Complete Electronics Self-Teaching Guide with Projects
Page 26
Figure 11.28 shows a function generator and oscilloscope attached to Circuit #2.
Figure 11.28
In Figure 11.29, the signal across the primary coil is represented by the upper waveform, and the signal across the resistor is represented by the lower waveform. Read the number of divisions for the peak-to-peak voltage of the upper waveform, and multiply it by the corresponding VOLTS/DIV setting to determine Vpp for the signal across the primary coil. Read the number of divisions for the peak voltage of the lower waveform, and multiply it by the corresponding VOLTS/DIV setting to determine Vp for the signal across the resistor.
Figure 11.29
As you measure Vpp and Vp, you may need to adjust the TIME/DIV, the VOLTS/DIV, and the vertical POSITION controls on the oscilloscope. The controls shown in Figure 11.30 are adjusted to measure Vpp in the primary coil and Vp across the load resistor for Circuit # 2.
Figure 11.30
Your values should be close to those shown in the following table.
Circuit # Vpp (Primary Coil) Vp (Load Resistor)
1 20 volts 4.4 volts
2 20 volts 8.8 volts
Vp across the load resistor for Circuit # 2 is twice the value of Vp for Circuit #1. That's because Circuit #1 uses a center-tapped secondary coil that supplies a 10 Vpp sine wave to the diodes, whereas the secondary coil in Circuit #2 supplies a 20 Vpp sine wave. In each circuit, the 0.6 to 0.7 voltage drop that occurs as the signal passes through each diode causes Vp across the load resistor to be slightly lower than half the Vpp supplied by the secondary coil.
What this chapter has explored to this point is how AC is turned into pulsating DC. In fact, rectified AC is often called pulsating DC. The next step in your understanding of power supplies is to learn how you turn pulsating DC into level DC.
Level DC (Smoothing Pulsating DC)
14 A basic power supply circuit can be divided into four sections, as shown in Figure 11.31.
Figure 11.31
Questions
A. If you use a center-tap transformer in a power supply, how many diodes would you need to produce a full-wave rectified output? _____
B. Will the power supply circuit shown in Figure 11.31 result in full- or half-wave rectification? _____
C. What type of output will the rectifier section of the power supply circuit shown in Figure 11.31 produce? _____
Answers
A. Two
B. Half-wave
C. Pulsating DC
15 The function of the smoothing section of a power supply circuit is to take the pulsating DC (PDC) and convert it to a “pure” DC with as little AC “ripple” as possible. The smoothed DC voltage, shown in the illustration on the right in Figure 11.32, is then applied to the load.
Figure 11.32
The load (which is “driven” by the power supply) can be a simple lamp or a complex electronic circuit. Whatever load you use, it requires a certain voltage across its terminals and draws a current. Therefore, the load has a resistance.
Usually, the voltage and current required by the load (and, hence, its resistance) are known, and you must design the power supply to provide that voltage and current.
To simplify the circuit diagrams, you can treat the load as a simple resistor.
Questions
A. What does the smoothing section of a power supply do? _____
B. What is connected to a power supply, and what can you treat it like? _____
Answers
A. The smoothing section converts the pulsating DC to a “pure” DC.
B. A load such as a lamp or an electronic circuit is connected to a power supply. In most cases, you can treat the load as you do a resistor.
16 Figure 11.33 shows a power supply circuit with a resistor as the load.
Figure 11.33
Questions
Look at the circuit shown in Figure 11.33 and answer the following questions:
A. What type of secondary coil is used? _____
B. What type of rectifier is used? _____
C. What components make up the smoothing section? _____
D. What output would you expect from the rectifier section? _____
Answers
A. A secondary coil with no center tap
B. A single-diode, half-wave rectifier
C. A resistor and two capacitors (R1, C1, and C2)
D. Half-wave pulsating DC
17 Figure 11.34 shows the output waveform from the rectifier portion of the power supply circuit shown in Figure 11.33.
Figure 11.34
This waveform is the input to the smoothing section of the power supply circuit. Use one of the DC pulses (shown in Figure 11.35) to analyze the effect of the smoothing section on the waveform.
