CHAPTER 2. Diode Applications

Transcription

1 CHAPTER 2 Diode Applications 1

2 Objectives Explain and analyze the operation of both half and full wave rectifiers Explain and analyze filters and regulators and their characteristics Explain and analyze the operation of diode limiting and clamping circuits Explain and analyze the operation of diode voltage multipliers Interpret and use a diode data sheet Troubleshoot simple diode circuits 2

3 Lecture s contents 2-1 The Basic DC Power Supply 2.2 Half-wave rectifier 2-3 Full-wave rectifier 2-4 Power supply filters and regulators 2-5 Diode limiting and clamping circuits 2-6 Voltage multipliers 2-7 Diode data sheet 2-8 Troubleshooting Summary 3

4 2-1 The Basic DC 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. 4

5 2-1 The Basic DC Power Supply(cont.) The basic function of a DC power supply is to convert an AC voltage to a constant DC voltage (AC DC) Eliminates the fluctuations - produce smooth dc voltage Maintains a constant dc voltage Either half or full-wave rectifier convert ac input voltage to a pulsating dc voltage. (a) Complete power supply with transformer, rectifier, filter, and regulator (b) Half-wave rectifier Fig. 2-1: Block diagram of a dc power supply with a load and rectifier. 5

6 2-1 The Basic DC Power Supply(cont.) The basic function of a DC power supply is to convert an AC voltage to a constant DC voltage (AC DC) Eliminates the fluctuations - produce smooth dc voltage Maintains a constant dc voltage Either half or full-wave rectifier convert ac input voltage to a pulsating dc voltage. (a) Complete power supply with transformer, rectifier, filter, and regulator (b) Half-wave rectifier Fig. 2-1: Block diagram of a dc power supply with a load and rectifier. 6

7 2-1 The Basic DC Power Supply(cont.) 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. Φ Fig.2-2: The general arrangement of a transformer 7

8 2-1 The Basic DC Power Supply(cont.) 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. + - N N + P S V ac Step-up V ac N N + P S V ac Step-down V ac N N + P S V ac Isolation V ac - (a) (b) (c) Fig.2-3 8

9 2-1 The Basic DC Power Supply(cont.) 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: N N sec pri V = V sec pri = 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 I I pri sec (2-1) 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 ). 9

10 2-2 2 Half-Wave Rectifiers Diode ability to conduct current in one direction and block current in other direction used in circuit called RECTIFIER (ac dc) Objective: Discuss the operation of half-wave rectifiers Describe a basic dc power supply & half-wave rectifications Determine the average value, V AVG of half-waves rectified voltage Discuss the effect of barrier potential, V P on a half-wave rectifier output Define Peak Inverse Voltage (PIV) Describe Transformer-couple half-wave rectifier 10

11 2-2 2 Half-Wave Rectifiers (cont.) A half wave rectifier(ideal) allows conduction for only 180 or half of a complete cycle. During first one cycle: -Vin goes positive diode FB conduct current -Vin goes negative diode RB no current- 0V The output frequency is the same as the input (same shape). (The Half-Wave Rectifier) ac source load resistor The average value V DC or V AVG : Vp V AVG = π -Measure on dc voltmeter Ideal diode model 11

12 2-2 2 Half-Wave Rectifiers (cont.) (Effect of the Barrier Potential on the Half-Wave Rectifier Output) Practical Diode barrier potential of 0.7V (Si) taken into account. During +ve half-cycle V in must overcome V potential for forward bias. Example 1: Calculate the peak o/p voltage, Vp(out)? The peak o/p voltage: V = V 0. 7 V p( out) p( in) 7 V p( out) = V p( in) = 5V 0.7 V = 4.30 V 0.7 V 12

13 2-2 2 Half-Wave Rectifiers (cont.) (Effect of the Barrier Potential on the Half-Wave Rectifier Output) Example 2: Sketch the output V 0 and determine the output level voltage for the network in above figure. 13

