Saturday, December 13, 2014

Solid-state power supplies : learning objectives ,the power transformer , the conventional full-wave rectifier and the bridge rectifier

SOLID-STATE POWER SUPPLIES
LEARNING OBJECTIVES


Upon completion of this chapter you will be able to:

  • Identify the various sections of a power supply.
  • State the purpose of each section of a power supply.
  • Describe the operation of the power supply from both a whole unit standpoint and from a subunit standpoint.
  • Describe the purpose of the various types of rectifier circuits used in power supplies.
  • Describe the purpose of the various types of filter circuits used in power supplies.
  • Describe the operation of the various voltage and current regulators in a power supply.
  • Describe the operation of the various types of voltage multipliers.
  • Trace the flow of ac and dc in a power supply, from the ac input to the dc output on a schematic diagram.
  • Identify faulty components through visual checks.
  • Identify problems within specific areas of a power supply by using a logical isolation method of troubleshooting.
  • Apply safety precautions when working with electronic power supplies.

In today's Navy all electronic equipment, both ashore and on board ship, requires a power supply. The discovery of the silicon diode and other solid-state components made possible the reduction in size and the increase in reliability of electronic equipment. This is especially important on board ship where space and accessibility to spare parts are a major concern.
In this chapter, you will read about the individual sections of the power supply, their components, and the purpose of each within the power supply.
THE BASIC POWER SUPPLY
View A of figure 4-1 shows the block diagram of a basic power supply. Most power supplies are made up of four basic sections: a TRANSFORMER, a RECTIFIER, a FILTER, and a REGULATOR.
Figure 4-1A. - Block diagram of a basic power supply.

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As illustrated in view B of figure 4-1, the first section is the TRANSFORMER. The transformer steps up or steps down the input line voltage and isolates the power supply from the power line. The RECTIFIER section converts the alternating current input signal to a pulsating direct current. However, as you proceed in this chapter you will learn that pulsating dc is not desirable. For this reason a FILTER section is used to convert pulsating dc to a purer, more desirable form of dc voltage.
Figure 4-1B. - Block diagram of a basic power supply.

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The final section, the REGULATOR, does just what the name implies. It maintains the output of the power supply at a constant level in spite of large changes in load current or input line voltages.
Now that you know what each section does, let's trace an ac signal through the power supply. At this point you need to see how this signal is altered within each section of the power supply. Later on in the chapter you will see how these changes take place. In view B of figure 4-1, an input signal of 115 volts ac is applied to the primary of the transformer. The transformer is a step-up transformer with a turns ratio of 1:3. You can calculate the output for this transformer by multiplying the input voltage by the ratio of turns in the primary to the ratio of turns in the secondary; therefore, 115 volts ac X 3 = 345 volts ac (peak-to-peak) at the output. Because each diode in the rectifier section conducts for 180 degrees of the 360-degree input, the output of the rectifier will be one-half, or approximately 173 volts of pulsating dc. The filter section, a network of resistors, capacitors, or inductors, controls the rise and fall time of the varying signal; consequently, the signal remains at a more constant dc level. You will see the filter process more clearly in the discussion of the actual filter circuits. The output of the filter is a signal of 110 volts dc, with ac ripple riding on the dc. The reason for the lower voltage (average voltage) will be explained later in this chapter. The regulator maintains its output at a constant 110-volt dc level, which is used by the electronic equipment (more commonly called the load).

Q.1 What are the four basic sections of a power supply?

Q.2 What is the purpose of the rectifier section?

Q.3 What is the purpose of the filter section?

Q.4 What is the purpose of the regulator section?


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THE POWER TRANSFORMER


In some cases a power supply may not use a transformer; therefore, the power supply would be connected directly to the source line voltage. This type of connection is used primarily because it is economical. However, unless the power supply is completely insulated, it presents a dangerous shock hazard to anyone who comes in contact with it. When a transformer is not being used, the return side of the ac line is connected to the metal chassis. To remove this potential shock hazard and to have the option of stepping up or stepping down the input voltage to the rectifier, a transformer must be used.

