i
Lecture Notes
Electric Control Devices and Applications
Xiaoping Liu
Department of Electrical Engineering
Lakehead University
ii
Contents
1 Circuit Analysis Using Laplace Transform
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . .
1.2 Review of Laplace Transform . . . . . . . . . . . . .
1.3 Differential Equations . . . . . . . . . . . . . . . . .
1.4 Partial Fractions . . . . . . . . . . . . . . . . . . . .
1.5 Electrical Element Models . . . . . . . . . . . . . . .
1.6 Analysis of Electrical Network by Laplace Transform
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1
1
2
4
4
6
6
2 Thyristors
2.1 Silicon-Controlled Rectifiers . . . . . . .
2.2 Gate-Turn-Off Thyristors . . . . . . . .
2.3 DIAC . . . . . . . . . . . . . . . . . . .
2.4 TRIAC: Bidirectional Triode Thyristors
2.5 TRIAC-DIAC Applications (Lab 2) . . .
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17
17
19
19
19
20
3 Rectifiers
3.1 Single-Phase Uncontrolled Half-Wave Rectifiers with R Loads . . . . . . .
3.2 Single-Phase Uncontrolled Half-Wave Rectifiers with RL Loads . . . . . .
3.3 Single-Phase Controlled Half-Wave Rectifiers with R Loads . . . . . . . .
3.3.1 Single-Phase Controlled Half-Wave Rectifiers with RL Loads . . .
3.3.2 Freewheeling Diodes . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.3 Single-Phase Uncontrolled Full-Wave Rectifiers with R Loads . . .
3.3.4 Single-Phase Controlled Full-Wave Rectifiers with R Loads . . . .
3.3.5 Single-Phase Controlled Full-Wave Rectifiers with RL Loads . . .
3.4 Three-Phase Uncontrolled Full-Wave Rectifiers with R Loads . . . . . . .
3.5 Three-Phase Controlled Full-Wave Rectifiers with Highly Inductive Loads
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21
21
23
24
26
28
29
30
33
35
38
4 AC Voltage Controllers
4.1 Single-Phase AC Voltage Controllers with R Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
41
5 DC
5.1
5.2
5.3
Choppers
Buck Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Boost Converters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Impulse-Commutated Chopper Circuit (for Lab #3) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
45
49
52
6 Inverter
6.1 Half-Bridge Inverter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Full-Bridge Inverter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
55
59
7 DC Drives
7.1 Basic Characteristics of DC Motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Single-Phase Full-Wave Converter Drive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
63
65
8 AC
8.1
8.2
8.3
69
69
72
73
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Drives
Induction Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Frequency Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Voltage and Frequency Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
0
Chapter 1
Circuit Analysis Using Laplace Transform
1.1
Introduction
Example
Consider the RL series circuit shown in Fig. 1.1. Assume that the current through the inductor is iL (0− ) = 1/L
when the switch is open. If the switch is closed at t = 0, then find i(t) for t > 0.
Solution
The current i(t) satisfies the following equation
i(t)R + L
di(t)
=0
dt
(1.1)
This is a first-order differential equation with constant coefficients, which can be solved by substituting i(t) = eλt and
solving the characteristic equation
R + Lλ = 0
for λ. λ is given by
λ=−
R
L
and thus the solution to the differential equation is given by
R
i(t) = Ce− L t
Figure 1.1: Circuit in t domain
1
2
CHAPTER 1. CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
C is an integration constant, which can be determined by the initial condition i(0) = iL (0− ) = 1/L, that is, C = 1/L.
Therefore, the current i(t) is
1 R
i(t) = e− L t
L
This differential equation can also be solved by using Laplace transform.
1.2
Review of Laplace Transform
Definition
Let f (t) be a given function defined for t ≥ 0. Then, its Laplace transform is defined as
Z ∞
F (s) = L{f (t)} =
e−st f (t)dt
0
which shows that the function f (t) in time domain is transformed to the function F (s) in s or complex frequency
domain by Laplace transform operation. F (s) and L{f (t)} is called the Laplace transform of f (t), and the original
function f (t) is called the inverse transform or inverse of F (s), denoted by L−1 {F (s)}, that is,
f (t) = L−1 {F (s)}
Properties
1. Linearity
L{af (t) + bg(t)} = aL{f (t)} + bL{g(t)}
Z
L{af (t) + bg(t)} =
= a
∞
e−st [af (t) + bg(t)]dt
Z ∞
∞
e−st f (t)dt + b
e−st g(t)dt
0Z
0
0
= aL{f (t)} + bL{g(t)}
(1.2)
2. First Shifting Theorem
L{eat f (t)} = F (s − a)
eat f (t) = L−1 {F (s − a)}
F (s − a)
=
Z
=
Z0 ∞
(1.3)
∞
e−(s−a)t f (t)dt
e−st [eat f (t)]dt
0
= L{eat f (t)}
3. Transform of Derivatives
L{f (n) (t)} = sn L{f (t)} − sn−1 f (0) − sn−2 f 0 (0) − · · · − sf (n−2) (0) − f (n−1) (0)
Case n = 1
L{f 0 (t)} =
=
Z
∞
Z0
∞
e−st f 0 (t)dt
0
e−st df (t)
(1.4)
1.2. REVIEW OF LAPLACE TRANSFORM
3
=
−st
∞
e f (t) 0 + s
Z
∞
e−st f (t)dt
0
= sL{f (t)} − f (0)
d(uv)
= vdu + udv
Z
Z
Z
=
d(uv) = vdu + udv
Z
= uv − vdu
uv
Z
(1.5)
udv
Case n = 2
L{f 00 (t)} = sL{f 0 (t)} − f 0 (0)
= s[sL{f (t)} − f (0)] − f 0 (0)
= s2 L{f (t)} − sf (0) − f 0 (0)
(1.6)
4. Transform of Integral
L
Z
t
f (τ )dτ
0Z
1
F (s)
s =
t
f (τ )dτ
= L−1
0
F (s)
which implies that L
nR
t
0
f (τ )dτ
o
= L{f (t)} = L
= sL
Z
= sL
Z0 t
(Z
f (τ )dτ
f (τ )dτ
t
1
F (s)
s
t
f (τ )dτ
0
−
Z
f (τ )dτ
5. Initial Value Theorem
f (0) = lim sF (s)
s→∞
6. Final Value Theorem
f (∞) = lim f (t) = lim sF (s)
7. Table of Lapace Transforms
unit impulse
unit step
unit ramp
unit acceleration
n-th order ramp
exponential
n-th order exponential
sine
cosine
damped sine
damped cosine
δ(t)
1(t)
t
t2
2
tn
n!
−at
e
tn −at
n! e
sin ωt
cos ωt
e−at sin ωt
e−at cos ωt
1
1
s
1
s2
1
s3
1
sn+1
1
s+a
1
(s+a)n+1
ω
s2 +ω 2
s
s2 +ω 2
ω
(s+a)2 +ω 2
s+a
(s+a)2 +ω 2
0 )
0
= 1s F (s).
s→0
(1.7)
0
0
t→∞
(1.8)
4
CHAPTER 1. CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
1.3
Differential Equations
Consider a second-order differential equation with constant coefficients:
00
0
y (t) + ay (t) + by(t)
= u(t),
y(0) = K1 ,
0
y (0) = K2
(1.9)
with constant a and b. Here u(t) is the input (driving force and current source or voltage source) applied to the
(mechanical and electrical) system and y(t) is the output (response) of the system. Laplace transform method involves
the following three steps:
1. Taking the transform on both sides of (1.9) gives
0
[s2 Y (s) − sy(0) − y (0)] + a[sY (s) − y(0)] + bY (s) = U (s)
which is called the subsidiary equation. Collecting Y -terms, we have
0
(s2 + as + b)Y (s) = (s + a)y(0) + y (0) + U (s)
2. Solving the subsidiary equation algebraically for Y (s) produces
Y (s)
0
=
[(s + a)y(0) + y (0)]Q(s) + U (s)Q(s)
(1.10)
where Q(s) is transfer function and is defined as
Q(s) =
1
s2 + as + b
0
If y(0) = y (0) = 0, then (1.10) becomes Y (s) = U (s)Q(s). Thus, Q(s) is the quotient
Q(s) =
L{output}
Y (s)
=
U (s)
L{input}
which explains the name of Q(s).
3. Using partial fractions, reduce (1.10) to a sum of terms whose inverses can be found from the table, so that the
solution y(t) = L−1 {Y (s)} of (1.9) is obtained.
Example
Solve the differential equation (1.1), that is,
i(t)R + L
di(t)
= 0,
dt
i(0) = 1/L
Taking the Laplace transform on both sides yields
I(s)R + LsI(s) − 1 = 0
Solving this equation gives
I(s) =
1
1
= LR
Ls + R
s+ L
Taking the inverse on both sides produces
i(t) =
1.4
1 −Rt
e L
L
Partial Fractions
The Solution Y (s) of a subsidiary equation of a differential equation usually comes out as a quotient of two polynomials,
Y (s) =
P (s)
Q(s)
In order to find y(t) by taking the inverse of Y (s), it is better to write Y (s) as a sum of partial fractions. The form
of the partial fractions depends on the types of factors in the product form of Q(s). We often encounter the following
cases.
1.4. PARTIAL FRACTIONS
5
1. Unrepeated factors (s − a1 )(s − a2 )
Y (s) =
A2
A1
+
s − a1
s − a2
Y (s) =
A2
A1
+
s − a (s − a)2
2. Repeated factors (s − a)2
3. Complex factors [s − (α + jβ)][s − (α − jβ)]
Y (s) =
As + B
As + B
=
[s − (α + jβ)][s − (α − jβ)]
(s − α)2 + β 2
Example Find a partial-fraction expansion for the transfer function
2s2 + 15s + 24
s(s + 2)(s + 5)
Solution Let
A1
A2
A3
2s2 + 15s + 24
=
+
+
s(s + 2)(s + 5)
s
s+2 s+5
Then, multiplying the common denominator s(s + 2)(s + 5) on both sides produces
2s2 + 15s + 24 = A1 (s + 2)(s + 5) + A2 s(s + 5) + A3 s(s + 2)
= A1 (s2 + 7s + 10) + A2 (s2 + 5s) + A3 (s2 + 2s)
= (A1 + A2 + A3 )s2 + (7A1 + 5A2 + 2A3 )s + 10A1
Equating coefficients of like powers of s gives
A1 + A2 + A3
7A1 + 5A2 + 2A3
10A1
= 2
= 15
= 24
Solving these equations yields A1 = 12/5, A2 = −1/3, and A3 = −1/15.
A1 , A2 , and A3 can also be determined by letting s = 0, s = −2, and s = −5. Let s = 0, we get
24 = 10A1
that is, A1 = 12/5. Let s = −2, we have
2 = −6A2
i.e., A2 = −1/3. Letting s = −5 gives
−1 = 15A3
or A3 = −1/15.
Therefore, the partial-fraction expansion is
2s2 + 15s + 24
1/3
12/5
1/15
−
=
−
s(s + 2)(s + 5)
s
s+2 s+5
Example Find a partial-fraction expansion for the transfer function
10s
(s + 1)(s2 + 4)
Solution Let
10s
A1
A2 s + A3
=
+
2
(s + 1)(s + 4)
s+1
s2 + 4
Then, multiplying the common denominator (s + 1)(s2 + 4) on both sides produces
10s = A1 (s2 + 4) + (A2 s + A3 )(s + 1)
= A1 (s2 + 4) + (A2 s2 + A2 s + A3 s + A3 )
= (A1 + A2 )s2 + (A2 + A3 )s + 4A1 + A3
6
CHAPTER 1. CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
Equating coefficients of like powers of s gives
A1 + A2
A2 + A3
4A1 + A3
= 0
= 10
= 0
Solving these equations yields A1 = −2, A2 = 2, and A3 = 8.
Therefore, the partial-fraction expansion is
2
2s + 8
10s
=−
+
(s + 1)(s2 + 4)
s + 1 s2 + 4
Example Find a partial-fraction expansion for the transfer function
(s2
Solution Let
4 − 2s
+ 1)(s − 1)2
A3
A4
A1 s + A2
4 − 2s
+
+
=
(s2 + 1)(s − 1)2
s2 + 1
s − 1 (s − 1)2
Then, multiplying the common denominator (s2 + 1)(s − 1)2 on both sides produces
4 − 2s = (A1 s + A2 )(s − 1)2 + A3 (s2 + 1)(s − 1) + A4 (s2 + 1)
= (A1 s + A2 )(s2 − 2s + 1) + A3 (s3 − s2 + s − 1) + A4 (s2 + 1)
= (A1 s3 + A2 s2 − 2A1 s2 − 2A2 s + A1 s + A2 ) + A3 (s3 − s2 + s − 1) + A4 (s2 + 1)
= (A1 + A3 )s3 + (A2 − 2A1 − A3 + A4 )s2 + (−2A2 + A1 + A3 )s + A2 − A3 + A4
Equating coefficients of like powers of s gives
A1 + A3
A2 − 2A1 − A3 + A4
−2A2 + A1 + A3
A2 − A3 + A4
= 0
A1
= 0
A4
= −2 A2
= 4
A3
= −A3
= −A3 − 1
=1
= −2
Solving these equations yields A1 = 2, A2 = 1, A3 = −2, and A4 = 1. Therefore, the partial-fraction expansion is
2
1
2s + 1
4 − 2s
−
+
= 2
(s2 + 1)(s − 1)2
s + 1 s − 1 (s − 1)2
1.5
Electrical Element Models
Resistor
v(t) = Ri(t)
V (s) = RI(s)
Capacitor
i(t) = C
dv(t)
dt
I(s) = CsV (s) − CVC (0)
or V (s) =
1
1
I(s) + VC (0)
Cs
s
Inductor
di(t)
V (s) 1
V (s) = LsI(s) − LiL (0) or I(s) =
+ iL (0)
dt
Ls
s
The circuit models in both t-domain and s-domain are shown in Fig. 1.2.
v(t) = L
1.6
Analysis of Electrical Network by Laplace Transform
1. Determine the initial conditions.
2. Draw equivalent circuit in s-domain.
3. Write loop voltage equations or node current equations.
4. Solve the resulting algebraic equations for unknown variables.
1.6. ANALYSIS OF ELECTRICAL NETWORK BY LAPLACE TRANSFORM
Figure 1.2: Element models in s domain
5. Reduce the equations for the unknowns to a sum of terms by partial fractions.
6. Take the inverses and find the expressions for the unknowns in time domain.
Example
Find the current in Fig. 1.3 by Laplace transform, assuming that all initial conditions are zero.
