1. 2.1 Power Supply Power supply is a group of circuits that convert the standard ac voltage (120 V, 60 Hz) provided by the wall outlet to constant dc voltage. The voltage produced is used to power all types of electronic circuits including: Consumer electronics (ex: radio, television, DVD, etc.) Computers Industrial controllers Most laboratory instrumentation systems and equipment The dc voltage level required depends on the application, but most applications require relatively low voltage. There are two basic types of power supplies: a linear power supply and a switching power supply. These components are described as follows: 1.A linear power supply is one that provides a constant current path between its input and its load. 2.A switching power supply provides an intermittent current path between its input and its output.

2. Fig. 2-1: Block diagram of a dc power supply with a load and rectifier. (a) Complete power supply with transformer, rectifier, filter, and regulator (b) Half-wave rectifier

3. 2.2 Transformer A transformer is a device that changes ac electric power at one voltage level to ac electric power at another voltage level through the action of a magnetic field. Simple transformer consist of: 1. Primary winding (input winding) 2. Secondary winding (output winding) 3. Magnetic core If the secondary has more turns than the primary, the output voltage across the secondary will be higher and the current will be smaller. If the secondary has fewer turns than the primary, the output voltage across the secondary will be lower and the current will be higher. The core has a function to concentrate the magnetic flux. Fig.2-2: The general arrangement of a transformer

4. There are three types of transformers: step-up, step-down, and isolation. These components are described as follows: 1.The step-up transformer provides a secondary voltage that is greater than the primary voltage. Ex: a step-up transformer may provides a 240 V ac output with a 120 V ac input. 2.The step-down transformer provides a secondary voltage that is less than the primary voltage. Ex: a step-down transformer may provides a 30 V ac output with a 120 V ac input. 3.An isolation transformer provides an output voltage that is equal to the input voltage. This type of transformer is used to isolate the power supply electrically from the ac power line. + - NPNP + - NSNS 120 V ac 240 V ac + - NPNP + - NSNS 120 V ac 30 V ac + - NPNP + - NSNS 120 V ac 1:2 4:11:1 Step-up Step-down Isolation (a)(b)(c) Fig.2-3

5. The turns ratio of a transformer is equal to the voltage ratio of the component and since, the voltage ratio is the inverse of the current ratio. By formula: where N Sec = the number of turns in the secondary N Pri = the number of turns in the primary V Sec = the secondary voltage V Pri = the primary voltage I Sec = the secondary current I Pri = the primary current By the equation (2-1) can be stated that: Step-down transformer secondary current is greater than its primary current (I Sec > I Pri ). Step-up transformer secondary current is less than its primary current (I Pri > I Sec ). (2-1)

6. 2.3 Half-Wave Rectifiers A rectifier is a diode circuits that converts the ac input voltage to a pulsating dc voltage. There are three basic types of rectifier circuits: 1. Half-wave rectifier 2. Full-wave rectifier 3. Bridge rectifier The half-wave rectifier is simply a diode that is placed in series between a transformer (or ac line input) and its load. When a half-wave rectifier is positioned as shown in Fig.2-4 (a), it eliminates the negative alternation of the input. And when positioned as shown in Fig.2-4 (b), it eliminates the positive alternation of the input. The direction of the diode determines whether the output from the rectifier is positive or negative: 1.When the diode points toward the load (R L ), the output from the rectifier will be positive. 2.When the diode points toward the source, the output from the rectifier will be negative.

