Complete Electronics Self-Teaching Guide with Projects
Page 5
17 Answer the following questions for another example.
Questions
A. Could a 3-watt silicon diode carry the current calculated for the germanium diode for problem 16? _____
B. What would be its safe current? _____
Answers
A. No, 10 amperes would cause a power dissipation of 7 watts, which would burn up the diode.
B.
Any current less than this would be safe.
18 The next several examples concentrate on finding the current through the diode. Look at the circuit shown in Figure 2.20.
Figure 2.20
The total current from the battery flows through R1, and then splits into I2 and ID. I2 flows through R2, and ID flows through the diode.
Questions
A. What is the relationship between IT, I2, and ID? _____
B. What is the value of VD? _____
Answers
A. Remember KCL, IT = I2 + ID _____
B. VD = 0.7 volt _____
19 To find ID, you need to go through the following steps because there is no way to find ID directly:
1. Find I2. This is done using VD = R2 × I2.
2. Find VR. For this, use VR = VS − VD (KVL again).
3. Find IT (the current through R1). Use VR = IT × R1.
4. Find ID. This is found by using IT = I2 + ID (KCL again).
To find ID in the circuit shown in Figure 2.21, go through these steps, and then check your answers.
Figure 2.21
Questions
A. I2 = _____
B. VR = _____
C. IT = _____
D. ID = _____
Answers
A.
B. VR = VS – VD = 5 volts – 0.7 volt = 4.3 volts
C.
D. ID = IT – I2 = 100 mA – 10 mA = 90 mA
20 For this problem, refer to your answers in problem 19.
Question
What is the power dissipation of the diode in problem 19? _____
Answer
P = VD × ID = (0.7 volt)(90 mA) = 63 milliwatts
21 To find the current in the diode for the circuit shown in Figure 2.22, answer the following questions in order.
Figure 2.22
Questions
A. I2 = _____
B. VR = _____
C. IT = _____
D. ID = _____
Answers
A.
B. VR = VS−VD = 1.6 – 0.3 = 1.3 volts
C.
D. ID = IT − I2 = 1.8 mA
If you want to take a break soon, this is a good stopping point.
Diode Breakdown
22 Earlier, you read that if the circuit in Project 2.1 was not working correctly, then the diode may be in backward. If you place the diode in the circuit backward—as shown on the right in Figure 2.23—then almost no current flows. In fact, the current flow is so small, it can be said that no current flows. The V-I curve for a reversed diode looks like the one shown on the left in Figure 2.23.
Figure 2.23
The V-I curve for a perfect diode would show zero current for all voltage values. But for a real diode, a voltage is reached where the diode “breaks down” and the diode allows a large current to flow. The V-I curve for the diode breakdown would then look like the one in Figure 2.24.
Figure 2.24
If this condition continues, the diode will burn out. You can avoid burning out the diode, even though it is at the breakdown voltage, by limiting the current with a resistor.
Question
The diode in the circuit shown in Figure 2.25 is known to break down at 100 volts, and it can safely pass 1 ampere without overheating. Find the resistance in this circuit that would limit the current to 1 ampere. _____
Figure 2.25
Answer
Because 1 ampere of current is flowing, then
23 All diodes break down when connected in the reverse direction if excess voltage is applied to them. The breakdown voltage (which is a function of how the diode is made) varies from one type of diode to another. This voltage is quoted in the manufacturer's data sheet.
Breakdown is not a catastrophic process and does not destroy the diode. If the excessive supply voltage is removed, the diode can recover and operate normally. You can use it safely many more times, provided the current is limited to prevent the diode from burning out.
A diode always breaks down at the same voltage, no matter how many times it is used.
The breakdown voltage is often called the peak inverse voltage (PIV) or the peak reverse voltage (PRV). Following are the PIVs of some common diodes:
Diode PIV
1N4001 50 volts
1N4002 100 volts
1N4003 200 volts
1N4004 400 volts
1N4005 600 volts
1N4006 800 volts
Questions
A. Which can permanently destroy a diode, excessive current or excessive voltage? _____
B. Which is more harmful to a diode, breakdown or burnout? _____
Answers
A. Excessive current. Excessive voltage cannot harm the diode if the current is limited.
B. Burnout. Breakdown is not necessarily harmful, especially if the current is limited.
Inside the Diode
At the junction of the N and P type regions of a diode, electrons from the N region are trapped by holes in the P region, forming a depletion region as illustrated in the following figure.
