by Earl Boysen
Define diode breakdown.
Differentiate between zeners and other diodes.
Determine when a diode can be considered “perfect.”
Understanding Diodes
1 Silicon and germanium are semiconductor materials used in the manufacture of diodes, transistors, and integrated circuits. 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. 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.
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?_____
Figure 2.1
Answer
Diagrams (b) and (e) only
2 In a diode, the P material is called the anode. The N material is called the cathode.
Question
Identify which part of the diode shown in Figure 2.2 is P material and which part is N material._____
Figure 2.2
Answer
The anode is P material; the cathode is N material.
3 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.
Figure 2.3
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.
Figure 2.4
Question
In a diode, does current flow from anode to cathode, or cathode to anode?_____
Answer
Current flows from anode to cathode.
4 In the circuit shown in Figure 2.5, an arrow shows the direction of current flow.
Figure 2.5
Questions
A. Is the diode connected correctly to permit current to flow?_____
B. Notice the way the battery and the diode connect. Is the anode at a higher or lower voltage than the cathode?_____
Answers
A. Yes.
B. 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.
5 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. Examine the way the diode is connected to the battery shown in Figure 2.6.
Figure 2.6
Question
Is the diode forward-biased? Give the reasons for your answer._____
Answer
No, it is not forward-biased. 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.
6 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.
Question
Draw a reverse-biased diode in the circuit shown in Figure 2.7._____
Figure 2.7
Answer
Your drawing should look something like Figure 2.8.
Figure 2.8
7 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.
Question
From your knowledge of basic electricity, what other component has zero voltage drop across its terminals in one condition and conducts no current in an alternative condition?_____
Answer
The mechanical switch. When closed, it has no voltage drop across its terminals, and when open, it conducts no current.
8 A forward-biased perfect diode can thus be compared to a closed switch. 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. No current flows through it, and the voltage difference between its terminals equals the supply voltage.
Question
Which of the switches shown in Figure 2.9 performs like a forward-biased perfect diode?_____
Figure 2.9
Answers
Switch (2) represents a closed switch and, like a forward-biased perfect diode, allows current to flow through it. There is no voltage drop across its terminals.
Project 2.1: The Diode
If you have access to electronic equipment, you may want to perform the simple project described in the next few problems. If this is the first time you have tried such a project, get help from an instructor or someone who is familiar with electronic projects.
When building electronic circuits, eventually you'll make a mistake (as all of us do), and sometimes those mistakes cause circuits to fry. If you smell hot electronic components, disconnect the battery from the circuit, and then check the circuit to determine what connections you should change.
When fixing a circuit, follow some simple safety rules. Do not try to rearrange connections with the battery connected because you may short leads together.
Also, don't touch bare wires with live electricity. Even with batteries, you have a chance of being burned or seriously injured. If your skin is wet, it forms an electrical connection with lower resistance, allowing more current to flow, potentially injuring you.
If you do not have access to equipment, do not skip this project. Read through the project, and try to picture or imagine the results. This is sometimes called “dry-labbing” the experiment. You can learn a lot from reading about this project, even though it is always better to actually perform the project. This advice also applies to the other projects in many of the following chapters.
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 2.10, 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.
Figure 2.10
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
One 330-ohm, 0.5-watt resistor
One 1N4001 diode
One breadboard
One 1 MΩ potentiometer
One terminal block
Step-by-Step Instructions
Set up the circuit as shown in Figure 2.11. 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.
Figure 2.11
Carefully check your ci
rcuit against Figure 2.11, 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.
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 330-ohm 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.
V (volts) I (mA)
7. Graph the points recorded in the table using the blank graph shown in Figure 2.12. Your curve should look like the one shown in Figure 2.10.
Figure 2.12
Expected Results
Figure 2.13 shows the breadboarded circuit for this project.
Figure 2.13
Figure 2.14 shows the test setup for this project.
