1. Chapter 2: Diode Applications

2. Load-Line Analysis (graphical solution)  The analysis of diode can follow one of two paths: using the actual characteristics or applying an approximate model for the device.  Load Line Analysis: is used to analyze diode circuit using its actual characteristics. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

3. Load-Line Analysis (graphical solution) A straight line is defined by the parameters of the network. It is called the load line because the intersection on the vertical axes is defined by the applied load R. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

4. Load-Line Analysis (graphical solution) The maximum I D equals E/R, and the maximum V D equals E. The point where the load line and the characteristic curve intersect is the Q-point, which identifies I D and V D for a particular diode in a given circuit. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

5. Example 2.1 For the given diode configuration and diode characteristics, determine: V DQ, I DQ and V R. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

6. Example 2.1 - Solution The load line is firstly drawn between V D =E=10 V and I D =E/R=10/0.5k=20mA. The intersection between the load line and characteristics defines the Q-point as V DQ =0.78 and I DQ =18.5mA. V R =I DQ R=(18.5mA)(1K)=18.5 V. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

7. Diode Configurations  The forward resistance of the diode is usually so small compared to the other series elements of the network that it can be ignored.  In general, a diode is in the “on” state if the current established by the applied sources is such that its direction matches that of the arrow in the diode symbol, and V D ≥0.7V for silicon, V D ≥0.3V for germanium, and V D ≥1.2V for gallium arsenide.  You may assume the diode is “on”, and then find the current in the diode. If the current flows into the positive terminal of the diode, then the assumption is right, otherwise, the diode is “off”. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

8. Series Diode Configurations Forward Bias Constants Silicon Diode: V D = 0.7 V Germanium Diode: V D = 0.3 V Analysis (for silicon) V D = 0.7 V (or V D = E if E < 0.7 V) V R = E – V D I D = I R = I T = V R / R= ( E-V D ) / R Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Equivalent circuit for the “on” diode

9. Series Diode Configurations Reverse Bias Diodes ideally behave as open circuits Analysis V D = E V R = 0 V I D = 0 A Equivalent circuit for the “off” diode Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

10. Example 2.4 Determine V D, V R and I D. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

11. Example 2.5 Determine V D, V R and I D. Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

12. Source Notation Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

13. Example 2.6 Determine V D, V R and I D. Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

14. Example 2.7 Determine V o and I D. The forward bias voltage for red LED is 1.8 V. Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

15. Example 2.8 Determine I D, V D2 and V o. Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

16. Example 2.9 Determine I, V 1,V 2 and V o Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

17. Parallel and Series-Parallel Configurations Example 2.10 Determine V O, I 1, I D1, and I D2 Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

18. Example 2.11 : Find the resistor R to ensure a current of 20 mA throughthe“on”diodeforthegivencircuit.Bothdiodeshave reverse breakdown voltage of 3V and average turn-on voltage of 2V. Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

19. Example 2.12 Determine the voltage V o. Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

20. Example 2.13 Determine the currents I 1, I 2 and I D2 Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Solution

21. Determine V o AND/OR Gates Example 2.14 Logic OR gate Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

22. AND/OR Gates: Example 2.15 Determinetheoutputlevelforthelogic AND gate Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Logic AND gate Solution

23. Diode Applications (a)Rectifiers (b)Clippers or Limiters (c)Clampers (d)Voltage Multipliers Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Diodes are used in many applications:

24. Sinusoidal Inputs: Half-Wave Rectification Half-wave Rectifier Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

25. Sinusoidal Inputs: Half-Wave Rectification Conduction region (0  T/2). Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.  For t= 0  T/2, the diode is on.  Diode is substituted with short-circuit equivalence for ideal diode (reduce complexity).

26. Sinusoidal Inputs: Half-Wave Rectification Nonconduction region (T/2  T).  For the periodT/2  T, the diode is off.  Diode is substituted with an open circuit. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

27. Sinusoidal Inputs: Half-Wave Rectification The DC output voltage is 0.318 V m, where V m = the peak AC voltage. V DC =0.318 V m Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

28. Sinusoidal Inputs: Half-Wave Rectification  The effect of using a silicon diode with V K =0.7 is shown.  The diode is “on” when the applied signal is at least 0.7 V.  v o = v i –v K  For V m >>V k : V DC ≈0.318 (V m -V K ) Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

29. a)Sketch dc output v o and determine the dc level of the output. b)Repeat (a) if the ideal diode is replaced by silicon diode. Example 2.16 Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

30. Example 2.16 - Solution (a) Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. (b) V DC =-0.318 V m V DC =-0.318 (20) V DC =-6.36 V V DC =-0.318 (V m - 0.7) V DC =-0.318 (19.3) V DC =-6.14 V

31. PIV (PRV) Because the diode is only forward biased for one-half of the AC cycle, it is also reverse biased for one-half cycle. It is important that the reverse breakdown voltage rating of the diode be high enough to withstand the peak, reverse-biasing AC voltage and avoid entering the Zener region. PIV (or PRV) > V m PIV = Peak inverse voltage PRV = Peak reverse voltage V m = Peak AC voltage Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

32. Full-Wave Rectification Half-wave: V dc = 0.318 V m Full-wave: V dc = 0.636 V m Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.  The rectification process can be improved by using a full-wave rectifier circuit.  Full-wave rectification produces a greater DC output:

33. Full-Wave Rectification – Bridge Network Network for the period 0  T/2 of the input voltage v i Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

34. Full-Wave Rectification – Bridge Network Conduction path for the negative region of v i Conduction path for the positive region of v i Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