Solution
Applying Laplace transform yields the equivalent circuit shown in Fig. 1.4:
By Kirchhoff’s voltage law, we get
I(s)
=
=
Now let
5
s+1
6+s+
8
s
=
5s
(s + 1)(s2 + 6s + 8)
5s
(s + 1)(s + 2)(s + 4)
5s
A1
A2
A3
=
+
+
(s + 1)(s + 2)(s + 4)
s+1 s+2 s+4
Multiplying the common denominator (s + 1)(s + 2)(s + 4) on both sides gives
5s = A1 (s + 2)(s + 4) + A2 (s + 1)(s + 4) + A3 (s + 1)(s + 2)
Let s = −1, we get −5 = 3A1 or A1 = −5/3.
Let s = −2, we have −10 = −2A2 or A2 = 5.
Let s = −4, we obtain −20 = 6A3 or A3 = −10/3.
Therefore,
I(s) = −
5
10/3
5/3
+
−
s+1 s+2 s+4
Taking inverse on both sides yields
5
10
i(t) = − e−t + 5e−2t − e−4t
3
3
Example
7
8
CHAPTER 1. CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
Figure 1.3: Circuit in t domain
Figure 1.4: Circuit in s domain
Consider the RLC parallel circuit shown in Fig. 1.5 with R = 3/4Ω, C = 1/3F , L = 1H, vC (0− ) = 2V , and
iL (0− ) = 0. The switch is closed at t = 0. Find v(t) for t > 0.
Solution
Taking Laplace to transform, we get the equivalent circuit in s domain as shown in Fig. 1.6, which is equivalent
to the circuit shown in Fig. 1.7 to the right by Norton Theorem.
By Kirchhoff’s current law, we get
I = IC + IR + IL
which is equivalent to
2C = sCV (s) +
1
1
V (s) +
V (s)
R
sL
Solving for V (s) produces
V (s)
=
=
=
=
2C
sC + R1 +
s
3
2
3
4
3
1
sL
+ 1s
2s
s2 + 4s + 3
2s
(s + 1)(s + 3)
+
Now, let us take a partial-fraction expansion. Set
A1
A2
2s
=
+
(s + 1)(s + 3)
s+1 s+3
Multiplying the common denominator (s + 1)(s + 3) on both sides gives
2s = A1 (s + 3) + A2 (s + 1)
Setting s = −1 gives A1 = −1, setting s = −3 gives A2 = 3. Therefore, we get
V (s) = −
1
1
+3
s+1
s+3
Taking inverse on both sides produces
v(t) = 3e−3t − e−t
Example
1.6. ANALYSIS OF ELECTRICAL NETWORK BY LAPLACE TRANSFORM
9
Figure 1.5: Circuit in t domain
Figure 1.6: Circuit in s domain
Figure 1.7: Norton equivalent circuit
Consider the electrical circuit shown in Fig. 1.8 with R = 30Ω, L = 10mH, C = 50µF , and
−50mA t ≤ 0
is (t) =
50mA
t>0
Find the current through the inductor by Laplace transform.
Solution
First let us determine the initial conditions by using the circuit in Fig. 1.9. Note that at t = 0− the circuit reaches
a steady-state, so iL (0− ) = 0 and
VC (0− ) = is (0− )R = (−50mA)(30Ω) = −1.5V
Applying Laplace transform gives the equivalent circuit in s-domain as shown in Fig. 1.10, which is equivalent to the
circuit in Fig. 1.11 by Thevenin theorem.
By Kirchhoff’s voltage law, I(s) can be found from
vC (0)
1
= RIs (s) −
I(s) R + sL +
sC
s
that is,
or
1 −1.5
1
I(s) 30 + 10 × 10 s +
= 30 × 50 × 10−3 −
50 × 10−6 s
s
s
20000
3
s
+
=
I(s) 30 +
100
s
s
−3
Multiplying 100s on both sides produces
I(s)(3000s + s2 + 2000000) = 300
that is,
I(s) =
300
300
=
2
(3000s + s + 2000000)
(s + 1000)(s + 2000)
10
CHAPTER 1. CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
Figure 1.8: Circuit in t domain
Figure 1.9: Circuit for initial conditions
Figure 1.10: Circuit in s domain
Figure 1.11: Thevenin quivalent circuit
1.6. ANALYSIS OF ELECTRICAL NETWORK BY LAPLACE TRANSFORM
Let
11
300
A1
A2
=
+
(s + 1000)(s + 2000)
s + 1000 s + 2000
Multiplying (s + 1000)(s + 2000) on both sides of the above equation gives
300 = A1 (s + 2000) + A2 (s + 1000)
Letting s = −1000 gives A1 = 0.3 and s = −2000 gives A2 = −0.3. Then, we get
I(s) =
0.3
0.3
−
s + 1000 s + 2000
Taking inverse on both sides gives
i(t) = 0.3e−1000t − 0.3e−2000t
Example
Consider the electrical circuit in Fig. 1.12 with R = 2Ω, L = 1H, C = 1F , and
4V
t≤0
vs (t) =
4e−t V t > 0
Find v0 (t) by Laplace transform.
Solution
First we need to determine the initial conditions. Note that the circuit in Fig. 1.13 reaches its steady-state
condition at t = 0− , that is, the current through the capacitor is zero and the voltage across the inductor is zero.
Therefore, we get
4V
vs (0− )
=
= 1A
2Ω + 2Ω
4Ω
−
−
= v0 (0 ) = iL (0 )(2Ω) = (1A)(2Ω) = 2V
iL (0− )
=
vC (0− )
Then, taking Laplace transform produces the equivalent circuit in s domain in Fig. 1.14.
By writing mesh equations for I1 and I2 , we have
0
=
0
=
4
1
2
− 2I1 − (I1 − I2 ) −
s+1
s
s
2 1
− (I2 − I1 ) − (s + 2)I2 + 1
s s
which are equivalent to
2
1
4
1
− − 2+
I1 + I2
s+1 s
s
s
2s + 1
2s − 2
1
−
I1 + I2
s(s
s
s
+ 1)
1 2s − 2
− (2s + 1)I1 + I2
s s+1
1
1
2
+ 1 + I1 − s + 2 +
I2
s
s
s
2+s 1
s2 + 2s + 1
+ I1 −
I2
s
s
s
1
(2 + s) + I1 − (s2 + 2s + 1)I2
s
0 =
=
=
0 =
=
=
that is,
0
0
2s − 2
− (2s + 1)I1 + I2
s+1
= (2 + s) + I1 − (s2 + 2s + 1)I2
=
From the second equation, I1 can be expressed as
I1 = (s2 + 2s + 1)I2 − (2 + s)
(1.11)
12
CHAPTER 1. CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
Figure 1.12: Circuit in t domain
Figure 1.13: Circuit for initial conditions
Figure 1.14: Circuit in s domain
1.6. ANALYSIS OF ELECTRICAL NETWORK BY LAPLACE TRANSFORM
Substituting this into the first equation gives
0=
2s − 2
− (2s + 1)[(s2 + 2s + 1)I2 − (2 + s)] + I2
s+1
that is,
0=
2s − 2
+ (2s + 1)(2 + s) − [(2s + 1)(s2 + 2s + 1) − 1]I2
s+1
Solving for I2 produces
I2
=
=
=
=
=
=
2s−2
s+1
+ (2s + 1)(2 + s)
(2s + 1)(s2 + 2s + 1) − 1
2s−2
2
s+1 + 2s + 5s + 2
2s3 + 5s2 + 4s
2s − 2 + (s + 1)(2s2 + 5s + 2)
(s + 1)(2s3 + 5s2 + 4s)
2s3 + 7s2 + 9s
(s + 1)(2s3 + 5s2 + 4s)
2s2 + 7s + 9
(s + 1)(2s2 + 5s + 4)
2s2 + 7s + 9
2(s + 1) s2 + 52 s + 2
As a result, the voltage V0 (s) is given by V0 (s) = 2I2 (s), that is,
V0 (s) =
2s2 + 7s + 9
(s + 1) s2 + 52 s + 2
Now we need to do partial-fraction expansion. Note that the second term in the denominator can be factorized as
√
√
[s − (−5/4 + j 7/4)][s − (−5/4 − j 7/4)]
which has complex factors. Hence, let
A2 s + A3
A1
2s2 + 7s + 9
=
+
s + 1 s2 + 52 s + 2
(s + 1) s2 + 52 s + 2
Multiplying the common denominator (s + 1) s2 + 52 s + 2 produces
5
2
2
2s + 7s + 9 = A1 s + s + 2 + (A2 s + A3 )(s + 1)
2
Let s = −1, we find A1 = 8. Then, we have
5
2s2 + 7s + 9 = 8 s2 + s + 2 + (A2 s + A3 )(s + 1)
2
that is,
2s2 + 7s + 9 = (8 + A2 )s2 + (20 + A2 + A3 )s + 16 + A3
Equating coefficients of like powers of s gives
2 = 8 + A2
7 = 20 + A2 + A3
9 = 16 + A3
Solving these equations for A2 and A3 gives A2 = −6 and A3 = −7. Therefore we get
V0 (s)
=
=
6s + 7
8
− 2 5
s + 1 s + 2s + 2
6s + 7
8
−
s + 1 s2 + 5 s + 5 2 + 2 −
2
4
5 2
4
13
14
CHAPTER 1. CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
Figure 1.15: Circuit in t domain
8
6s + 7
−
2
7
s+1
s + 54 + 16
5
6 s+ 4 +7−
8
−
2
s+1
s+ 5 + 7
=
=
4
30
4
16
s + 54
2 √ 2
s + 54 + 47
=
8
−6
s+1
+
7
2
4
√
√ 2
7 s + 5 2 +
7
4
4
√
Taking inverse on both sides gives
vo (t)
√ !
7
t
= 8e − 6e
cos
4
√ !
2 −5t
7
4
+ √ e
t
sin
4
7
−t
− 54 t
Example
√ Consider the half-wave rectifier circuit with an inductive R − L load, as shown in Fig. 1.15. vi = Vmax sin ωt =
2V sin ωt. Find the current i(t) through the load.
Solution
During the positive half cycle of the input voltage, the diode is forward biased and so current flow commences as
the supply voltage goes positive. The presence of L delays the current change. The current continues to flow at the
end of the positive half cycle because of this inductance. The diode remains on and the load sees the negative supply
voltage until the current decays to zero. This process can be analyzed mathematically as follows.
By writing Kirchhoff voltage equation, we get
vi = VL + VR
that is,
Vmax sin ωt = L
di(t)
+ Ri(t)
dt
Note that i(0) = 0. Taking the Laplace transform on both sides of the above equation gives
ωVmax
= sLI(s) + RI(s)
s2 + ω 2
Solving for I(s) produces
I(s) =
Let a =
R
L.
(s2
ωVmax
ωVmax
=
2
2
+ ω )(R + sL)
L(s + ω 2 )(s +
R
L)
Then, it follows that
I(s) =
L(s2
ωVmax
ωVmax
1
=
2
2
2
+ ω )(s + a)
L (s + ω )(s + a)
1.6. ANALYSIS OF ELECTRICAL NETWORK BY LAPLACE TRANSFORM
Let
15
1
As + B
C
= 2
+
(s2 + ω 2 )(s + a)
(s + ω 2 ) s + a
Multiplying the common denominator (s2 + ω 2 )(s + a) yields
= (As + B)(s + a) + C(s2 + ω 2 )
= As2 + Aas + Bs + Ba + Cs2 + Cω 2
= (A + C)s2 + (Aa + B)s + Ba + Cω 2
1
By equating coefficients of like powers of s, we get
A+C
Aa + B
Ba + Cω 2
= 0
= 0
= 1
The first equation gives C = −A and the second gives B = −aA. Substituting both C = −A and B = −aA into the
third equation produces
−a2 A − ω 2 A = 1
that is,
A=−
1
So, we have C = a2 +ω
2 and B =
Therefore, I(s) is given by
1
a2 + ω 2
a
a2 +ω 2 .
I(s)
=
=
=
=
(
)
1
a
1
ωVmax − a2 +ω2 s + a2 +ω2
a2 +ω 2
+
L
s2 + ω 2
(s + a)
ωVmax
s
a
1
−
+
+
L(a2 + ω 2 ) s2 + ω 2
s2 + ω 2
(s + a) s
a
1
ωVmax
− 2
+ 2
+
2
2
2
2)
s
+
ω
s
+
ω
(s
+
a)
L( R
+
ω
L2
s
a
1
ωLVmax
ω
−
+
+
R2 + (ωL)2 )
s2 + ω 2
ω s2 + ω 2
(s + a)
Taking the inverse transform gives
i(t)
=
=
=
p
with Z = R2 + (ωL)2 .
Now let
o
ωLVmax n
a
−at
sin
ωt
+
e
−
cos
ωt
+
R2 + (ωL)2 )
ω
o
ωLVmax n a
−at
sin
ωt
−
cos
ωt
+
e
R2 + (ωL)2 ) ω
o
ωLVmax n a
−at
sin
ωt
−
cos
ωt
+
e
Z2
ω
a
sin ωt − cos ωt = A sin(ωt − φ)
ω
Then, we have
a
sin ωt − cos ωt = A sin ωt cos φ − A cos ωt sin φ
ω
By letting ωt = 0 and ωt = π2 , we get
−1 = −A sin φ
a
= A cos φ
ω
which implies that
tan φ =
and
A2 = 1 +
ω
ωL
=
a
R
a2
R2
(ωL)2 + R2
Z2
=1+
=
=
2
2
2
ω
(ωL)
(ωL)
(ωL)2
16
CHAPTER 1. CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
Hence,
A=
Z
ωL
Therefore, we have derived the following.
i(t)
ωLVmax Z
−at
sin(ωt
−
φ)
+
e
Z 2 ωL
ωL −at
Vmax ωL Z
sin(ωt − φ) +
e
Z Z ωL
Z
Vmax
ωL −at
sin(ωt − φ) +
e
Z
Z
=
=
=
Recall that A sin φ = 1 with A =
Z
ωL .
