# 4. Semiconductor Diodes

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30 Οκτ 2013 (πριν από 4 χρόνια και 8 μήνες)

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4
-
1

4. Semiconductor Diodes

Introduction

So far we have looked at only so
-
called passive electronic devices: capacitors and resistors respond to
voltages applied across them by accumulating charge or passing currents, resp
ectively. In the next two
labs, we’ll take a look at two
active

devices which behave very differently depending upon what voltages
are applied to them. The
diode

essentially acts as a one
-
way switch controlled by voltage. For one
polarity of voltage, i
f the voltage across the diode is greater than a threshold value (often ~0.6 Volts), it
conducts current with essentially no resistance. If the voltage is below that value, or has the opposite
polarity, the diode acts as an open switch and conducts no cur
rent. The
transistor

differs from the other
devices we’ve considered so far in that it has three leads or connections. The voltage applied between
two of these leads controls whether current can be conducted between two others. Although the
transistor a
lso acts as a switch, it does not merely shift between fully off and fully on. Its importance lies
in the fact that a relatively low
-
power voltage supply can control the flow of a more powerful current
over a range of values. It is this switching and dec
ision
-
making property of diodes and transistors, which
makes all of modern day electronics, including computers and telecommunications, possible.

The study of semiconductors and the devices made from them falls under the category of solid state
physics.
In this experiment, we will work with one useful device: the diode. Although we will discuss
the theory briefly, you will mainly be expected to be able to understand how diodes function in circuits.

Energy Bands

The electrons of an isolated atom have d
iscrete allowed energies that we call energy levels. The Pauli
Exclusion Principle states that at most two electrons can occupy any allowed energy level. For example,
Figure 1 shows schematically the energy levels for a Lithium atom. The vertical scale
is associated with
increased energy. The dots denote electrons occupying a given level. In order to minimize energy the
electrons fill the levels from the bottom up.

Electron

Energy

Figure 1: Energy levels of an isolated a
tom

When two isolated atoms are brought close together their electric fields interact and cause a splitting
of the energy levels. Each original energy level splits into two, one going slightly higher and the other

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Semiconductor Diodes

going lower (see Figure 2). (A detaile
d explanation of the cause of this will have to be postponed until a
quantum mechanics course.) In a crystalline solid, where many atoms exist close together, this effect is
multiplied many times over. Each energy level spreads out over a small continuou
s range of energies
called an
energy band.

far apart

both close together

many atoms close together

Figure 2: Energy level splitting due to atomic interaction

If the o
riginal energy levels were spaced closely together the associated energy bands might overlap,
resulting in a larger continuous band. Sometimes, however, the energy levels are spaced far enough apart
that the bands don't overlap. This “gap” between bands
is what provides the interesting physics of
semiconductor devices. Remember that energy level diagrams represent allowed or accessible states that
an electron may occupy. A gap between bands indicates a
forbidden energy range

for electrons.

Conduction a
nd Valence Bands

Conceptually, we can imagine gradually filling up the electronic states with electrons until all of them
are accommodated, even though solids aren't really made that way. In order to minimize the energy, the
lowest states of the system f
ill first. Each band can hold 2 electrons for each atom in the crystal, since each
band is derived from one atomic energy level for each atom, and each atomic energy level can hold two
electrons (one spin up and one spin down). The
highest fully occupied

band is called the valence band
.
The next band above that, which may be partially filled, is called the conduction band. For an electron to
participate in conduction it must be able to gain energy in small amounts from the applied electric field,
i.e.
,
there must be empty levels close in energy to that occupied by the electron. Thus the valence band
electrons are immobile and cannot contribute to the conductivity, whereas electrons in the conduction
band can contribute to conduction (as the name suggest
s).

Partially Populated Conduction Band: Conductors

Figure 3 diagrams a material for which the conduction band is partly occupied by electrons. If an external
electric field is applied to this material, some of the electrons can gain a small amount of en
ergy and jump
to a higher state in the previously unoccupied section of the conduction band. Thus a material of this sort
responds to the application of an electric field with a large current flow. This is the typical metallic
behavior.

Semiconductor Diodes

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3

Figure 3: The
energy bands for a conductor

Empty Conduction Band: Insulators

If the conduction band is completely empty (Figure 4) the material acts like an electrical insulator.
Electrons in the valence band cannot gain enough energy to jump over the forbidden regi
on into the
conduction band. In addition, the electrons in the valence band cannot move through the solid to create a
current because there are no empty states in the valence band for a traveling electron to occupy. Since no
current can flow, the materia
l is an insulator.

