Boise State University
Department of Electrical and Computer Engineering
ECE 472 – Power Electronics
Lecture #7: Single-Phase Rectifiers
iD
Breakdown
Region
Reverse−Biased
Forward−Biased
Region
Region
vD
Shockley Diode Equation:
(
vD
)
iD = Is e nVT − 1
where
iD is the diode current directed from anode to cathode
Is is the reverse leakage current
vD is the diode voltage from anode to cathode
VT is the thermal voltage defined below and
n is an ideality factor (also known as a quality factor or emission coefficient). The ideality factor
n varies from 1 to 2 depending on the fabrication process and semiconductor material.
The thermal voltage vT is approximately equal to 25.85 mV or about 26 mV at 300 o K which is
equal to 27o C, a temperature close to the “room temperature” commonly used in device simulation
software. At any other temperature T ,
VT
=
kT
q
where
k = 1.380648813 × 10−23 JK−1 is the Boltzmann constant
T is the absolute temperature of the p-n junction in o K and
q = 1.60217646 × 10−19 is the elementary charge of an electron.
1
Diode Models:
Imax
VD =0.7 V
VD =0.7 V
+ −
+ −
iD
Imax
iD
RD
Imax
iD
1
−Vbr
−Vbr
vD
(a) Ideal Model
RD
−Vbr
VD vD
(b) CVD Model
2
VD
(c) PWL Model
vD
Single-Phase Rectifier with RL Load
When the diode conducts:
Vm sin ωt = Ri(t) + L
di
,
dt
i(0) = 0
General Solution:
i(t) = ih (t) + ip (t)
Homogeneous Solution:
dih
R
= − ih =⇒ ih (t) = Ce−Rt/L
dt
L
Particular Solution: Assume that
ip (t) = A sin ωt + B cos ωt
solves the differential equation, that is,
dip
dt
Substituting this sinusoidal steady-state solution into the above differential equation,
Vm sin ωt = Rip (t) + L
Vm sin ωt = R(A sin ωt + B cos ωt) + L(ωA cos ωt − ωB sin ωt)
=⇒ Vm sin ωt = (RA − ωLB) sin ωt + (RB + ωLA) cos ωt
Solving for A and B,
{
RA − ωLB = Vm
ωLA + RB = 0
A =
B =
Vm −ωL 0
R R −ωL ωL
R R V m ωL 0 R −ωL ωL
R =⇒ A =
=⇒ B =
R2
−
RVm
+ (ωL)2
R2
ωLVm
+ (ωL)2
ip (t) = A sin ωt + B cos ωt
(
)
√
A
B
2
2
√
√
=
A +B
sin ωt +
cos ωt
A2 + B 2
A2 + B 2
)
(
√
B
=
A2 + B 2 sin ωt + tan−1
A
)
(
Vm
ωL
−1
= √ 2
sin ωt − tan
R
R + (ωL)2
Vm
sin (ωt − ϕ)
=
Z
3
where
√
Z =
R2 + (ωL)2
ϕ = tan−1
ωL
R
Alternative Derivation of the Particular Solution:
^
Im
^
Vm =Vm
0o
R
+
−
jωL
Iˆm =
V̂m
R + jωL
Vm ̸ 0o
R2 + (ωL)2 ̸ tan−1 ωL/R
Vm
ωL
̸
= √ 2
− tan−1
2
R
R + (ωL)
=⇒ ip (t) = |Iˆm | sin(ωt + ̸ Iˆm )