7. Fig.2-4: Half-wave rectifiers.

8. 2.3.1 Load Voltage and Load Current When we take the value of V F (barrier potential) into account, the peak out voltage (peak load voltage) is found as V p(sec) is the peak secondary voltage which is the peak input voltage (V p(in) ) of the transformer and is found as where N sec /N pri = the ratio of transformer secondary turns to primary turns V p(pri) = the peak primary voltage of a transformer (2-2) (2-3)

9. Fig.2-5: Ideal half-wave rectifier operation. The source voltage are given as rms values as follows: (2-4)

10. 2.3.2 Average Voltage and Current The value of average voltage (V AVG ) for an ac (or other) waveform is the value that would be measured with a dc voltmeter. For a half-wave rectifier, V AVG is found as: where V p = the peak value of the voltage. Fig.2-6 : Average value of the half-wave rectified signal. (2-5)

11. We can also convert a peak current to an average current. The value of the average current (I AVG ) for an ac waveform is the value that would be measured with a dc ammeter. The value of I AVG can be calculated in one of two ways: 1. We can determine the value of V AVG and then use Ohm’s law as follows: 2. We can calculate the value of I p. Then, we can convert this peak value to average form using equations similar to those we used to convert V p to V AVG. The current forms of these equations are: (2-6) (2-7)

12. 2.3.3 Peak Inverse Voltage (PIV) The maximum reverse bias that will be applied to a diode in a given circuit is called the peak inverse voltage (PIV) of the circuit. For the half-wave rectifier, the PIV is found as When the diode is reverse biased (Fig.2-5 (b)), no voltage is dropped across the load. Therefore, all of V sec is dropped across the diode. (2-8)

13. Ex. 2-1: The fuse shown in Fig. 2-7 is used to limit the current in the primary of the transformer. Assuming that the fuse limits the value of I pri to 1 A, what is the limit on the value of the secondary current if the transformer has a turns ratio of 1:4? Ans.: 250 mA Ex. 2-2: A circuit like the one shown in Fig.2-7 has a turns ratio of 1:5 and a fuse that limits the primary current to 750 mA. Calculate the maximum allowable value of I sec. Ans.: 150 mA Ex. 2-3: Determine the peak load voltage for the circuit shown in Fig.2-7. The ratio of secondary turns to primary turns is 5:1. Ans.: 33.3 V Ex. 2-4: A half-wave rectifier has values of N pri = 10, N sec = 1, and V p(pri) = 180 V. What is the peak load voltage for the circuit? Ans.: 17.3 V Ex. 2-5: Using the results of ex. 2-3 and 2-4, what are the peak load current, the value of V AVG and the value of I AVG for the circuit if the load current of the circuit is 120 Ω?

14. Figure 2-7. F 1 1 A

15. 2.4 Full-Wave Rectifiers There are two types of full-wave rectifiers: center-tapped and bridge full-wave rectifier. 2.4.1 The Center-Tapped Full-Wave Rectifier A center-tapped rectifier is a type of rectifier that uses two diodes connected to the center of the secondary winding of a center-tapped transformer, as shown in Fig. 2-8. The input voltage is coupled through the transformer to the center-tapped secondary. Half of the total secondary voltage appears between the center tap and each end of the secondary winding as shown. Fig.2-8: A center-tapped full-wave rectifier.

16. The number of positive alternations that make up the full-wave rectified voltage is twice that of the half-wave voltage for the same time interval, as illustrated by Fig. 2-9. Therefore, the average load voltage for the full-wave rectifier is found as: Fig. 2-9: Typical rectifier waveforms. toto Input signal Half-wave rectifier output Full-wave rectifier output (2-9)

17. Fig. 2-30: The operation of the center-tapped full-wave rectifier during the one complete cycle of the input signal.

18. During a positive half-cycle of the input voltage, the polarities of the secondary voltages are shown in Fig. 2-30 (a). This condition causes diode D 1 is forward-biased, and diode D 2 is reverse-biased. The current path is through D 1 and the load resistor R L, as indicated. During a negative half-cycle of the input voltage, the polarities of the secondary voltages are shown in Fig. 2-30 (b). This condition reverse-biases diode D 1, and forward- biases diode D 2. The current path is through D 2 and R L, as indicated. Because the current through the load resistor is in the same direction, so the output voltage always has the same polarity. Assuming the diode is an ideal model, the peak output voltage (value of peak load voltage, V p(load) ) can be approximated as: The load voltage is approximately equal to half the secondary voltage because the transformer is center tapped. The voltage from either end of a center-tapped transformer to the center tap is always half the total secondary voltage. (2-10)

