ELEC1 CH2 DIODES

Électronics 1 - Chapitre 2
DIODES
1. UNDERSTANDING DIODES
MATERIALS
Silicon and germanium are semiconductor materials
used in the manufacture of diodes, transistors, and
integrated circuits.
DOPING
Semiconductor material is refined to an extreme level of
purity, and then minute, controlled amounts of a specific
impurity are added (a process called doping). Based on
which impurity is added to a region of a semiconductor
crystal, that region is said to be N type or P type.
In addition to electrons (which are negative charge carriers
used to conduct charge in a conventional conductor),
semiconductors contain positive charge carriers called
holes.
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DOPING (continued)
The impurities added to an N type region increases the
number of electrons capable of conducting charge,
whereas the impurities added to a P type region increase
the number of holes capable of conducting charge.
DIODE JUNCTION
When a semiconductor chip contains an N doped region
adjacent to a P doped region, a diode junction (often called
a PN junction) is formed.
Diode junctions can also be made with either silicon or
germanium. However, silicon and germanium are never
mixed when making PN junctions.
QUESTION
Which diagrams in Figure 2.1 show diode junctions?
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FIGURE 2.1
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2. ANODE / CATHODE
In a diode, the P material is called the
The N material is called the cathode : Figure 2.2
anode.
CURRENT FLOW
Diodes are useful because electric current can flow through
a PN junction in one direction only. Figure 2.3 shows the
direction in which the current flows.
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DIODE SYMBOL
Figure 2.4 shows the circuit symbol for a diode. The arrowhead
points in the direction of current flow. Although the anode and
cathode are indicated here, they are not usually indicated in
circuit diagrams. In a diode, current flows from anode to
cathode
In the circuit shown in Figure 2.5, an arrow shows the direction
of current flow.
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CONNECT THE DIODE IN A CIRCUIT
Forward-biased
In the circuit shown in Figure 2.5, The anode connects to the
positive battery terminal, and the cathode connects to the
negative battery terminal. Therefore, the anode is at a
higher voltage than the cathode.
When the diode is connected so that the current flows, it
is forward-biased. In a forward biased diode, the anode
connects to a higher voltage than the cathode, and current
flows.
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Reverse-biased
In the circuit shown in Figure 2.6, The cathode is connected to
the positive battery terminal, and the anode is connected to the
negative battery terminal. Therefore, the cathode is at a higher
voltage than the anode.
When the cathode is connected to a higher voltage level than
the anode, the diode cannot conduct. In this case, the diode is
reverse-biased.
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Perfect diode
In many circuits, the diode is often considered to be a perfect
diode to simplify calculations.
A perfect diode has zero voltage drop in the forward direction
and conducts no current in the reverse direction.
A forward-biased perfect diode can
be compared to a
closed switch, Figure 2.7 (2). It has no voltage drop across its
terminals, and current flows through it.
A reverse-biased perfect diode can be compared to an
open switch, Figure 2.7 (1). No current flows through it, and the
voltage difference between its terminals equals the supply
voltage.
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PROJECT 2 : DIODE
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OBJECTIVE
The objective of this project is to plot the V-I curve
(also called a characteristic curve) of a diode,
which shows how current flow through the diode
varies with the applied voltage.
As shown in Figure 2.8, the I-V curve for a diode
demonstrates that if low voltage is applied to a
diode, current does not flow. However, when the
applied voltage exceeds a certain value, the
current flow increases quickly.
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FIGURE 2.8 : the I-V curve for a diode
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GENERAL INSTRUCTIONS
While the circuit is set up, measure the current for
each voltage value. As you perform the project,
observe how much more rapidly the current
increases for higher applied voltages.