Figure 2.14
Compare your measurements with the ones shown in the following table:
V (volts) I (mA)
0.44 0.00
0.46 0.01
0.50 0.06
0.52 0.11
0.55 0.23
0.58 0.49
0.60 0.92
0.63 1.74
0.68 4.86
0.72 15.1
0.73 20.9
0.74 25.2
Further reductions in the resistance below the 330 included in the circuit causes little increase in the voltage but produces large increases in the current.
Figure 2.15 shows the V-I curve generated using the measurements shown in the preceding table.
Figure 2.15
The V-I curve (or diode characteristic curve) is repeated in Figure 2.16 with three important regions marked on it.
Figure 2.16
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.
9 The knee voltage is also a limiting voltage. That is, it is the highest voltage that can be obtained across the diode in the forward direction.
Questions
A. Which has the higher limiting voltage, germanium or silicon?_____
B. What happens to the diode resistance at the limiting or knee voltage?_____
Answers
A. Silicon, with a limiting voltage of 0.7 volt, is higher than germanium, which has a limiting voltage of only 0.3 volt.
B. It changes from high to low.
Note You use these knee voltages in many later chapters as the voltage drop across the PN junction when it is forward-biased.
10 Refer back to the diagram of resistance regions shown in Figure 2.16.
Question
What happens to the current when the voltage becomes limited at the knee?_____
Answer
It increases rapidly.
11 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.
Questions
A. Why are imperfect diodes specified here?_____
B. Would you use a high or low resistance to prevent excessive current?_____
Answers
A. All diodes are imperfect, and the 0.3 or 0.7 voltage values are only approximate. In fact, in some later problems, it is assumed that the voltage drop across the diode, when it is conducting, is 0 volts. This assumes, then, that as soon as you apply any voltage above 0, current flows in an ideal resistor. (That is, the knee voltage on the V-I curve for an ideal diode is 0 volts.)
B. Generally, use a high resistance. However, you can calculate the actual resistance value given the applied voltage and the maximum current the diode can withstand.
12 Calculate the current through the diode in the circuit shown in Figure 2.17 using the steps in the following questions.
Figure 2.17
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, which was discussed in Chapter 1, “DC Review and Pre-Test.” VR = _____
C. Calculate the current through the resistor. Use I = VR/R.
I = _____
D. Finally, determine the current through the diode. I =
Answers
You should have written these values:
A. 0.7 volt
B. VR = VS – VD = 5 volts – 0.7 volt = 4.3 volts
C.
D. 4.3 mA (In a series circuit, the same current runs through each component.)
13 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.
Questions
A. Calculate the current through the silicon diode, as shown in Figure 2.18.
Figure 2.18
VD = _____
VR = VS – VD = _____
_____
ID = _____
B. Calculate the current through the perfect diode, as shown in Figure 2.18.
VD = _____
VR = VS – VD = _____
_____
ID = _____
Answers
A. 0.7 volt; 9.3 volt; 9.3 mA; 9.3 mA
B. 0 volt; 10 volt; 10 mA; 10 mA
14 The difference in the values of the two currents found in problem 13 is less than 10 percent of the total current. That is, 0.7 mA is less than 10 percent of 10 mA. Many electronic components have a plus or minus 5 percent tolerance in their nominal values. This means that a 1 k resistor can be anywhere from 950 ohms to 1,050 ohms, meaning that the value of current through a resistor can vary plus or minus 5 percent.
Because of a slight variance in component values, calculations are often simplified if the simplification does not change values by more than 10 percent. Therefore, a diode is often assumed to be perfect when the supply voltage is 10 volts or more.
Questions
A. Examine the circui
t in Figure 2.19. Is it safe to assume that the diode is perfect? _____
Figure 2.19
B. Calculate the current through the diode. _____
Answers
A. Yes, it can be considered a perfect diode.
B. I = 10 mA
15 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 × 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.
Question
Assume a current of 2 amperes is flowing through a silicon diode. How much power is being dissipated? _____
Answer
16 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.
Question
Suppose the maximum power rating of a germanium diode is 3 watts. What is its highest safe current? _____
Answer