19. (2-11) Effect of the Turns Ratio on the Output Voltage Using the practical diode model, in the any case, the peak load voltage for a full-wave rectifier is always one-half of the peak secondary voltage less the diode drop (due to the barrier potential): If the transformer’s turns ratio is 1, the output voltage equals half the peak primary voltage less the voltage drop, as shown in Fig. 2-31. Fig.2-31: Center-tapped full-wave rectifier with a transformer turns ratio of 1. (2-12)

20. In order to obtain an output voltage with a peak equal to the input peak, a step- transformer with a turns ratio of n = 2 must be used, as shown in Fig. 2-32. Meaning that the total peak secondary voltage is twice the peak primary voltage (V p(sec) = 2V p(pri) ), so the voltage across each half of the secondary is equal to V p(pri). Fig.2-32: Center-tapped full-wave rectifier with a transformer turns ratio of 2. (2-13)

21. Peak Inverse Voltage When one of the diodes in a full-wave rectifier is reverse-biased as shown in Fig. 2-33 where D 2 is assumed to be reverse-biased, the maximum anode voltage of D 1 is and the maximum anode voltage of D 2 is (2-14) (2-15)

22. Fig.2-33: Full-wave rectifier PIV (D 1 is forward-biased and D 2 is reverse-biased.

23. Thus, the peak inverse voltage across diode D 2 is Since D 1 is assumed to be forward-biased, its cathode is at the same voltage as its anode minus diode drop; this is also the voltage on the cathode of D 2. By formula, Because the peak load voltage supplied by the full-wave rectifier is approximately half the secondary voltage. By formula, or (2-16) (2-17) (2-18)

24. Then, by substitution, the peak inverse voltage across either diode is 2.4.2 The Bridge Full-Wave Rectifier A bridge rectifier is a type of rectifier that uses four diodes connected as shown in Fig. 2-34. The bridge rectifier is the most commonly used full-wave rectifier for several reasons: 1.It does not require the use of a center-tapped transformer and therefore can be coupled directly to the ac power line if desired. 2.When connected to a transformer with the same secondary voltage, it produces nearly twice the peak output voltage of the conventional full-wave rectifier. This results in a higher dc output voltage from the supply. (2-19)

25. Fig.2-34: Bridge rectifier operation. (a) During the positive half-cycle of the input, D 1 and D 2 are forward-biased and conduct current. D 3 and D 4 are reverse-biased. (b) During the negative half-cycle of the input, D 3 and D 4 are forward-biased and conduct current. D 1 and D 2 are reverse-biased.

26. If the center-tapped full-wave rectifier produces its output by alternating conduction between two diodes, the bridge full-wave rectifier alternates conduction between two diode pairs. When the input half-cycle is positive, the transformer has the polarity as shown in Fig. 2-34(a), causing diodes D 1 and D 2 are forward-biased and conduct the current in the direction shown. A voltage is developed across R L that looks like the positive half of the input cycle. During the time, diodes D 3 and D 4 are reverse-biased. When the input half-cycle is negative, the transformer has the polarity as shown in Fig. 2-34(b). Diodes D 3 and D 4 now are forward-biased and conduct the current in the same direction through R L. During the time, diodes D 1 and D 2 are reverse-biased. A full-wave rectified output voltage appears across R L as a result of this action. Bridge Output Voltage Assuming the diodes in the bridge as shown in Fig. 2-35 to be ideal (in other word, the diode drops are neglected), the rectifier has a peak output voltage of (2-20)

27. Fig. 2-35: Bridge operation during a positive half-cycle of the primary and secondary voltages for a ideal diodes.

28. If the voltage drops across the two conducting diodes as shown in Fig. 2-36 are taken into account, the output voltage is The 1.4 V in the equation represents the sum of the diode voltage drops. Note that in the bridge rectifier, two diodes are always in series with the load resistor during both the positive and negative half-cycles. Fig. 2-36: Bridge operation during a positive half-cycle of the primary and secondary voltages for a practical diodes. (2-21)

