then, observe
and sketch the output voltage vo. How does this output voltage compare to
that in step 2?
Diode clippers may be used in pairs to perform double-ended limiting at
two independent levels. Fig. 8.9 shows a double-diode clipper that limits
at two independent levels. Set VR1 = 3 V and VR2 = 2 V. Test this clipping
circuit with a sinusoidal input that has a peak amplitude of 5 V and a
frequency of 1 kHz. Observe the output voltage Vo and sketch it. Please
note that this will require the use of both the positive and negative
variable DC supplies. These outputs are labeled SUPPLY+ and SUPPLYon the protoboard.
Comment on the diode circuits studied in this experiment. Where might
they be used in?
Fig. 8.9
Full-wave Rectification:
For many electronic circuits, DC supply voltages are required but only AC voltages are
available. Then, the required DC voltages are obtained from the AC voltages by rectification
and filtering. In the rectification process, an AC current is converted to a time-varying current
that flows in a single direction, and so it is a time-varying DC current. Subsequent filtering
“smoothes out” the variations to produce an almost constant dc current and voltage.
The diode bridge rectifier circuit is shown in Fig. 8.10. The operation can be understood
by considering the polarity of the applied voltage Vi at the top node a, with respect to the bottom
node b. When this voltage is positive, diodes D1 and D2 are forward biased and therefore
conduct, thereby causing the output voltage Vo to be equal to the input voltage Vi. (For
simplicity, the small voltage drops across the conducting diodes are neglected.) When, however,
Vi is negative, node b is positive with respect to node a and thus diodes D3 and D4 conduct,
making Vo = -Vi. Since, however, Vi is negative, Vo is positive. Thus, Vo is always positive and
is equal to the magnitude of Vi. That is, Vo = | Vi | .
Fig. 8.10: Diode Bridge Full-Wave Rectifier
Suppose that the input voltage is Vi = 10 sin t V. Then, if VT = 0, the input and output
voltage waveforms will be as shown in Fig. 10.2 by the dotted curve. Actually, though, for a real
bridge rectifier, the output waveform will be shifted down by 2VT because of the diode voltage
drops. This is shown by the solid curve in the bottom figure 8.11. Also, because of this shift, the
output waveform will be zero for a short time between the rectified half cycles. This type of
rectification is called full-wave rectification because both half cycles of a cycle of the input
contribute to the output.
Fig. 8.11
Now, suppose that in addition to a sinusoid, the input voltage contains a DC component,
as can be obtained from the sources shown in Fig. 8.12. What then is the output voltage Vo?
Fig. 8.12
Again, the key to understanding the circuit operation is the polarity of the voltage drop
from node a to b, which is
Vab = 10 sin t + 3 V.
When Vab is positive, diodes D1 and D2 conduct. When this voltage is negative, only diodes D3
and D4 conduct. Thus, the input and output voltage waveforms are as shown in Fig. 8.13.
Fig. 8.13
Finally, suppose in the circuit of Fig. 8.10, that a capacitor is placed across the load
resistor. What would the Vo waveform be then? The answer is that when Vi increases for the
first time to its positive peak, the capacitor would charge to this peak value minus 2VT. Then,
when Vi started to decrease, the capacitor could not discharge through the diodes because that
would require reverse diode currents. The capacitor would, however, start discharging through
the load resistor RL. Then, if the RLC time constant was much greater than the duration of a half
cycle of the sinusoidal input voltage, the capacitor voltage would not decrease as fast as the
sinusoidal voltage, and thus would reverse bias the diodes. The diodes would remain reverse
biased until the input sinusoidal voltage (minus 2VT) exceeded the capacitor voltage. In the
meantime, though, the load voltage would be the slow exponentially decaying capacitor voltage
as is shown in Fig. 8.14. Clearly, the resulting load voltage is smoother than a full-wave
rectified sinusoid.
Fig. 8.14
Construct the circuit shown in Fig. 8.15. Be careful to connect the diodes so that
they conduct in the directions indicated. Diode casings usually have a stripe at
one end to designate the end corresponding to the bar in the diode circuit symbol
(the cathode end). The 2.2-k resistor is included to limit the current to a safe
value. Apply a 2-V peak, 60-Hz sine wave and, using an oscilloscope, observe
the voltage across the a-b terminals. Examine the waveform, including amplitude
and time values.