Hence
sin φ =
ωL
1
=
A
Z
Therefore, the current takes the form of
i(t)
=
o
R
Vmax n
sin(ωt − φ) + sin φe− L t
Z
Now let us calculate the time at which the current decays to zero. To this end, assume that the current will become
zero at ωt = β, that is,
o
R β
Vmax n
sin(β − φ) + sin φe− L ω = 0
Z
The solution to this equation has no analytical form, so we have to use the numerical method to solve for β.
This differential equation can also be solved as follows:
Let i(t) = if (t) + in (t) where if (t) is force response or steady-state response and in (t) is natural response or
zero-state response.
if (t) is determined by the phasor analysis as
if (t) =
Vmax
vi (t)
=
sin(ωt − φ)
Z
Z
p
where Z = R2 + (Lω)2 and φ = tan−1 Lω
R .
L
− τt
with τ = R
.
in (t) is determined by in (t) = Ae
Then, we have
t
Vmax
sin(ωt − φ) + Ae− τ
i(t) =
Z
where A is determined from the initial condition i(0) = 0, i.e.
A=
Vmax
sin φ
Z
Chapter 2
Thyristors
A thyristor is a four-layer seniconductor device of PNPN structure with three PN-junctions. It has three terminals:
anode, cathode, and gate.
There are several members of the thyristor family. The more commonly used thyristors are
1. silicon-controlled rectifiers (SCRs)
2. gate-turn-off (GTO) thyristors
3. bi-directional triode thyristors (TRIACs)
2.1
Silicon-Controlled Rectifiers
The SCR is a four-layer PNPN semiconductor device. A dc voltage is applied to the SCR across the anode and cathode
through a resistor RA and another dc voltage source is connected to the gate through a switch S and a resistor RG ,
as shown in Fig. 2.1. Assume that the switch is open initially.
In order to see how the SCR works, let us divide the middle PN regions along the dotted line, and model the SCR
by a PNP transistor and an NPN transistor (two transistor model).
S is open initially
In this case, there is no gate current, only small leakage current flows from anode to cathode and no conduction
can take place. The SCR is said to be in the forward blocking or off-state condition, and the leakage current is known
as off-state or forward leakage current ID .
If the anode-to-cathode voltage VAK is increased to a sufficiently large value VBO , the reversed-biased junction will
break. This is known as avalanche break down and VBO is called forward breakdown voltage. A large forward anode
current flows from anode to cathode and the SCR is in a conducting state or on state.
When the cathode voltage is positive with respect to the anode, the SCR will be in the reverse blocking state or
off-state and a reverse leakage current would flow through the device. Reverse breakdown will take place if the reverse
voltage reaches reverse breakdown voltage (VRBO ).
S is closed
There is a current iG flowing into the gate, Q2 will conduct and a current from the collector of Q2 to emitter of
Q2 will flow. This is drawn from the base of Q1 which causes Q1 to conduct. Now the main current from Q1 is fed
into the base of Q2 which holds Q2 on. Hence, if the gate current is removed, conduction process continues and the
current iA flows from anode to cathode.
Figure 2.1: SCR test circuit
17
18
CHAPTER 2. THYRISTORS
In the on-state, the voltage drop across the SCR is very small, typically, 1V. The anode current is limited by the
external resistance RL . The anode current must be more than a value known as latching current IL , which is the
minimum anode current required to maintain the thyristor in the on-state immediately after a SCR has been turned
on and the gate signal has been removed.
Once a SCR conducts, it behaves like a conducting diode and keep on-state until the forward anode current is
reduced below a level known as the holding current IH , which is the minimum anode current to maintain the SCR in
the on-state. IH is in the order of milliamperes and less than IL .
SCR Turn-On
A SCR can be turned on by the following methods.
1. Gate current. If an SCR is forward biased, the injection of gate current by applying positive gate voltage
between the gate and cathode terminals would turn on the SCR. As the gate current is increased, the forward
blocking voltage is decreased.
2. High voltage. If the forward anode-to-cathode voltage is greater than VBO , the SCR will be turned on.
However, this type of turn-on may be destructive and should be avoided.
3. Light. If light is allowed to strike the junctions of a SCR, the elctron-hole pairs will increase; and the SCR may
be turned on. The light-activated thyristors are turned on by allowing light to strike the silicon wafers.
4. Thermals. If the temperature of a SCR is high, the SCR may be turned on. This type of turn-on is normally
avoided because it may cause thermal runaway.
5. dv/dt. The SCR may be turned on by high rate of rise of the anode-to-cathode voltage, which can cause damage
of the SCR and should be avoid.
SCR Turn-Off
A SCR is normally switched on by applying a pulse of gate signal. Once the SCR is turned on and the output
requirements are satisfied, it is usually necessary to turn it off. The turn-off means that the forward conduction of the
SCR has ceased and the reapplication of a positive voltage to the anode will not cause current flow without applying
the gate signal. SCRs can be turned off by reducing the anode current to be lower than their holding currents.
Commutation is the process of turning off a SCR and it normally causes transfer of current flow to other parts of the
circuit. There are many techniques to commutate a SCR. However, these can be broadly classified as two types:
1. Natural commutation
2. Forced commutation
Natural Commutation
If the source (or input) voltage is ac, the SCR current goes through a natural zero, and a reverse voltage appears
across the SCR. The device is then automatically turned off due to the natural behavior of the source voltage. This
is known as natural commutation or line commutation.
The figures 3.5 and 3.6 shows the circuit arrangement for natural commutation and the voltage and current
waveforms with a delay angle α.
The delay angle α is defined as the angle between the zero crossing of the input voltage and the instant the SCR
is fired. Note that a train of gate current pulses is applied to the gate terminal. The SCR is triggered and turned on
by the gate current pulses whenever the voltage across anode and cathode is positive. On the other hand, the SCR is
switched off whenever the applied voltage goes negative.
The natural commutation is applied in ac voltage controllers, phase-controllerd rectifier, and cycloconverters.
Forced Commutation
In circuits where the signal to be controlled will never have a negative period, the SCR must be switched off by
using an external circuit, which is known as commutation circuit. This is achieved by diverting the anode current flow.
This technique is called forced commutation and normally applied in dc-dc converters (choppers) and dc-ac converters
(inverters).
The circuit for the forced commutation is given in fig. 2.2.
When T1 is fired, the load R1 is connected to the supply voltage Vs , and at the same time the capacitor C is
charged to Vs through the other load with R2 . When T2 is fired, the capacitor is then placed across T1 and the load
R2 is connected to the supply voltage Vs . T1 is reverse biased and is turned off. Once T1 is switched off, the capacitor
voltage is reversed to −Vs through R1 , T2, and the supply. If T1 is fired again, SCR2 is turned off and the cycle is
repeated. Normally, the two SCRs conduct with equal time intervals.
2.2. GATE-TURN-OFF THYRISTORS
19
Figure 2.2: Forced commutation circuit
(a) T1 is fired at t = t1 . Redefining the time origin at t1 . Assume that the capacitor has been charged to Vs in
the previous commutation, that is, VC (0) = −Vs . By Kirchhof voltage law,
vs (t) = Ri(t) + vc (t) = RC
dvc (t)
+ vc (t)
dt
which has a solution of
t
vc (t) = Vs − 2Vs e− RC
(b) T2 is fired at t = t2 . Redefining the time origin at t = t2 , vc (t) satisfies
vs (t) = −Ri(t) − vc (t) = RC
dvc (t)
+ vc (t),
dt
vC (0) = Vs
which has a solution of
t
vc (t) = −Vs + 2Vs e− RC
2.2
Gate-Turn-Off Thyristors
The GTO is yet another four layer device with 3 therminals.
Although the symbol is different, operation is similar to SCR and two transistor analogy still applies. However,
the gate can be used to turn the device on and off. A positive gate current pulse will switch the GTO on, while a
negative gate current pulse will switch the GTO off. Since the GTO can be turned off by a short negative pulse to its
gate, it has advantage over SCRs: elimination of commutating components in forced commutation.
2.3
DIAC
A DIAC is a device containing five semiconductor layers (PNPNP) that behaves like two PN diodes connected back to
back. It can conduct in either direction once the breakover voltage is exceeded. It turns on when the applied voltage
in either direction exceeds VBO . Once it is turned on, a DIAC remains on until its current falls below IH .
The name DIAC is derived from the word diode with AC applications.
2.4
TRIAC: Bidirectional Triode Thyristors
A TRIAC is a device that behaves like two SCRs connected back to back with a common gate lead. It can conduct in
either direction once its breakover voltage is exceeded. The breakover voltage in a TRIAC decreases with increasing
gate current in just the same manner as it does in an SCR, except that a TRIAC responds to either positive or negative
pulses at its gate. Once it is turned on, a TRIAC remains on until its current falls below IH .
TRI implies ”three terminal device” and AC implies AC application.
Because a single TRIAC can conduct in both directions, it can replace a more complex pair of back-to-back SCRs
in many AC control circuits. However, TRIACs generally switch more slowly then SCRs, and are available only at
lower power ratings. As a result, their use is largely restricted to low- to medium-power applications in 50Hz or 60Hz
circuits, such as light dimmer circuits.
20
CHAPTER 2. THYRISTORS
Figure 2.3: TRIAC-DIAC application circuit
2.5
TRIAC-DIAC Applications (Lab 2)
Although DIACs can be used as the main power switch device, they are merely always used in the gate circuit of the
TRIAC circuit. Since VBO is accurately known, it can provide an accurate firing (Triggering) voltage to the TRIAC.
The circuit for light dimmer control will be given in fig. 2.3. An adjustable resistance R, together with a capacitor
C, makes a single-element phase shift network. When the voltage across C reaches VBO of the DIAC, the DIAC is
turned on and C is discharged through the DIAC and TRIAC gate. The discharging current triggers the TRIAC into
the conduction mode for the remainder of that half cycle. Triggering is in the 1st quadrant and 3rd quadrant modes
of the circuit. This circuit has many small range applications, such as light dimmer control, heater and fan speed
control.
vC (t) satisfies the following differential equation
vs (t) = RC
dvC (t)
+ vC (t)
dt
with vs (t) = Vm sin ωt and vC (0) = 0. Assume that
vC (t) = vCf (t) + vCn (t)
Then,
vCf (t) = √
Vm ωC
sin(ωt − φ)
1 + ω 2 C 2 R2
1
with φ = tan−1 [ ωRC
]. vCn (t) satisfies
0 = RC
dvC (t)
+ vC (t)
dt
t
which has a solution of the form vCn (t) = Ae− RC with A determined by vC (0) = 0, that is,
0= √
Vm ωC
sin(−φ) + A
1 + ω 2 C 2 R2
So
A= √
Therefore, we get
vCf (t) = √
Vm ωC
sin(φ)
1 + ω 2 C 2 R2
t
Vm ωC
sin(ωt − φ) + sin φe− RC
1 + ω 2 C 2 R2
Chapter 3
Rectifiers
A rectifier is used to convert AC voltages to DC voltages. It is a type of AC-DC converter. Depending on the source
voltages, the rectifiers are classified into single-phase and three phase. Depending on whether the output voltages can
be adjusted or not, the rectifiers are classified into uncontrolled and controlled.
3.1
Single-Phase Uncontrolled Half-Wave Rectifiers with R Loads
Positive half-cycle: D is forward biased and conducts. The voltage across D is zero and the output voltage is the
same as the source voltage.
Negative half-cycle: D is reverse biased and turns off, making the current zero. The output voltage is zero and
the voltage across the diode is the same as the source voltage.
vo (t)
vD (t)
=
=
vs (t) = Vm sin ωt
0
0 ≤ ωt ≤ π
π ≤ ωt ≤ 2π
0
vs (t) = Vm sin ωt
0 ≤ ωt ≤ π
π ≤ ωt ≤ 2π
The average output voltage is calculated by
Vo
and the average output current is Io =
The rms output voltage is
Z π
1
=
Vm sin ωtd(ωt)
2π 0
Vm
π
[cos ωt]0
= −
2π
Vm
=
π
Vo
R.
Vrms
=
s
s
1
2π
Vm2
2π
Z
π
(Vm sin ωt)2 d(ωt)
0
Z
π
1 − cos 2ωt
d(ωt)
2
0
s
π
1
Vm2
=
ωt − sin 2ωt
4π
2
0
=
=
Vm
2
and the rms output current Irms = Vrms
R .
The DC output power is Pdc = Vo Io and the AC output power is Pac = Vrms Irms . The efficiency of the rectifier is
dc
η=P
Pac .
21
22
CHAPTER 3. RECTIFIERS
Figure 3.1: Single-Phase Uncontrolled Half-Wave Rectifiers with R Loads
v
s
π
2π
3π
4π
ωt
π
2π
3π
4π
ωt
π
2π
3π
4π
ωt
vo,io,is
vD
Figure 3.2: Single-Phase Uncontrolled Half-Wave Rectifiers with R Loads
3.2. SINGLE-PHASE UNCONTROLLED HALF-WAVE RECTIFIERS WITH RL LOADS
23
Figure 3.3: Single-Phase Uncontrolled Half-Wave Rectifiers with RL Loads
v ,i ,i
s o s
π
β
2π
3π
2π+β
4π
ωt
π
β
2π
3π
2π+β
4π
ωt
π
β
2π
3π
2π+β
4π
ωt
π
β
2π
3π
2π+β
4π
ωt
v
o
VD
V
L
Figure 3.4: Single-Phase Uncontrolled Half-Wave Rectifiers with RL Loads
The power factor of the circuit is calculated from
PF =
3.2
Vrms Irms
Pac
=
=
S
Vs,rms Is,rms
Vm
2 Irms
V
√m Irms
2
√
=
2
2
Single-Phase Uncontrolled Half-Wave Rectifiers with RL Loads
As the source voltage goes through zero, becoming positive, the diode becomes forward biased. The current flow in
the circuit satisfies
di(t)
= vs (t) = Vm sin ωt, i(0) = 0
Ri(t) + L
dt
The solution to the above equation is
i(t) =
i
ωt
Vm h
sin(ωt − θ) + sin θe− ωτ
Z
p
and τ =
where Z = R2 + (ωL)2 , θ = tan−1 ωL
R
It is important to note the following facts:
L
R.