29. Peak Inverse Voltage When V sec has the polarity as shown in Fig. 2-37(a), D 1 and D 2 are forward-biased that the current flows through D 1 and D 2 as shorts (ideal model). As shown, D 3 and D 4 are connected across the transformer secondary and thus, D 3 and D 4 have a peak inverse voltage equal to the peak secondary voltage. Since the output voltage is ideally equal to the secondary voltage, If the voltage drops of the forward-biased diodes are included as shown in Fig. 2- 36(b), the peak inverse voltage across each reverse-biased diode in terms of V p(out) is (2-22) (2-23)

30. Fig.2-37: Peak inverse voltages across diode D 3 and D 4 in a bridge rectifier during the positive half-cycle of the secondary voltage.

32. 2.5 Filters and Regulators Filter is a circuit implemented with capacitor that follows the rectifier in a power supply. Filters are used to reduce the fluctuation in the rectified output voltage and produces a constant dc output voltage as shown in Fig. 2-38. The small amount of fluctuation in the filter output voltage is called ripple. Fig.2-38: The effects of filtering on the output of a half-wave rectifier.

33. The amount of ripple in the output voltage from a given filter depends on rectifier used, filter component values, and load resistance. Too much ripple in the output can have different adverse effects, depending on the application of the power supply. For example: In an audio amplifier, excessive ripple can produce an “annoying hum” at 60 or 120 Hz, depending on the type rectifier used. In video circuits, excessive ripple can produce “video hum bars” in the picture. In the digital circuits, excessive ripple can result in “erroneous outputs” from logic gates. Therefore, filtering is necessary to provide power and biasing for proper operation of electronic circuits.

34. 2.5.1 Capacitive Filter Capacitive filter is simply a capacitor connected in parallel with the load resistance or connected from the rectifier output to ground, as shown in Fig.2-39. During the positive first quarter-cycle of the input, the diode is forward-biased, allowing the capacitor charges rapidly, as illustrated in Fig.2-39(a). When the input begins to go negative, the diode is reverse-biased, and the capacitor slowly discharges through the load resistance (Fig.2-39(b)). As the output from the rectifier drops below the charged voltage of the capacitor, the capacitor acts as the voltage source for the load. During first quarter of the next cycle, as illustrated in part (c), the diode will again become forward-biased when the input voltage exceeds the capacitor voltage. Capacitor Fig.2-39: Basic capacitive filter. Current indicates charging or discharging of the capacitor

35. Ripple Voltage Ripple voltage is the fluctuation in the capacitor voltage due to the difference between the charge and discharge times. The difference between the charge and discharge times is caused by two distinct RC time constant in the circuit. One time constant is found as: (2-24) where R and C are the total circuit resistance and capacitance, respectively. Since it takes five time constants for a capacitor to charge or discharge fully, this time period (T) can be found as: (2-25)

36. The discharge path for the capacitor is through the resistor as shown in Fig. 2-40(b). For this circuit, the time constant is found as: and the total capacitor discharge time is found as: For example, refer to Fig. 2-40(a), the capacitor charges through the diode. Assuming that diode has a forward resistance of 5 Ω, so the time constant for the circuit is found as: and the total capacitor charge time is found as:

37. (a) Charge circuit (b) Discharge circuit Fig.2-40: The basic capacitive filter.

38. Fig.2-41: Comparison of ripple voltages for half-wave and full-wave rectified voltages with the same filter capacitor and load and derived from the same sinusoidal input voltage. The shorter time between peaks of the rectifier output voltage causes the capacitor reduces the more fluctuation. Therefore, the full-wave rectified voltage has a smaller ripple than the half-wave rectified voltage for the same load resistance and capacitor values (Fig. 2-41).