Fig. 8.15
Place a 0.1-F capacitor across (in parallel with) RL, and reexamine the voltage waveform across the a,b nodes.
Add a 1.0 V offset voltage to the function generator. This has the
same effect as the configuration in Fig. 8.16. Using the
oscilloscope, again observe the voltage waveform across the a,b
Repeat step 2.
Fig. 8.16
Time Averages and RMS Values
In electrical circuits, one frequently encounters periodically varying current and voltage
waveforms. If the waveform x(t) repeats itself every T seconds, then
x(t) = x(t + NT)
where N is an integer and T is called the period. Figures 8.17 (a)-(d) show several periodic
Figure 8.17
(a) sinusoid, (b) triangular, (c) square, (d) half-wave rectified waveforms
The average value of a periodic waveform is defined mathematically as
Xave =  x(t)  = {0T x(t) dt}/T.
The integration represents the area under the curve during one period. Thus, it can be seen that
the average of a sinusoid is zero.
Prelab Question:
Determine the time averages of waveforms in Figures 8.17(b), (c), (d).
The Root-Mean Square of a Periodic Waveform:
We saw earlier that the time average of one of the most important periodic waveforms, the
sinusoid, is zero. However, the average itself is of little importance. Since power absorbed or
provided by any element in a circuit depends on the square of the current or voltage, average
power can be computed by knowing the time average (mean) of the square of the periodic
 x2(t)  = 0T x2(t) dt/T.
The root-mean square (RMS) value is simply the square root of the mean of the square.
Xrms =   x2(t) 
Note: AC voltmeters and ammeters measure the RMS value, not the peak value. The RMS value
is sometimes also referred to as the effective value.
Prelab Questions:
Show that the RMS value of the sinusoid x(t) = A cos(t + ) is A2.
Hint: Use trigonometric identity: 2(cos)2 = 1 + cos2
in evaluating the integral.
Determine the RMS value of the waveform shown in Figure 8.17(d).
What would be the RMS value of a full-wave rectified waveform?
Give your answer in terms of the amplitude of the input sinusoid.
Ripple Filtering:
The full-wave rectified waveform shown in the bottom curve of Figure 8.11 presents two
1. The time-average value (the DC value) of the output is much lower than the amplitude of the
input sinusoid.
2. The output oscillates around the DC value, an undesirable feature for most applications
(Imagine it running the DC motor in a portable shaver).
The full-wave rectified waveform can be considered as made up of two parts: the DC part and
the oscillating part. The oscillating part is called the ripple.
v0(t) = VDC + vripple(t).
where VDC = < vo(t) >
The ripple can be reduced (filtered) by using an appropriate capacitor connected across the load
resistor RL (refer to Figure 8.15 for the load resistor). This is explained in the text preceding
Figure 8.14. The filtered waveform is shown in Figure 8.14, along with the unfiltered waveform
in Figure 8.11. The filtered waveform is much smoother.
DC Value of the Filtered Waveform and Ripple Factor:
At t = T/4, the capacitor is fully charged (to the peak voltage of the input sinusoid, Vp).
Then, it discharges via the load resistor with a time constant  = RLC. Since the capacitor is
chosen to ensure that  >> T, the exponential decay of the capacitor voltage continues almost
until t = 3T/4. Therefore, in the interval T/4  t 3T/4, the waveform is given by
v0(t) = Vp exp{ (t  T/4)/}.
The minimum occurs at t = 3T/4 and is given by
Vmin  Vp exp(  T/2)  Vp (1  T/2)
The ripple factor (rf) and the DC voltage are given by the following equations:
rf = (Vp  Vmin)/VDC
VDC = (Vp + Vmin)/2 = Vp(1  T/4) = Vp(1  1/4fRLC)
Prelab Question:
Determine the ripple factor for f = 60 Hz, RL = 1 M, and C = 0.1 F.
For the circuit in Fig 8.15 with a capacitor added in parallel to the load resistor:
Measure the peak value and the minimum value in the rectified output.
Measure the average voltage (VDC) of the rectified output using the Oscilloscope.
Determine the ripple factor.
Compare the DC voltage and ripple factor with the respective theoretical values.