View A of figure 4-2 shows the schematic diagram for a STEP-UP transformer; view B shows a STEP-DOWN transformer; and, view C shows a STEP-UP, CENTER-TAPPED transformer. The step-up and step-down transformers were discussed in earlier NEETS modules, so only the center-tapped transformer will be mentioned in this chapter. The primary purpose of the center-tapped transformer is to provide two equal voltages to the conventional full-wave rectifier.
Figure 4-2A. - Common types of transformers. STEP-UP

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Figure 4-2B. - Common types of transformers. STEP-DOWN

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Figure 4-2C. - Common types of transformers. CENTER-TAPPED
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THE RECTIFIER
From previous discussions, you should know that rectification is the conversion of an alternating current to a pulsating direct current. Now let's see how the process of RECTIFICATION occurs in both a half-wave and a full-wave rectifier.

The Half-Wave Rectifier

Since a silicon diode will pass current in only one direction, it is ideally suited for converting alternating current (ac) to direct current (dc). When ac voltage is applied to a diode, the diode conducts ONLY ON THE POSITIVE ALTERNATION OF VOLTAGE; that is, when the anode of the diode is positive with respect to the cathode. This simplest type of rectifier is the half-wave rectifier. As shown in view A of figure 4-3, the half- wave rectifier uses only one diode. During the positive alternation of input voltage, the sine wave applied to the diode makes the anode positive with respect to the cathode. The diode then conducts, and current (I) flows from the negative supply lead (the secondary of the transformer), through the milliammeter, through the diode, and to the positive supply lead. As indicated by the shaded area of the output waveform in view B, this current exists during the entire period of time that the anode is positive with respect to the cathode (in other words, for the first 180 degrees of the input sine wave).
Figure 4-3A. - Simple half-wave rectifier. HALF-WAVE RECTIFIER

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Figure 4-3B. - Simple half-wave rectifier. OUTPUT WAVEFORM

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During the negative alternation of input voltage (dotted polarity signs), the anode is driven negative and the diode cannot conduct. When conditions such as these exist, the diode is in cutoff and remains in cutoff for 180 degrees, during which time no current flows in the circuit. The circuit current therefore has the appearance of a series of positive pulses, as illustrated by the shaded areas on the waveform in view B. Notice that although the current is in the form of pulses, the current always flows in the same direction. Current that flows in pulses in the same direction is called PULSATING DC. The diode has thus RECTIFIED the ac input voltage.

Rms, Peak, and Average Values

View A of figure 4-4 is a comparison of the rms, peak, and average values of the types of waveforms associated with the half-wave rectifier. Ac voltages are normally specified in terms of their rms values. Thus, when a 115-volt ac power source is mentioned in this chapter, it is specifying the rms value of 115 volts ac. In terms of peak values,

Erms = Epeak X .707

The peak value is always higher than the rms value. In fact,

Epeak = Erms X 1.414

therefore, if the rms value is 115 volts ac, then the peak value must be:

Epeak = Erms X 1.414

Epeak = 115 volts ac X 1.414

Epeak = 162.6 volts

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The average value of a sine wave is 0 volts. View B of figure 4-4 shows how the average voltage changes when the negative portion of the sine wave is clipped off. Since the wave form swings positive but never negative (past the "zero-volt" reference line), the average voltage is positive. The average voltage (Eavg) is determined by the equation:

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Figure 4-4B. - Comparison of Epeak to E avg in a half-wave rectifier. Eavg WAVEFORM

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Ripple Frequency
The half-wave rectifier gets its name from the fact that it conducts during only half the input cycle. Its output is a series of pulses with a frequency that is the same as the input frequency. Thus when operated from a 60-hertz line, the frequency of the pulses is 60 hertz. This is called RIPPLE FREQUENCY.

Q.5 What is the name of the simplest type of rectifier which uses one diode?

Q.6 If the output of a half-wave rectifier is 50-volts peak, what is the average voltage?

Q.7 In addition to stepping up or stepping down the input line voltage, what additional purpose does the transformer serve?


