211-4SemiconductorDiodes_08

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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, well take a look at two activedevices which behave very differently depending upon what

voltages are applied to them. The diodeessentially 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 transistordiffers from the other

devices weve 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 thetransistor 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

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4-2 Semiconductor Diodes

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

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


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Semiconductor Diodes 4-3

of this sort responds to the application of an electric field with a large current flow. This is the typical

metallic behavior.

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 ofelectrons 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.


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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 anintrinsic semiconductor (a process called doping). We will use silicon as an example.

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

extraelectron to the crystal. This extra electron (often called a donorelectron) is easily excited into

the conduction band as a freely roaming current carrier.

III IV V

5B 6C 7N Group III: -1e-: acceptor p type

13Al 14Si 15P

31Ga 32Ge 33As Group V: +1e-: donor n type

49In 50Sn 51Sb

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 stealsan 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 positivecharge.

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.


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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. Thisfigure 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 theimmediate 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.


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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 chargeswhich 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


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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 wouldnt 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 diodes health would be useful for troubleshooting

purposes.

1) Your handheldDigital Multimeter contains a built in diode tester. (Note: we have noticed

the diode testers on the plug-in DMMs 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


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directions. Test a good diode and then test one from the dead diodes bin. (You need not

write down anything on this.)

Experiment 4-2: Current-Voltage curve of a diode1) 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 couldnt simply connect Ch. 1 to the top

wire (as its shown), Ch. 2 to the middle wire, and the scope ground to the bottom wire, i.e.

explain why we cant 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.


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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-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 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 kas before, and C = 10F.

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 fractionsof the p-p applied

voltage, since they are proportional to the input. How do your results for the maximum

voltage compare with expectations?

5) Alsotry 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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Experiment 4-4: Full-wave rectifier

A more efficient power supply would utilize the current on bothparts 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 10F 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.)


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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 Thevenins Equivalent Circuit theorem; the circuit is behaving as if it has

no internal resistance. Why does Thevinins 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 RLbetween 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 RLis

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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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

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