1

Chap. 3 Diodes

•Simplest semiconductor device

•Nonlinear

•Used in power supplies

•Voltage limiting circuits

2

3.1 Ideal Diodes

Forward bias(on)

Reverse bias(off)

3

I-V characteristics of an ideal diode

4

Ideal diode operation

on onoff off

5

Ideal diode operation

diode on

diode off

6

24

12

on off on off

Vin = 24 sint

Vout

Ideal diode operation

Diode conducts when 24 sint = 12

sint = 12/24 t = 30

30

7

Exercise 3.4(a)

5V2.5K Find I and V

Assume diode is on.

V = 0, I = 5V/ 2.5K

I = 2mA, implies diode is on.

Correct assumption5V2.5K

I

+V-

I

8

Exercise 3.4(b)

5V2.5K Find I and V

Assume diode is off.

VD = - 5, ID = 0

implies diode is off.

Correct assumption

V = 5, ID = 0

5V2.5K

I

+V-

9

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/1KmA.

Check assumption,VD1 = 0, onVD2 = -1, offVD3 = -2, offCorrect assumption (old-style OR gate)

+

V

-

I

+3

+2

+1

Find I and V

10

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.

11

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

12

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.

13

Prob. 3.9(b)

Assume both diodes are on.

10V = (10K)I1

I1 = 10V/10K = I1 = 1mA

0 = (5K)I2 - 10V, I2 = 2mA

Current in D2 = I2 = 2mA, onCurrent in D1 = I1 - I2 = -1mA, offDoes not match assumption; start over.

Are the diodes on or off?

I1

I2

14

Prob. 3.9(b)

Assume D1 off and D2 on.

10V = (10K)I + (5K)I -10V

20V = (15K)II = 20V/15K = 1.33mA

Current in D2 = I = 1.33mA, onVoltage across D1 10V - 10K(1.33mA) = -3.33V, off

Matches assumption; done.

Are the diodes on or off?

I

15

I-V characteristics of an ideal diode

16

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.

17

Prob. 3.10(b)

Assume diode on.

15V = (10K)I1 + (10K)(I1- I2)

15 = (20K)I1 - (10K)I2 1

0 = (10K)(I2- I1) + (10K)(I2- I3)

0 = -(10K)I1 + (20K)I2 - (10K)I3 2

0 = (10K)( I3- I2) + (10K)I3 + 10

-10 = -(10K)I2 + (20K)I3 3

Is the diode on or off?

I1

I3

I2

Put 3 into 2. -5 = -(10K)I1 + (15K)I2, Put 1 into this equation, solve for I2.

I2 = 0.875mA, Current through diode is negative! Diode can’t be on.

18

Prob. 3.10(b)

Assume diode off.

15V = (10K)I1 + (10K)I1

I1 = 0.75mA

I2 = 0

0 = (10K)I3 + (10K)I3 + 10I3 = -0.5mA

I1

I3

I2

Find V1. V1 = (10K)I1 = 7.5V

Find V2. V2 = -(10K)I3 = 5V

Voltage across diode is V2 - V1 = -2.5V, diode is off

V1 V2

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3.2 Real diodesCharacteristics of a real diode

Forward biasReverse bias

breakdown

20

Reverse bias region

•A small current flows when the diode is reversed bias, IS

•IS is called the saturation or leakage current

•IS 1nA

•-VZ is the reverse voltage at which the diode breaks down.

•VZ is the Zener voltage in a Zener diode (controlled breakdown).

•Otherwise, VZ is the peak inverse voltage (PIV)

IS

21

Forward bias region

•For Silicon diodes, very little current flows until V 0.5V

•At V 0.7V, the diode characteristics arenearly vertical

•In the vicinity of V 0.7V, a wide range ofcurrent may flow.

•The forward voltage drop of a diode is oftenassumed 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

22

3.4 Analysis of diode circuits(Simplified diode models) p. 159-162

•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

23

Constant-voltage drop modelI-V characteristics

•A straight line is used to represent the fast-rising characteristics.•Resistance of diode when slope is vertical is zero.

24

Constant-voltage drop model

I-V characteristics and equivalent circuit

0.7V

0.7V

+

-

25

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

26

Constant-voltage drop with resistor model

I-V characteristics and equivalent circuit

0.7V

0.7V

50

+

-

27

Prob. 3.9(b) (using constant voltage-drop model)

Assume both diodes are on.

10V = (10K)I1 + 0.7I1 = 9.3V/10K = I1 = 0.93mA

0 = -0.7 + 0.7 + (5K)I2 - 10V, I2 = 2mA

Current in D2 = I2 = 2mA, onCurrent in D1 = I1 - I2 = -1.07mA, offDoes not match assumption; start over.

Are the diodes on or off?

I1

I2

28

Prob. 3.9(b) (using constant voltage-drop model)

Assume D1 off and D2 on.

10V = (10K)I + 0.7 + (5K)I -10V

19.3V = (15K)II = 19.3V/15K = 1.29mA

Current in D2 = I = 1.29mA, onVoltage across D1 10V - 10K(1.29mA) = -2.9V, off

Matches assumption; done.

Are the diodes on or off?

I

29

Prob. 3.10(b) (using constant voltage-drop model)

Assume diode on.

15V = (10K)I1 + (10K)(I1- I2)

15 = (20K)I1 - (10K)I2 1

0 = (10K)(I2- I1) - 0.7 + (10K)(I2- I3)

0.7 = -(10K)I1 + (20K)I2 - (10K)I3 2

0 = (10K)( I3- I2) + (10K)I3 + 10

-10 = -(10K)I2 + (20K)I3 3

Is the diode on or off?

I1

I3

I2

Put 3 into 2. -4.3 = -(10K)I1 + (15K)I2, Put 1 into this equation, solve for I2.

I2 = 0.91mA, Current through diode is negative! Diode can’t be on.

30

Prob. 3.10(b) (using constant voltage-drop model)

Assume diode off.

15V = (10K)I1 + (10K)I1

I1 = 0.75mA

I2 = 0

0 = (10K)I3 + (10K)I3 + 10I3 = -0.5mA

I1

I3

I2

Find V1. V1 = (10K)I1 = 7.5V

Find V2. V2 = -(10K)I3 = 5V

Voltage across diode is V2 - V1 = -2.5V, diode is off

V1 V2

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3.7 Rectifier circuits

Block diagram of a dc power supply

32

Half-wave rectifier

•Simple•Wastes half the input

33

Full-wave rectifierVS > 0

VS < 0

•Current goes through load in same direction for + VS. •VO is positive for + VS.

•Requires center-tap transformer

34

Full-wave rectifier

•Entire input waveform is used

35

Bridge rectifier

•A type of full-wave rectifier•Center-tap not needed•Most popular rectifier

VS > 0 D1, D2 on; D3, D4 off

VS < 0 D3, D4 on; D1, D2 off

36

Bridge rectifier

•VO is 2VD less than VS

37

Filter

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

38

Filter

Diode on

Diode off

•When the diode is off, the capacitor discharges.•Vo = Vpexp(-t/RC)•Assuming t T, and T=1/f

VP - Vr = Vpexp(-1/fRC) half-wave rectifier (t T) VP - Vr = Vpexp(-1/2fRC) full-wave rectifier (t T/2)