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The Conventional Full-Wave Rectifier

A full-wave rectifier is a device that has two or more diodes arranged so that load current flows in the same direction during each half cycle of the ac supply.
A diagram of a simple full-wave rectifier is shown in figure 4-5.
The transformer supplies the source voltage for two diode rectifiers, D1 and D2. This power transformer has a center-tapped, high-voltage secondary winding that is divided into two equal parts (W1 and W2). W1 provides the source voltage for D1, and W2 provides the source voltage for D2. The connections to the diodes are arranged so that the diodes conduct on alternate half cycles.
During one alternation of the secondary voltage, the polarities are as shown in view A. The source for D2 is the voltage induced into the lower half of the secondary winding of the transformer (W2). At the specific instant of time shown in the figure, the anode voltage on D2 is negative, and D2 cannot conduct. Throughout the period of time during which the anode of D2 is negative, the anode of D1 is positive. Since the anode of D1 is positive, it conducts, causing current to flow through the load resistor in the direction shown by the arrow.
Figure 4-5A. - Full-wave rectifier. POSITIVE ALTERNATION

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View B shows the next half cycle of secondary voltage. Now the polarities across W1 and W2 are reversed. During this alternation, the anode of D1 is driven negative and D1 cannot conduct. For the period of time that the anode of D1 is negative, the anode of D2 is positive, permitting D2 to conduct. Notice that the anode current of D2 passes through the load resistor in the same direction as the current of D1 did. In this circuit arrangement, a pulse of load current flows during each alternation of the input cycle. Since both alternations of the input voltage cycle are used, the circuit is called a FULL-WAVE RECTIFIER.
Figure 4-5B. - Full-wave rectifier. NEGATIVE ALTERNATION

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Now that you have a basic understanding of how a full-wave rectifier works, let's cover in detail a practical full-wave rectifier and its waveforms.
A Practical Full-Wave Rectifier
A practical full-wave rectifier circuit is shown in view A of figure 4-6. It uses two diodes (D1 and D2) and a center-tapped transformer (T1). When the center tap is grounded, the voltages at the opposite ends of the secondary windings are 180 degrees out of phase with each other. Thus, when the voltage at point A is positive with respect to ground, the voltage at point B is negative with respect to ground. Let's examine the operation of the circuit during one complete cycle.
Figure 4-6. - Practical full-wave rectifier.

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During the first half cycle (indicated by the solid arrows), the anode of D1 is positive with respect to ground and the anode of D2 is negative. As shown, current flows from ground (center tap), up through the load resistor (RL), through diode D1 to point A. In the transformer, current flows from point A, through the upper winding, and back to ground (center tap). When D1 conducts, it acts like a closed switch so that the positive half cycle is felt across the load (RL).
During the second half cycle (indicated by the dotted lines), the polarity of the applied voltage has reversed. Now the anode of D2 is positive with respect to ground and the anode of D1 is negative. Now only D2 can conduct. Current now flows, as shown, from ground (center tap), up through the load resistor (RL), through diode D2 to point B of T1. In the transformer, current flows from point B up through the lower windings and back to ground (center tap). Notice that the current flows across the load resistor (RL) in the same direction for both halves of the input cycle.

View B represents the output waveform from the full-wave rectifier. The waveform consists of two pulses of current (or voltage) for each cycle of input voltage. The ripple frequency at the output of the full-wave rectifier is therefore twice the line frequency.

The higher frequency at the output of a full-wave rectifier offers a distinct advantage: Because of the higher ripple frequency, the output is closely approximate to pure dc. The higher frequency also makes filtering much easier than it is for the output of the half-wave rectifier.
In terms of peak value, the average value of current and voltage at the output of the full-wave rectifier is twice as great as that at the output of the half-wave rectifier. The relationship between the peak value and the average value is illustrated in figure 4-7. Since the output waveform is essentially a sine wave with both alternations at the same polarity, the average current or voltage is 63.7 percent (or 0.637) of the peak current or voltage.
Figure 4-7. - Peak and average values for a full-wave rectifier.

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As an equation:
Where:
Emax = The peak value of the load voltage pulse
Eavg = 0.637 X Emax (the average load voltage)
Imax = The peak value of the load current pulse
Iavg = 0.637 X Imax (the average load current)

Example: The total voltage across the high-voltage secondary of a transformer used to supply a full-wave rectifier is 300 volts. Find the average load voltage (ignore the drop across the diode).

