Chap. 3 Diodes
•Simplest semiconductor device
•Nonlinear
•Used in power supplies
•Voltage limiting circuits
0
3.1 Ideal Diodes
Forward bias
(on)
Reverse bias
(off)
1
I-V characteristics of an ideal diode
2
Ideal diode operation
on
off
on
off
3
Ideal diode operation
diode on
diode off
4
Ideal diode operation
Vin = 24 sinwt
24
12
Vout
on
off
on
off
30
Diode conducts when 24 sinwt = 12
sinwt = 12/24
wt = 30
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Exercise 3.4(a)
I
5V
+
V
-
2.5KW
I
Find I and V
Assume diode is on.
V = 0, I = 5V/ 2.5KW
I = 2mA, implies diode is on.
5V
2.5KW
Correct assumption
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Exercise 3.4(b)
I
5V
2.5KW
+
V
-
Find I and V
Assume diode is off.
VD = - 5, ID = 0
implies diode is off.
5V
2.5KW
Correct assumption
V = 5, ID = 0
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Exercise 3.4(e)
Find I and V
+3
(Start with largest voltage)
+2
Assume D1 on,
then D2 will be off, and D3 will be off
+1
I
+
V = 3V, and I = 3V/1KW = 3mA.
V
Check assumption,
VD1 = 0, on
VD2 = -1, off
VD3 = -2, off
Correct assumption (old-style OR gate)
-
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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.
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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
10
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.
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Prob. 3.9(b)
Are the diodes on or off?
Assume both diodes are on.
10V = (10K)I1
I1 = 10V/10K = I1 = 1mA
I1
0 = (5K)I2 - 10V, I2 = 2mA
I2
Current in D2 = I2 = 2mA, on
Current in D1 = I1 - I2 = -1mA, off
Does not match assumption; start over.
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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.
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I-V characteristics of an ideal diode
14
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.
15
Prob. 3.10(b)
Is the diode on or off?
Assume diode on.
I1
I3
I2
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
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.
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Prob. 3.10(b)
Assume diode off.
V1
I1
15V = (10K)I1 + (10K)I1
I1 = 0.75mA
V2
I3
I2 = 0
I2
0 = (10K)I3 + (10K)I3 + 10
I3 = -0.5mA
Find V1. V1 = (10K)I1 = 7.5V
Find V2. V2 = -(10K)I3 = 5V
Voltage across diode is V2 - V1 = -2.5V, diode is off
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3.2 Real diodes
Characteristics of a real diode
breakdown
Reverse bias
Forward bias
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Reverse bias region
•A small current flows when the diode is
reversed bias, IS
•IS is called the saturation or leakage current
IS
•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)
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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
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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
21
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.
22
Constant-voltage drop model
I-V characteristics and equivalent circuit
+
-
0.7V
0.7V
23
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.
24
Constant-voltage drop with resistor model
I-V characteristics and equivalent circuit
+
0.7V
-
 50W
0.7V
25
Prob. 3.9(b) (using constant voltage-drop model)
Are the diodes on or off?
Assume both diodes are on.
10V = (10K)I1 + 0.7
I1 = 9.3V/10K = I1 = 0.93mA
I1
0 = -0.7 + 0.7 + (5K)I2 - 10V, I2 = 2mA
I2
Current in D2 = I2 = 2mA, on
Current in D1 = I1 - I2 = -1.07mA, off
Does not match assumption; start over.
26
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 + 0.7 + (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.
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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)I2
I1
1
I3
I2
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
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.
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Prob. 3.10(b) (using constant voltage-drop model)
Assume diode off.
V1
I1
15V = (10K)I1 + (10K)I1
I1 = 0.75mA
V2
I3
I2 = 0
I2
0 = (10K)I3 + (10K)I3 + 10
I3 = -0.5mA
Find V1. V1 = (10K)I1 = 7.5V
Find V2. V2 = -(10K)I3 = 5V
Voltage across diode is V2 - V1 = -2.5V, diode is off
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3.7 Rectifier circuits
Block diagram of a dc power supply
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Half-wave rectifier
•Simple
•Wastes half the input
31
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
32
Full-wave rectifier
•Entire input waveform is used
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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
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Bridge rectifier
•VO is 2VD less than VS
35
Filter
•Capacitor acts as a filter.
•Vi charges capacitor as Vi increases.
•As Vi decreases, capacitor supplies current to load.
36
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)
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