●
PARTS LIST
❑ One 9 V battery
❑ One snap battery connector
❑ One multimeter set to measure current (mA)
❑ One multimeter set to measure DC voltage
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PARTS LIST (continued)
❑ One 330-ohm, 0.5-watt resistor
❑ One 1N4001 diode
❑ One breadboard
❑ One 1 MΩ potentiometer
❑ One terminal block
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STEP-BY-STEP INSTRUCTIONS
Set up the circuit as shown in Figure 2.9. The
circled “A” designates a multimeter set to
measure current, and the circled “V” designates
a multimeter set to measure DC voltage. If you
have some experience in building circuits, this
schematic (along with the previous parts list)
should provide all the information you need to
build the circuit. If you need a bit more help in
building the circuit, look at the photos of the
completed circuit in the “Expected Results”
section.
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FIGURE 2.9
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Carefully check your circuit against Figure 2.9,
especially the direction of the battery and the diode.
The diode has a band at one end. Connect the lead at
the end of the diode with the band to the ground bus
on the breadboard.
After you check your circuit, follow these steps, and
record your measurements in the blank table following
the steps:
1. Set the potentiometer to its highest value. This sets the
voltage applied to the diode to its lowest possible value.
2. Measure and record the voltage applied to the diode.
3. Measure and record the current.
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4. Adjust the potentiometer slightly to give a higher voltage.
5. Measure and record the new values of voltage and current.
6. Repeat steps 4 and 5 until the lowest resistance of the
potentiometer is reached, taking as many readings as possible.
This results in the highest voltage and current readings for this
circuit. At this point, the potentiometer resistance is zero ohms,
and the current is limited to approximately 27 mA by the 330ohm resistor.
This resistor is included in the circuit to avoid overheating the
components when the potentiometer is set to zero ohms. If your
circuit allows currents significantly above this level as you adjust
the potentiometer, something is wrong. You should disconnect
the battery and examine the circuit to see if it were connected
incorrectly. If V gets large—above 3 or 4 volts—and I remains
small, then the diode is backward. Reverse it and start again.
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FIGURE 2.10
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7. Graph the points recorded in the table using the blank
graph shown in Figure 2.11. Your curve should look like the
one shown in Figure 2.8.
FIGURE 2.11
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●
EXPECTED RESULTS
Figure 2.12 shows the breadboarded circuit for this
project.
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FIGURE 2.12
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Figure 2.13 shows the test setup for this project.
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CHAPTER 2 (continued)
The V-I curve (or diode characteristic curve) is repeated in
Figure 2.14 with three important regions marked on it.
The most important region is the knee region. This is not a
sharply defined changeover point, but it occupies a narrow
range of the curve where the diode resistance changes
from high to low.
The ideal curve is shown for comparison.
For the diode used in this project, the knee voltage is about
0.7 volt, which is typical for a silicon diode. This means (and
your data should verify this) that at voltage levels below 0.7
volt, the diode has such a high resistance that it limits the
current flow to a low value. This characteristic knee voltage is
sometimes referred to as a threshold voltage.
If you use a germanium diode, the knee voltage is
approximately 0.3 volt.
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Figure 2.14
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NOTE
For any given diode, the knee voltage is not exactly 0.7
volt or 0.3 volt. Rather, it varies slightly. But when using
diodes in practice (that is, imperfect diodes), you can
make two assumptions :
■ The voltage drop across the diode is either 0.7 volt or
0.3 volt.
■ You can prevent excessive current from flowing through
the diode by using the appropriate resistor in series with
the diode.
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Current through the diode
Calculate the current through the diode in the circuit shown
in Figure 2.15 using the steps in the following questions.
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QUESTIONS
A. The voltage drop across the diode is known. It is 0.7 volt for
silicon and 0.3 volt for germanium. (“Si” near the diode means it
is silicon.) Write down the diode voltage drop.
VD =
B. Find the voltage drop across the resistor. This can be
calculated using VR = VS - VD. This is taken from KVL.
VR =
C. Calculate the current through the resistor. Use I = VR / R.
I=
D. Finally, determine the current through the diode.
ID =
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QUESTIONS
A. The voltage drop across the diode is known. It is 0.7 volt for
silicon and 0.3 volt for germanium. (“Si” near the diode means it
is silicon.) Write down the diode voltage drop.