When the electrons are trapped by holes, they can no longer move, which is what produces this depletion region that contains no mobile electrons. The static electrons give this region the properties of an electrical insulator.
You apply a forward-biased voltage to a diode by applying negative voltage to the N region and positive voltage to the P region. Electrons in the N region are repelled by the negative voltage, pushing more electrons into the depletion region. However, the electrons in the N region are also repelled by the electrons already in the depletion region. (Remember that like charges repel each other.) When the forward-biased voltage is sufficiently high (0.7 volts for silicon diodes, and 0.3 volts for germanium diodes), the depletion region is eliminated. As the voltage is raised further, the negative voltage, in repelling electrons in the N region, pushes them into the P region, where they are attracted by the positive voltage.
This combination of repulsive and attractive forces allows current to flow, as illustrated in the following figure.
Note that an interesting aspect of diodes is that while the N and P regions of the diode have mobile charges, they do not have a net charge. The mobile charges (electrons in N regions and holes in P regions) are simply donated by the impurity atoms used to dope the semiconductor material. The impurity atoms have more electrons in the N region and less in the P region than are needed to bond to the adjacent silicon or germanium atoms in the crystalline structure. However, because the atoms have the same number of positive and negative charges, the net charge is neutral.
In a similar way, metal is a conductor because it has more electrons than are needed to bond together the atoms in its crystalline structure. These excess electrons are free to move when a voltage is applied, either in a metal or an N type semiconductor, which makes the material electrically conductive. Electrons moving through the P region of the diode jump between the holes, which physicists model as the holes moving. In a depletion region, where electrons have been trapped by holes, there is a net negative charge.
You apply a reverse-biased voltage to a diode by applying positive voltage to the N type region and negative voltage to the P type region. In this scenario, electrons are attracted to the positive voltage, which pulls them away from the depletion region. No current can flow unless the voltage exceeds the reverse breakdown voltage.
Diodes are marked with one band indicating the cathode (the N region) end of the diode. The band on the diode corresponds to the bar on the diode symbol, as shown in the following figure. Use this marking to orient diodes
in a circuit.
The part number is marked on diodes. However, the part numbers don't tell you much about the diode. For example, the 1N4001 is a silicon diode that can handle a peak reverse voltage of 50 volts, whereas the 1N270 is a germanium diode that can handle a peak reverse voltage of 80 volts. Your best bet is to refer to the manufacturer's data sheet for the diode peak reverse voltage and other characteristics. You can easily look up data sheets on the Internet. Also, you can find links to the data sheets for the components used in this book on the website at www.buildinggadgets.com/index_datasheets.htm.
The Zener Diode
24 Diodes can be manufactured so that breakdown occurs at lower and more precise voltages than those just discussed. These types of diodes are called zener diodes, so named because they exhibit the “Zener effect”—a particular form of voltage breakdown. At the zener voltage, a small current flows through the zener diode. This current must be maintained to keep the diode at the zener point. In most cases, a few milliamperes are all that is required. Figure 2.26 shows the zener diode symbol and a simple circuit.
Figure 2.26
In this circuit, the battery determines the applied voltage. The zener diode determines the voltage drop (labeled Vz) across it. The resistor determines the current flow. Zeners are used to maintain a constant voltage at some point in a circuit.
Question
Why are zeners used for this purpose, rather than ordinary diodes? _____
Answer
Because zeners have a precise breakdown voltage.
25 Examine an application in which a constant voltage is wanted—for example, a lamp driven by a DC generator. In this example, when the generator turns at full speed, it puts out 50 volts. When it runs more slowly, the voltage can drop to 35 volts. You want to illuminate a 20-volt lamp with this generator. Assume that the lamp draws 1.5 amperes. Figure 2.27 shows the circuit.
Figure 2.27
You need to determine a suitable value for the resistance. Follow these steps to find a suitable resistance value:
1. Find RL, the lamp resistance. Use the following formula:
2. Find VR. Use VS = VR + VL.
3. Find R. Use the following formula:
Questions
Work through these steps, and write your answers here.
A. RL = _____
B. VR = _____
C. R = _____
Answers
A.
B.
C.
26 Assume now that the 20-ohm resistor calculated in problem 25 is in place, and the voltage output of the generator drops to 35 volts, as shown in Figure 2.28. This is similar to what happens when a battery gets old. Its voltage level decays and it will no longer have sufficient voltage to produce the proper current. This results in the lamp glowing less brightly, or perhaps not at all. Note, however, that the resistance of the lamp stays the same.
