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0 Chap. 3 Diodes Simplest semiconductor device Nonlinear Used in power supplies Voltage limiting circuits.

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Presentation on theme: "0 Chap. 3 Diodes Simplest semiconductor device Nonlinear Used in power supplies Voltage limiting circuits."— Presentation transcript:

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2 0 Chap. 3 Diodes Simplest semiconductor device Nonlinear Used in power supplies Voltage limiting circuits

3 1 3.1 Ideal Diodes Forward bias (on) Reverse bias (off)

4 2 I-V characteristics of an ideal diode

5 3 Ideal diode operation on off

6 4 Ideal diode operation diode on diode off

7 on off V in = 24 sin  t V out Ideal diode operation Diode conducts when 24 sin  t = 12 sin  t = 12/24  t = 30  30 

8 6 Exercise 3.4(a) 5V 2.5K  Find I and V Assume diode is on. V = 0, I = 5V/ 2.5K  I = 2mA, implies diode is on. Correct assumption 5V 2.5K  I +V-+V- I

9 7 Exercise 3.4(b) 5V 2.5K  Find I and V Assume diode is off. V D = - 5, I D = 0 implies diode is off. Correct assumption V = 5, I D = 0 5V 2.5K  I +V-+V-

10 8 Exercise 3.4(e) (Start with largest voltage) Assume D1 on, then D2 will be off, and D3 will be off V = 3V, and I = 3V/1K  mA. Check assumption, V D1 = 0, on V D2 = -1, off V D3 = -2, off Correct assumption (old-style OR gate) +V-+V- I Find I and V

11 9 3.6 Zener diodes Designed to break down at a specific voltage Used in power supplies and voltage regulators When a large reverse voltage is reached, the diode conducts. Vz is called the breakdown, or Zener voltage.

12 10 Typical use of Zener diode The Zener diode will not usually conduct, it needs Vs > 12.5V to break down Assume Vs fluctuates or is noisy If Vs exceeds 12.5V, the diode will conduct, protecting the load

13 11 Solving ideal diode problems (determining if the diode is on or off) Assume diodes are on or off. Perform circuit analysis, find I & V of each diode. Compare I & V of each diode with assumption. Repeat until assumption is true.

14 12 Prob. 3.9(b) Assume both diodes are on. 10V = (10K)I 1 I 1 = 10V/10K = I 1 = 1mA 0 = (5K)I V, I 2 = 2mA Current in D2 = I 2 = 2mA, on Current in D1 = I 1 - I 2 = -1mA, off Does not match assumption; start over. Are the diodes on or off? I1I1 I2I2

15 13 Prob. 3.9(b) Assume D1 off and D2 on. 10V = (10K)I + (5K)I -10V 20V = (15K)I I = 20V/15K = 1.33mA Current in D2 = I = 1.33mA, on Voltage across D1 10V - 10K(1.33mA) = -3.33V, off Matches assumption; done. Are the diodes on or off? I

16 14 I-V characteristics of an ideal diode

17 15 Solving ideal diode problems (determining if the diode is on or off) Assume diodes are on or off. Perform circuit analysis, find I & V of each diode. Compare I & V of each diode with assumption. Repeat until assumption is true.

18 16 Prob. 3.10(b) Assume diode on. 15V = (10K)I 1 + (10K)(I 1 - I 2 ) 15 = (20K)I 1 - (10K)I = (10K)(I 2 - I 1 ) + (10K)(I 2 - I 3 ) 0 = -(10K)I 1 + (20K)I 2 - (10K)I = (10K)( I 3 - I 2 ) + (10K)I = -(10K)I 2 + (20K)I 3 3 Is the diode on or off? I1I1 I3I3 I2I2 Put 3 into = -(10K)I 1 + (15K)I 2, Put 1 into this equation, solve for I 2. I 2 = 0.875mA, Current through diode is negative! Diode can’t be on.