VD =
B. Find the voltage drop across the resistor. This can be
calculated using VR = VS - VD. This is taken from KVL.
VR =
C. Calculate the current through the resistor. Use I = VR / R.
I=
D. Finally, determine the current through the diode.
ID =
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NOTE
In practice, when the battery voltage is 10 volt or
above, the voltage drop across the diode is often
considered to be 0 volt, instead of 0.7 volt.
The assumption here is that the diode is a perfect
diode, and the knee voltage is at 0 volts, rather than
at a threshold value that must be exceeded. As
discussed later, this assumption is often used in
many electronic designs.
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POWER DISSIPATION
When a current flows through a diode, it causes heating and
power dissipation, just as with a resistor. The power formula
for resistors is P = V x I. This same formula can be applied
to diodes to find the power dissipation.
To calculate the power dissipation in a diode, you must first
calculate the current as shown previously. The voltage drop
in this formula is assumed to be 0.7 volt for a silicon diode,
even if you considered it to be 0 volts when calculating the
current.
For example, a silicon diode has 100 mA flowing through it.
Determine how much power the diode dissipates.
P = (0.7 volt)(100 mA) = 570 mW
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NOTE
Diodes are made to dissipate a certain amount of power, and
this is quoted as a maximum power rating in the manufacturer’s
specifications of the diode.
Assume a silicon diode has a maximum power rating of
2 watts. How much current can it safely pass?
Provided the current in the circuit does not exceed this, the
diode cannot overheat and burn out.
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QUESTIONS
Look at the circuit shown in Figure 2.16. To find ID, you need to
go through the following steps because there is no way to find ID
directly :
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DIODE BREAKDOWN
If you place the diode in the circuit backward—as shown on the
right in Figure 2.17—then almost no current flows. In fact, the
current flow is so small, it can be said that no current flows. The
V-I curve for a reversed diode looks like the one shown on the
left in Figure 2.17.
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The V-I curve for a perfect diode would show zero current for all
voltage values. But for a real diode, a voltage is reached where the
diode “breaks down” and the diode allows a large current to flow. The
V-I curve for the diode breakdown would then look like the one in
Figure 2.18.
If this condition continues, the diode will burn out. You can avoid
burning out the diode, even though it is at the breakdown voltage, by
limiting the current with a resistor.
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QUESTION
The diode in the circuit shown in Figure 2.19 is known to break
down at 100 volts, and it can safely pass 1 ampere without
overheating. Find the resistance in this circuit that would limit
the current to 1 ampere.
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ANSWER
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Breakdown voltage
All diodes break down when connected in the reverse direction if
excess voltage is applied to them. The breakdown voltage
(which is a function of how the diode is made) varies from one
type of diode to another. This voltage is quoted in the
manufacturer’s data sheet.
Breakdown is not a catastrophic process and does not destroy
the diode. If the excessive supply voltage is removed, the diode
can recover and operate normally. You can use it safely many
more times, provided the current is limited to prevent the diode
from burning out.
A diode always breaks down at the same voltage, no matter how
many times it is used.
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The breakdown voltage is often called the peak inverse
voltage (PIV) or the peak reverse voltage (PRV). Following
are the PIVs of some common diodes :
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INSIDE THE DIODE
At the junction of the N and P type regions of a diode, electrons from
the N region are trapped by holes in the P region, forming a depletion
region as illustrated in the following figure.
When the electrons are trapped by holes, they can no longer move,
which is what produces this depletion region that contains no mobile
electrons. The static electrons give this region the properties of an
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electrical insulator.
Forward-biased
You apply a forward-biased voltage to a diode by applying negative
voltage to the N region and positive voltage to the P region.
Electrons in the N region are repelled by the negative voltage, pushing
more electrons into the depletion region. However, the electrons in the
N region are also repelled by the electrons already in the depletion
region. (Remember that like charges repel each other.)