1. The diode remains forward biased longer than π.
24
CHAPTER 3. RECTIFIERS
2. The point when the diode turns off is when the current i(t) reaches zero. The first positive value of ωt that
results in zero current is called the extinction angle, β, which can be determined by
i(β/ω) =
or
i
β
Vm h
sin(β − θ) + sin θe− ωτ = 0
Z
β
sin(β − θ) + sin θe− ωτ = 0
There is no closed-form solution for β, and some numerical method is required to solve the equation above.
3. The inductor voltage is negative when the current is decreasing (vL (t) = L di(t)
dt ).
3.3
Single-Phase Controlled Half-Wave Rectifiers with R Loads
The output voltage of an uncontrolled rectifier is fixed if the source and load parameters are established. Many
applications, such as DC motor drives, need variable DC voltages. One way to control the output voltages of rectifiers
is to replace diodes with SCRs. The rectifiers built with SCRs are called controlled rectifiers.
Two conditions must be met before an SCR can conduct: (1) the SCR must be forward biased and (2) a current
must be applied to the gate of the SCR.
Unlike the diode, the SCR will not begin to conduct as soon as the source becomes positive. Conduction is delayed
until a gate current is applied, which is the basis for using the SCR as a means of control. Once the SCR is conducting,
the gate current can be removed and the SCR remains on until the current goes to zero.
3.3. SINGLE-PHASE CONTROLLED HALF-WAVE RECTIFIERS WITH R LOADS
25
Figure 3.5: Single-Phase Controlled Half-Wave Rectifiers with R Loads
vs
α
π
2π 2π+α
3π
4π
ωt
α
π
2π 2π+α
3π
4π
ωt
α
π
2π 2π+α
3π
4π
ωt
vo,io,is
vSCR
Figure 3.6: Single-Phase Controlled Half-Wave Rectifiers with R Loads
Positive half-cycle: The SCR is forward biased. It does not conduct until it is fired. Before the SCR conducts,
the output voltage is zero and the voltage across the SCR is the same as the source. Suppose it is fired at ωt = α.
Then it begins to conduct at ωt = α and the voltage across the SCR drops to zero. The output voltage is the same
as the source voltage.
Negative half-cycle: The SCR is reverse biased and it is turned off. The voltage across the SCR is the same as
the source and the output voltage becomes zero.
The voltages across the load and SCR are
vs (t) = Vm sin ωt α ≤ ωt ≤ π
vo (t) =
0
otherwise
0
α ≤ ωt ≤ π
vSCR (t) =
vs (t) = Vm sin ωt otherwise
Delay angle or firing angle is the angle after the input voltage starts to go positive until the SCR is fired.
The average output voltage is calculated from
Z π
1
Vo =
Vm sin ωtd(ωt)
2π α
Vm
π
[cos ωt]α
= −
2π
Vm
(1 + cos α)
=
2π
26
CHAPTER 3. RECTIFIERS
and the average output current is Io =
The rms output voltage is
Vo
R.
Vrms
=
s
s
1
2π
Vm2
2π
Z
π
(Vm sin ωt)2 d(ωt)
α
Z
π
1 − cos 2ωt
d(ωt)
2
α
s
π
1
Vm2
ωt − sin 2ωt
=
4π
2
α
r
Vm
α sin 2α
=
1− +
2
π
2π
=
and the rms output current Irms =
The power factor is
PF
Vrms
R .
=
=
Pac
Vrms Irms
=
=
S
Vs,rms Is,rms
√ r
2
α sin 2α
1− +
2
π
2π
Vm
2
q
1−
α
π
+
sin 2α
2π Irms
V
√m Irms
2
Note that α = 0 gives Vo = Vπm and Vrms = V2m , which are the same as those for an uncontrolled half-wave rectifier.
Example: Design a circuit to produce an average voltage of 40V across a 100Ω load resistor from a 120V rms
60Hz AC source. Determine the power absorbed by the resistance and the power factor.
Solution: If an uncontrolled half-wave rectifier is used, the average voltage would be
√
120 2
Vm
Vo =
=
= 54V
π
π
Therefore, some means of reducing the average resistor voltage to the design specification of 40V must be formed. A
series resistance or inductance could be added to an uncontrolled rectifier, or a controlled rectifier could be used. The
controlled rectifier has the advantage of not altering the load or introducing losses, so it is selected for this design.
In order to get an output voltage of 40V, the required delay angle is
2πVo
2π × 40
√ − 1 = 61.2◦ = 1.07rad
α = cos−1
− 1 = cos−1
Vm
120 × 2
With this angle, the rms output voltage is
r
√ r
120 2
Vm
α sin 2α
1.07 sin(2 × 61.2◦)
Vrms =
=
+
= 75.6V
1− +
1−
2
π
2π
2
π
2π
The load power is
Pac =
2
75.62
Vrms
=
= 57.1W
R
100
The power factor is
√ r
√ r
2
2
α sin 2α
1.07 sin(2 × 61.2◦)
=
+
= 0.63
1− +
1−
PF =
2
π
2π
2
π
2π
3.3.1
Single-Phase Controlled Half-Wave Rectifiers with RL Loads
The analysis of this circuit is similar to that of the uncontrolled half-wave rectifier with the RL load.
Now assume that the delay angle is α. Then the SCR will be turned on at ωt = α and conduct until the current
in the circuit reaches zero. The current can be found from
Ri(t) + L
di(t)
= vs (t) = Vm sin ωt,
dt
i(α/ω) = 0
3.3. SINGLE-PHASE CONTROLLED HALF-WAVE RECTIFIERS WITH R LOADS
27
Figure 3.7: Single-Phase Controlled Half-Wave Rectifiers with RL Loads
v ,i ,i
s o s
α
π
β
2π 2π+α
3π
2π+β
4π
ωt
α
π
β
2π 2π+α
3π
2π+β
4π
ωt
α
π
β
2π 2π+α
3π
2π+β
4π
ωt
α
π
β
2π 2π+α
3π
2π+β
4π
ωt
vo
VSCR
V
L
Figure 3.8: Single-Phase Controlled Half-Wave Rectifiers with RL Loads
28
CHAPTER 3. RECTIFIERS
Figure 3.9: Single-Phase Uncontrolled Half-Wave Rectifiers with RL Loads and Freewheeling Diode
Figure 3.10: Single-Phase Controlled Half-Wave Rectifiers with RL Loads and Freewheeling Diode
The solution to the above equation is
i(t) =
i
α−ωt
Vm h
sin(ωt − θ) − sin(α − θ)e ωτ
Z
p
L
where Z = R2 + (ωL)2 , θ = tan−1 ωL
.
and τ = R
R
The extinction angle is defined as the angle at which the current returns to zero, as in the case of the uncontrolled
rectifier, that is,
i
α−β
Vm h
i(β/ω) =
sin(β − θ) − sin(α − θ)e ωτ = 0
Z
or
α−β
sin(β − θ) − sin(α − θ)e ωτ = 0
There is no closed-form solution for β, and some numerical method is required to solve the equation above.
The angle β − α is called the conduction angle, γ = β − α
3.3.2
Freewheeling Diodes
Recall that output voltages from both uncontrolled and controlled rectifiers with RL loads contain negative parts,
which result in smaller average output voltages. The average output voltages can be increased by making β = π,
which is possible by adding freewheeling diodes to the rectifiers as shown in the figures.
Positive half-cycle: D is forward biased and Df
Positive half-cycle: SCR is forward biased and
reverse biased. The output voltage is the same as
is triggered at ωt = α. Df is reverse biased. The
the source voltage.
output voltage is the same as the source voltage.
Negative half-cycle: Df is forward biased and D
Negative half-cycle: Df is forward biased and
reverse biased. The output voltage is zero.
SCR reverse biased. The output voltage is zero.
The output voltage is given by
The output voltage is given by
vs (t) 0 ≤ t ≤ π
vs (t) α ≤ t ≤ π
vo (t) =
vo (t) =
0
π ≤ t ≤ 2π
0
otherwise
The average output voltage is Vo = Vπm , which is the
same as that for the uncontrolled half-wave rectifier
with R load.
m
The average output voltage is Vo = V2π
(1 + cos α),
which is the same as that for the controlled half-wave
rectifier with R load.
3.3. SINGLE-PHASE CONTROLLED HALF-WAVE RECTIFIERS WITH R LOADS
29
Figure 3.11: Single-Phase Uncontrolled Full-Wave Rectifiers with R Loads
v
s
π
2π
3π
4π
ωt
π
2π
3π
4π
ωt
π
2π
3π
4π
ωt
π
2π
3π
4π
ωt
v ,i
o o
vD1,2
v
D3,4
Figure 3.12: Single-Phase Uncontrolled Full-Wave Rectifiers with R Loads
3.3.3
Single-Phase Uncontrolled Full-Wave Rectifiers with R Loads
Positive half-cycle: Both D1 and D2 are forward biased and D3 and D4 reverse biased. The output voltage is equal
to the source voltage. The voltages across D1 and D2 are zero and the voltages across D3 and D4 are −vs (t).
Negative half-cycle: Both D3 and D4 are forward biased and D1 and D2 reverse biased. The output voltage is
−vs (t). The voltages across D1 and D2 are the source voltages and the voltages across D3 and D4 are zero.
The voltages across the load and diodes are given by
0 ≤ ωt ≤ π
vs (t) = Vm sin ωt
vo (t) =
−vs (t) = −Vm sin ωt π ≤ ωt ≤ 2π
0
0 ≤ ωt ≤ π
vD1,2 (t) =
vs (t) = Vm sin ωt π ≤ ωt ≤ 2π
−vs (t) = −Vm sin ωt 0 ≤ ωt ≤ π
vD3,4 (t) =
0
π ≤ ωt ≤ 2π
The average output voltage is calculated by
Vo
Z π
2
Vm sin ωtd(ωt)
2π 0
2Vm
π
[cos ωt]0
= −
2π
2Vm
=
π
=
30
CHAPTER 3. RECTIFIERS
and the average output current is Io =
The rms output voltage is
Vo
R.
Vrms
=
s
s
2
2π
Z
π
(Vm sin ωt)2 d(ωt)
0
Z
2Vm2 π 1 − cos 2ωt
d(ωt)
2π 0
2
s
π
1
Vm2
=
ωt − sin 2ωt
2π
2
0
=
=
Vm
√
2
and the rms output current Irms = Vrms
R .
The DC output power is Pdc = Vo Io and the AC output power is Pac = Vrms Irms . The efficiency of the rectifier is
dc
η=P
Pac .
The power factor of the circuit is calculated from
Pac
Vrms Irms
PF =
=
=
S
Vs,rms Is,rms
V
√m Irms
2
V
√m Irms
2
=1
Example: A full-wave bridge rectifier is supplied by a sinusoidal voltage source of 110V rms. The load resistance
is 100Ω. Determine the efficiency, power factor of the rectifier, and the peak inverse voltage (PIV) of the diode D1 .
Solution:
√
2 × 110 2
2Vm
=
= 99V
Vo =
π
π
99
Vo
Io =
=
= 0.99A
R
100
Pdc = Vo Io = 99 × 0.99 = 98W
√
Vm
110 2
Vrms = √ = √
= 110V
2
2
110
Vrms
=
= 1.1A
Irms =
R
100
Pac = Vrms Irms = 110 × 1.1 = 121W
Pdc
98
η =
=
= 0.81
Pac
121
PF = 1
√
P IV = Vm = 110 × 2 = 155.6V
3.3.4
Single-Phase Controlled Full-Wave Rectifiers with R Loads
Positive half-cycle: Both SCR1 and SCR2 are forward biased and SCR3 and SCR4 reverse biased. The output
voltage is still zero, the voltages across SCR1 and SCR2 are 12 vs (t) and the voltages across SCR3 and SCR4 are
− 21 vs (t). When SCR1 and SCR2 are fired at ωt = α, SCR1 and SCR2 conduct until the current becomes zero at
ωt = π. The output voltage is equal to the source voltage. The voltages across SCR1 and SCR2 are zero and the
voltages across SCR3 and SCR4 are −vs (t).
Negative half-cycle: Both SCR1 and SCR2 are reverse biased and SCR3 and SCR4 forward biased. The output
voltage is still zero, the voltages across SCR1 and SCR2 are 12 vs (t) and the voltages across SCR3 and SCR4 are
− 21 vs (t). When SCR3 and SCR4 are fired at ωt = π + α, SCR3 and SCR4 conduct until the current becomes zero
at ωt = 2π. The output voltage is equal to −vs (t). The voltages across SCR1 and SCR2 are vs (t) and the voltages
across SCR3 and SCR4 are zero.
The voltages across the load and diodes are given by

α ≤ ωt ≤ π
 vs (t) = Vm sin ωt
−vs (t) = −Vm sin ωt π + α ≤ ωt ≤ 2π
vo (t) =

0
otherwise
3.3. SINGLE-PHASE CONTROLLED HALF-WAVE RECTIFIERS WITH R LOADS
31
Figure 3.13: Single-Phase Controlled Full-Wave Rectifiers with R Loads
vs
α
π
π+α
2π 2π+α
3π 3π+α
4π
ωt
α
π
π+α
2π 2π+α
3π 3π+α
4π
ωt
π
π+α
2π 2π+α
3π 3π+α
4π
ωt
π
π+α
2π 2π+α
3π 3π+α
4π
ωt
v ,i
o o
vSCR1,2
α
v
SCR3,4
α
Figure 3.14: Single-Phase Controlled Full-Wave Rectifiers with R Loads
32
CHAPTER 3. RECTIFIERS
vSCR1,2 (t)
vSCR3,4 (t)

α ≤ ωt ≤ π
 0
vs (t) = Vm sin ωt
π + α ≤ ωt ≤ 2π
=
 1
vs (t) = 12 Vm sin ωt otherwise
 2
α ≤ ωt ≤ π
 −vs (t) = −Vm sin ωt
0
π + α ≤ ωt ≤ 2π
=
 1
− 2 vs (t) = − 21 Vm sin ωt otherwise
The average output voltage is calculated from
Vo
and the average output current is Io =
The rms output voltage is
Z π
2
=
Vm sin ωtd(ωt)
2π α
Vm
π
[cos ωt]α
= −
π
Vm
(1 + cos α)
=
π
Vo
R.
Vrms
=
s
s
2
2π
Vm2
π
Z
π
(Vm sin ωt)2 d(ωt)
α
Z
π
1 − cos 2ωt
d(ωt)
2
α
s
π
1
Vm2
=
ωt − sin 2ωt
2π
2
α
r
Vm
α sin 2α
= √
1− +
π
2π
2
=
and the rms output current Irms =
The power factor is
PF
Vrms
R .
