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Lecture 26
Diode Models, Circuits
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Piecewise-Linear Diode Models
v  Ra i  Va
Va
1
i
v
Ra
Ra
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Piecewise-Linear Diode Models
1 v

m i
1.6V  0.6V

100 mA  0 mA
1V

 10 
0.1 A
1 v

m i
 7.2V  6.0V

 100 mA  0 mA
 1.2V

 12 
 0.1 A
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):
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.9
Find the equivalent circuit if RL= 10 K and 1 K
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.9
For RL= 10 K:
(10 K)( 2 K)
 1.67 K
10 K  2 K
10 K
VT  voc 
10V  8.33V
10 K  2 K
RT 
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:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.9
Vt  iRt  6  i12
Vt  6
8.66  6
2.66
i


 1.58 mA
Rt  12 1,667  12 1,679
vo  6V  (12 )(1.58 mA)  6.018V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.9
For RL= 1 K:
(1K)( 2 K)
 0.667 K
1K  2 K
1K
VT  voc 
10V  3.33V
1K  2 K
RT 
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 voc=
3.33V
Diode=open circuit
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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.10
For segment A:
1 v
2V


 400 
m i 5mA
intercept  0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.10
Intercept 1.5 V
For segment B:
1 v 0.5V


 100 
m i 5mA
intercept  1.5V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.10
-5.5V
For segment C:
1 v
2V


 800 
m i 2.5mA
intercept  -5.5V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Simple Piecewise-Linear Diode
Equivalent Circuit
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.
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.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Half-Wave Rectifier with Smoothing
Capacitor
The charge removed from
the capacitor in one cycle:
Q  I LT  VC  Vr C
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Half-Wave Rectifier with Smoothing
Capacitor
I LT
Q  I LT  VC  Vr C  C 
Vr
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=VM. For the half wave rectifier with a
smoothing capacitor PIV=2VM
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:
I LT
C
2Vr
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Full-Wave Rectifier
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Clipper Circuit
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Clipper Circuit
+5V
vin  iR  v A  6  0  v A  vin  6  iR  1  off
 vin  iR  9  v B  0  v B  vin  9  iR  14  off
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Clipper Circuit
+7V
vin  iR  v A  6  0  v A  vin  6  iR  1  on
 vin  iR  9  v B  0  v B  vin  9  iR  16  off
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Clipper Circuit
-8V
vin  iR  v A  6  0  v A  vin  6  iR  14  off
 vin  iR  9  v B  0  v B  vin  9  iR  1  off
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Clipper Circuit
-10V
vin  iR  v A  6  0  v A  vin  6  iR  16  off
 vin  iR  9  v B  0  v B  vin  9  iR  1  on
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Clipper Circuit
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 VZ
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vo = vin
For vin>10V D1 is on and D2 is in reverse breakdown, then
vout = 9.4+0.6 = 10V
For vin<-1.8V D3, D4 and D5 are on and vout = 3(-0.6) = -1.8V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.14
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.14
When -5V < vin < 5V both diodes are off and vo = vin
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.14
When vin > 5V D6 is on and D7 is in reverse breakdown.
Then vin = (1k)i +0.6V+4.4V+ (1k)i = 5+2000i
vin  5
i
2000
and vo = 5+1000i = 0.5vin+2.5
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.14
When vin < -5V D6 is in reverse breakdown and D7 is on
Then vin = (1k)i -0.6V-4.4V+ (1k)i= -5+2000i
vin  5 and v = -5+1000i = 0.5v -2.5
i
o
in
2000
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 10.14
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.