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# C hapter 2 diode applications

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### C hapter 2 diode applications

1. 1. DIODEAPPLIC CHAPTER 2ATIONS MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
2. 2. TOPIC OUTLINES  LOAD LINE ANALYSIS SERIES DIODE CONFIGURATION PARALLEL AND SERIES-PARALLEL CONFIGURATION AND/OR GATES HALF WAVE RECTIFICATION FULL WAVE RECTIFICATION CLIPPERS CLAMPERS ZENER DIODES VOLTAGE MULTIPLIER CIRCUITSMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
3. 3. Load Line Analysis• The load line plots all possible current (ID) conditions for all voltages applied to the diode (VD) in a given circuit. E/R is the maximum ID and E is the maximum VD.• Where the load line and the characteristic curve intersect is the Q point, which specifies a particular ID and VD for a given circuit.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
4. 4. How to determine the Q point of a system?• Identify diode model• Using Kirchoff’s Law : – Set VD = 0V (horizontal line) – Set ID = 0A (vertical line) – Obtain VDQ, IDQ from the graph intersection ( Q-point)MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
5. 5. Diode ApproximationApproximate model notationIn Forward Bias: SiSilicon Diode: VD = 0.7VGermanium Diode: VD = 0.3V GeIn Reverse Bias:Both diodes act like opens VD = source voltage and ID =0A Ideal model notation VD = 0V and ID = 0AMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
6. 6. Diode in DC Series Circuit: Forward Bias The diode is forward bias. • VD = 0.7V (or VD = E if E <0.7V) • VR = E – VD • ID = IR = VR /RMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
7. 7. Diode in DC Series Circuit: Reverse Bias The diode is reverse biased. • VD = E • VR = 0V • ID = IR = IT = 0AMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
8. 8. • An open circuit can have any voltage across its terminals, but the current is always 0A.• A short circuit has a 0V drop across its terminals, but the current is limited only by the surrounding network.• Source notation :MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
9. 9. ExampleFor the series diode configuration below, employing the diodecharacteristics of figure below, determine VDQ, IDQ and VR.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
10. 10. SolutionStep1: Find the maximum ID. VD = 0V→ ID = IR= E/RStep 2: Find the maximum VD. ID =0A → E = VD + IDRStep 3: Plot the load lineStep 4 : Find the intersection between the load line and thecharacteristic curve. This is the Q-pointStep 5: Checking !!!! MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
11. 11. Example Determine ID, VD2 and Vo for the circuit.Remember, the combination ofshort circuit in series with anopen circuit always results inan open circuit and ID=0A. MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
12. 12. Example Determine I, V1, V2 and VoMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
13. 13. Series – Parallel ConfigurationsSolve this circuit like any Series/Parallelcircuit, knowing VD = 0.7V (or up to 0.7V) inforward bias and as an open in reverse bias. VD1 = VD2 = Vo =0 .7V VR = 9.3VDiodes in parallel are used to limit current:IR = E – VD = 10V -0 .7V = 28mA R 0.33kID1 = ID2 = 28mA/2 = 14mA MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
14. 14. Example Determine the resistance R for the network when I=200mA. Si SiMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
15. 15. Example Determine the currents I1, I2, and ID2 for the networkMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
16. 16. Diodes in AC Circuits •Inputs: -Sinusoidal waveform -Square wave •This circuit is called half-wave rectifier, which generate waveform vo that will have an average value of particular use in the ac-to-dc conversion process. •The diode only conducts when it is in forward bias, therefore only half of the AC cycle passes through the diode.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
17. 17. Half-Wave Rectification•The diode that employed in the rectification process is typically referred to rectifier.•The diode only conducts for one-half of the AC cycle. The remaining half is either all positive or all negative. This is a crude AC to DC conversion.•The DC Voltage out of the diode : Vdc = 0.318Vm where Vm = the peak voltageMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
18. 18. Peak Inverse Voltage (PIV)• Because the diode is only forward biased for one-half of the AC cycle, it is then also off for one-half of the AC cycle. It is important that the reverse breakdown voltage rating of the diode be high enough to withstand the peak AC voltage. – PIV (PRV) > Vm• PIV = Peak Inverse Voltage; PRV = Peak Reverse Voltage• Vm = Peak AC VoltageMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
19. 19. Full-Wave Rectification: Bridge Network• The dc level obtained from a sinusoidal input can be improved 100% using a process called full-wave rectification.• The most familiar network is bridge configuration with 4 diodes. Vdc = 0.636 VmMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
20. 20. Operation of the Bridge Rectifier CircuitFor the positive half of the AC cycle: For the negative half of the AC cycle: MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
21. 21. Determining Vo for silicon diodes in the bridge configuration The effect of using a silicon diode with VD=0.7 is demonstrated in below figure. The dc level has change to:MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
22. 22. Example Determine the output waveform for the network below and calculate the output dc level.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
23. 23. Solution Conduction path for the +ve regionMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
24. 24. Solution Conduction path for the -ve regionMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
