Module 7—Introduction to Solid-State Devices and Power Supplies

Chapter 4: Pages 4-11 through 4-20

Pages i - ix, 1-1 to 1-10, 1-11 to 1-20, 1-21 to 1-30, 1-31 to 1-40, 1-41 to 1-47, 2-1 to 2-10, 2-11 to 2-20, 2-21 to 2-30,

2-31 to 2-40, 2-41 to 2-50, 2-51 to 2-54, 3-1 to 3-10, 3-11 to 3-20, 3-21 to 3-30, 3-31 to 3-40, 3-41 to 3-50, 3-51 to 3-54,

4-1 to 4-10, 4-11 to 4-20, 4-21 to 4-30, 4-31 to 4-40, 4-41 to 4-50, 4-51 to 4-62, Index

Figure 4-8.—Bridge rectifier.

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 R

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 R

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

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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

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

Q11. What is the main disadvantage of a conventional full-wave rectifier?

Q12. What main advantage does a bridge rectifier have over a conventional full-wave rectifier?

While the output of a rectifier is a pulsating dc, most electronic circuits require a substantially pure dc for proper operation. This type of output is provided by single or multisection filter circuits placed between the output of the rectifier and the load.

There are four basic types of filter circuits:

· Simple capacitor filter

· LC choke-input filter

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· LC capacitor-input filter (pi-type)

· RC capacitor-input filter (pi-type)

The function of each of these filters will be covered in detail in this chapter.

Filtering is accomplished by the use of capacitors, inductors, and/or resistors in various combinations. Inductors are used as series impedances to oppose the flow of alternating (pulsating dc) current. Capacitors are used as shunt elements to bypass the alternating components of the signal around the load (to ground). Resistors are used in place of inductors in low current applications.

Let's briefly review the properties of a capacitor. First, a capacitor opposes any change in voltage. The opposition to a change in current is called capacitive reactance (X

From the formula, you can see that if frequency or capacitance is increased, the X

To obtain a steady dc output, the capacitor must charge almost instantaneously to the value of applied voltage. Once charged, the capacitor must retain the charge as long as possible. The capacitor must have a short charge time constant (view A). This can be accomplished by keeping the internal resistance of the power supply as small as possible (fast charge time) and the resistance of the load as large as possible (for a slow discharge time as illustrated in view B).

Figure 4-10A.—Capacitor filter. FAST CHARGE

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Figure 4-10B.—Capacitor filter. SLOW DISCHARGE

From your earlier studies in basic electricity, you may remember that one time constant is defined as the time it takes a capacitor to charge to 63.2 percent of the applied voltage or to discharge to 36.8 percent of its total charge. This action can be expressed by the following equation:

**t = RC**

Where: R represents the resistance of the charge or discharge path

And: C represents the capacitance of the capacitor.

You should also recall that a capacitor is considered fully charged after five RC time constants. Refer to figure 4-11. You can see that a steady dc output voltage is obtained when the capacitor charges rapidly and discharges as slowly as possible.

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Figure 4-11.—RC time constant.

In filter circuits the capacitor is the common element to both the charge and the discharge paths. Therefore, to obtain the longest possible discharge time, you want the capacitor to be as large as possible. Another way to look at it is: The capacitor acts as a short circuit around the load (as far as the ac component is concerned), and since

the larger the value of the capacitor (C), the smaller the opposition (X

Now let's look at inductors and their application in filter circuits. Remember, AN INDUCTOR OPPOSES ANY CHANGE IN CURRENT. In case you have forgotten, a change in current through an inductor produces a changing electromagnetic field. The changing field, in turn, cuts the windings of the wire in the inductor and thereby produces a counter electromotive force (CEMF). It is the CEMF that opposes the change in circuit current. Opposition to a change in current at a given frequency is called inductive reactance (X

Mathematically,

**X _{L} = 2πfL**

If frequency or inductance is increased, the XL increases. Since inductors are placed in series with the load (as shown in figure 4-12), the larger the XL, the larger the ac voltage developed across the load.

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Figure 4-12.—Voltage drops in an inductive filter.

Now refer to figure 4-13. When the current starts to flow through the coil, an expanding magnetic field builds up around the inductor. This magnetic field around the coil develops the CEMF that opposes the change in current. When the rectifier current decreases, as shown in figure 4-14, the magnetic field collapses and again cuts the turns (windings) of wire, thus inducing current into the coil. This additional current merges with the rectifier current and attempts to keep it at its original level.

Figure 4-13.—Inductive filter (expanding field).

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Figure 4-14.—Inductive filter (collapsing field).

Now that you have read how the components in a filter circuit react to current flow from the rectifier, the different types of filter circuits in use today will be discussed.

Q13. If you increase the value of the capacitor, will the X

The simple capacitor filter is the most basic type of power supply filter. The application of the simple capacitor filter is very limited. It is sometimes used on extremely high-voltage, low-current power supplies for cathode-ray and similar electron tubes, which require very little load current from the supply. The capacitor filter is also used where the power-supply ripple frequency is not critical; this frequency can be relatively high. The capacitor (C1) shown in figure 4-15 is a simple filter connected across the output of the rectifier in parallel with the load.