(
)
Vm
ωL
−1
= √ 2
sin ωt − tan
R
R + (ωL)2
Vm
=
sin(ωt − ϕ)
Z
=
√
Complete Solution:
Vm
sin(ωt − ϕ)
Z
Vm
i(0) = 0 = C +
sin(−ϕ) =⇒ C
Z
Vm sin ϕ −Rt/L Vm
i(t) =
e
+
sin(ωt − ϕ)
Z
Z
]
Vm [
=
sin(ωt − ϕ) + sin ϕe−Rt/L
Z
i(t) = Ce−Rt/L +
=
4
Vm
sin ϕ
Z
Alternate Forms of the Complete Solution:
i(t) =
i(t) =
i(θ) =
Vm
Z
Vm
Z
Vm
Z
[
]
sin(ωt − ϕ) + sin ϕe−Rt/L ,
0 ≤ t ≤ to
[
]
sin(ωt − ϕ) + sin ϕe−ωt/(ωL/R) ,
0 ≤ t ≤ to
]
[
sin(θ − ϕ) + sin ϕe−θ/ tan ϕ ,
0 ≤ θ ≤ β
Extinction Angle:
i(β) =
]
Vm [
sin(β − ϕ) + sin ϕe−β/ tan ϕ
Z
= 0
0 = sin(β − ϕ) + sin ϕe−β/ tan ϕ
Case #1: Pure Resistor(ϕ = 0o )
0 = sin(β − 0) + sin(0)e−β/ tan 0 =
sin β =⇒ β = π = 180o
Case #2: Pure Inductor (ϕ = 90o )
0 = sin(β − 90o ) + sin(90o )e−β/ tan 90
o
− cos β + 1 =⇒
=
cos β = 1 =⇒ β = 360o
General Case: Resistive-Inductive Load (0 ≤ ϕ ≤ 90o )
sin(β − ϕ) = − sin ϕe−β/ tan ϕ
[
π − (β − ϕ) = − sin−1 sin ϕe−β/ tan ϕ
]
[
β = π + ϕ + sin−1 sin ϕe−β/ tan ϕ
]
Gauss Algorithm:
[
β (k+1) = π + ϕ + sin−1 sin ϕe−β
(k) / tan ϕ
]
,
β (0) = π
Example:
R = 10 Ω
L = 15.3 mH =⇒ X = ωL = 120π × 0.0153 ∼
= 5.768 Ω
5.768
X
= tan−1
= tan−1 0.5768 ∼
ϕ = tan−1
= 29.976o ∼
= 30.0o
R
10
Answers:
β (0) = π
[
β (1) = π + ϕ + sin−1 sin ϕe−β
(0) / tan ϕ
β (2) = π + ϕ + sin−1 sin ϕe−β
(1) / tan ϕ
β (3) = π + ϕ + sin−1 sin ϕe−β
(2) / tan ϕ
[
[
]
]
]
∼
= 3.6669 rad
∼
= 3.6656 rad
∼
= 3.6656 rad ∼
= 210.0o = 180o + ϕ
5
Solving for the Extinction Angle β:
% MATLAB script for solving the following nonlinear equation
%
0 = sin(beta - phi) + sin(phi)*exp(-beta/tan(phi))
% using an iterative method (Gauss and Newton methods)
% Gauss method
beta = pi;
for m = 0:90
phid(m+1) = m;
phi = phid(m+1)*pi/180;
for n = 1:10
beta = pi + phi + asin(sin(phi)*exp(-beta/tan(phi)));
end
betad(m+1) = beta*180/pi;
end
%plot(phid,betad),grid
% Newton method
beta = pi;
for m = 0:90
phi = phid(m+1)*pi/180;
for n = 1:10
F = sin(beta-phi) + sin(phi)*exp(-beta/tan(phi));
FP = cos(beta-phi) - cos(phi)*exp(-beta/tan(phi));
betanew = beta - F/FP;
beta = betanew;
end
betad1(m+1) = beta*180/pi;
end
plot(phid,betad,’r’,phid,betad1,’.b’),grid
xlabel(’Power Factor Angle \phi [deg]’)
ylabel(’Extinction Angle \beta [deg]’)
print -deps betasol
6
360
340
320
Extinction Angle β [deg]
300
280
260
240
220
200
180
0
10
20
30
40
50
60
Power Factor Angle φ [deg]
7
70
80
90