14 2-2Half 2Half-Wave Rectifiers (cont.) - Peak inverse voltage (PIV) is the maximum voltage across the diode when it is in reverse bias. [Peak Inverse Voltage (PIV)] The diode must be capable of withstanding this amount of voltage. PIV =V p(in) 14

15 2-2 2 Half-Wave Rectifiers (cont.) (Half-Wave Rectifier with Transformer-Coupled Input Voltage) Transformers are often used for voltage change and isolation. The turns ratio, n of the primary to secondary determines the output versus the input. to couple ac input to the rectifier V sec = nv pri The advantages of transformer coupling: 1) allows the source voltage to be stepped up or down 2) the ac source is electrically isolated from the rectifier, thus prevents shock hazards in the secondary circuit. n = N sec N pri If n>1, Vsec is greater than Vpri. If n<1, Vsec is less than Vpri. If n=1, Vsec= Vpri. Vp ( out) = Vp (sec) 0. 7V PIV =V p(sec) 15

16 Example 3: Determine the peak value of output voltage as shown in below Figure. 16

17 2-3 Full-Wave Rectifiers (Introduction) Objective: Explain & Analyze the operation of Full-Wave Rectifier. Discuss how full wave rectifier differs from half-wave rectifier Determine the average value Describe the operation of center-tapped & bridge. Explain effects of the transformers turns ratio PIV Comparison between center-tapped & bridge. 17

18 2-3 Full-Wave Rectifiers (cont.) (Introduction) A full-wave rectifier allows current to flow during both the positive and negative half cycles or the full 360º whereas half-wave rectifier allows only during one-half of the cycle. The no. of +ve alternations is twice the half wave for the same time interval The output frequency is twice the input frequency. The average value the value measured on a dc voltmeter V AVG = 2V π p Twice output 63.7% of Vp 18

19 2-3 Full-Wave Rectifiers (i - The Center-Tapped Full-Wave Rectifier) This method of rectification employs two diodes connected to a secondary center-tapped transformer. The i/p voltage is coupled through the transformer to the center-tapped secondary. Coupled input voltage 19

20 2-3 Full-Wave Rectifiers (cont.) (i - The Center-Tapped Full-Wave Rectifier) +ve half-cycle input voltage (forward-bias D1 & reverse-bias D2)-the current pass through the D1 and RL -ve half-cycle input voltage (reverse-bias D1 & forward-bias D2)-the current pass through D2 and RL The output current on both portions of the input cycle same direction through the load. The o/p voltage across the load resistors full-wave rectifiers 20

21 2-3 Full-Wave Rectifiers (cont.) (i - The Center-Tapped Full-Wave Rectifier) -Effect of the Turns Ratio on the Output Voltage- If n=1, V p(out) =V p(pri) - 0.7V 2 V p(sec) =V p(pri) V = sec 2V pri If n=2, V 2V sec = pri ( out) = p( pri) V p V 0.7 In any case, the o/p voltage is always one-half of the total secondary voltage minus the diode drop (barrier potential), no matter what the turns ratio. Vsec V out = 0. 7V 2 21

22 2-3 Full-Wave Rectifiers (cont.) (i - The Center-Tapped Full-Wave Rectifier) -Peak Inverse Voltage (PIV)- V p V p (sec) (sec) Maximum anode voltage: D 1 = + D2 = 2 2 D1: forward-bias its cathode is at the same voltage of its anode minus diode drop; This is also the voltage on the cathode of D2. PIV across D2 : Vp(sec) Vp(sec) PIV = 0.7V 2 2 = V 0.7V V p( out) p(sec) We know that Vp(sec) = 0.7V 2 V = 2V + 1.4V p(sec) Thus; PIV p( out) = 2 V out ) p( V 22