Figure 11.35
As the voltage level of the DC pulse rises to its peak, the capacitor C1 is charged to the peak voltage of the DC pulse.
When the input DC pulse drops from its peak voltage back to 0 volts, the electrons stored on capacitor C1 discharge through the circuit. This maintains the voltage across the load resistor at close to its peak value, as shown in Figure 11.36. The DC pulse to the right of the diode stays at the peak voltage, even though Vin drops to zero.
Figure 11.36
Question
What discharge path is available for the capacitor C1? _____
Answer
The diode is not conducting, so the capacitor cannot discharge through the diode. The only possible discharge path is through R1 and the load RL.
18 If no further pulses pass through the diode, the voltage level drops as the capacitor discharges, resulting in the waveform shown in Figure 11.37.
Figure 11.37
However, if another pulse passes through the diode before the capacitor is discharged, the resulting waveform looks like that shown in Figure 11.38.
Figure 11.38
The capacitor discharges only briefly before the second pulse recharges it to peak value. Therefore, the voltage applied to the load resistor drops only a small amount.
Applying further pulses can produce this same recharging effect again and again. Figure 11.39 shows the resulting waveform.
Figure 11.39
The waveform in Figure 11.39 has a DC level with an AC ripple, which varies between Vp and Vx. If you choose values of C1, R1, and RL that produce a discharge time constant for C1 equal to about 10 times the duration of an input pulse, Vx will be approximately 80 percent of Vp.
If the discharge time you select is greater than 10 times the duration of an input pulse, the smoothing effect minimizes the AC ripple. A time constant of 10 times the pulse duration results in practical design values that are used throughout this chapter.
Note The smoothing section of a power supply circuit is sometimes referred to as a low-pass filter. Though such a circuit can function as a low-pass filter, in the case of a power supply circuit converting AC to DC, it is the release of electrons by the capacitor that is primarily responsible for leveling out the pulsating DC. For that reason, this discussion uses the term smoothing section.
Question
Estimate the average DC output level of the waveform shown in Figure 11.39. _____
Answer
Approximately 90 percent of Vp
19 The output from the secondary coil of the circuit shown in Figure 11.40 is a 28 Vrms, 60 Hz sine wave. For this circuit, you need 10 volts DC across the 100-ohm load resistor.
Figure 11.40
Question
What is the peak voltage out of the rectifier? _____
Answers
The transformer secondary coil delivers 28 Vrms, so
or about 40 volts.
20 Figure 11.41 shows the waveform after the diode has rectified the sine wave for the halfway rectifier circuit shown in Figure 11.40.
Figure 11.41
Question
Calculate the duration of one pulse. _____
Answer
60 Hz represents 60 cycles (that is, wavelengths) in 1 second. Therefore, one wavelengt
h lasts for 1/60 second.
1/60 second = 1000/60 milliseconds = 16.67 ms
Therefore, the duration of a pulse, which is half a wavelength, is 8.33 ms.
21 The average DC voltage at point B in the circuit shown in Figure 11.40 is approximately 90 percent of the peak value of the sine wave from the secondary coil, or VB = 0.9 × 40 volts = 36 volts. R1 and RL act as a voltage divider to reduce the 36-volt DC level to the required 10 volts DC at the output.
Question
Using the voltage divider formula, calculate the value of R1 that will result in 10 volts DC across the 100-ohm load resistor. _____
Answer
Therefore, R1 = 260 ohms
22 Figure 11.42 shows the half-wave rectifier circuit with the 260-ohm value you calculated for R1.
Figure 11.42
Now, choose a value for C1 that produces a discharge time through the two resistors equal to 10 times the input wave duration.