Figure 4: The energy bands for an insulator

Semiconductors

Semiconductors are a special case of insulators in which the forbidden region between the valence
and conduction bands is relatively small (about 0.5 to 1.5 eV). In this ca
se an extremely small number of
electrons are excited across the energy gap by thermal excitation and occupy states in the conduction
band. These excited electrons can respond to the applied fields but since their number is comparatively
few, the material

as a whole is not a good conductor, so we call it a semiconductor. The conductivity of a
semiconductor is very sensitive to temperature since it depends on thermally excited electrons.

Doping

In general, the amount of current that a semiconductor can c
arry is not enough to make a useful
device. Most commercial semiconductors are made by introducing small amounts of impurities to an
intrinsic semiconductor (a process called doping). We will use silicon as an example.

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Semiconductor Diodes

Silicon (Si) is an intrinsic semic
onductor, but in its natural state it conducts very poorly. Silicon is a
group IV element on the periodic chart and has four electrons in its outermost shell. When silicon is
doped with arsenic (As), a group V element, the arsenic atoms replace silicon a
toms at a small number of
points on the crystal lattice. Since arsenic has 5 electrons in its outer shell, it adds a loosely
-
bound
“extra”electron to the crystal. This extra electron (often called a “donor”electron) is easily excited into the
conduction
band as a freely roaming current carrier.

III

IV

V

5
B

6
C

7
N

Group III:
-
1e
-
: acceptor Ñ p type

13
Al

14
Si

15
P

31
Ga

32
Ge

33
As

Group V: +1e
-
: donor Ñ n type

49
In

50
Sn

51
Sb

Figure 5: Section of Periodic Table

Silicon can also be doped with

an element from group III of the periodic table, such as gallium (Ga).
In this case the impurity has only three electrons in its outermost shell so there is a deficiency of one
electron at every point where a gallium atom replaces a silicon atom. This i
s called an “acceptor site”
since the gallium would very much like to have a fourth electron to complete its bonds. The gallium often
“steals”an electron from a neighboring silicon atom leaving a “hole” or empty state in the valence band of
the silicon.
This “hole” is free to roam around in the valence band and effectively acts as a positive charge
carrier.

Holes move through the crystal lattice in the same way spaces between cars move in a traffic jam: the
cars (electrons) move forward to fill up the s
paces (holes) in front of them, only to create another space
behind them. The holes move in the direction opposite the electrons, hence the “effective” positive
charge.

Doping with group V elements results in an n
-
type semiconductor since the charge carr
iers (electrons)
are negative. On the other hand, doping with group III elements produces a p
-
type semiconductor
because the effective charge carriers (holes) are positive.

Semiconductor Diodes

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5

Diodes

Figure 6a: distribution of
mobile

charge car
riers in a p
-
n junction before (spontaneous) charge diffusion

Figure 6b: distribution of
excess

charge in a p
-
n junction before (spontaneous) charge diffusion. This
figure shows that each side starts out electrically neutral
, because the negative cha
rges due to electrons
are exactly balanced by the positively charged nuclei in both the n
-

and p
-
type regions.

Diodes are formed by producing a piece of semiconductor that is p
-
type at one end and n
-
type at the
other such as Figure 6a. Although electron
s and holes are free to roam in each section, the material is
electrically neutral.

However, in less than a nanosecond, some of the free electrons will diffuse into the p
-
type, and an
equal number of holes will diffuse into the n
-
type. As the electrons a
nd holes diffuse across the junction
they recombine and “'eliminate” each other. The end result is a lack of mobile charge carriers in the
immediate vicinity of the junction (see Figure 7a). In addition, the region of either side of the junction is
no
longer electrically neutral so a “built
-
in” electric field is established, as shown in Figure 7b. Since the
junction region in Figure 7a is devoid of free charge carriers, it will have a low conductivity and high
resistance. Another way to see this is th
at the built
-
in electric field opposes the motion of any holes from
the p
-
type side which are trying to move to the right and any electrons from the n
-
type side which are
trying to move to the left.

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Semiconductor Diodes

Figure 7a: Diffusion of mobile charge in a p
-
n junct
ion after charge diffusion.

Figure 7b: Diffusion of excess charge in a p
-
n junction after charge diffusion. The negative charges which
have diffused out of the n
-
type side into the p
-
type side leave a net negative charge there, and vice
-
versa.
This
charge distribution causes the “built
-
in” electric field shown near the junction.

Reverse Biased Junction

Figure 8 shows a p
-
n junction attached to a voltage source such that the positive terminal is connected
to the n
-
type semiconductor. The electric f
ield due to the applied voltage source

to the built
-
in field.
(Note that the conductivity of the p
-

and n
-
type regions away from the junction is greater than that of the
junction region so the potential varies mainly in the proximity of the junction.
) Hence, the addition of the
second field even further opposes the motion of any holes from the p
-
type side which are trying to move
to the right and any electrons from the n
-
type side which are trying to move to the left.