39. Ripple Factor The ripple factor (r) is an indication of the effectiveness of the filter and defined as: where, V r(pp) is the peak-to-peak ripple voltage and V DC is the dc (average) value of the filter’s output voltage (Fig.2-42). Fig.2-42: V r and V DC determine the ripple factor. (2-26)

40. Fig. 2-43. The lower the ripple factor, the better the filter. The ripple factor (or the amplitude of the ripple voltage) at the output of a filter can be lowered by increasing the value of filter capacitance or increasing the load resistance, as illustrated in Fig. 2-43:

41. For a full-wave rectifier with a capacitive filter, approximations for the peak-to-peak ripple voltage, V r(pp), and dc value of the filter output voltage, V DC, are given in the following equations: V p(rect) = the unfiltered peak rectified voltage. (2-27) (2-28)

42. 2.5.2 Surge Current Before the switch in Fig.2-44 is closed, the filter capacitor is uncharged. At the instant the switch is closed, voltage is connected to the bridge and the uncharged capacitor acts as a short circuit. As a result, the current is initially limited only by the resistance of the transformer secondary and the bulk resistance of the diode. Since these resistances are usually very low, the initial current tends to be extremely high. This high initial current is referred to as surge current. Fig.2-44: Surge current in a capacitive filter.

43. The value of surge current can be calculated as follows: where, V p(sec) = the peak secondary voltage R W = the resistance of the secondary windings R B = the diode bulk resistance (2-29)

44. 2.5.3 Voltage Regulators A voltage regulator is connected to the output of a filtered rectifier and maintains a constant output voltage (or current) despite changes in the input, the load current, or the temperature. The regulator reduces the ripple to negligible amount. Most regulators are integrated circuits and have three terminals: an input terminal, an output terminal, and a reference (or adjust) terminal. Three-terminal regulators designed for fixed output voltages require only external capacitors to complete the regulation portion of the power supply, as shown in Fig.2-45. Fig.2-45: A voltage regulator with input and output capacitors.

45. Fig.2-46: A basic +5 V regulated power supply

46. 2.5.4 Percent Regulation The regulation can be stated in a percentage in terms of input (line) regulation or load regulation. Line regulation specifies how much change occurs in the output voltage for a given change in the input voltage. It is mathematically defined as a ratio of a change in output voltage for a corresponding change in the input voltage expressed as a percentage. Load regulation specifies how much change occurs in the output voltage over a certain range of load current values, usually from minimum current (no load, NL) to maximum current (full load, FL). It can be mathematically determined with the following formula: (2-30) (2-31)

47. 2.6 Clippers Clipper is a diode circuit that is used to limit or clip the positive part of the input voltage. Clipper is often referred to as a limiter. 2.6.1 Series Clippers Each series clipper contains a diode that is positioned in series with a load resistor, as shown in Fig.2-47 (a) and (b). Fig.2-47: Two basic clipper configurations. + - + - RLRL + - + - RLRL I = 0 (a) Negative series clipper(b) Positive series clipper

48. The operating principles of the series clipper are as follows: 1. When the diode in a negative series clipper is forward biased by the input signal, it conducts, and the load voltage is found as 2. When the diode in the negative series clipper is reverse biased by the input signal, it does not conduct. Therefore, A negative series clipper and its associated waveforms are shown in Fig.2-48(a). and 3. The positive series clipper operates in the same fashion. The only differences are: a.The output voltage polarities are reversed. b.The current directions through the circuit are reversed. A positive series clipper and its associated waveforms are shown in Fig.2-48(b). (2-32) (2-33) (2-34)

49. Comparing Fig.2-4 and 2-49, we can see that the operation of the series clipper is identical to that of the half-wave rectifier. Fig.2-48: Series clipper operation