Questions
A. How long should the discharge time constant be for the circuit in Figure 11.42? Refer to problems 18 and 20. _____
B. Given the time constant, calculate the value of C1. _____
Answers
A. The time constant should be 10 times the pulse duration (8.33 ms), so
B. τ = 10 × 8.33 ms = 83.3 ms or 0.083 seconds
C. τ = R × C = (R1 + RL) × C1 = 360 × C1
D. Therefore, 0.0833 = 360 × C1, or C1 = 230 μF
23 Figure 11.43 shows voltage waveforms at various points in the half-wave rectifier circuit.
Figure 11.43
Questions
A. What happens to the DC output voltage between points B and C in this circuit? _____
B. What happens to the waveform between points A and C in the circuit? _____
Answers
A. The voltage has been reduced from 36 volts to 10 volts.
B. The waveform has changed from pulsating DC to a 10-volt DC level with an AC ripple.
24 In most cases, the level of the AC ripple is still too high, and further smoothing is required. Figure 11.44 shows the portion of the half-wave rectifier circuit that forms a voltage divider using R1, and the parallel combination of RL and C2. This voltage divider reduces the AC ripple and the DC voltage level.
Figure 11.44
Choose a value for C2 that causes the capacitor's reactance (XC2) to be equal to or less than one-tenth of the resistance of the load resistor. C2, R1, and R2 form an AC voltage divider. As discussed in problem 26 of Chapter 6, “Filters,” choosing such a value for C2 simplifies the calculations for an AC voltage divider circuit containing a parallel resistor and capacitor.
Questions
A. What should the value of XC2 be? _____
B. What is the formula for the reactance of a capacitor? _____
C. What is the frequency of the AC ripple? _____
D. Calculate the value of the capacitor C2. _____
Answers
A. XC2 = RL/10 = 100/10 = 10 Ω or less
B.
C. 60 Hz. This is identical to the frequency of the sine wave output from the transformer's secondary coil.
D. Solving the reactance formula for C results in the following:
25 Figure 11.45 shows the half-wave rectifier circuit with all capacitor and resistor values.
Figure 11.45
Because XC2 is one-tenth of RL, you can ignore RL in AC voltage divider calculations. Figure 11.46 shows the resulting AC voltage divider circuit.
Figure 11.46
Questions
A. What is the peak-to-peak voltage at the input to the AC voltage divider? _____
B. Find the AC ripple output across RL using the AC voltage divider formula discussed in problem 26 of Chapter 6. _____
Answers
A. Vpp = Vp − Vx = 40 volts − 32 volts = 8 VPP
B.
Note This result means that the addition of C2 lowers the AC ripple shown by curve C in Figure 11.43, with peak values of 11.11 and 8.89, to values of 10.155 and 9.845 volts. This represents a lower ripple at the output. Hence, C2 aids the smoothing of the 10 volts DC at the output.
26 You can apply the calculations you performed for a half-wave rectifier circuit in the last few problems to a full-wave rectifier circuit. In the next few problems, you calculate the values of R1, C1, and C2 required to provide 10 volts DC across a 100-ohm load for a full-wave rectifier circuit with a 28 Vrms sine wave supplied by the secondary coil of a transformer.
Figure 11.47 shows the output waveform from the rectifier section of the circuit.
Figure 11.47
Figure 11.48 shows the waveform that results from using a smoothing capacitor.
Figure 11.48
If the discharge time constant C1 is 10 times the period of the waveform, Vx is approximately 90 percent of Vp. The average DC level is approximately 95 percent of Vp.
Questions
A. What is the average DC level for the half-wave rectifier at point B in Figure 11.43? _____
B. What is the average DC level for the waveform in Figure 11.48 given that Vp = 40 volts? _____
C. Why does a full-wave rectifier have a higher average DC level than a half-wave rectifier? _____
Answers
A. 36 volts, which is 90 percent of Vp.
B. 38 volts, which is 95 percent of Vp.
C. The slightly higher values occur because the capacitor does not discharge as far with full-wave rectification, and, as a result, there is slightly less AC ripple. Therefore, Vx is higher and the average DC level is higher.
