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ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 26 Diode Models, Circuits.

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Presentation on theme: "ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 26 Diode Models, Circuits."— Presentation transcript:

1 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 26 Diode Models, Circuits

2 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Piecewise-Linear Diode Models

3 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Piecewise-Linear Diode Models

4 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Piecewise-Linear Diode Models Use the piecewise-linear diode model for the circuit shown in (a):

5 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.9 Find the equivalent circuit if R L = 10 K  and 1 K 

6 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.9 For R L = 10 K  :

7 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.9 Since the diode is reverse biased by more than -6 V, we use the diode model for line segment C:

8 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.9

9 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.9 For R L = 1 K  :

10 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.9 Since the diode is reverse biased by 3.3V, we use the diode model for line segment B, where the diode is an open circuit. The output voltage is then just v oc = 3.33V Diode=open circuit

11 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.10 Find a circuit model for each line segment shown in the figure

12 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.10 For segment A:

13 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.10 For segment B: Intercept 1.5 V

14 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.10 For segment C: -5.5V

15 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Simple Piecewise-Linear Diode Equivalent Circuit

16 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Half-Wave Rectifier with Resistive Load The diode is on during the positive half of the cycle. The diode is off during the negative half of the cycle.

17 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Half-Wave Rectifier used to Charge a Battery The diode is on when the signal voltage exceeds the battery voltage, charging the battery. The diode is off when the signal voltage is less than the battery voltage.

18 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. The charge removed from the capacitor in one cycle: Half-Wave Rectifier with Smoothing Capacitor

19 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Half-Wave Rectifier with Smoothing Capacitor

20 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Peak Inverse Voltage An important aspect of rectifier circuits is the peak inverse voltage (PIV) across the diodes. For the half wave rectifier with a resistive load PIV=V M. For the half wave rectifier with a smoothing capacitor PIV=2V M

21 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Full-Wave Rectifier The capacitance required for a full-wave rectifier is given by:

22 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Full-Wave Rectifier

23 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Clipper Circuit

24 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Clipper Circuit +5V

25 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. +7V Clipper Circuit

26 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. -8V Clipper Circuit

27 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. -10V Clipper Circuit

28 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Clipper Circuit

29 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Clipper Circuits using Zener Diodes Zener diode provides a reference voltage of V Z

30 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.14 All diodes are off for -1.8V < v < 10V  v o = v in For v in >10V D 1 is on and D 2 is in reverse breakdown, then v out = 9.4+0.6 = 10V For v in <-1.8V D 3, D 4 and D 5 are on and v out = 3(-0.6) = -1.8V

31 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.14

32 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.14 When -5V < v in < 5V both diodes are off and v o = v in

33 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.14 When v in > 5V D 6 is on and D 7 is in reverse breakdown. Then v in = (1k  )i +0.6V+4.4V+ (1k  )i = 5+2000i and v o = 5+1000i = 0.5v in +2.5

34 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.14 When v in < -5V D 6 is in reverse breakdown and D 7 is on Then v in = (1k  )i -0.6V-4.4V+ (1k  )i= -5+2000i and v o = -5+1000i = 0.5v in -2.5

35 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 10.14

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