23 2-2 2 Full-Wave Rectifiers (cont.) (ii - The Bridge Full-Wave Rectifier) It employs four diodes arranged such that current flows in the direction through the load during each half of the cycle. When Vin +ve, D1 and D2 FB and conduct current. A voltage across R L looks like +ve half of the input cycle. During this time, D3 and D4 are RB. When Vin ve, D3 and D4 are FB and conduct current. D1 and D2 are RB. Used 4 diode: 2 diode in forward 2 diode in reverse Without diode drop (ideal diode): V = V p( out) p(sec) 2 diode always in series with load resistor during +ve and ve half cycle. With diode drop (practical diode): Vp( out) = Vp(sec) 1. 4V 23

24 2-2 2 Full-Wave Rectifiers (cont.) (ii - The Bridge Full-Wave Rectifier) For ideal diode, PIV = Vp(out) For each diode, PIV = V p( out ) V PIV = V p(out) 0V (ideal diode) PIV = V p( out) V Note that in most cases we take the diode drop into account. 24

25 2-3 Power Supply Filters And Regulators (introduction) Objective: Explain & Analyze the operation & characteristic of power supply filters & Regulators Explain the purpose of a filter Describe the capacitor-input input filter Define ripple voltage & calculate the ripple voltage Discuss surge current in capacitor-input input filter Discuss voltage regulation & integrated circuit regulator 25

26 2-4 Power Supply Filters And Regulators (cont.) (introduction) Power Supply Filters To reduce the fluctuations in the output voltage of half / full-wave rectifier produces constant-level dc voltage. It is necessary electronic circuits require a constant source to provide power & biasing for proper operation. Filters are implemented with capacitors. Regulators Voltage regulation in power supply done using integrated circuit voltage regulators. To prevent changes in the filtered dc voltage/ to fix output dc voltage due to variations in input voltage or load. 26

27 2-4 Power Supply Filters And Regulators (cont.) (introduction) In most power supply 60 Hz ac power line voltage is converted to constant dc voltage. 60Hz pulsating dc output must be filtered to reduce the large voltage variation. Small amount of fluctuation in the filter o/p voltage - ripple ripple 27

28 2-4 Power Supply Filters And Regulators (cont.) Capacitive Filter Capacitive filter is simply a capacitor connected in parallel with the load resistance or connected from the rectifier outpu to ground, as shown in Fig (Capacitor-Input Filter) load 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). capacitor 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. 28

29 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: τ = RC 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: T = 5 RC = 5τ 29

30 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: τ = ( 5Ω)(100 µf ) = 500 µs and the total capacitor charge time is found as: T = ( 5)(500 µ s) = 2. 5ms 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: τ = ( 1kΩ)(100 µ F) = 100ms and the total capacitor discharge time is found as: T = ( 5)(100 ms) = 500 ms 30

31 (a) Charge circuit (b) Discharge circuit Fig.2-40: The basic capacitive filter. 31

32 2-4Power Supply Filters And Regulators (cont.) (Capacitor-Input Filter) Ripple Voltage: the variation in capacitor voltage due to the charging and discharging times. The advantage of a full-wave rectifier over a half-wave is quite clear. The capacitor can more effectively reduce the ripple when the time between peaks is shorter. Easier to filter -shorted time between peaks. -smaller ripple. 32

33 2-4 Power Supply Filters And Regulators (cont.) Ripple factor: indication of the effectiveness of the filter r = V r ( pp) V DC (Capacitor-Input Filter) [half-wave rectifier] V r(pp) = peak to peak ripple voltage; V DC = V AVG = average value of filter s output voltage. Lower ripple factor better filter [can be lowered by increasing the value of filter capacitor or increasing the load resistance] For the full-wave rectifier: V V r( pp) DC = V AVG 1 fr C L V = 1 p( rect ) 1 2 fr L C V p( rect ) V p(rect) = unfiltered peak. 33