35. Full-Wave Rectification – Bridge Network The DC level is now twice that of half wave rectifier=2(0.318V m ) Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. V DC =0.636V m

36. Full-Wave Rectification – Bridge Network If silicon diode is used, v i –V K – v o - V K =0 v o = v i - 2V K Vo max= Vm - 2VKVo max= Vm - 2VK Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. For V m >>2V k : V DC ≈0.636 (V m - 2V K )

37. Full-Wave Rectification Center Tapped Transformer Rectifier V DC =0.636V m Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

38. Center Tapped Transformer Rectifier Network conditions for the positive region of v i Network conditions for the negative region of v i Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

39. Summary of Rectifier Circuits V m = peak of the AC voltage. In the center tapped transformer rectifier circuit, the peak AC voltage is the transformer secondary voltage to the tap. Rectifier Ideal V DC Realistic V DC Half Wave Rectifier V DC = 0.318 V m V DC = 0.318( V m – 0.7) Bridge Rectifier V DC = 0.636 V m V DC = 0.636( V m – 2(0.7)) Center-Tapped Transformer Rectifier V DC = 0.636 V m V DC = 0.636( V m – 0.7 ) Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

40. Determine the output waveform for the network and calculate the output dc level. Solution Example 2.17 Network for the positive region of v i Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

41. Example 2.17 - Solution Resulting output Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. v o =(1/2) v i V omax =(1/2) V imax =(1/2) 10=5V V DC =0.636(5V)=3.18 V. For the negative part the roles of the diodes are interchanged and v o appears as shown in figure.

42. Full Wave Rectifier with Smoothing Capacitor (AC to DC Converter) Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

43. Diode Clippers Clippers are networks that employ diodes to “clip” away a portion of an input signal without distorting the remaining part of the applied waveform. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

44. Biased Clippers Adding a DC source in series with the clipping diode changes the effective forward bias of the diode. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

45. Determining v o for the diode in the “on” state. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

46. Determine the output waveform for the network. Example 2.18 Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Solution

47. Determine the output waveform for the network. Example 2.19 Solution Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

48. Parallel Clippers The diode in a parallel clipper circuit“clips” any voltage that forward bias it. DC biasing can be added in series with the diode to change the clipping level. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

49. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Example 2.20 Determine v o for the network shown. Solution

50. Determine v o for the network if silicon diode is used. Solution Example 2.21 Determining v o for the diode in the “on” state. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

51. Summary of Clipper Circuits more… Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

52. Summary of Clipper Circuits Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

53. Clampers A clamper is a network constructed of a diode, a resistor, and a capacitor that shifts a waveform to a different DC level. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

54. Clampers Diode “on” and the capacitor charging to V volts. Determining v o with the diode “off.” R is chosen such that the discharge period 5τ=5RC is much larger than the period T/2→T, and the capacitor is assumed to hold onto all its charge. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

55. Clampers Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

56. Determine v o for the network. Example 2.22 Solution Determining v o and V C with the diode in the “on” state. Determining v o with the diode in the “off” state. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

57. Example 2.22 Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

58. Summary of Clamper Circuits Clamping circuits with ideal diodes (5  = 5RC >> T/2). Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

59. Zener Diodes A Zener diode is a type of diode that permits current not only in the forward direction like a normal diode, but also in the reverse direction if the voltage is larger than the breakdown voltage known as "Zener voltage“ (V Z ). Common Zener voltages are between 1.8 V and 200 V. Zener diode is used as regulator. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

60. Zener Diodes Approximate equivalent circuits for the Zener diode in the three possible regions of application. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

61. Example 2.24 Determine the reference voltages provided by the network which uses a white LED (4V) to indicate power is on. What is the power delivered to the LED and to the 6 V Zener diode. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

62. Basic Zener Regulator Determining the state of the Zener diode. Substituting the Zener equivalent for the “on” situation Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Remove Zener diode from network. Calculate V across open circuit. If V ≥Vz, Zener diode is on. If V <Vz, Zener diode is off.

63. Example 2.26 (a)For the Zener diode network, determine V L, V R, I Z and P Z. (b)Repeat part (a) with R L =3 kΩ. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

64. Example 2.26 - Solution (a) determine V L, V R, I Z and P Z. (R L =1.2 kΩ) V=8.73 V (<10) I Z =0 Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

65. Example 2.26 - Solution (b)determine V L, V R, I Z and P Z. (R L =3 kΩ) Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

66. Voltage Doubler This half-wave voltage doubler’s output can be calculated by: V out = V C2 = 2V m where V m = peak secondary voltage of the transformer Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

67. Voltage Doubler Positive Half-Cycle Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. oD 1 conducts oD 2 is switched off oCapacitor C 1 charges to V m Negative Half-Cycle oD 1 is switched off oD 2 conducts oCapacitor C 2 charges to 2V m V out = V C2 = 2V m (a) positive half-cycle; (b) negative half-cycle.

68. Voltage Doubler Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

69. Practical Applications (AC to DC Converter) Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

70. Practical Applications Battery Charger Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

71. Practical Applications Battery Charger Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

72. Practical Applications - Polarity insurance (a) Polarity protection for an expensive, sensitive piece of equipment; (b) correctly applied polarity; (c) application of the wrong polarity. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

73. Practical Applications - Polarity Detector Polarity detector using diodes and LEDs. Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

74. Practical Applications – Exit sign using LEDS Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

75. Practical Applications – AC Regulator Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

76. Practical Applications – Square-Wave Generator Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.