Vrms Irms
Pac
=
=
=
S
Vs,rms Is,rms
r
α sin 2α
=
1− +
π
2π
V
√m
2
q
1−
α
π
+
sin 2α
2π Irms
V
√m Irms
2
Note that α = 0 gives Vo = 2Vπm and Vrms = V√m2 , which are the same as in the case of an uncontrolled full-wave
rectifier.
Example: A full-wave bridge controlled rectifier has an AC input of 120V rms at 60Hz and 20Ω load resistor. The
delay angle is α = 40◦ . Determine the average current in the load, the power absorbed by the load, efficiency and the
source volt-amperes.
Solution:
√
120 2
Vm
Vo =
(1 + cos α) =
(1 + cos 40◦ ) = 95.4V
π
π
95.4
Vo
Io =
=
= 4.77A
R
20
Pdc = Vo Io = 95.4 × 4.77 = 455W
r
√ r
120 2
sin 2 × 40
Vm
α sin 2α
40
= √
+
= 116V
Vrms = √
1− +
1−
π
2π
180
2π
2
2
116
Vrms
Irms =
=
= 5.8A
R
20
Pac = Vrms Irms = 116 × 5.8 = 673W
Pdc
455
η =
= 0.68
=
Pac
673
S = Vs,rms Is,rms = Vs,rms Irms = 120 × 5.8 = 696V A
3.3. SINGLE-PHASE CONTROLLED HALF-WAVE RECTIFIERS WITH R LOADS
33
Figure 3.15: Single-Phase Controlled Full-Wave Rectifiers with RL Loads
3.3.5
Single-Phase Controlled Full-Wave Rectifiers with RL Loads
Discontinuous Current:
Starting the analysis at ωt = 0 with zero load current, i.e. i(0) = 0, SCR1 and SCR2 are forward biased and SCR3
and SCR4 are reverse biased as the source voltage becomes positive. Assume that gate signals are applied to SCR1
and SCR2 at ωt = α, turning SCR1 and SCR2 on. Then the voltage across the load is vo (t) = vs (t) = Vm sin ωt, and
the load current is determined from
Ri(t) + L
di(t)
= vs (t) = Vm sin ωt,
dt
i(α/ω) = 0
The solution to the above equation is
i(t) =
where Z =
p
R2 + (ωL)2 , θ = tan−1
ωL
R
i
α−ωt
Vm h
sin(ωt − θ) − sin(α − θ)e ωτ
Z
,τ=
L
R,
and β is determined by
sin(β − θ) − sin(α − θ)e
α−β
ωτ
=0
If β < π + α, then the current remains zero until ωt = π + α, when the gate signals are applied to SCR3 and
SCR4 , which are then forward biased and begin to conduct. This mode of operation is called discontinuous current.
Continuous Current:
If the load current is still positive at ωt = π + α when the gate signals are applied to SCR3 and SCR4 in the
preceding analysis, SCR3 and SCR4 are turned on and SCR1 and SCR2 are forced off. Since the initial condition for
the current in the second half-cycle is not zero, the current expression derived above is no longer valid. The current
in this mode is given by
α−ωt
2
Vm
ωτ
sin(ωt − θ) −
i(t) =
sin(α
−
θ)e
π
Z
1 − e− ωτ
The boundary between continuous and discontinuous current is when β ≥ π + α. The current at ω = π + α must
be greater than zero for continuous current operation, that is,
i
α−π−α
Vm h
π+α
=
i
sin(π + α − θ) − sin(α − θ)e ωτ
ω
Z
π Vm =
− sin(α − θ) − sin(α − θ)e− ωτ
Z
Vm π =
sin(θ − α) + sin(θ − α)e− ωτ
Z
Vm
π sin(θ − α) 1 + e− ωτ
=
Z
≥ 0
which implies that θ − α ≥ 0, that is, θ ≥ α.
In summary, if θ ≥ α, the current is continuous; otherwise discontinuous.
Example: The controlled full-wave bridge rectifier has a source of 120V rms at 60Hz, R = 10Ω, L = 20mH, and
α = 60◦ .
1. Check the continuity of the current.
34
CHAPTER 3. RECTIFIERS
vs,io
Discontinuous Mode
α
π
β π+α
2π
2π+α
3π
3π+α
2π+β
4π
ωt
α
π
β π+α
2π
2π+α
3π
3π+α
2π+β
4π
ωt
α
π
β π+α
2π
2π+α
3π
3π+α
2π+β
4π
ωt
α
π
β π+α
2π
2π+α
3π
3π+α
2π+β
4π
ωt
α
π
β π+α
2π
2π+α
3π
3π+α
2π+β
4π
ωt
vo
VSCR1,2
VSCR3,4
VL
Figure 3.16: Single-Phase Controlled Full-Wave Rectifiers with RL Loads: Discontinuous Current
v ,i
s o
Continuous Mode
α
π
π+α
2π 2π+α
3π 3π+α
4π
ωt
α
π
π+α
2π 2π+α
3π 3π+α
4π
ωt
π
π+α
2π 2π+α
3π 3π+α
4π
ωt
α
π
π+α
2π 2π+α
3π 3π+α
4π
ωt
α
π
π+α
2π 2π+α
3π 3π+α
4π
ωt
vo
VSCR1,2
α
VSCR3,4
VL
Figure 3.17: Single-Phase Controlled Full-Wave Rectifiers with RL Loads: Continuous Current
3.4. THREE-PHASE UNCONTROLLED FULL-WAVE RECTIFIERS WITH R LOADS
35
2. Determine an expression for the load current.
3. Check the continuity of the current if L = 100mH.
Solution:
1.
p
p
R2 + (ωL)2 = 102 + (2π × 60 × 0.02)2 = 12.5Ω
ωL
2π × 60 × 0.02
−1
−1
= tan
= tan
= 0.646rad = 37◦
R
10
Z
=
θ
The current is discontinuous because of α > θ.
2. Note that ωτ =
ωL
R
=
2π×60×0.02
10
i(t)
= 0.754rad and α = 60◦ = 1.047rad. Then
=
=
=
3. Note that θ = tan−1
3.4
ωL
R
= tan−1
i
α−ωt
Vm h
sin(ωt − θ) − sin(α − θ)e ωτ
Z √
i
1.047−ωt
120 2 h
sin(ωt − 37◦ ) − sin(60◦ − 37◦ )e 0.754
12.5
ωt
13.6 sin(ωt − 37◦ ) − 21.2e− 0.754
2π×60×0.1
10
= 75◦ > α. So the current is continuous.
Three-Phase Uncontrolled Full-Wave Rectifiers with R Loads
Suppose three-phase AC voltages are applied to the three-phase uncontrolled full-wave bridge rectifier, which are 120◦
out of phase and given by
van (t)
= Vm sin ωt
vbn (t)
= Vm sin(ωt −
vcn (t)
2π
)
3
2π
)
= Vm sin(ωt +
3
The line voltages are given by
√
vab (t)
vbc (t)
vca (t)
π
)
6
√
π
= vbn − vcn = 3Vm sin(ωt − )
2
√
5π
)
= vcn − van = 3Vm sin(ωt +
6
= van − vbn =
3Vm sin(ωt +
Some basic observations about the circuit are as follows:
1. Only one diode in the top half of the bridge may conduct at any instant. The diode with its anode connected to
the phase voltage which is highest at the instant conducts.
2. Only one diode in the bottom half of the bridge may conduct at any instant. The diode with its cathode
connected to the phase voltage which is lowest at the instant conducts.
3. D1 and D4 cannot conduct at the same time. Similarly, D3 and D6 cannot conduct simultaneously, nor can D5
and D2 .
4. The output voltage across the load is one of line-to-line voltages of the source.
Time Interval
[0, π6 ]
[ π6 , 3π
6 ]
5π
[ 3π
,
6
6 ]
5π 7π
[6, 6]
9π
[ 7π
6 , 6 ]
9π 11π
[6, 6 ]
Conducting Diodes
D5 and D6
D6 and D1
D1 and D2
D2 and D3
D3 and D4
D4 and D5
Output Voltages
vo (t) = vcb (t) = vcn (t) − vbn (t)
vo (t) = vab (t) = van (t) − vbn (t)
vo (t) = vac (t) = van (t) − vcn (t)
vo (t) = vbc (t) = vbn (t) − vcn (t)
vo (t) = vba (t) = vbn (t) − van (t)
vo (t) = vca (t) = vcn (t) − van (t)
36
CHAPTER 3. RECTIFIERS
Figure 3.18: Three-Phase Uncontrolled Full-Wave Rectifiers with R Loads
v
v
an
v
bn
cn
π
D5,D6
D6,D1
v
D1,D2
v
ab
ac
D2,D3
v
bc
2π
D3,D4
v
ba
D4,D5
v
ca
ωt
D5,D6
v
cb
v ,i
o o
π/6
π/2
5π/6
7π/6
9π/6
11π/6
ωt
π/6
π/2
5π/6
7π/6
9π/6
11π/6
ωt
π/6
π/2
5π/6
7π/6
9π/6
11π/6
ωt
i
a
id1
Figure 3.19: Three-Phase Uncontrolled Full-Wave Rectifiers with R Loads
3.4. THREE-PHASE UNCONTROLLED FULL-WAVE RECTIFIERS WITH R LOADS
37
5. The output voltage has a frequency of 6ω, where ω is the frequency of the source.
The average output voltage is
Vo
=
=
=
Z π2 √
6
π
3Vm sin ωt +
d(ωt)
2π π6
6
√
π
3 3Vm h
π i 2
− cos ωt +
π
6 π6
√
3 3Vm
π
The rms output voltage is
Vrms
=
=
s
s
6
2π
Z
9Vm2
π
π
2
h√
π
6
Z
π
2
π
6
π i2
3Vm sin ωt +
d(ωt)
6
1 − cos 2 ωt + π6
d(ωt)
2
v
u
π
u 9V 2 1
π 2
= t m ωt − sin 2 ωt +
2π
2
6 π
6
s
√
3 9 3
+
= Vm
2
4π
and the rms output current is Irms = Vrms
R .
Each diode conducts one-third of the time, resulting
ID,avg
=
ID,rms
=
The source current is given by
Is,rms =
The apparent power from the source is
1
Io
3
1
√ Irms
3
r
2
Irms
3
√
S=
3VL−L,rms Is,rms
Example: A three-phase bridge rectifier has a purely resistive load of R. The rectifier delivers Io = 60A at an
output voltage of Vo = 280.7V and the source frequency is 60Hz. Determine the efficiency, the peak inverse voltage
(PIV) of each diode, and the apparent power from the source.
Solution:
Vm
=
Vrms
=
R
=
Pdc
=
Pac
=
η
=
P IV
=
Vo π
280.7π
√ = √ = 169.7V
3 3
3 3
s
s
√
√
3 9 3
3 9 3
+
= 169.7
+
= 280.9
Vm
2
4π
2
4π
Vo
280.7
= 4.68Ω
=
Io
60
Vo Io = 280.7 × 60 = 16842W
2
280.92
Vrms
=
= 16860W
R
4.68
Pdc
16842
= 99.89%
=
Pac
16860
√
√
3Vm = 3 × 169.7 = 293.9V
38
CHAPTER 3. RECTIFIERS
√
VL−L,rms
Is,rms
S
3.5
√ Vm
√
169.7
3Vphase,rms = 3 √ = 3 × √ = 208V
2
2
r
r
r
2
2 Vrms
2 280.9
Irms =
=
×
= 49A
=
3
3 R
3
4.68
√
√
=
3VL−L,rms Is,rms = 3 × 208 × 49 = 17653V A
=
Three-Phase Controlled Full-Wave Rectifiers with Highly Inductive
Loads
Suppose three-phase AC voltages are applied to the three-phase controlled full-wave bridge rectifier, which are 120◦
out of phase and given by
van (t)
= Vm sin ωt
vbn (t)
= Vm sin(ωt −
vcn (t)
2π
)
3
2π
)
= Vm sin(ωt +
3
The line voltages are given by
√
π
)
6
√
π
vbc (t) = vbn − vcn = 3Vm sin(ωt − )
2
√
5π
vca (t) = vcn − van = 3Vm sin(ωt +
)
6
SCRs do not conduct until gate signals are applied while forward biased. The delay angle is referenced from where
the SCR would begin to conduct if it were a diode. The delay angle is the interval between when the SCR becomes
forward biased and when the gate signal is applied. SCR i will be fired at ωt = π6 + (i−1)π
+ α for i = 1, 2, 3, 4, 5, 6.
3
The output voltage across the load is one of line-to-line voltages of the source.