So, very little current flows.

Figure 8: Distribution of mobile charges in a reverse biased diode

Semiconductor Diodes

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7

Forward Biased Junction

Suppose, on the other hand, that the voltage supply was connected with the positive terminal wired
to the p
-
type semiconductor. The electric field due to the vo
ltage source will now be in the direction
opposite to the built
-
in field. Now, the opposition is reduced holes from the p
-
type side which are trying
to move to the right and electrons from the n
-
type side which are trying to move to the left. Since these

are the directions that the applied voltage is trying to push these charges, current can flow fairly easily, at
least once the applied voltage is big enough to mostly overcome the opposition from the built
-
in field.
Thus the p
-
n junction provides an inte
resting device that conducts current in only one direction.

In fact, if you literally connected a battery in the forward
-
bias direction directly across a diode, so
much current would flow that the diode would burn out! To avoid this, you would need to ad
d a
“current
-
limiting resistor”, as shown below.

Figure 9: Circuit diagram of a forward biased diode, with resistor added to limit the current.

Experimental Procedure

Experiment 4
-
1: Diode Tester

Sometimes, despite precautio
ns, too much current passes through a diode and it "burns out". This
"burn out" is not very flashy; in fact, you wouldn’t know that anything happened except for the fact that
your circuit would malfunction. The diode would look the same as before and you

would be wondering
what went wrong. Obviously, a quick test of a diode’s health would be useful for troubleshooting
purposes.

1)

Your
handheld

Digital Multimeter contains a built in diode tester.
(Note: we have noticed the
diode testers on the plug
-
in

DMM’s sometimes do not work!)
Rotate the function switch to
the diode symbol and connect the red lead and black leads in the forward bias direction
across the diode. The DMM is now applying a current of a few mA through the diode, and
displaying the vol
tage needed to reach this current. For a 'healthy' diode, it should read
about 600mV, indicating that only a modest voltage is needed to make current flow in the
forward direction. A burnt out diode will either read 0.0 mV or the open circuit voltage
(“
OL”). When the diode is connected reversed biased the meter will read the open circuit
voltage. To fully test the diode, you need to check both the forward and reverse bias
directions. Test a good diode and then test one from the “dead diodes” bin. (Yo
u need not
write down anything on this.)

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Semiconductor Diodes

Experiment 4
-
2: Current
-
Voltage curve of a diode

1)

Build the circuit in Figure 10 to plot the
I
-
V

(current versus voltage) curve for a diode using
the X
-
Y mode on the scope.

Pre
-
lab question 1:

Calculate the v
oltage across the resistor which corresponds to a current of
50mA (the maximum current rating of the diode).

TURN UP THE GENERATOR VOLTAGE SLOWLY

so that you do not exceed this value (on
the Y
-
axis of the scope) or operate for longer than a second at t
his value.

2)

Use Channel 2 to measure the voltage across the resistor as an indirect method to obtain the
current. Note where the ground of the circuit is located. Set the scope to DC mode. Because
channel 2 is connected in the direction opposite to c
hannel 1, you should press the CH2
INVERT button on your oscilloscope.

Figure 10: Circuit to display
I
-
V

diagram of diode.

Pre
-
lab question 2:

We want Ch. 1 to display the voltage across the diode and Ch. 2 to display
the
voltage across the resistor. Explain why we couldn’t simply connect Ch. 1 to the top wire
(as it’s shown), Ch. 2 to the middle wire, and the scope ground to the bottom wire, i.e. explain
why we can’t interchange the connections for scope ground and Ch. 2
.

3)

Set the function generator to produce a triangle wave. Put the oscilloscope in X
-
the instructor to check your I
-
V curve and if necessary help you to adjust the gains for a good
display.

Semiconductor Diodes

4
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9

At about what voltage does the diode start to

conduct?

The diode maintains a nearly
constant forward voltage (V
f
) for a wide range of forward currents once this voltage is
exceeded.

Sketch the I
-
V curve you observe in your lab notebook.

Make sure you label axes and give
units.

How much re
verse current is there,
e.g.

at
-
5 V?

To answer this question, you may need to
change to a larger resister,
e.g.

10k, since the reverse current is small.
How effective is your
diode at only letting current pass in one direction?

Experiment 4
-
3: Half
-
w
ave rectifier

You will now use a function generator and a diode to build a half
-
wave rectifier (circuit diagrams in
Figure 11) to eliminate the negative part of an oscillatory signal.

1)

Assemble the circuit in Figure 11a, with
R

=

2 k

, and an input
signal of about 10 V peak
-
to
-
peak (p
-
p). Monitor the voltage across the resistor with the oscilloscope. (It is best when
possible to connect the
ground

lead of the scope to the
negative

side of the generator.)

2)

Explain how the AC signal from the func
tion generator is “rectified” by the diode.