Solution: Since the total secondary voltage (ES) is 300 volts, each diode is supplied one-half of this value, or 150 volts. Because the secondary voltage is an rms value, the peak load voltage is:

Emax = 1.414 X ES
Emax = 1.414 X 150
Emax = 212 volts
The average load voltage is:
Eavg = 0.637 X Emax
Eavg = 0.637 X 212
Eavg = 135 volts
NOTE: If you have problems with this equation, review the portion of NEETS, module 2, that pertain to this subject.
As you may recall from your past studies in electricity, every circuit has advantages and disadvantages. The full-wave rectifier is no exception. In studying the full-wave rectifier, you may have found that by doubling the output frequency, the average voltage has doubled, and the resulting signal is much easier to filter because of the high ripple frequency. The only disadvantage is that the peak voltage in the full-wave rectifier is only half the peak voltage in the half-wave rectifier. This is because the secondary of the power transformer in the full-wave rectifier is center tapped; therefore, only half the source voltage goes to each diode.

Fortunately, there is a rectifier which produces the same peak voltage as a half-wave rectifier and the same ripple frequency as a full-wave rectifier. This circuit, known as the BRIDGE RECTIFIER, will be the subject of our next discussion.

Q.8 What was the major factor that led to the development of the full-wave rectifier?

Q.9 What is the ripple frequency of a full-wave rectifier with an input frequency of 60 Hz?

Q.10 What is the average voltage (Eavg) Output of a full-wave rectifier with an output of 100 volts peak?


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The Bridge Rectifier

When four diodes are connected as shown in figure 4-8, the circuit is called a BRIDGE RECTIFIER. The input to the circuit is applied to the diagonally opposite corners of the network, and the output is taken from the remaining two corners.
Figure 4-8. - Bridge rectifier.

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One complete cycle of operation will be discussed to help you understand how this circuit works. We have discussed transformers in previous modules in the NEETS series and will not go into their characteristics at this time. Let us assume the transformer is working properly and there is a positive potential at point A and a negative potential at point B. The positive potential at point A will forward bias D3 and reverse bias D4. The negative potential at point B will forward bias D1 and reverse

bias D2. At this time D3 and D1 are forward biased and will allow current flow to pass through them; D4 and D2 are reverse biased and will block current flow. The path for current flow is from point B through D1, up through RL, through D3, through the secondary of the transformer back to point B. This path is indicated by the solid arrows. Waveforms (1) and (2) can be observed across D1 and D3.

One-half cycle later the polarity across the secondary of the transformer reverses, forward biasing D2 and D4 and reverse biasing D1 and D3. Current flow will now be from point A through D4, up through RL, through D2, through the secondary of T1, and back to point A. This path is indicated by the broken arrows. Waveforms (3) and (4) can be observed across D2 and D4. You should have noted that the current flow through RL is always in the same direction. In flowing through RL this current develops a voltage corresponding to that shown in waveform (5). Since current flows through the load (RL) during both half cycles of the applied voltage, this bridge rectifier is a full-wave rectifier.
One advantage of a bridge rectifier over a conventional full-wave rectifier is that with a given transformer the bridge rectifier produces a voltage output that is nearly twice that of the conventional full-wave circuit. This may be shown by assigning values to some of the components shown in views A and B of figure 4-9. Assume that the same transformer is used in both circuits. The peak voltage developed between points X and Y is 1000 volts in both circuits. In the conventional full-wave circuit shown in view A, the peak voltage from the center tap to either X or Y is 500 volts. Since only one diode can conduct at any instant, the maximum voltage that can be rectified at any instant is 500 volts. Therefore, the maximum voltage that appears across the load resistor is nearly - but never exceeds - 500 volts, as a result of the small voltage drop across the diode. In the bridge rectifier shown in view B, the maximum voltage that can be rectified is the full secondary voltage, which is 1000 volts. Therefore, the peak output voltage across the load resistor is nearly 1000 volts. With both circuits using the same transformer, the bridge rectifier circuit produces a higher output voltage than the conventional full-wave rectifier circuit.
Figure 4-9A. - Comparison of a conventional and bridge full-wave rectifier. CONVENTIONAL FULL-WAVE RECTIFIER

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Figure 4-9B. - Comparison of a conventional and bridge full-wave rectifier. FULL-WAVE BRIDGE RECTIFIER

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Q.11 What is the main disadvantage of a conventional full-wave rectifier?

Q.12 What main advantage does a bridge rectifier have over a conventional full-wave rectifier?


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