50. 2.6.2 Shunt Clippers Each shunt clipper contains a diode that is positioned in parallel with a load resistor, as shown in Fig.2-49 (a) and (c). The operating principles of the shunt clipper are as follows: As the input signal goes positive (Fig.2-49(a)), the diode is forward biased and conducts current. With the diode conducting, the voltage across the diode equals the diode forward voltage (V F ). Since the diode and load are in parallel, the output voltage also equals the diode forward voltage. By formula: Fig.2-49: Two basic shunt clipper configurations. (a) Positive shunt clipper. The diode is forward-biased during the positive alternation (above 0.7 V). (b) Positive shunt clipper. The diode is reverse-biased during the negative alternation (below 0.7 V). When the input signal goes back below 0.7 V (Fig.2-49(b)), the diode becomes reverse biased and acts as an open circuit. The output voltage looks like the negative part of the input voltage, but with a magnitude determined by the voltage divider formed by R 1 and R L, as follows: + - (2-35)

51. If the diode is turned around, as in Fig. 2-50, the negative part of the input signal is clipped off. When the diode is forward-biased during the negative part of the input signal, the voltage across the diode is held to - 0.7 V. When the input signal goes above -0.7 V, the diode is no longer forward-biased; and the output voltage is proportional to the input voltage. If R 1 is small compared to R L, then (2-36) (2-37) Fig.2-50: Negative shunt clipper. The diode is forward-biased during the negative alternation and reverse-biased during the positive alternation.

52. 2.6.3 Biased Clippers/Limiters A biased clipper uses a dc bias source, V Bias, in series with the diode to limit an output voltage to certain level, as shown in Fig. 2-51. The positive-biased clipper (Fig.2- 51(a)) clips the input signal at the values of V Bias + 0.7 V before the diode will become forward-biased and conduct. Once the diode starts to conduct, the voltage at point A is limited to V Bias + 0.7 V so that all input signal above this level is clipped off. The negative-biased clipper (Fig.2- 51(b)) works in the same fashion, but it clips the input signal at the values of -V Bias - 0.7 V. (a) A positive clipper or limiter (b) A negative clipper or limiter Fig.2-51: Biased shunt clipper or limiter

53. By turning the diode around, the positive clipper can be modified to limit the output voltage to the portion of the input voltage waveform above V Bias - 0.7 V, as shown in Fig.2-52(a). Similarly, the negative clipper can be modified to limit the output voltage to the portion of the input voltage waveform below –V Bias + 0.7 V, as shown in part (b). Fig. 2-52.

54. 2.6.4 Voltage-Divider Bias The bias voltage sources can be replaced by a resistive voltage divider that derives the desired bias voltage from the dc supply voltage, as illustrated in Fig.2-53. The bias voltage is calculated by using the voltage-divider formula as follows: Fig. 2–53: Diode clippers implemented with voltage–divider bias. (a) A positive clipper(b) A negative clipper(b) A variable positive clipper (2-38)

55. Fig.2-54: The input/output characteristics of the positive clamper circuit 2.7 Clampers (DC Restorers) Clamper is a diode circuit designed to shift a waveform either above or below a given reference voltage without distorting the waveform. There are two types of clampers: the positive clamper and the negative clamper. 1. A positive clamper shifts its input waveform so that the negative peak of the waveform is equal to the clamper dc reference voltage. For example: Fig. 2-54 shows what happens when a 20 V pp sin wave is applied to a positive clamper with a dc reference of 0 V. The input and output waveforms have the value of 20 V pp. However, the clamper output waveform has the positive peak of +20 V and the negative peak of 0 V. The positive clamper has shifted the entire waveform so that its negative peak is equal to the circuit’s dc reference voltage.

