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

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

2. I-V characteristics of an ideal diode

3. Ideal diode operation on
off

4. Ideal diode operation
diode on
diode off

5. Ideal diode operation sinwt = 12/24 Vin = 24 sinwt 24 12 Vout
on off on off
30
Diode conducts when 24 sinwt = 12
sinwt = 12/24
wt = 30

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

7. Exercise 3.4(b) I + 2.5KW V 5V - 2.5KW 5V Find I and V
Assume diode is off.
VD = - 5, ID = 0
implies diode is off.
Correct assumption
V = 5, ID = 0
2.5KW 5V

8. Exercise 3.4(e) Find I and V +3 (Start with largest voltage)
Assume D1 on,
then D2 will be off, and D3 will be off
V = 3V, and I = 3V/1KW = 3mA.
Check assumption,
VD1 = 0, on
VD2 = -1, off
VD3 = -2, off
Correct assumption (old-style OR gate) +2 +1 + V - I

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.

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

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.

12. Prob. 3.9(b) Are the diodes on or off? Assume both diodes are on.
10V = (10K)I1
I1 = 10V/10K = I1 = 1mA
0 = (5K)I2 - 10V, I2 = 2mA
Current in D2 = I2 = 2mA, on
Current in D1 = I1 - I2 = -1mA, off
Does not match assumption; start over. I1 I2

13. Prob. 3.9(b) Are the diodes on or off? I 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.

14. I-V characteristics of an ideal diode

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.

16. Prob. 3.10(b) Is the diode on or off? Assume diode on.
15V = (10K)I1 + (10K)(I1- I2)
15 = (20K)I1 - (10K)I
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)I I3 I1 I2
Put 3 into = -(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.

17. Prob. 3.10(b) Assume diode off. 15V = (10K)I1 + (10K)I1 I1 = 0.75mA
Find V1. V1 = (10K)I1 = 7.5V
Find V2. V2 = -(10K)I3 = 5V
Voltage across diode is V2 - V1 = -2.5V, diode is off

18. 3.2 Real diodes Characteristics of a real diode breakdown Reverse bias
Forward bias

19. 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

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

21. 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

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.

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

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

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

26. Prob. 3.9(b) (using constant voltage-drop model)
Are the diodes on or off?
Assume both diodes are on.
10V = (10K)I
I1 = 9.3V/10K = I1 = 0.93mA
0 = (5K)I2 - 10V, I2 = 2mA
Current in D2 = I2 = 2mA, on
Current in D1 = I1 - I2 = -1.07mA, off
Does not match assumption; start over. I1 I2

27. Prob. 3.9(b) (using constant voltage-drop model)
Are the diodes on or off? I
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.

28. Prob. 3.10(b) (using constant voltage-drop model)
Is the diode on or off?
Assume diode on.
15V = (10K)I1 + (10K)(I1- I2)
15 = (20K)I1 - (10K)I
0 = (10K)(I2- I1) (10K)(I2- I3)
0.7 = -(10K)I1 + (20K)I2 - (10K)I3 2
0 = (10K)( I3- I2) + (10K)I3 + 10
-10 = -(10K)I2 + (20K)I I3 I1 I2
Put 3 into = -(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.

29. 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 + 10
I3 = -0.5mA V1 V2 I3 I1 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

30. 3.7 Rectifier circuits
Block diagram of a dc power supply

31. Half-wave rectifier
Simple
Wastes half the input

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

33. Full-wave rectifier
Entire input waveform is used

34. Bridge rectifier VS > 0 D1, D2 on; D3, D4 off
A type of full-wave rectifier
Center-tap not needed
Most popular rectifier

35. Bridge rectifier
VO is 2VD less than VS

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

37. Filter Diode off Diode on
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)