25. 25. Full-Wave Rectification: Center-Tapped TransformerA second popular full-wave rectifier with only two diodes butrequiring a center-tapped transformer to establish the inputsignal across each section of the secondary of the transformer. Two diodes and a center-tapped transformer are required. VDC = 0.636(Vm) for ideal diode Note that Vm here is the transformer secondary voltage to the tap.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
26. 26. Operation of the Center–Tapped Transformer Rectifier Circuit For the positive half of the AC cycle During the positive cycle of vi applied to the primary of the Transformer the network will appear as shown in figure. D1 assumes the short-circuit equivalent and D2 the open-circuit equivalent, as determined by the secondary voltages and the resulting current directions.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
27. 27. For the negative half of the AC cycleDuring the negative cycle of vi, reversing the roles of the diodes(D2 is short-circuit) but maintaining the same polarity for thevoltage across the load resistor R.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
28. 28. Animation of center-tapped transformer rectifierMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
29. 29. Example Show the voltage waveform across the secondary winding and across R when an input sinusoidal is applied to the primary winding.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
30. 30. SolutionThe transformer turns ratio = 0.5.The total peak secondary voltageis,Vp(sec) = nVp(pri) =0.5(100)=50V.There is a 25 V peak across each ofthe secondary with respect toground.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
31. 31. Rectifier Circuit Summary Note: Vm = peak of the AC voltage. Be careful, in the center tapped transformer rectifier circuit the peak AC voltage is the transformer secondary voltage to the tap.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
32. 32. Clippers• Clippers or diode limiting is a diode network that have the ability to “clip”(cut short/crop) off a portion on the input signal without distorting the remaining part of the alternating waveform.• Clippers are used to eliminate amplitude noise or to fabricate new waveforms from an existing signal.• Simplest form of diode clipper- one resistor and a diode• Depending on the orientation of the diode, the positive or negative region of the applied signal is clipped off.• 2 general of clippers: a) Series clippers b) Parallel clippers• Series Clippers – The series configuration is defined as one where the diode is in series with the load. – A half-wave rectifier is the simplest form of diode -clipper-one resistor and diode.• Parallel Clippers – The parallel configuration has the diode in a branch parallel to the load.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
33. 33. Series Clipper • Diodes “clip” a portion of the AC wave. • The diode “clips” any voltage that does not put it in forward bias. That would be a reverse biasing polarity and a voltage less than 0.7V for a silicon diode. Any type of signals can be applied to a clipperMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
34. 34. Analysis steps for series clippersThere is no general procedure for analyzing series clippersnetwork but there are some things one can do to give theanalysis some direction. 1. Take careful note of where the output voltage is defined. 2. Try to develop an overall sense of the response by simply noting the “pressure” established by each supply and the effect it will have on the conventional current direction through the diode. 3. Determine the applied voltage (transition voltage) that will result in a change of state for the diode from the “off” to the “on” state. 4. It is often helpful to draw the output waveform directly below the applied voltage using the same scales for the horizontal axis and the vertical axis. Series clipper with dc supply examplesMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
35. 35. Series clipper with dc supply By adding a DC source to the circuit, the voltage required to forward bias the diode can be changed.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
36. 36. Series clipper example• Positive region of Vi - turn the diode ON.• Negative region of Vi - turn the diode OFF.• Vi > V to turn ON the diode.• In general, diode is open circuit (OFF state) and short circuit (ON state)• For Vi > V the Vo = Vi – V• For Vi = V the Vo= 0 V MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
37. 37. Example 1 Determine the output waveform for the network below:MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
38. 38. Solution (continued)MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
39. 39. Example 2 Repeat previous example for the square-wave input.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
40. 40. Solution (continued): - ve region  OFF stateMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
41. 41. Parallel Clipper • By taking the output across the diode, the output is now the voltage when the diode is not conducting. • A DC source can also be added to change the diode’s required forward bias voltage. Parallel clipper exampleMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
42. 42. Example 2 Determine the Vo and sketch the output waveform for the below networkMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
43. 43. SolutionMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
44. 44. Solution (continued)MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
45. 45. Solution (continued)MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
46. 46. Example 2 Repeat the previous example using a silicon diode with VD=0.7 V SolutionMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
47. 47. Solution (continued) For input voltages greater than 3.3 V the diode  open circuit and Vo=Vi. For input voltages less than 3.3 V the diode  short circuit and the network result as shown belowMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
48. 48. Series Clippers (Ideal Diode) SummaryMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
49. 49. Parallel Clippers (Ideal Diode) SummaryMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