56. 2. A negative clamper shifts its input waveform so that the positive peak of the waveform is equal to the clamper dc reference voltage. For example: Fig. 2-55 shows what happens when a 20 Vpp sin wave is applied to a negative clamper with a dc reference of 0 V. In this case, The clamper output waveform has the positive peak of 0 V and the negative peak of –20 V. The negative clamper has shifted the entire waveform so that its negative peak is equal to the circuit’s dc reference voltage. Fig.2-55: The input/output characteristics of the negative clamper circuit

57. Clamper Operation The clamper is similar to a shunt clipper; the difference is added capacitor in the clamper, as illustrated in Fig. 2-56. For the circuit in Fig. 2-56(a), the diode is forward biased and it charges the capacitor. Thus, the charging time constant is found as: (a) Capacitor charge circuit Reverse -biased + − (b) Capacitor discharge circuit Fig.2-56: Clamper charge and discharge where R D is bulk resistance of the diode and C is capacitance of the capacitor. The total charge time is: (2-39) (2-40)

58. When the diode is reverse biased, the capacitor starts to discharge through the resistor, as shown in Fig. 2-56(b). Therefore, the discharge time constant is found as: and the total discharge time is found as: (2-41) (2-42) The effect of the clamping action is shown in Fig. 2-56. The capacitor retains a charge approximately equal to the input peak less the diode drop so that it acts as a battery. Fig.2-57: Positive clamper operation

59. If the diode is turned around, a negative dc voltage is added to the input voltage to produce the output voltage as shown in Fig. 2-57: Fig.2-57: Negative clamper operation

60. Ex. 2-6: Assume the circuit in Fig.2-40 has an input of 170 V pk, a turns ratio of 2:1, and values of R W = 0.5 Ω and R B = 8 Ω. What is the initial surge current for the circuit? Ex. 2-7: A negative shunt clipper has a values of R L = 510 Ω and R 1 = 100 Ω. If the input voltage is +15 V pk, what is the peak load voltage?

61. 2.8 Voltage Multipliers A voltage multiplier is a circuit providing a dc output voltage that is a multiple of its peak input voltage. When a voltage multiplier increases a peak input voltage by a given factor, the peak input current is decreased by approximately the same factor. The typical application of a voltage multiplier is to supply the high-voltage, low-current input required to operate the cathode-ray tube in a television and to operate the particle accelerators. There are several types of the voltage multiplier: Voltage doublers : half-wave voltage doubler and full-wave voltage doubler. Voltage tripler Voltage quadrupler

62. Fig.2–58: Half-wave voltage doubler operation. V p is the peak secondary voltage. 2.8.1 Voltage Doubler Voltage doubler is a voltage multiplier with a multiplication factor of two. Half-Wave Voltage Doubler Half-wave voltage doubler consist of two diodes and two capacitors, as shown in Fig.2-58. During the positive half-cycle of the secondary voltage (in Fig.2-58(a)), diode D 1 is forward-biased and D 2 is reverse-biased. Capacitor C 1 charges until its plate-to-plate voltage equals the peak value of the secondary voltage less the diode drop. During the negative half-cycle, D 2 is forward-biased and D 1 is reverse-biased (in part b). AC voltage source and C 1 charge C 2 until its plate-to-plate voltage is approximately 2 V p.

63. Applying Kirchhoff’s law around the loop as shown in part (b), the voltage across C 2 is Neglecting the diode drop of D 2, V C1 = V p. Therefore, Input and output waveforms for a half-wave voltage doubler are shown in Fig. 2-59. Fig. 2-59.

64. Full-Wave Voltage Doubler The full-wave voltage doubler closly resembles the half-wave voltage doubler, as shown in Fig.2-60. During the positive half-cycle of the secondary voltage, D 1 is forward-biased and C 1 charges until its plate-to-plate voltage is approximately equal to V p. During the negative half-cycle, D 2 is forward-biased and C 2 charges until its plate-to- plate voltage is approximately equal to V p. Since C 1 and C 2 are in series, the total output voltage across the two capacitors is 2 V p. Fig.2–60: Full-wave voltage doubler operation.

65. 2.8.2 Voltage Tripler Voltage tripler is variation on the half-wave voltage doubler. It is created by the addition of another diode-capacitor section to the half-wave voltage doubler, as shown in Fig.2- 61. Fig. 2–61: Voltage tripler. On the positive half-cycle of the secondary voltage, C 1 charges through D 1 until its plate-to-plate voltage is approximately equal to V p. During the negative half-cycle, C 2 charges through D 2 until its plate-to-plate voltage is approximately equal to 2V p. During next positive half-cycle, C 3 charges through D 3 until its plate-to-plate voltage is approximately equal to 2Vp. The voltage across C 1 and C 3 add up to 3V p.