34 2-4 Power Supply Filters And Regulators (cont.) (Capacitor-Input Filter) Surge Current in the Capacitor-Input Filter: When you first turn on power supply, the filter capacitor has no accumulated charge to oppose V p(sec). For first instant, the capacitor appears as a short circuit, thus the current through the diodes can momentarily be quite high. To reduce risk of damaging the diodes, a surge current limiting resistor is placed in series with the filter and load. The min. surge Resistor values: R surge = V p(sec) 1.4 I FSM V I FSM = forward surge current rating specified on diode data sheet. 34

35 2-4 Power Supply Filters And Regulators (cont.) (IC Regulators) Connected to the output of a filtered & maintains a constant output voltage (or current) despite changes in the input, load current or temperature. Combination of a large capacitor & an IC regulator inexpensive & produce excellent small power supply Popular IC regulators have 3 terminals: (i) input terminal (ii) output terminal (iii) reference (or adjust) terminal Type number: 78xx (xx refer to output voltage) i.e 7805 (output voltage +5.0V); 7824 (output voltage +24V) 35

36 2-4 Power Supply Filters And Regulators (cont.) (IC Regulators) Regulation is the last step in eliminating the remaining ripple and maintaining the output voltage to a specific value. Typically this regulation is performed by an integrated circuit regulator. There are many different types used based on the voltage and current requirements. Connected to the output of filtered rectifier output Bridge-full wave Gnd rectifier filter regulators Fig : A basic +5.0V regulated power supply 36

37 2-4 Power Supply Filters And Regulators (cont.) Percent Regulation (Percent Regulations) 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. Line regulation V OUT = x100% V IN 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: VNL V Load regulation= VFL FL x100% 37

38 Objectives: 2-5 Diode Limiting & Clamping Circuits (Introduction) Analyze the operation of diode limiting, clamping circuit, voltage multipliers and interpret and use diode data sheet. Determine V of biased limiter & used voltage-divider bias to set limiting level. Discuss voltage doublers, triplers & quadruples. Identify V & current ratings. Determine the electrical characteristics of a diode. Analyze graphical data Select an appropriate diode for a given set of specifications. 38

39 2-5 Diode Limiting & Clamping Circuits (Diode Limiters) Diode limiters/clippers that limits/clips the portion of signal voltage above or below certain level. Limiting circuits limit the positive or negative amount of an input voltage to a specific value. 4 basic clipper configuration: Negative series clipper diode is in series with its load Positive series clipper Negative shunt clipper diode is in parallel with its load Positive shunt clipper 39

40 2-5 Diode Limiting & Clamping Circuits (Diode Limiters) Clipper configuration 40

41 2-5 Diode Limiters (series clipper/limiter) Negative series clipper Diode is forward-biased during +ve alternation of i/p signal. Diode is reverse-biased when i/p signal is ve. Eliminates negatives alternation of its i/p. Positive series clipper Diode is reverse-biased during +ve alternation of i/p signal. Diode is forward-biased when i/p signal is ve. Eliminates positive alternation of its input. 41

42 2-5Diode Limiters (shunt clipper/limiter) Positive shunt clipper Forward-biased diode when i/p is +ve cycle. Reverse-biased diode when i/p is in ve cycle. o/p signal is limit/clip to +0.7V during +ve cycle of i/p signal. Negative shunt clipper Reverse-biased diode act as open cct during +ve cycle. Forward-biased diode act as short cct during ve cycle. o/p signal is limit/clip to -0.7V during ve cycle of i/p signal. 42

43 2-5 Diode Limiting & Clamping Circuits (Diode Limiters) Series clipper When diode in ve series clipper is FB, load voltage is: V L =V in 0.7V When diode is RB, doesn t conduct, so: V L = 0V +ve series clipper operates the same. The only differences are: O/p voltage polarities are reversed. V L =-V in + 0.7V Current direction through cct are reversed. Shunt clipper For veshunt, when i/p signal +vecycle, diode is RB (open circuit), thus: V out R = R 1 + L R V During vecycle, diode is FB, load voltage is equal to diode forward voltage. V L = -V F = -0.7V For +veshunt, o/p voltage and current direction are reversed. L in 43