When the forward-biased voltage is sufficiently high (0.7 volts for
silicon diodes, and 0.3 volts for germanium diodes), the depletion
region is eliminated.
As the voltage is raised further, the negative voltage, in repelling
electrons in the N region, pushes them into the P region, where they
are attracted by the positive voltage.
This combination of repulsive and attractive forces allows current to
flow, as illustrated in the following figure.
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This combination of repulsive and attractive forces
allows current to flow, as illustrated in the following
figure.
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Note that an interesting aspect of diodes is that while the N
and P regions of the diode have mobile charges, they do not
have a net charge.
The mobile charges (electrons in N regions and holes in P
regions) are simply donated by the impurity atoms used to
dope the semiconductor material.
The impurity atoms have more electrons in the N region and
less in the P region than are needed to bond to the adjacent
silicon or germanium atoms in the crystalline structure.
However, because the atoms have the same number of
positive and negative charges, the net charge is neutral.
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In a similar way, metal is a conductor because it has more
electrons than are needed to bond together the atoms in its
crystalline structure. These excess electrons are free to
move when a voltage is applied, either in a metal or an N
type semiconductor, which makes the material electrically
conductive.
Electrons moving through the P region of the diode jump
between the holes, which physicists model as the holes
moving. In a depletion region, where electrons have been
trapped by holes, there is a net negative charge.
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Reverse-biased
You apply a reverse-biased voltage to a diode by
applying positive voltage to the N type region and
negative voltage to the P type region.
In this scenario, electrons are attracted to the
positive voltage, which pulls them away from the
depletion region.
No current can flow unless the voltage exceeds the
reverse breakdown voltage.
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Diode symbol
Diodes are marked with one band indicating the cathode (the N
region) end of the diode. The band on the diode corresponds to
the bar on the diode symbol, as shown in the following figure.
Use this marking to orient diodes in a circuit.
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Part number
The part number is marked on diodes. However, the part
numbers don’t tell you much about the diode.
For example, the 1N4001 is a silicon diode that can handle
a peak reverse voltage of 50 volts, whereas the 1N270 is a
germanium diode that can handle a peak reverse voltage of
80 volts.
Your best bet is to refer to the manufacturer’s data sheet for
the diode peak reverse voltage and other characteristics.
You can easily look up data sheets on the Internet. Also,
you can find links to the data sheets for the components
used on the website at :
www.buildinggadgets.com/index_datasheets.htm.
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3. THE ZENER DIODE
Diodes can be manufactured so that breakdown occurs at
lower and more precise voltages than those just discussed.
These types of diodes are called zener diodes, so named
because they exhibit the “Zener effect”—a particular form of
voltage breakdown.
At the zener voltage, a small current flows through the
zener diode.
This current must be maintained to keep the diode at the
zener point. In most cases, a few milliamperes are all that
is required.
Figure 2.20 shows the zener diode symbol and a simple
circuit.
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FIGURE 2.20
In this circuit, the battery determines the applied voltage. The
zener diode determines the voltage drop (labeled Vz) across it.
The resistor determines the current flow. Zeners are used to
maintain a constant voltage at some point in a circuit.
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APPLICATION
Examine an application in which a constant voltage is
wanted—for example, a lamp driven by a DC generator.
In this example, when the generator turns at full speed, it
puts out 50 volts. When it runs more slowly, the voltage can
drop to 35 volts.
You want to illuminate a 20-volt lamp with this generator.
Assume that the lamp draws 1.5 amperes.
Figure 2.21 shows the circuit.
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FIGURE 2.21
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A)
You need to determine a suitable value for the resistance.
Follow these steps to find a suitable resistance value:
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B) Assume now that the 20-ohm resistor calculated is in place,
and the voltage output of the generator drops to 35 volts, as
shown in Figure 2.21. This is similar to what happens when a
battery gets old. Its voltage level decays and it will no longer
have sufficient voltage to produce the proper current. This
results in the lamp glowing less brightly, or perhaps not at all.
Note, however, that the resistance of the lamp stays the same.