19 17 Prob. 3.10(b) Assume diode off. 15V = (10K)I 1 + (10K)I 1 I 1 = 0.75mA I 2 = 0 0 = (10K)I 3 + (10K)I I 3 = -0.5mA I1I1 I3I3 I2I2 Find V 1. V 1 = (10K)I 1 = 7.5V Find V 2. V 2 = -(10K)I 3 = 5V Voltage across diode is V 2 - V 1 = -2.5V, diode is off V1V1 V2V2

20 Real diodes Characteristics of a real diode Forward biasReverse bias breakdown

21 19 Reverse bias region A small current flows when the diode is reversed bias, I S I S is called the saturation or leakage current I S  1nA -V Z is the reverse voltage at which the diode breaks down. V Z is the Zener voltage in a Zener diode (controlled breakdown). Otherwise, V Z is the peak inverse voltage (PIV) ISIS

22 20 Forward bias region For Silicon diodes, very little current flows until V  0.5V At V  0.7V, the diode characteristics are nearly vertical In the vicinity of V  0.7V, a wide range of current may flow. The forward voltage drop of a diode is often assumed to be V = 0.7V Diodes made of different materials have different voltage drops V  0.2V - 2.4V Almost all diodes are made of Silicon, LEDs are not and have V  1.4V - 2.4V

23 Analysis of diode circuits (Simplified diode models ) p Ideal diode Constant-voltage drop model Constant-voltage drop model with resistor All use assumptions because actual diode characteristics are too difficult to use in circuit analysis

24 22 Constant-voltage drop model I-V characteristics A straight line is used to represent the fast-rising characteristics. Resistance of diode when slope is vertical is zero.

25 23 Constant-voltage drop model I-V characteristics and equivalent circuit 0.7V + -

26 24 Constant-voltage drop with resistor model A straight line with a slope is used to represent the fast-rising characteristics. Resistance of diode is 1/slope. I-V characteristics

27 25 Constant-voltage drop with resistor model I-V characteristics and equivalent circuit 0.7V  50  + -

28 26 Prob. 3.9(b) (using constant voltage-drop model) Assume both diodes are on. 10V = (10K)I I 1 = 9.3V/10K = I 1 = 0.93mA 0 = (5K)I V, I 2 = 2mA Current in D2 = I 2 = 2mA, on Current in D1 = I 1 - I 2 = -1.07mA, off Does not match assumption; start over. Are the diodes on or off? I1I1 I2I2

29 27 Prob. 3.9(b) (using constant voltage-drop model) Assume D1 off and D2 on. 10V = (10K)I (5K)I -10V 19.3V = (15K)I I = 19.3V/15K = 1.29mA Current in D2 = I = 1.29mA, on Voltage across D1 10V - 10K(1.29mA) = -2.9V, off Matches assumption; done. Are the diodes on or off? I

30 28 Prob. 3.10(b) (using constant voltage-drop model) Assume diode on. 15V = (10K)I 1 + (10K)(I 1 - I 2 ) 15 = (20K)I 1 - (10K)I = (10K)(I 2 - I 1 ) (10K)(I 2 - I 3 ) 0.7 = -(10K)I 1 + (20K)I 2 - (10K)I = (10K)( I 3 - I 2 ) + (10K)I = -(10K)I 2 + (20K)I 3 3 Is the diode on or off? I1I1 I3I3 I2I2 Put 3 into = -(10K)I 1 + (15K)I 2, Put 1 into this equation, solve for I 2. I 2 = 0.91mA, Current through diode is negative! Diode can’t be on.

31 29 Prob. 3.10(b) (using constant voltage-drop model) Assume diode off. 15V = (10K)I 1 + (10K)I 1 I 1 = 0.75mA I 2 = 0 0 = (10K)I 3 + (10K)I I 3 = -0.5mA I1I1 I3I3 I2I2 Find V 1. V 1 = (10K)I 1 = 7.5V Find V 2. V 2 = -(10K)I 3 = 5V Voltage across diode is V 2 - V 1 = -2.5V, diode is off V1V1 V2V2

32 Rectifier circuits Block diagram of a dc power supply

33 31 Half-wave rectifier Simple Wastes half the input

34 32 Full-wave rectifier V S > 0 V S < 0 Current goes through load in same direction for + V S. V O is positive for + V S. Requires center-tap transformer

35 33 Full-wave rectifier Entire input waveform is used

36 34 Bridge rectifier A type of full-wave rectifier Center-tap not needed Most popular rectifier V S > 0 D 1, D 2 on; D 3, D 4 off V S < 0 D 3, D 4 on; D 1, D 2 off

37 35 Bridge rectifier V O is 2V D less than V S

38 36 Filter Capacitor acts as a filter. Vi charges capacitor as Vi increases. As Vi decreases, capacitor supplies current to load.

39 37 Filter Diode on Diode off When the diode is off, the capacitor discharges. V o = V p exp(-t/RC) Assuming t  T, and T=1/f V P - V r = V p exp(-1/fRC) half-wave rectifier (t  T) V P - V r = V p exp(-1/2fRC) full-wave rectifier (t  T/2)


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