Time Interval
SCR fired Conducting SCRs Output Voltages
π
1π
[ π6 + 0π
+
α,
+
+
α]
SCR1
SCR6 and SCR1 vo (t) = vab (t) = van (t) − vbn (t)
3
6
3
π
2π
[ π6 + 1π
+
α,
+
+
α]
SCR
SCR1 and SCR2 vo (t) = vac (t) = van (t) − vcn (t)
2
3
6
3
π
3π
[ π6 + 2π
+
α,
+
+
α]
SCR
SCR2 and SCR3 vo (t) = vbc (t) = vbn (t) − vcn (t)
3
3
6
3
π
4π
[ π6 + 3π
+
α,
+
+
α]
SCR
SCR3 and SCR4 vo (t) = vba (t) = vbn (t) − van (t)
4
3
6
3
π
4π
π
5π
[ 6 + 3 + α, 6 + 3 + α] SCR5
SCR4 and SCR5 vo (t) = vca (t) = vcn (t) − van (t)
π
6π
[ π6 + 5π
+
α,
+
+
α]
SCR
SCR5 and SCR6 vo (t) = vcb (t) = vcn (t) − vbn (t)
6
3
6
3
The average output voltage is
Z π2 +α √
6
π
3Vm sin ωt +
Vo =
d(ωt)
2π π6 +α
6
√
π i π2 +α
3 3Vm h
− cos ωt +
=
π
6 π6 +α
√
3 3Vm
cos α
=
π
The rms output voltage is
s
Z π2 +α h√
6
π i2
Vrms =
3Vm sin ωt +
d(ωt)
2π π6 +α
6
s
Z π2 +α
1 − cos 2 ωt + π6
9Vm2
d(ωt)
=
π
π
2
6 +α
v
u
π +α
u 9V 2 1
π 2
= t m ωt − sin 2 ωt +
2π
2
6 π +α
6
s
√
3 9 3
+
cos 2α
= Vm
2
4π
vab (t)
= van − vbn =
3Vm sin(ωt +
3.5. THREE-PHASE CONTROLLED FULL-WAVE RECTIFIERS WITH HIGHLY INDUCTIVE LOADS
Figure 3.20: Three-Phase Controlled Full-Wave Rectifiers with RL Loads
v
v
an
v
bn
cn
π
SCR5,6
SCR6,1
v
v
SCR1,2
v
ab
SCR2,3
v
ac
2π
SCR3,4
v
bc
SCR4,5
v
ba
ωt
v
ca
cb
o
π/6
π/6+α
π/2
π/2+α
5π/6
5π/6+α
7π/6
7π/6+α
9π/6
9π/6+α
11π/6
ωt
ωt
ia
π/6
π/2
5π/6
7π/6
9π/6
11π/6
ωt
π/6
π/2
5π/6
7π/6
9π/6
11π/6
ωt
π/6
π/2
5π/6
7π/6
9π/6
11π/6
ωt
i
SCR1
i
o
Figure 3.21: Three-Phase Controlled Full-Wave Rectifiers with RL Loads
39
40
CHAPTER 3. RECTIFIERS
Each SCR conducts one-third of the time, resulting
ISCR,avg
=
ISCR,rms
=
The source current is given by
Is,rms =
The apparent power from the source is
r
1
Io
3
1
√ Irms
3
2
Irms
3
√
S=
3VL−L,rms Is,rms
Chapter 4
AC Voltage Controllers
An AC voltage controller is a converter that converts an AC voltage to an AC voltage with different magnitude
and frequency. Such a converter is also called an AC-AC converter. The most common applications of AC voltage
controllers are light-dimmer circuits and speed control of induction motors.
4.1
Single-Phase AC Voltage Controllers with R Loads
Positive half-cycle: S1 is forward biased and S2 is reverse biased. Before S1 conducts, the output voltage is zero
and the voltage across the SCRs is the same as the source voltage. Suppose S1 is fired at ωt = α. Then, the output
voltage is equal to the input voltage and the voltage across the SCRs is zero.
At ωt = π, io (t) = 0A, S1 is turned off because of natural commutation.
Negative half-cycle: S2 is forward biased and S1 is reverse biased. Before S2 conducts, the output voltage is
zero and the voltage across the SCRs is the same as the source voltage. Suppose S2 is fired at ωt = π + α. Then, the
output voltage is equal to the input voltage and the voltage across the SCRs is zero.
At ωt = 2π, io (t) = 0A, S2 is turned off because of natural commutation.
The voltages across the load and the SCRs are given by

 vs (t) = Vm sin ωt α ≤ ωt ≤ π
vs (t) = Vm sin ωt π + α ≤ ωt ≤ 2π
vo (t) =

0
otherwise

α ≤ ωt ≤ π
 0
0
π + α ≤ ωt ≤ 2π
vSCR1,2 (t) =

vs (t) = Vm sin ωt otherwise
The average output voltage and current are zero. The rms output voltage is calculated by
s
Z π
2
(Vm sin ωt)2 d(ωt)
Vrms =
2π α
s
Z
Vm2 π 1 − cos 2ωt
=
d(ωt)
π α
2
s
π
1
Vm2
=
ωt − sin 2ωt
2π
2
α
r
Vm
α sin 2α
= √
1− +
π
2π
2
and the rms output current is Irms = Vrms
R .
The average and rms SCR currents are given by
ISCR,rms
=
ISCR,avg
=
1
√ Irms
2
Vm
(1 + cos α)
2πR
41
42
CHAPTER 4. AC VOLTAGE CONTROLLERS
Figure 4.1: Single-Phase AC Voltage Controllers with R Loads
v
s
α
π
π+α
2π
2π+α
3π
3π+α
4π
ωt
α
π
π+α
2π
2π+α
3π
3π+α
4π
ωt
α
π
π+α
2π
2π+α
3π
3π+α
4π
ωt
vo,io,is
vSCR
Figure 4.2: Single-Phase AC Voltage Controllers with R Loads
4.1. SINGLE-PHASE AC VOLTAGE CONTROLLERS WITH R LOADS
43
The power factor is
PF
Pac
Vrms Irms
=
=
=
S
Vs,rms Is,rms
r
α sin 2α
=
1− +
π
2π
V
√m
2
q
1−
α
π
+
sin 2α
2π Irms
V
√m Irms
2
Example: The single-phase AC voltage controller has a 120V rms 60Hz source. The load resistance is 15Ω. The
delay angle is 88◦ . Determine the rms output voltage, the rms source current, the rms and average currents through
SCRs, and the power factor.
Solution:
r
r
Vm
sin 2 × 88◦
Vm
α sin 2α
88
√
√
Vrms =
=
+
= 86.7V
1− +
1−
π
2π
180
2π
2
2
86.7
Vrms
Irms =
=
= 5.78A
R
15
1
1
ISCR,rms = √ Irms = √ × 5.78 = 4.09A
2
2
√
120 2
Vm
(1 + cos α) =
(1 + cos 88◦ ) = 1.86A
ISCR,avg =
2πR
2π × 15
r
r
sin 2 × 88◦
α sin 2α
88
== 1 −
+
= 0.72
1− +
PF =
π
2π
180
2π
44
CHAPTER 4. AC VOLTAGE CONTROLLERS
Midterm Review
Circuit Analysis by Laplace Transform:
i(t) = C
dv(t)
dt
I(s) = CsV (s) − CVC (0)
or V (s) =
1
1
I(s) + VC (0)
Cs
s
di(t)
V (s) 1
V (s) = LsI(s) − LiL (0) or I(s) =
+ iL (0)
dt
Ls
s
Single-Phase Rectifiers With R Load
UnControlled ←←
Full-Wave Controlled
→→ Half-Wave Controlled
m
Vo = 2Vπm
α = 0 Vo = Vπm (1q
+ cos α)
÷2
Vo = V2π
(1q
+ cos α)
√
sin(2α)
sin(2α)
V
V
V
α
m
m
m
Vrms = √2
α = 0 Vrms = √2 1 − π + 2π
÷ 2 Vrms = 2 1 − α
π +
2π
q
q
√
√
sin(2α)
sin(2α)
α
2
α
P F =1
α = 0 P F = 1 − π + 2π
÷ 2 P F = 2 1 − π + 2π
p
L
2
2
Single-Phase Rectifier With RL Load: Z = R + (ωL) , θ = tan−1 ωL
R ,τ = R
Half Wave hControlled:
i
α−ωt
io (t) = VZm sin(ωt − θ) − sin(α − θ)e ωτ , α ≤ ωt ≤ β
v(t) = L
Full-Wave Controlled-Discontinuous
Mode:
h
i θ < α,
α−ωt
Vm
io (t) = Z sin(ωt − θ) − sin(α − θ)e ωτ , α ≤ ωt ≤ β
Full-Wave Controlled-Continuous
Mode: θ ≥ α, i
h
α−ωt
Vm
2
ωτ
io (t) = Z sin(ωt − θ) −
, α ≤ ωt ≤ π + α
− π sin(α − θ)e
ωτ
1−e
Three-Phase Rectifiers
Controlled
with Highly L Load
√
Vo = 3 π 3 Vm
qcos α
→→→→
α = 0
√
Vrms = Vm 32 + 94π3 cos 2α
α = 0
ISCR,avg = 13 Io
SCR → D
ISCR,rms = √13 Irms
SCR → D
q
Is,rms = 23 Irms
→→→→
Single-Phase AC Voltage
Controller:
q
q
Vo = 0, Vrms =
ISCR,avg =
Vm
2πR
V√m
2
1−
α
π
+
(1 + cos α) ,
sin(2α)
2π , P F
=
ISCR,rms =
Uncontrolled
with R Load
√
Vo = 3 π 3 Vm
q
Vrms = Vm 32 +
ID,avg = 13 Io
ID,rms = √13 Irms
q
Is,rms = 23 Irms
1−
√1 Irms
2
α
π
+
sin(2α)
2π ,
√
9 3
4π
→→
α=0
UnControlled
Vo = Vπm
α=0
Vrms =
α=0
PF =
Vm
2
√
2
2
Chapter 5
DC Choppers
A DC chopper converts directly from DC to DC voltage and is also known as a DC to DC converter. A chopper can
be considered as DC equivalent to an AC transformer with a continuously variable ratio. Like a transformer, it can be
used to step-down or step-up a DC voltage source. Step-down chopper circuit is called buck converter, while step-up
converter is referred to as boost converter. Choppers are widely used for DC power supplies, DC motor drives, and so
on.
5.1
Buck Converter
The figure in the next page shows a buck converter circuit. The chopper switch in this circuit can be implemented by
using a power BJT, power MOSFET, GTO, and so on.
Analysis of the buck converter begins by making the following assumptions:
1. The circuit is operating in the steady state.
2. The inductor current is continuous (always positive)
3. The capacitor is large enough so that the output voltage is held constant at Vo .
4. The switching frequency is f and the switching period is T = 1/f . The switch is closed for time DT and open
for time (1 − D)T with D duty cycle, which is defined as a ratio of the on-time and period.
5. The components are ideal.
The operation of the buck converter can be divided into two modes.
Mode 1: 0 ≤ t ≤ DT
In this mode, the switch is closed and the freewheeling diode is reverse biased. The equivalent circuit for this mode
is given in next page.
The inductor current iL is determined by
vL = Vs − Vo = L
diL
dt
that is,
Vs − Vo
diL
=
dt
L
Suppose that iL (0) = Imin . Then, integrating both sides of the above differential equation gives
Z
t
0
diL
ds =
dt
Z
t
0
Vs − Vo
ds
L
As a result,
Vs − Vo
t + Imin
L
which means that the inductor current increases linearly with time t.
Mode 2: DT ≤ t ≤ T
iL (t) =
45
46
CHAPTER 5. DC CHOPPERS
Figure 5.1: Buck converter
Figure 5.2: Buck converter: Mode 1
Figure 5.3: Buck converter: Mode 2
v
L
Vs−Vo
DT
T
t
DT
T
t
DT
T
t
−V
o
iL
I
max
IR
Imin
i
C
Imax−IR
Imin−IR
Figure 5.4: Buck converter
5.1. BUCK CONVERTER
47
In this mode, the switch is open and the freewheeling diode is forced to conduct by the inductor current. The
equivalent circuit for this mode is given in last page.
The inductor current iL is determined by
diL
vL = −Vo = L
dt
that is,
Vo
diL
=−
dt
L
Suppose that iL (DT ) = Imax . Then, integrating both sides of the above differential equation gives
Z
t
DT
diL
ds = −
dt
Z
t
DT
Vo
ds
L
As a result,
Vo
(t − DT ) + Imax
L
which means that the inductor current decreases linearly with time t.
Under the steady state conditions, the inductor current at the end of Mode 1 is the same as that at the beginning
of Mode 2, that is,
Vs − Vo
DT + Imin = Imax
(5.1)
L
and the inductor current at the end of Mode 2 is the same as that at the beginning of Mode 1, that is,
iL (t) = −
−
Vo
(T − DT ) + Imax = Imin
L
(5.2)
which implies that
Imax − Imin =
Vo (1 − D)T
L
(5.3)
Adding (5.1) to (5.2) gives
Vo
Vs − Vo
DT − (T − DT ) = 0
L
L
Solving this equation for Vo yields
Vo = DVs
which means that the output voltage depends only on the input voltage and duty cycle.
In order to determine the inductor current, we need to find Imin and Imax . To this end, the average currents in
the circuit satisfy the following equation:
Vo
IL = IC + IR =
R
Note that the average capacitor current IC must be zero for the steady state.
With IL determined, Imax and Imin are given by
(1 − D)T
Imax − Imin
1
= Vo
+
Imax = IL +
2
R
2L
(1 − D)T
Imax − Imin
1
= Vo
−
Imin = IL −
2
R
2L
It is important to note that the above analysis is valid under the assumption of continuous inductor current, which
means that Imin ≥ 0. If Imin < 0, then the freewheeling diode is off and so the inductor current will remain zero until
the switch is closed again.
In order to guarantee the continuity of the inductor current, the inductance L and the switching period T must be
appropriately selected so that
(1 − D)T
1
−
≥0
Imin = Vo
R
2L
that is,
L≥
R(1 − D)T
2
48
CHAPTER 5. DC CHOPPERS
or
T ≤
2L
R(1 − D)
which imply that
Lmin =
or
Tmax =
R(1 − D)T
2
2L
R(1 − D)
R(1 − D)
2L
The peak-to-peak ripple voltage at output is determined by
fmin =
∆Vo =
Vo (1 − D)
Vo (1 − D)T 2
=
8LC
8LCf 2
Example: A buck converter is supplied by a DC source of 48V. It produces an output voltage of 18V across a
10Ω load resistor. Assume that the capacitor is large enough so that the output voltage is kept constant.
1. Determine the duty cycle D.
2. Find the minimum inductor size Lmin if the switching frequency is 40kHz.
3. Calculate the inductor current if L = 1.25Lmin and f=40kHz.
4. Select C so that the peak-to-peak output ripple voltage
∆Vo
Vo
does not exceed 0.5% if L = 1.25Lmin and f=40kHz.
5. Compute the minimum switching frequency fmin if L = 1.25Lmin.
Solution
1. D =
Vo
Vs
2. Lmin =
=
18
48
= 0.375
R(1−D)
2f
=
10×(1−0.375)
2×40×103
= 78µH
3.