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QUESTIONS
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ANSWER
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C) In many applications, a lowering of voltage across the
lamp (or some other component) may be unacceptable. You
can prevent this by using a zener diode, as shown in the
circuit in Figure 2.22.
If you choose a 20-volt zener (that is, one that has a 20-volt drop
across it), then the lamp always has 20 volts across it, no matter
what the output voltage is from the generator (provided, of
course, that the output from the generator is always above 20
volts).
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QUESTIONS
Say that the voltage across the lamp is constant, and the
generator output drops.
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ZENER ACTION
To make this circuit work and keep 20 volts across the lamp
at all times, you must find a suitable value of R.
This value should allow sufficient total current to flow to
provide 1.5 amperes required by the lamp, and the small
amount required to keep the diode at its zener voltage.
To do this, you start at the “worst case” condition. (“Worst
case” design is a common practice in electronics. It is used
to ensure that equipment can work under the most adverse
conditions.)
The worst case here would occur when the generator puts
out only 35 volts.
Figure 2.23 shows the paths that current would take in this
circuit.
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Figure 2.23 shows the paths that current would take in this
circuit.
Find the value of R that enables 1.5 amperes to flow through the
lamp. How much current can flow through the zener diode? You
can choose any current you like, provided it is above a few
milliamperes, and provided it does not cause the zener diode to
burn out. In this example, assume that the zener current IZ is 0.5
amperes.
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QUESTIONS
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ZENER ACTION (continued)
Now, take a look at what happens when the generator
supplies 50 volts, as shown in Figure 2.24.
Because the lamp still has 20 volts across it, it can still draw only
1.5 amperes. But the total current and the zener current change.
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QUESTIONS
Although the lamp voltage and current remain the same, the
total current and the zener current both changed.
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QUESTIONS
The increase in IT flows through the zener diode and
not through the lamp.
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POWER DISSIPATED
The power dissipated by the zener diode changes as the
generator voltage changes.
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PROJECT 3 :
The Zener Diode Voltage Regulator
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OBJECTIVE
The objective of this project is to measure the
voltage applied to the lamp, and the current
through the lamp for different supply voltages,
demonstrating the use of a zener diode to
provide a steady voltage and current to a
lamp when the supply voltage changes.
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OBJECTIVE
The objective of this project is to measure the
voltage applied to the lamp, and the current
through the lamp for different supply voltages,
demonstrating the use of a zener diode to
provide a steady voltage and current to a lamp
when the supply voltage changes.
GENERAL INSTRUCTIONS
While the circuit is set up, measure the
lamp current, zener diode current, and
supply voltage as the voltage from the
9-volt battery drops.
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PARTS LIST
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STEP-BY-STEP INSTRUCTIONS
Set up the circuit shown on Figure 2.25. The
circled “A” designates a multimeter set to
measure current, and the circled “V” designates
a multimeter set to measure DC voltage. If you
have some experience in building circuits, this
schematic (along with the previous parts list)
should provide all the information you need to
build the circuit. If you need a bit more help in
building the circuit, look at the photos of the
completed circuit in the “Expected Results”
section.
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STEP-BY-STEP INSTRUCTIONS (continued)
FIGURE 2.25
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STEP-BY-STEP INSTRUCTIONS (continued)
Carefully check your circuit against the diagram, especially
the direction of the battery and the diode. The diode has a
band at one end. Connect the lead at the end of the diode
without the band to the ground bus on the breadboard.
After you check your circuit, follow these steps, and record
your measurements in the blank table following the steps:
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STEP-BY-STEP INSTRUCTIONS (continued)
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EXPECTED RESULTS
Figure 2.26 shows the breadboarded circuit for this project.
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Figure 2.27 shows the test setup for this project.
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Complete your measurements like the ones shown in the
following table.
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NOTE
You must see in the recorded datas that,
even though the supply voltage dropped by
approximately 15 percent, the lamp current
stayed roughly constant, showing less than
a 1 percent drop.
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