66. 2.8.3 Voltage Quadrupler Voltage quadrupler is also variation on the half-wave voltage doubler. It is produced by the addition of still another diode-capacitor section, as shown in Fig.2-62. Fig. 2–62: Voltage quadrupler. C 4 charges through D 4 until its plate-to-plate voltage is approximately equal to 2V p on a negative half-cycle. The voltage across C 1 and C 3 now add up to 4V p.

67. Summary: Half-wave voltage doubler provides a dc output voltage that is approximately twice its peak input voltage. Full-wave voltage doubler provides a dc output voltage that is approximately twice its peak input voltage. Voltage tripler produces a dc output voltage that is approximately three times its peak input voltage. Voltage quadrupler produces a dc output voltage that is approximately four times its peak input voltage.

68. 2.9 The Diode Datasheet Datasheet gives detailed information on a device as maximum ratings, electrical characteristics, mechanical data, and graphs of various parameter. Thus, it can be used properly in a given application. Diode datasheet is commonly divided into two categories: 1. Data given in table form: (a) absolute maximum ratings, (b) thermal characteristics and (c) electrical characteristics. 2. Characteristics shown in graphical form: (a) forward current derating curve, (b) forward characteristics curve, (c) nonrepetitive surge current and (d) reverse characteristics.

69. Fig. 2–63: A selection of rectifier diodes based on maximum ratings of I O, I FSM, and V RRM.

71. 2.10 Troubleshooting Power supply faults may occur in the transformer, rectifier, filter, or regulator. 2.10.1 Transformer Faults The transformer in a power supply develop one of several possible faults: 1.A shorted primary or secondary winding. 2.An open primary or secondary winding. 3.A short between the primary and secondary winding and the transformer frame. In most cases, a shorted primary or secondary winding will cause the fuse to blow. 2.10.2 Rectifier Faults Half-wave rectifier is the easiest rectifier to troubleshoot. If the diode in the rectifier shorts, the output will be a sine wave. If the diode opens, the resulting symptom is an output of 0 V, as illustrated in Fig. 2-64.

72. Fig. 2–64: The effect of an open diode in a half-wave rectifier is an output of 0 V.

73. In full-wave rectifier, a shorted diode will cause the power supply fuse to blow. Full- wave center-tapped rectifier is shown in Fig.2-65. If either of two diodes is open, the output voltage will have twice the normal ripple voltage and the ripple frequency will go from 120 to 60 Hz. Fig. 2–66: Effect of an open diode in a bridge rectifier.

74. The symptoms for open diode in the bridge rectifier are the same as those for the center-tapped full-wave rectifiers, as shown in Fig.2-66. In the case, we have more diodes that must be tested. Fig. 2–66: Effect of an open diode in a bridge rectifier.

75. 2.10.3 Filter Faults Capacitors store an electrical charge, and they can retain that charge even after the power switch has been turned off or the ac plug has been disconnected. This is a danger to the technician and also, the test equipment can be damaged. As a safety precaution, capacitors should be discharged by shorting the terminals with a shorting tool before taking any measurements. Three types of defects of a filter capacitor are illustrated in Fig.2-67: If the filter capacitor for a full-wave rectifier opens, the output is a full-wave rectified voltage. If the filter capacitor shorts, the fuse will blow and the output is 0 V. When the filter capacitor shorts, this effectively shorts out the transformer secondary, causing excessive secondary current. If the filter capacitor leaks, the leakage resistance will reduce the time constant and allow the capacitor to discharge more rapidly than the normal. This results in an increase in the ripple voltage on the output. This fault is rare.

76. Fig. 2–67: Effects of a faulty filter capacitor.