44 Question 4: What would you expect to see displayed on an oscilloscope connected across RL in the limiter shown below. 44

45 Solution question 4 The diode is forward biased and conducts when input voltage goes below -0.7V. So, for ve limiter, the peak output voltage across RL is: RL 1.0k Ω Vp ( out) = Vp ( in) 10V = 9. 09V R1 R = + L 1.1k Ω The waveform is shown below: 45

46 2.5 Diode Limiting & Clamping Circuits (cont.) (Diode Limiters) Biased Limiters : Use dc biasing source, V BIAS to set limit on the circuit output voltage. This allow the circuit to clip input waveform at values other than diode forward voltage, 0.7V. In each circuit, bias voltage is in series with shunt diode. As a result, the diode conducts and clips the i/p waveform when signal voltage equals sum of V F and V BIAS. 2 type of biased limiter: positive-biased limiter negative-biased limiter Positive limiter The voltage at point A must equal V BIAS +0.7V before diode become FB and conduct. Once diode begin to conduct, voltage at point A is limited to V BIAS +0.7V, so all i/p voltage above this level is clipped off. Negative limiter Voltage at point A must go below V BIAS - 0.7V to forward-bias the diode and initiate limiting action. A positive limiter A negative limiter 46

47 Voltage-Divider Bias: 2-5 Diode Limiting & Clamping Circuits (cont.) (Diode Limiters) The bias voltage source used to illustrate the basic operation of diode limiters can be replace by a resistive voltage divider that derives the desired bias voltage from dc Vsupply. VBIAS set by the resistor values according to the voltage-divider formula: V R BIAS = V + SUPPLY R 2 R The desired amount of limitation can be attained by a power supply or voltage divider. The amount clipped can be adjusted with different levels of VBIAS. The bias resistor << R1- the forward current through the diode will not effect V BIAS 47

48 48

49 2-5 Diode Limiting & Clamping Circuits (cont.) (Diode Limiter Applications) Half-wave rectifier Circuit alters the shape of ac signal and change it to pulsating dc. Transient protection circuit Transient abrupt current or voltage spike in short duration. Many digital circuits have i/p that cannot tolerate voltage fall outside a specified range which can cause serious damage. A clipper can be used to protect these circuits. AM detector Eliminate ve portion of i/p waveform, so capacitor charges and discharges at rate of peak i/p variations. This provides a signal at load that is a reproduction of i/p signal. 49

50 Example 5: 1. Sketch the output voltage waveform as shown in the circuit combining a positive limiter with negative limiter in Figure V -15V 6V 6V Figure

51 Example 5 (cont.): 2. A student construct the circuit as shown in Figure 5-2. Describe the output voltage waveform on oscilloscope CH2. +20V -20V +15V CH2 Figure

52 2.7 Clampers (DC Restorers) 2-5 Diode Limiting & Clamping Circuits (cont.) (Diode Clampers) 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 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.

53 2-5 Diode Limiting & Clamping Circuits (cont.) (Diode Clampers) 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 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.

54 Clamper Operation The clamper is similar to a shunt clipper; the difference is added capacitor in the clamper, as illustrated in Fig 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 τ = R D C τ (2-39) where R D is bulk resistance of the diode and C is capacitance of the capacitor. The total charge time is: + Reversebiased T Charg e = 5 R D C (2-40) (b) Capacitor discharge circuit Fig.2-56: Clamper charge and discharge

55 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: τ = R C L (2-41) and the total discharge time is found as: T Discharg e = 5 R C L (2-42) The effect of the clamping action is shown in Fig 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 55