L =
IL
Imax
Imin
1.25Lmin = 1.25 × 78 = 97.5µH
18
Vo
=
= 1.8A
=
R 10
(1 − D)
(1 − 0.375)
1
1
= 3.24A
+
= 18 ×
+
= Vo
R
2Lf
10 2 × 97.5 × 10−6 × 40 × 103
(1 − D)
(1 − 0.375)
1
1
= 0.36A
−
= 18 ×
−
= Vo
R
2Lf
10 2 × 97.5 × 10−6 × 40 × 103
Note that
T
=
DT
Vs − Vo
L
Vo
L
=
=
=
1
= 0.000025s
40 × 103
0.375 × 0.000025 = 0.0000094s
48 − 18
= 307692
97.5 × 10−6
18
= 184615
97.5 × 10−6
So the inductor current is given by
Vs −Vo
0 ≤ t ≤ DT
L t + Imin
iL (t) =
− VLo (t − DT ) + Imax DT ≤ t ≤ T
307692t + 0.36
0 ≤ t ≤ 0.0000094s
=
−184615(t − 0.0000094) + 3.24 0.0000094s ≤ t ≤ 0.000025s
5.2. BOOST CONVERTERS
4. Note that
∆Vo
Vo
=
1−D
8LCf 2
49
≤ 0.005. Then,
Cmin =
1−D
1 − 0.375
=
= 100µF
0.005 × 8Lf 2
0.005 × 8 × 97.5 × 10−6 × (40 × 103 )2
5.
fmin =
5.2
10(1 − 0.375)
R(1 − D)
=
= 32kHz
2L
2 × 97.5 × 10−6
Boost Converters
The figure in the next page shows a boost converter circuit. The chopper switch in this circuit can be implemented
by using a power BJT, power MOSFET, GTO, and so on.
Analysis of the boost converter begins by making the following assumptions:
1. The circuit is operating in the steady state.
2. The inductor current is continuous (always positive)
3. The capacitor is large enough so that the output voltage is held constant at Vo .
4. The switching frequency is f and the switching period is T = 1/f . The switch is closed for time DT and open
for time (1 − D)T with D duty cycle, which is defined as a ratio of the on-time and period.
5. The components are ideal.
The operation of the boost converter can be divided into two modes.
Mode 1: 0 ≤ t ≤ DT
In this mode, the switch is closed and the freewheeling diode is reverse biased by Vo . The equivalent circuit for
this mode is given in next page.
The inductor current iL is determined by
diL
vL = Vs = L
dt
that is,
diL
Vs
=
dt
L
Suppose that iL (0) = Imin . Then, integrating both sides of the above differential equation gives
Z
t
0
diL
ds =
dt
Z
t
0
Vs
ds
L
As a result,
Vs
t + Imin
L
which means that the inductor current increases linearly with time t.
Mode 2: DT ≤ t ≤ T
In this mode, the switch is open and the freewheeling diode is forward biased to provide a path for inductor current.
The equivalent circuit for this mode is given in last page.
The inductor current iL is determined by
iL (t) =
vL = Vs − Vo = L
diL
dt
that is,
diL
Vs − Vo
=
dt
L
Suppose that iL (DT ) = Imax . Then, integrating both sides of the above differential equation gives
Z
t
DT
diL
ds =
dt
Z
t
DT
Vs − Vo
ds
L
50
CHAPTER 5. DC CHOPPERS
Figure 5.5: Boost converter
Figure 5.6: Boost converter: Mode 1
Figure 5.7: Boost converter: Mode 2
vL
Vs
DT
T
t
DT
T
t
DT
T
t
DT
T
t
Vs−Vo
iL
I
max
IL
Imin
iD
I
max
IL
Imin
Imax−IR
I
−IR
min
iC
−I
R
Figure 5.8: Boost converter
5.2. BOOST CONVERTERS
51
As a result,
Vs − Vo
(t − DT ) + Imax
L
Under the steady state conditions, the inductor current at the end of Mode 1 is the same as that at the beginning
of Mode 2, that is,
Vs
DT + Imin = Imax
(5.4)
L
which implies that
Vs DT
(5.5)
Imax − Imin =
L
and the inductor current at the end of Mode 2 is the same as that at the beginning of Mode 1, that is,
iL (t) =
Vs − Vo
(T − DT ) + Imax = Imin
L
(5.6)
Adding (5.4) to (5.6) gives
Vs − Vo
Vs
DT +
(T − DT ) = 0
L
L
Solving this equation for Vo yields
Vs
1−D
which means that the output voltage is higher than the input voltage.
The average inductor current IL is determined as follows. Note that the power supplied by the source is the same
as the power absorbed by the load due to the assumptions, that is,
2
Vs
2
1−D
V
Po = o =
= Ps = IL Vs
R
R
Vo =
Solving this equation for IL gives
IL =
Vs
(1 − D)2 R
With IL determined, Imax and Imin are given by
Imax
Imin
DT
Imax − Imin
1
= Vs
+
2
(1 − D)2 R
2L
Imax − Imin
DT
1
= IL −
= Vs
−
2
(1 − D)2 R
2L
= IL +
Note that the above analysis is valid under the assumption of continuous inductor current, which means that
Imin ≥ 0. In order to guarantee the continuity of the inductor current, the inductance L and the switching period T
must be appropriately selected so that
DT
1
Imin = Vs
−
≥0
(1 − D)2 R
2L
that is,
D
1
−
≥0
2
(1 − D) R 2Lf
or
L≥
D(1 − D)2 R
2f
which imply that
D(1 − D)2 R
2f
The peak-to-peak ripple voltage at output is determined by
Lmin =
∆Vo =
Vo D
Vo DT
=
RC
RCf
Example: A boost converter is supplied by a DC source of 12V. It produces an output voltage of 30V across a
50Ω load resistor. Assume that the capacitor is large enough so that the output voltage is kept constant.
52
CHAPTER 5. DC CHOPPERS
1. Determine the duty cycle D.
2. Find the minimum inductor size Lmin if the switching frequency is 25kHz.
3. Calculate the inductor current if L = 120µH and f=25kHz.
4. Select C so that the peak-to-peak output ripple voltage
∆Vo
Vo
does not exceed 1% if L = 120µH and f=25kHz.
Solution
2. Lmin =
Vs
Vo
=1−
12
30
= 0.6
D(1−D)2 R
2f
=
0.6×(1−0.6)2 ×50
2×25×103
1. D = 1 −
= 96µH
3.
IL
Imax
Imin
Vs
12
=
= 1.5A
(1 − D)2 R
(1 − 0.6)2 × 50
D
0.6
1
1
= 2.7A
+
= 12 ×
+
= Vs
(1 − D)2 R 2Lf
(1 − 0.6)2 × 50 2 × 120 × 10−6 × 25 × 103
D
0.6
1
1
= 0.3A
−
= 12 ×
−
= Vs
(1 − D)2 R 2Lf
(1 − 0.6)2 × 50 2 × 120 × 10−6 × 25 × 103
=
Note that
T
=
DT
Vs
L
Vs − Vo
L
=
=
=
1
= 0.00004s
25 × 103
0.6 × 0.00004 = 0.000024s
12
= 100000
120 × 10−6
12 − 30
= −150000
120 × 10−6
So the inductor current is given by
Vs
0 ≤ t ≤ DT
L t + Imin
iL (t) =
Vs −Vo
(t
−
DT
)
+
I
max DT ≤ t ≤ T
L
100000t + 0.3
0 ≤ t ≤ 0.000024s
=
−150000(t − 0.000024) + 2.7 0.000024s ≤ t ≤ 0.00004s
4. Note that
∆Vo
Vo
=
D
RCf
≤ 0.01. Then,
Cmin =
5.3
D
0.6
= 48µF
=
0.001Rf
0.01 × 50 × 25 × 103
Impulse-Commutated Chopper Circuit (for Lab #3)
At the beginning, trigger T2 and leave T1 off. D is reverse biased. C charges through T2 and R to Vs according to the
equation
dvc
+ vc
Vs = RC
dt
which has a solution of
t
vc (t) = Vs 1 − e− τ , τ = RC
The capacitor current is
Vs − t
e τ
R
When ic (t) becomes zero, T2 is turned off. The load voltage is
ic (t) =
vo (t) = Vs − vc
5.3. IMPULSE-COMMUTATED CHOPPER CIRCUIT (FOR LAB #3)
53
Mode 1: T1 is fired.
vc = Vs is applied across T2 so that T2 is cut off. The output voltage is
v0 (t) = Vs
At the same time, D is forward biased by vc (t). C discharges through T1 , D and L according to equation
0 = vc + L
dic
d2 vc
= vc + LC 2
dt
dt
which has a solution of
vc (t) = Vs cos
The capacitor current ic (t) is given by
ic (t) = −Vs
r
√
C
sin
L
1
t
LC
√
1
t
LC
√
When t = tr = π LC, the capacitor current becomes zero, which forces D off, and the voltage across the capacitor is
kept at -Vs because D prevent ic (t) from becoming negative. Now it is ready to fire T2 again.
Mode 2: T2 is fired.
vc = −Vs is applied across T1 so that T1 is turned off. C charges through T2 and R to Vs according to the equation
Vs = RC
which has a solution of
dvc
+ vc
dt
t
vc (t) = Vs 1 − 2e− τ
The capacitor current is
ic (t) =
2Vs − t
e τ
R
The time tof f when vc (t) becomes zero is given by
tof f =0
vc (tof f ) = Vs 1 − 2e− τ
that is, tof f = τ ln 2 = 0.69RC. In order to turn off T1 successfully, the turn off time for T1 must be less than tof f .
The time tc for C to recharge to Vs is given by tc = 5 ∗ τ . Now it is ready to√fire T1 again.
In order to make the chopper work, T1 must be on at least for time tr = π LC and T2 must be on at least for
time tc = 5 ∗ τ .
When ic (t) becomes zero, T2 is turned off. The load voltage is
vo (t) = Vs − vc
54
CHAPTER 5. DC CHOPPERS
Chapter 6
Inverter
The function of an inverter is to convert a DC input voltage to a symmetrical AC output voltage of desired magnitude
and frequency. The output voltage could be at a fixed or variable frequency. A variable voltage can be obtained by
varying the gain of the inverter, which is accomplished by pulse-width-modulation (PWM) control within the inverter,
Inverter gain =
AC output voltage
DC input voltage
The output waveforms are non-sinusoidal for practical inverters. Ideal inverters will produce sinusoidal voltages. Square
wave voltages may be acceptable for low and medium power applications, and low distorted sinusoidal waveforms for
high power applications.
The harmonic contents of output voltages can be minimized by switching techniques with the use of high-speed
power semiconductor devices.
Inverters are widely used in industrial applications, for example, variable speed AC motor drives, induction heating,
standby power supplies, un-interruptible power supplies, and so on.
Inverters are classified as single phase and three phase. This course will focus on single phase only.
6.1
Half-Bridge Inverter
A pulse with a period of T and duty cycle of 0.5, ig (t), is applied to the base of Q1 , while it is applied to the base of
Q2 through a NOT gate. So, Q1 and Q2 are turned on and off alternately, each 50% of the period, with one transistor
on while the other is off.
Mode 1: 0 ≤ t ≤ 0.5T
In this mode, either Q1 or D1 is turned on and Q2 and D2 are off. The output voltage is
vo = Vdc
Now assume that io (0) = Imin < 0. Then, D1 is turned on and Q1 is still off at t = 0. The load current is determined
by
dio
vo = Vdc = Rio + L
dt
Solution to this equation takes the form of
t
Vdc
+ Ae− τ
io (t) =
R
with τ =
L
R.
The constant A is determined by the initial condition io (0) = Imin , that is,
io (0) =
Vdc
+ A = Imin
R
Solving the above equation for A gives
A = Imin −
Vdc
R
Therefore, the load current in this mode is
io (t) =
Vdc
Vdc − t
+ Imin −
e τ
R
R
55
56
CHAPTER 6. INVERTER
Figure 6.1: Half-bridge inverter
Vdc
v
o
T/2
−V
dc i
Imax o
t
1
T/2 T/2+t
t
T/2
1
T
t
T
t
T
t
T
t
T
t
T
t
T
t
T
t
I
min i
S1
I
max
1
I
min i
Imax S2
T/2
I
min i
D1
I
max
T/2+t
1
t
1
I
max
i
D2
T/2+t
T/2
I
max
1
i
Q1
t
T/2
1
I
max
i
Q2
T/2+t
1
Figure 6.2: Half-bridge inverter
6.1. HALF-BRIDGE INVERTER
57
This current increases from a negative value, Imin , to the maximum value Imax , which is reached at t = 0.5T , that is
Vdc
Vdc − 0.5T
+ Imin −
e τ
Imax = io (0.5T ) =
(6.1)
R
R
i
0.5T
0.5T
Vdc h
(6.2)
=
1 − e− τ + Imin e− τ
R
When it goes through zero and become positive, D1 is cut off, Q1 is turned on and begins to conduct. The time when
io (t) becomes zero is determined by
Vdc
Vdc − t1
0 = io (t1 ) =
+ Imin −
e τ
(6.3)
R
R
Mode 2: 0.5T ≤ t ≤ T
In this mode, either Q2 or D2 is turned on and Q1 and D1 are off. The output voltage is
vo = −Vdc
Note that io (0.5T ) = Imax > 0. Then, D2 is turned on and Q2 is still off at t = 0.5T . The load current is determined
by
dio
vo = −Vdc = Rio + L
dt
Solution to this equation takes the form of
io (t) = −
t−0.5T
Vdc
+ Be− τ
R
The constant B is determined by the initial condition io (0.5T ) = Imax , that is,
io (0.5T ) = −
Vdc
+ B = Imax
R
Solving the above equation for B gives
B = Imax +
Vdc
R
Therefore, the load current in this mode is
io (t) = −
Vdc
Vdc − t−0.5T
τ
+ Imax +
e
R
R
This current decreases from a positive value, Imax , to the minimum value Imin , which is negative and reached at
t = T , that is,
Vdc
Vdc − 0.5T
Imin = io (T ) = −
+ Imax +
e τ
(6.4)
R
R
i
0.5T
0.5T
Vdc h
(6.5)
= −
1 − e− τ + Imax e− τ
R
When it goes through zero and become negative, D2 is cut off, Q2 is turned on and begins to conduct. The time when
io (t) becomes zero is determined by
Vdc
Vdc − t2 −0.5T
τ
(6.6)
+ Imax +
e
0 = io (t2 ) = −
R
R
Adding (6.5) to (6.2) gives
Imax + Imin = (Imax + Imin )e−
So Imin = −Imax . Substituting it to (6.2) yields
T
Imax =
Vdc 1 − e− 2τ
T
R 1 + e− 2τ
0.5T
τ
58
CHAPTER 6. INVERTER
With Imax and Imin , it follows from (6.3) and (6.6) that
t1 = τ ln
t2
2
T
1 + e− 2τ
0.5T + t1
=
In summary, currents through the devices in the circuit is given below.
