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, respectively. 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, if 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 current. 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 also 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 decision-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 discrete 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 atom
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
4-1
4-2
Semiconductor Diodes
going lower (see Figure 2). (A detailed 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 continuous 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 original 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 and 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 fill 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 suggests).
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 energy 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
4-3
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 region 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 material 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 case 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 carry 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.
4-4
Semiconductor Diodes
Silicon (Si) is an intrinsic semiconductor, 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 atoms 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
5B
6C
7N
13Al
14Si
15P
31Ga
32Ge
33As
49In
50Sn
51Sb
Group III: -1e-: acceptor Ñ p type
Group V: +1e-: donor Ñ n type
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 is 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 spaces (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 carriers (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
4-5
Diodes
Figure 6a: distribution of mobile charge carriers 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 charges 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 electrons 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 and 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 that 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.
4-6
Semiconductor Diodes
Figure 7a: Diffusion of mobile charge in a p-n junction 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 field due to the applied voltage source adds 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
4-7
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 voltage 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 interesting 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 add 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 precautions, 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 voltage 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. (You need not
write down anything on this.)
4-8
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 voltage 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 this 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 channel 1, you should press the CH2
INVERT button on your oscilloscope.
R = 100 
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-Y mode. Ask
the instructor to check your I-V curve and if necessary help you to adjust the gains for a good
display.
Semiconductor Diodes
4-9
• At about what voltage does the diode start to conduct? The diode maintains a nearly
constant forward voltage (Vf) 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 reverse 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-wave 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-topeak (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 function 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, since 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
4-10
Semiconductor Diodes
Experiment 4-4: Full-wave rectifier
A more efficient power supply would 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
positive 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 – Building a radio
Experiment 4-6: Zener diode (OPTIONAL)
A Zener diode behaves like a normal diode when it is forward biased. However, when reversed
biased, the Zener will start to conduct current once a specified voltage (Zener Voltage, Vz) is exceeded,
but the voltage across the diode will not change significantly. The constancy of the reverse 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 dissipated 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 takes 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: Zener 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 Zener? To avoid any risk to the pot, don't set it for less than
100 ohms.
2)
Test the circuit by measuring VL (and hence IL) for a few (e.g. 5) values of RL 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 VL constant as RL 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.
4-12
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 should deliver a constant voltage (Vout = Vz) over a
wide range of output currents.
Figure 15: Regulated Power Supply