56 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 56

57 2-5 Diode Limiting & Clamping Circuits (cont.) A Clamper Application: (Diode Clampers) A clamping circuit is often used in TV receivers as a dc restorer. The incoming composite video signal is normally processed through capacitively coupled amplifiers that eliminate the dc component, thus losing black and white reference levels and the blanking level. Before applied to the picture tube, these reference level must be restored. 57

58 2-6 Voltage Multiplier (Introduction) Use clamping action to increase peak rectified voltages without necessary to increase input transformer s voltage rating. Multiplication factors: two, three or four. Three types of voltage multipliers: * Voltage doubler - Half wave voltage doubler - Full wave voltage doubler * Voltage tripler * Voltage Quadrupler Voltage multipliers are used in high-voltage, low current applications, i.e TV receivers. 58

59 Half-wave voltage Doubler: 2-6 Voltage Multiplier (cont.) (Voltage Doubler) Clamping action can be used to increase peak rectified voltage. Once C 1 and C 2 charges to the peak voltage they act like two batteries in series, effectively doubling the voltage output. The current capacity for voltage multipliers is low. PIV = 2V p By applying Kirchhoff s Law at (b): V V + V = ~ approximately 2V p (neglecting diode drop D 2 ) C 2 p C1 Half-wave voltage doubler operation. V p is the peak secondary voltage. 59

60 Full-wave voltage doubler: 2-6 Voltage Multiplier (cont.) (Voltage Doubler) Arrangement of diodes and capacitors takes advantage of both positive and negative peaks to charge the capacitors giving it more current capacity. forward-bias output charges forward-bias charges Secondary voltage positive Secondary voltage negative 60

61 2-6 Voltage Multiplier (cont.) (Voltage Tripler & Voltage Quadrupler) Voltage triplers and quadruplers utilize three and four diode capacitor arrangements, respectively. Voltage tripler and quadrupler gives output 3V p and 4V p, respectively. Tripler output is taken across C 1 and C 3, thus V out = 3V p Quadrupler output is taken across C 2 and C 4, thus V out = 4V p PIV for both cases: PIV = 2V p Voltage Triple Voltage Quadruple 61

62 2-7 The Diode Data Sheet (Introduction) The data sheet for diodes and other devices gives detailed information about specific characteristics such as the various maximum current and voltage ratings, temperature range, and voltage versus current curves (V-I characteristic). It is sometimes a very valuable piece of information, even for a technician. There are cases when you might have to select a replacement diode when the type of diode needed may no longer be available. These are the absolute max. values under which the diode can be operated without damage to the device. 62

63 2-7 The Diode Data Sheet (cont.) (Maximum Rating) Rating Peak repetitive reverse voltage Working peak reverse voltage DC blocking voltage Nonrepetitive peak reverse voltage rms reverse voltage Symbol V RRM V RWM V R 1N4001 1N4002 1N4003 UNIT V V RSM V V R(rms) V Average rectified forward current (single-phase, resistive load, 60Hz, T A = 75 o C I o 1 A Nonrepetitive peak surge current (surge applied at rated load conditions) I FSM 30 (for 1 cycle) A Operating and storage junction temperature range T j, T stg -65 to +175 o C 63

64 2-7 The Diode Data Sheet (cont.) (Maximum Rating) FIGURE 2-56 A selection of rectifier diodes based on maximum ratings of I O, I FSM, and I RRM 64 RRM.

65 2-8Troubleshooting (introduction) Objective: Troubleshoot diode circuits using accepted techniques. Discuss the relationship between symptom & cause, power check, sensory check, component replacement method and discuss the signal tracing technique in the three variations. Fault analysis. Our study of these devices and how they work leads more effective troubleshooting. Efficient troubleshooting requires us to take logical steps in sequence. Knowing how a device, circuit, or system works when operating properly must be known before any attempts are made to troubleshoot. The symptoms shown by a defective device often point directly to the point of failure. There are many different methods for troubleshooting. We will discuss a few. 65