50. 50. Clampers • A clamper is a network constructed of a diode, resistor, and a capacitor that shifts a waveform to a different dc level without changing the appearance of the applied signal. • Clamping networks have a capacitor connected directly from input to output with a resistive element in parallel with the output signal. The diode is also in parallel with the output signal but may or may not have a series dc supply as an added element.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
51. 51. Element of the clamper circuit • Magnitude of R and C must be appropriate to ensure г=RC where the time constant is large enough and capacitor may not discharge during the time interval while diode is not conducting. • We will assume that all practical purposes the diode will fully charge or discharge in 5 time constant. A diode in conjunction with a capacitor can be used to “clamp” an AC signal to a specific DC level.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
52. 52. •The input signal can be any type of waveform: - sine, square, triangle wave, etc. •You can adjust the DC camping level with a DC source. Clampers exampleMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
53. 53. There is a sequence of steps that can be applied tohelp make the analysis straight forward.1. Start the analysis by examining the response of the portion of the input signal that will forward bias the diode. If the diode is reverse bias, skip the analysis for that interval time, and start analysis for the next interval time.2. During the period that the diode is in the “on” state, assume that the capacitor will charge up instantaneously to a voltage level determined by the surrounding network.3. Assume that during the period when the diode is in the “off” state the capacitor holds on to its established voltage level.4. Throughout the analysis, maintain a continual awareness of the location and defined polarity for vo to ensure that the proper levels are obtained.5. Check that the total swing of the output matches that off the input.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
54. 54. Clampers SummaryMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
55. 55. Example Determine Vo for the below network.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
56. 56. Solution• f=1000Hz, so a period of 1ms or interval 0.5ms between each level.• Define the period that the diode is start to conduct (t1~t2), which is the V0=5V.• Determine VC from the Kirchoff’s Law – VC=20V+5V=25V• When in the positive input, we will find V0=35V(outside loop)• Time constant, г =RC = 10ms, total discharge time = 50ms where is large enough before the capacitor is discharge during interval t2~t3MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
57. 57. Output waveformMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
58. 58. Zener Diode The state of the diode must be determined followed by a substitution of the appropriate model and a determination of the unknown quantities of the network. For the off state as a defined by a voltage less than Vz but greater than 0V. The Zener equivalent is the open circuit.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
59. 59. Vi and R FixedThe applied dc voltage is fixed, as the load resistor.The analysis :1. Determine the state of the Zener diode by removing it from the network and calculating the voltage across the resulting open circuit.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
60. 60. 2. Substitute the appropriate equivalent circuit and solve for the desired unknowns. - For the on state diode, the voltages across parallel elements must be the same. VL=VZ The Zener diode current is determined by KCL: IZ = IR – IL The power dissipated by the Zener diode is determined by: PZ = VZ IZ - For the off state diode, the equivalent circuit is open- circuit.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
61. 61. Fixed Vi, Variable RL• Due to the offset voltage Vz, there is a specific range of resistor values (and therefore load current) which will ensure that the Zener is in the on state.• Too small RL  VL < Vz  Zener diode will be in the off state• To determine the min RL that will turn the Zener diode on :• Any load resistance value greater than the RL min will ensure that the Zener diode is in the on state and the diode can be replaced by its Vz source equivalent.• The max IL MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
62. 62. • Once the diode is in the on state, the voltage across R remains fixed at:• Iz is limited to IZM as provided on the data sheet, it does affect the range of RL and therefore IL.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
63. 63. Fixed RL, Variable Vi• For fixed values of RL, the voltage Vi must be sufficiently large to turn the Zener diode on. The min turn-on voltage Vi=Vi min :• The max value of Vi is limited by the max Zener current IZM. IRmax=IZM+IL• Since IL is fixed at VZ/RL and IZM is the max value of IZ, the max Vi is defined by: Vi max =VRmax+Vz Vi max=IRmaxR+VzMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
64. 64. Zener Diode ExamplesMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
65. 65. Example Determine the network to find the range of RL and IL to maintained VRL at 10V.MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
66. 66. Solution:MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
67. 67. Voltage Multiplier Circuit Half Wave voltage doublerMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
68. 68. Voltage Multiplier Circuit Full Wave voltage doublerMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
69. 69. Voltage Multiplier Circuit Voltage Tripler and QuadruplerMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
70. 70. Practical Application • Rectification • Protective Configuration • Polarity Insurance • Controlled Battery-Powered Backup • Polarity Detector • AC Regulator and Square Wave GeneratorMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN
71. 71. End of Chapter 2 Thank youMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN