Semiconductor Diodes4-1
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, 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.
ElectronEnergy /
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 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.
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.
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 / V5B / 6C / 7N / Group III: -1e-: acceptor Ñ p type
13Al / 14Si / 15P
31Ga / 32Ge / 33As / Group V: +1e-: donor Ñ 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 “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.
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.
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
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.)
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.
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.
• 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 k as 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 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
Experiment 4-4: Full-wave rectifier