Figure 2.28
Questions
A. Find the total current flowing. Use the following formula:
B. Find the voltage drop across the lamp. Use VL = IT × RL.
VL = _____
C. Have the voltage and current increased or decreased? _____
Answers
A.
B.
C. Both have reduced in value.
27 In many applications, a lowering of voltage across the lamp (or some other component) may be unacceptable. You can prevent this by using a zener diode, as shown in the circuit in Figure 2.29.
Figure 2.29
If you choose a 20-volt zener (that is, one that has a 20-volt drop across it), then the lamp always has 20 volts across it, no matter what the output voltage is from the generator (provided, of course, that the output from the generator is always above 20 volts).
Questions
Say that the voltage across the lamp is constant, and the generator output drops.
A. What happens to the current through the lamp? _____
B. What happens to the current through the zener? _____
Answers
A. The current stays constant because the voltage across the lamp stays constant.
B. The current decreases because the total current decreases.
28 To make this circuit work and keep 20 volts across the lamp at all times, you must find a suitable value of R. This value should allow sufficient total current to flow to provide 1.5 amperes required by the lamp, and the small amount required to keep the diode at its zener voltage. To do this, you start at the “worst case” condition. (“Worst case” design is a common practice in electronics. It is used to ensure that equipment can work under the most adverse conditions.) The worst case here would occur when the generator puts out only 35 volts. Figure 2.30 shows the paths that current would take in this circuit.
Figure 2.30
Find the value of R that enables 1.5 amperes to flow through the lamp. How much current can flow through the zener diode? You can choose any current you like, provided it is above a few milliamperes, and provided it does not cause the zener diode to burn out. In this example, assume that the zener current IZ is 0.5 amperes.
Questions
A. What is the total current through R?
IR = _____
B. Calculate the value of R.
R = _____
Answers
A.
B.
A different choice of IZ here would produce another value of R.
29 Now, take a look at what happens when the generator supplies 50 volts, as shown in Figure 2.31.
Figure 2.31
Because the lamp still has 20 volts across it, it can still draw only 1.5 amperes. But the total current and the zener current change.
Questions
A. Find the total current through R.
IR = _____
B. Find the zener current.
IZ = _____
Answers
A.
B. IZ = IR − IL = 4 − 1.5 = 2.5 amperes
30 Although the lamp voltage and current remain the same, the total current and the zener current both changed.
Questions
A. What has happened to IT (IR)? _____
B. What has happened to IZ? _____
Answers
A. IT has increased by 2 amperes.
B. IZ has increased by 2 amperes.
The increase in IT flows through the zener diode and not through the lamp.
31 The power dissipated by the zener diode changes as the generator voltage changes.
Questions
A. Find the power dissipated when the generator voltage is 35 volts. _____
B. Now, find the power when the generator is at 50 volts. _____
Answers
A. PZ = V × I = (20 volts) (0.5 ampere) = 10 watts
B. PZ = V × I = (20 volts) (2.5 ampere) = 50 watts
If you use a zener diode with a power rating of 50 watts or more, it does not burn out.
32 Use Figure 2.32 to answer the following question.
Figure 2.32
Question
For the circuit shown in Figure 2.32, what power rating should the zener diode have? The current and voltage ratings of the lamp are given. _____
Answer
At 24 volts, assuming a zener current of 0.5 ampere:
At 60 volts:
PZ = (15 volts)(2.8 amperes) = 42 watts
Project 2.2: The Zener Diode Voltage Regulator
Objective
The objective of this project is to measure the voltage applied to the lamp, and the current through the lamp for different supply voltages, demonstrating the use of a zener diode to provide a steady voltage and current to a lamp when the supply voltage changes.
General Instructions
While the circuit is set up, measure the lamp current, zener diode current, and supply voltage as the voltage from the 9-volt battery drops.
Parts List
One 9-volt battery
One battery snap connector
Two multimeters set to measure
current (mA)
One multimeter set to measure DC voltage
One 56-ohm, 0.5-watt resistor
One 1N4735A zener diode
One breadboard
One lamp rated for approximately 25 mA at 6 volts. (Part # 272-1140 from Radio Shack is a good fit for this project.)
Two terminal blocks
Step-by-Step Instructions
Set up the circuit shown on Figure 2.33. 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 2.33
Carefully check your circuit against the diagram, 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 without 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. Measure and record the supply voltage.
2. Measure and record the lamp current.
3. Measure and record the zener current.
4. Wait 30 minutes.
5. Measure and record the new values of voltage and current.
6. Repeat steps 4 and 5 four times.