27 You can use the method for calculating the value of R1 for a half-wave rectifier (see problem 21) to calculate the value of R1 for a full-wave rectifier.
Question
Calculate the value of R1 when RL = 100 Ω, Vin = 38 volts, and the required voltage across RL is 10 volts. _____
Answer
Therefore, R1 = 280 ohms.
28 You can also use the method for calculating the value of C1 for a half-wave rectifier (see problem 22) to calculate the value of C1 for a full-wave rectifier.
Question
Calculate the value of C1. _____
Answer
With a time constant of τ = 83.3 ms. and a discharge resistance of R1 + RL = 380 ohms, C1 = 220 μF.
29 You can use the voltage divider equation to find the amount of AC ripple across the load resistor for a full-wave rectifier with R1 = 280 Ω and RL = 100 Ω. For Vp = 40 volts, the calculation results in 10.52 volts. For Vx = 36 volts, the calculation results in 9.47 volts. Therefore, the voltage levels at the load resistor vary between 10.52 volts and 9.47 volts, with an average DC level of 10 volts. You can reduce the AC ripple by adding a second capacitor in parallel with the load resistor.
Questions
Use the method for calculating the value of C2 for the half-wave rectifier in problem 24.
A. Calculate the reactance of the second capacitor (C2). _____
B. Calculate the value of C2. (The frequency of the AC ripple for the full-wave rectifier is 120 Hz.) _____
Answers
A. The reactance should be one-tenth (or less) of the load resistance. Therefore, it should be 10 ohms or less.
B.
30 The AC ripple at the first smoothing capacitor ranges from 36 volts to 40 volts. The AC ripple at the load ranges from 9.47 volts to 10.52 volts when there is only one capacitor in the circuit.
Question
Calculate the upper and lower values of the AC ripple at the output if you use a second capacitor with a value of 135 μF in parallel with the load resistor. You can use the same formulas as those for the half-wave rectifier in problem 25. XC2 is 10 Ω from problem 29; R1 = 280 from problem 27; and AC Vin = Vp − Vx = (40 volts − 36 volts) = 4 Vpp. _____
Answer
The result of approximately 0.14 Vpp means that the output will now vary from 10.07 to 9.93 volts. This shows that the second capacitor lowers the ripple significantly. The AC ripple is less than half of the ripple shown f
or the half-wave rectifier in problem 25. In other words, a full-wave rectifier produces a smoother DC output than a half-wave rectifier.
31 Figure 11.49 shows a full-wave rectifier circuit with an output voltage of 5 volts across a 50 Ω load resistor. Use the following steps to calculate the values of the other components.
Figure 11.49
Questions
A. What are Vp, Vx, and the DC level at the first capacitor? _____
B. Calculate the value of R1 required to make the DC level at the output 5 volts. _____
C. Calculate the value of C1. _____
D. Calculate the value of C2. _____
E. What is the amount of AC ripple at the output? _____
F. Draw the final circuit showing the calculated values. Use a separate sheet of paper for your drawing.
Answers
A.
The DC level is 95 percent of Vp, which is 8.46 volts.
B. About 35 ohms.
C. 980 μF.
D. Using XC2 = 5 ohms and 120 Hz, C2 = 265 μF.
E. At the input to the smoothing section, the AC variation is 8.91 to 8.02, or 0.89 Vpp. Using the AC voltage divider equation with R1 = 35 ohms and XC2 = 5 ohms, AC Vout equals approximately 0.13 Vpp. Therefore, the AC variation at the output is 5.065 to 4.935 volts, a small AC ripple.
F. See Figure 11.40.
Figure 11.50
Using the simple procedure shown here always produces a working power supply circuit. This is not the only design procedure you can use for power supplies, but it is one of the simplest and most effective.