66 2-8 Troubleshooting (cont.) (Troubleshooting Techniques) Here are some helpful troubleshooting techniques: Power Check: Sometimes the obvious eludes the most proficient troubleshooters. Check for fuses blown, power cords plugged in, and correct battery placement. Sensory Check: : What you see or smell may lead you directly to the failure or to a symptom of a failure. Component Replacement: Educated guesswork in replacing components is sometimes effective. Signal Tracing: Look at the point in the circuit or system where you first lose the signal or incorrect signal. 66

67 Signal tracing techniques: 2-8 Troubleshooting (cont.) (Troubleshooting Techniques) Input to output Output to input 67

68 2-8 Troubleshooting (cont.) (Fault Analysis) Can be applied when you measure an incorrect voltage at a test point using signal tracing and isolate the fault to a specific circuit. Example 1: Effect of an Open Diode in a Half- Wave Rectifier: - Zero o/p voltage - Open diode breaks the current path from transformer secondary winding to the filter and load resistor no load current. Other faults: open transformer winding, open fuse, or no input voltage. 68

69 2-8 Troubleshooting (cont.) (Fault Analysis) Example 2: Effect of an Open Diode in a Full-Wave Rectifier: -The effect of either of two diodes is open diode, the o/p voltage will have large than normal ripple voltage at 60 Hz rather than at 120 Hz. - Another fault open in one of the halves of the transformer secondary winding. - Open diode give same symptom to bridge full-wave rectifier. (See Figure 2-63) 69

70 2-8 Troubleshooting (cont.) (Fault Analysis) Example 3: Effect of a Shorted Diode in a Full-Wave Rectifier: Fuse should blow cause by short circuit D1,D4 will probably burn open. 70

71 2-8 Troubleshooting (cont.) (Fault Analysis) Example 4: Effect of a fault filter capacitor: Open o/p is full-wave rectified voltage Shorted the o/p is 0V Leaky increase the ripple voltage on the o/p Example 5: Effect of a Faulty Transformer: Open primary/secondary winding of a transformer 0V o/p 71

72 2-8 Troubleshooting (cont.) (The complete Troubleshooting Process) The complete troubleshooting process: (i) Identify the symptom(s). (ii) Perform a power check (iii) Perform a sensory check (iv) Apply a signal tracing technique. (v) Apply fault analysis (vi) Use component replacement to fix the problem. 72

73 Summary The basic function of a power supply to give us a smooth ripple free DC voltage from an AC voltage. Half-wave rectifiers only utilize half of the cycle to produce a DC voltage. Transformer Coupling allows voltage manipulation through its windings ratio Full-Wave rectifiers efficiently make use of the whole cycle. This makes it easier to filter. The full-wave bridge rectifier allows use of the full secondary winding output whereas the center-tapped full wave uses only half. Filtering and Regulating the output of a rectifier helps keep the DC voltage smooth and accurate 73

74 Summary Limiters are used to set the output peak(s) to a given value. Clampers are used to add a DC voltage to an AC voltage. Voltage Multipliers allow a doubling, tripling, or quadrupling of rectified DC voltage for low current applications. The Data Sheet gives us useful information and characteristics of device for use in replacement or designing circuits. Troubleshooting requires use of common sense along with proper troubleshooting techniques to effectively determine the point of failure in a defective circuit or system. 74

75 Solution 2: The peak output voltage for the circuit is: V p(out) = V p(in) 0.7V = = 4.30 V 75

76 Solution 3: V p(pri) = V p(in) = 156 V The peak secondary voltage is: V p(sec) = nv p(pri) = 0.5 (156 V) = 78 V The rectified peak output voltage is: V p(out) = V p(sec) 0.7V = = 77.3 V where V p(sec) is the input to the rectifier. 76

77 Solution 5: V -6.7V 77

78 Solution 5 (cont.): 2. VBIAS = R = 10.31V 2 R + 3 R 3 V SUPPLY = V V V 78