3)

Now put a low pass filter on the output

of your circuit by adding a capacitor, as shown in
Figure 11b. This converts your AC signal to a DC voltage with some "ripple" remaining (if
the frequency isn't too
low). We suggest that you let R=2 k

as before, and
C

=

10

F.

4)

Measure the
maximum and the mean

value of the voltage, and the p
-
p amplitude of the
ripple

at a frequency of 60 Hz. It is best to express these as
fractions

of the p
-
p applied
voltage, s
ince they are proportional to the input.
How do your results for the maximum
voltage compare with expectations?

5)

Also

try a smaller capacitor and note the amplitude of the ripple in that case as well.
Is the
filter behaving as you would expect?

Explain.

You have successfully built a DC power
supply from an AC source. Congrtatulations!

(a)

(b)

Figure 11: Half wave rectifier

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Semiconductor Diodes

Experiment 4
-
4
: Full
-
wave rectifier

A more efficient power supply wou
ld utilize the current on
both

parts of the cycle. To do that, build
the full
-
wave rectifier shown in Figure 12. The AC signal is now fully "rectified".

1)

Explain how this circuit works by tracing the current paths (a) when the applied voltage is
pos
itive and (b) when it is negative.

2)

Build the circuit with a 10

F capacitor in parallel with the resistor to eliminate most of the AC
ripple, leaving a nearly constant DC voltage. Compare the mean voltage for this circuit
(again expressed as a fraction of the p
-
p input) to what you found for the half
-
wave rectifier.

Do you see why this circuit is a better DC power supply?

Figure 12: Full wave rectifier

Experiment 4
-
5: To be
demonstrated

in lab

Experiment 4
-
6: Zener diode (OPTIONAL)

A Zener diode behaves like a norma
l diode when it is forward biased. However, when reversed
biased, the Zener will start to conduct current once a specified voltage (Zener Voltage,
V
z
) is exceeded,
but the voltage across the diode will not change significantly. The constancy of the rever
se voltage is the
reason why the device is useful, as we shall see. (Note: All diodes will break down and start to conduct if
the reverse bias voltage is high, but for the Zener this process is repeatable and does not harm the diode
provided the maximum
power rating is not exceeded.)

1)

Using the same method as in experiment 4
-
2, plot the I
-
V curve of the zener diode
.
However, before turning on the power CALCULATE THE MAXIMUM ALLOWED
CURRENT for the circuit, given that the maximum
power

that can be d
issipated in the Zener
is 400mW. (To do this, remember that
P
=
I V
.)

Semiconductor Diodes

4
-
11

2)

Record the value at which the zener diode begins to conduct in the reverse
-
biased direction.

Experiment 4
-
7: Zener Voltage regulator circuit (OPTIONAL)

The circuit in Figure 13 t
akes advantage of the reverse breakdown voltage of the Zener diode to form
a regulated voltage supply. The Zener diode maintains a constant reverse
-
biased voltage for a wide
range of currents. The result is an output voltage that remains constant as the
delivered current varies.
Obviously, this “contradicts” Thevenin’s Equivalent Circuit theorem; the circuit is behaving as if it has no
internal resistance. Why does Thevinin’s Theorem not apply to this circuit?

Figure 13: Ze
ner regulator

1)

Construct the circuit in Figure 13. Before turning on the DC voltage, you need to consider
whether this circuit will keep you within the 1/4 W limit for the resistors and the 400mW
limit for the Zener. To do this, suppose that the pot
resistance is high so it doesn't draw any
current. Then all the current goes through the Zener. How much will that current be (given
the known Zener voltage for the diode you have been given? How much power will be
dissipated in the resistor and the Zen
er? To avoid any risk to the pot, don't set it for less than
100 ohms.

2)

Test the circuit

by measuring
V
L

(and hence
I
L
) for a few (e.g. 5) values of
R
L

between 100
ohms and 1000 ohms. You'll have to disconnect the pot when you adjust it to check its
resistance with the DMM. Over what range of load currents does this circuit maintain a
constant voltage output?

3)

Make sure that you understand how the
currents are changing to keep
V
L
constant as
R
L

is
varied. Ask your instructor if you're unsure. Congratulations! You have created a DC to DC
converter, a device that can be used to provide a constant DC voltage to a circuit independent
of variations in

the input voltage or the load.

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-
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Semiconductor Diodes

Experiment 4
-
8 (OPTIONAL, TO READ ONLY): Regulated power supply

The full
-
wave rectifier from part 4
-
4 and the Zener regulator from part 4
-
6 can be used to build a
regulated power supply Figure 15. This power supply shoul
d deliver a constant voltage (
V
out

=

V
z
) over a
wide range of output currents.

Figure 15: Regulated Power Supply