Figure 4-15.—Full-wave rectifier with a capacitor filter.

When this filter is used, the RC charge time of the filter capacitor (C1) must be short and the RC discharge time must be long to eliminate ripple action. In other words, the capacitor must charge up fast, preferably with no discharge at all. Better filtering also results when the input frequency is high; therefore, the full-wave rectifier output is easier to filter than that of the half-wave rectifier because of its higher frequency.

For you to have a better understanding of the effect that filtering has on E

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output waveforms in figure 4-16 represent the unfiltered and filtered outputs of the half-wave rectifier circuit. Current pulses flow through the load resistance (R

Figure 4-16A.—Half-wave rectifier with and without filtering. UNFILTERED

Figure 4-16B.—Half-wave rectifier with and without filtering. FILTERED

The value of the capacitor is fairly large (several microfarads), thus it presents a relatively low reactance to the pulsating current and it stores a substantial charge.

The rate of charge for the capacitor is limited only by the resistance of the conducting diode which is relatively low. Therefore, the RC charge time of the circuit is relatively short. As a result, when the pulsating voltage is first applied to the circuit, the capacitor charges rapidly and almost reaches the peak value of the rectified voltage within the first few cycles. The capacitor attempts to charge to the peak value of the rectified voltage anytime a diode is conducting, and tends to retain its charge when the

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rectifier output falls to zero. (The capacitor cannot discharge immediately.) The capacitor slowly discharges through the load resistance (R

The rate of discharge of the capacitor is determined by the value of capacitance and the value of the load resistance. If the capacitance and load-resistance values are large, the RC discharge time for the circuit is relatively long.

A comparison of the waveforms shown in figure 4-16 (view A and view B) illustrates that the addition of C1 to the circuit results in an increase in the average of the output voltage (E

Now, let's consider a complete cycle of operation using a half-wave rectifier, a capacitive filter (C1), and a load resistor (RL). As shown in view A of figure 4-17, the capacitive filter (C1) is assumed to be large enough to ensure a small reactance to the pulsating rectified current. The resistance of RL is assumed to be much greater than the reactance of C1 at the input frequency. When the circuit is energized, the diode conducts on the positive half cycle and current flows through the circuit, allowing C1 to charge. C1 will charge to approximately the peak value of the input voltage. (The charge is less than the peak value because of the voltage drop across the diode (D1)). In view A of the figure, the charge on C1 is indicated by the heavy solid line on the waveform. As illustrated in view B, the diode cannot conduct on the negative half cycle because the anode of D1 is negative with respect to the cathode. During this interval, C1 discharges through the load resistor (R

Figure 4-17A.—Capacitor filter circuit (positive and negative half cycles). POSITIVE HALF-CYCLE

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Figure 4-17B.—Capacitor filter circuit (positive and negative half cycles). NEGATIVE HALF-CYCLE

Since practical values of C1 and RL ensure a more or less gradual decrease of the discharge voltage, a substantial charge remains on the capacitor at the time of the next half cycle of operation. As a result, no

current can flow through the diode until the rising ac input voltage at the anode of the diode exceeds the voltage on the charge remaining on C1. The charge on C1 is the cathode potential of the diode. When the potential on the anode exceeds the potential on the cathode (the charge on C1), the diode again conducts, and C1 begins to charge to approximately the peak value of the applied voltage.

After the capacitor has charged to its peak value, the diode will cut off and the capacitor will start to discharge. Since the fall of the ac input voltage on the anode is considerably more rapid than the decrease on the capacitor voltage, the cathode quickly become more positive than the anode, and the diode ceases to conduct.

Operation of the simple capacitor filter using a full-wave rectifier is basically the same as that discussed for the half-wave rectifier. Referring to figure 4-18, you should notice that because one of the diodes is always conducting on. either alternation, the filter capacitor charges and discharges during each half cycle. (Note that each diode conducts only for that portion of time when the peak secondary voltage is greater than the charge across the capacitor.)

Figure 4-18.—Full-wave rectifier (with capacitor filter).

Another thing to keep in mind is that the ripple component (E) of the output voltage is an ac voltage and the average output voltage (E

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Introduction to Matter, Energy, and Direct Current, Introduction to Alternating Current and Transformers, Introduction to Circuit Protection, Control, and Measurement, Introduction to Electrical Conductors, Wiring Techniques, and Schematic Reading, Introduction to Generators and Motors, Introduction to Electronic Emission, Tubes, and Power Supplies, Introduction to Solid-State Devices and Power Supplies, Introduction to Amplifiers, Introduction to Wave-Generation and Wave-Shaping Circuits, Introduction to Wave Propagation, Transmission Lines, and Antennas, Microwave Principles, Modulation Principles, Introduction to Number Systems and Logic Circuits, Introduction to Microelectronics, Principles of Synchros, Servos, and Gyros, Introduction to Test Equipment, Radio-Frequency Communications Principles, Radar Principles, The Technician's Handbook, Master Glossary, Test Methods and Practices, Introduction to Digital Computers, Magnetic Recording, Introduction to Fiber Optics