(
t
Vdc
+ Imin − VRdc e− τ
0 ≤ t ≤ 0.5T
R
− t−0.5T
io (t) =
Vdc
Vdc
τ
0.5T ≤ t ≤ T
− R + Imax + R e
io (t) 0 ≤ t ≤ 0.5T
is1 (t) =
0
0.5T ≤ t ≤ T
0
0 ≤ t ≤ 0.5T
is2 (t) =
−io (t) 0.5T ≤ t ≤ T
−io (t) 0 ≤ t ≤ t1
iD1 (t) =
0
t1 ≤ t ≤ T
io (t) 0.5T ≤ t ≤ 0.5T + t1
iD2 (t) =
0
otherwise
io (t) t1 ≤ t ≤ 0.5T
iQ1 (t) =
0
otherwise
−io (t) 0.5T + t1 ≤ t ≤ T
iQ2 (t) =
0
otherwise
The rms output voltage is
s
Vrms =
1
0.5T
Z
0.5T
2 dt = V
Vdc
dc
0
and the rms output current is
s
Irms =
1
0.5T
Z
0.5T
0
2
Vdc − t
Vdc
τ
+ Imin −
e
dt
R
R
Both average output current and voltage are zero.
Example: A half-bridge inverter has the following data: Vdc = 300V , R = 10Ω, L = 0.05H, and f = 60Hz. Find
1. peak load current
2. time of current zero crossing
3. average transistor current
4. average diode current
Solution
1.
τ
=
T
=
Imax
=
0.05
L
=
= 0.005s
R
10
1
1
=
= 0.01667s
f
60
T
Vdc 1 − e− 2τ
T
R 1 + e− 2τ
0.01667
=
=
300 1 − e− 0.01
10 1 + e− 0.01667
0.01
20.47A
6.2. FULL-BRIDGE INVERTER
2.
t1 = τ ln
59
2
T
1 + e− 2τ
= 0.005 × ln
2
1 + e−
0.01667
0.01
= 0.0026s
3. The load current during the first half cycle is given by
300 − t
Vdc
Vdc − t
300
t
τ
io (t) =
+ Imin −
+ −20.47 −
=
e
e 0.005 = 30 − 50.47e− 0.005
R
R
10
10
So, the average transistor current is
Z
Z 0.00833
t
1 0.5T
1
io (t)dt =
[30 − 50.47e− 0.005 ]dt
IC =
T t1
0.01667 0.0026
i0.00833
h
1
t
=
30t − 54.47 × (−0.005)e− 0.005
0.01667
0.0026
i
h
0.00833
0.0026
1
=
30(0.00833 − 0.0026) − 54.47 × (−0.005) e− 0.005 − e− 0.005
0.01667
= 4.18A
4. The average diode current is calculated as follows:
Z
Z 0.0026
t
1 t1
1
ID =
−io (t)dt = −
[30 − 50.47e− 0.005 ]dt
T 0
0.01667 0
i0.0026
h
1
t
= −
30t − 54.47 × (−0.005)e− 0.005
0.01667
0
i
h
0.0026
1
0
= −
30(0.0026 − 0) − 54.47 × (−0.005) e− 0.005 − e− 0.005
0.01667
= 1.46A
6.2
Full-Bridge Inverter
A pulse with a period of T and duty cycle of 0.5, ig (t), is applied to the bases of Q1 and Q2 , while it is applied to the
bases of Q3 and Q4 through a NOT gate. So, Q1 and Q2 are turned on and Q3 and Q4 off during the first half cycle,
while Q1 and Q2 are off and Q3 and Q4 on during the second half cycle
Mode 1: 0 ≤ t ≤ 0.5T
In this mode, Q1 or D1 and Q2 or D2 are turned on and Q3 , D3 , Q4 and D4 are off. The output voltage is
vo = Vdc
Similar to the half-bridge inverter, the load current is determined by
vo = Vdc = Rio + L
Solution to this equation is given by
with τ =
L
R
dio
dt
Vdc
Vdc − τt
io (t) =
+ Imin −
e
R
R
and
T
Imin = −
Vdc 1 − e− 2τ
T
R 1 + e− 2τ
When the current io (t) is negative, both D1 and D2 conduct, otherwise both Q1 and Q2 conduct.
Mode 2: 0.5T ≤ t ≤ T
In this mode, Q3 or D3 and Q4 or D4 are turned on and Q1 , D1 , Q2 and D2 are off. The output voltage is
vo = −Vdc
Similar to the half-bridge inverter, the load current is determined by
vo = −Vdc = Rio + L
dio
dt
60
CHAPTER 6. INVERTER
Figure 6.3: Full-bridge inverter
v
o
V
dc
−Vdc
Imax
T/2
T
t
T/2
T
t
T/2
T
t
T/2
T
t
T/2
T
t
i
o
I
min
Imax
iS1,2
I
min
I
i
S3,4
max
I
min
I
i
s
max
I
min
Figure 6.4: Full-bridge inverter
6.2. FULL-BRIDGE INVERTER
61
Solution to this equation is given by
io (t) = −
Vdc − t−0.5T
Vdc
τ
+ Imax +
e
R
R
with
T
Imax =
Vdc 1 − e− 2τ
T
R 1 + e− 2τ
When the current io (t) is positive, both D3 and D4 conduct, otherwise both Q3 and Q4 conduct.
In summary, currents through the devices in the circuit is given below.
(
−t
Vdc
Vdc
τ
0 ≤ t ≤ 0.5T
R + Imin − R e
io (t) =
t−0.5T
Vdc
Vdc
0.5T ≤ t ≤ T
− R + Imax + R e− τ
0 ≤ t ≤ 0.5T
io (t)
is (t) =
−io (t) 0.5T ≤ t ≤ T
io (t) 0 ≤ t ≤ 0.5T
is1 (t) = is2 (t) =
0
0.5T ≤ t ≤ T
0
0 ≤ t ≤ 0.5T
is3 (t) = is4 (t) =
−io (t) 0.5T ≤ t ≤ T
The rms output voltage is
s
Vrms =
1
0.5T
Z
0.5T
0
2 dt = V
Vdc
dc
and the rms output current is
s
Irms =
1
0.5T
Z
0.5T
0
2
Vdc − t
Vdc
+ Imin −
e τ
dt
R
R
Both average output current and voltage are zero.
Example: A full-bridge inverter has the following data: Vdc = 100V , R = 10Ω, L = 0.025H, and f = 60Hz.
Determine
1. an expression for the load current
2. the power absorbed by the load
3. average current through the source
Solution
1.
τ
=
T
=
Imax
=
L
0.025
=
= 0.0025s
R
10
1
1
=
= 0.01667s
f
60
T
Vdc 1 − e− 2τ
T
R 1 + e− 2τ
0.01667
=
=
io (t)
=
=
=
100 1 − e− 0.005
10 1 + e− 0.01667
0.005
9.31A
(
−t
Vdc
Vdc
τ
0 ≤ t ≤ 0.5T
R + Imin − R e
t−0.5T
Vdc
Vdc
0.5T ≤ t ≤ T
− R + Imax + R e− τ
(
t
100
− 0.0025
− 100
0 ≤ t ≤ 0.00833
10 + −9.31
10 e t−0.00833
100
100
− 10 + 9.31 + 10 e− 0.0025
0.00833 ≤ t ≤ 0.01667
(
t
10 − 19.31e− 0.0025
0 ≤ t ≤ 0.00833
t−0.00833
− 0.0025
−10 + 19.31e
0.00833 ≤ t ≤ 0.01667
62
CHAPTER 6. INVERTER
2. The rms load current is
s
s
Z
Z 0.00833
2 0.5T
2
t
2
Irms =
[io (t)] dt =
[10 − 19.31e− 0.0025 ]2 dt
T 0
0.01667 0
r
i
h
2
t
2t
=
100 − 38.62e− 0.0025 + 19.312e− 0.0025 dt
0.01667
s
0.00833
2
0.0025 − 2t
t
100t + 38.62 × 0.0025e− 0.0025 − 19.312 ×
e 0.0025
=
0.01667
2
0
s
2
0.0025 − 2×0.00833
0.00833
−
2
0.0025
0.0025
0.83 + 38.62 × 0.0025 e
=
− 1 − 19.31 ×
−1
e
0.01667
2
= 6.64A
Po
2
= Irms
R = 6.642 × 10 = 441W
3. The average source current is calculated as follows:
Is
Z 0.00833
t
2
=
io (t)dt =
[10 − 19.31e− 0.0025 ]dt
0.01667
0
0
i0.00833
h
2
t
=
10t − 19.13 × (−0.0025)e− 0.0025
0.01667
0
i
h
0.00833
2
=
10(0.00833 − 0) − 19.13 × (−0.0025) e− 0.0025 − 1
0.01667
= 4.41A
2
T
Z
0.5T
Chapter 7
DC Drives
7.1
Basic Characteristics of DC Motors
The equivalent circuit for a separately excited dc motor is shown in Fig. 7.1. When a separately excited motor is
excited by a field current if and an armature current ia flows into the armature circuit, the motor develops a back
emf and a torque to balance the load torque at a certain speed. The equations describing the characteristics of a
separately excited motor can be determined as follows:
The instantaneous field current if can be found from
vf = Rf if + Lf
dif
dt
and the instantaneous armature current ia can be determined by
va = Ra ia + La
dia
+ ea
dt
The motor back emf ea is given by
ea = Kv ωif
and the torque developed by the motor is
τd = Kt if ia
which must be equal to the load torque
τd = J
dω
+ Bω + τL
dt
where
ω
B
Kv
Kt
La
Lf
Ra
Rf
τL
J
=
=
=
=
=
=
=
=
=
=
motor speed, rad/s
viscous friction constant, N · m/rad/s
voltage constant, V/A-rad/s
torque constant (Kt = Kv )
armature circuit inductance, H
field circuit inductance, H
armature circuit resistance, Ω
field circuit resistance, Ω
Load torque, N · m
inertia, Kg · m2 .
Under steady state conditions, the time derivatives become zero. Therefore, the steady-state average quantities
are
Vf
Va
Ea
τd
=
=
=
=
=
Rf If
Ra Ia + Ea
Kv ωIf
Kt If Ia
Bω + τL
63
(7.1)
64
CHAPTER 7. DC DRIVES
Figure 7.1: DC motor equivalent circuit
The developed power is given by
Pd = τd ω
The relationship between If and Ea is given by magnetization curve or characteristics of the motor. According to
the steady-state equations derived above, the speed of a separately excited motor can be determined as
ω=
Va − Ra Ia
Va − Ra Ia
=
Kv If
Kv Vf /Rf
(7.2)
which implies that the motor speed can be varied by
1. controlling the armature voltage Va , known as voltage control;
2. controlling the field current If , known as field control;
3. torque demand, which corresponds to an armature current Ia for a fixed field current If .
Base Speed The speed corresponding to the rated armature voltage, rated field current, and rated armature current
is called base speed.
Example 14-1
A 15hp 220V 2000rpm separately excited dc motor controls a load requiring a torque of τL = 45N · m at a speed
of 1200rpm. The field circuit resistance is Rf = 147Ω, the armature circuit resistance is Ra = 0.25Ω, and the voltage
constant of the motor is Kv = 0.7032V /A − rad/s. The field voltage is Vf = 220V . The viscous friction and no-load
losses are negligible. The armature current may be assumed continuous and ripple free. Determine
1. the back emf Ea ;
2. the required armature voltage Va ;
3. the rated armature current of the motor.
Solution
1.
ω
If
Ea
2π × 1200
2πn
=
= 125.66rad/s
60
60
Vf
220
= 1.497A
=
=
Rf
147
= Kv ωIf = 0.7032 × 125.66 × 1.497 = 132.28V
=
7.2. SINGLE-PHASE FULL-WAVE CONVERTER DRIVE
65
Figure 7.2: DC motor equivalent circuit
2.
Ia
Va
τd
45
= 42.75A
=
Kt If
0.7032 × 1.497
= Ra Ia + Ea = 0.25 × 42.75 + 132.28 = 142.97V
=
3. Since 1hp is equal to 746W,
Irated =
7.2
Prated
15 × 746
=
= 50.87A
Vrated
220
Single-Phase Full-Wave Converter Drive
In order to control the speed of the motor, we need to adjust the armature supply voltage Va or the field supply voltage
Vf , which implies that we need variable dc voltage supplies. As we know, controlled rectifiers provide a variable dc
output voltage from a fixed ac voltage, whereas choppers can provide a variable dc voltage from a fixed dc voltage.
According to variable dc supplies, dc drives can be classified into three types:
1. Single-phase drives
(a)
(b)
(c)
(d)
Single-phase
Single-phase
Single-phase
Single-phase
half-wave converter drives (up to 0.5kW)
semi-converter drives (up to 15kW)
full-wave converter drives (up to 15kW)
dual converter drives (up to 15kW)
2. Three-phase drives
(a)
(b)
(c)
(d)
Three-phase
Three-phase
Three-phase
Three-phase
half-wave converter drives (up to 40kW)
semi-converter drives (up to 115kW)
full-wave converter drives (up to 1500kW)
dual converter drives (up to 1500kW)
3. Chopper drives
We shall consider only single-phase full-wave converter drive here due to time limitations.
The connection of single-phase full-converter drive is shown in Fig. 7.2. Both armature and field circuits are
supplied by single-phase full-wave converters.
Recall that a single-phase full-wave converter provides an output voltage
Va =
2Vm
cos αa ,
π
0 ≤ αa ≤ π
Vf =
2Vm
cos αf ,
π
0 ≤ αf ≤ π
for the armature circuit, and
for the field circuit.
Example 14-3 The speed of a separately excited dc motor is controlled by a one-phase full-wave converter. The field
circuit is also controlled by a full converter and the field current is set to the maximum possible value. The AC supply
voltage to the armature and field converters is one-phase, 440V, 60Hz. The armature resistance is Ra = 0.25Ω, the
field circuit resistance is Rf = 175Ω, and the motor voltage constant is Kv = 1.4V /A − rad/s. The armature current
corresponding to the load demand is Ia = 45A. The viscous friction and no-load losses are negligible. The inductances
of the armature and field circuits are sufficient to make the armature and field currents continuous and ripple-free. If
the delay angle of the armature converter is αa = 60◦ and the armature current is Ia = 45A, determine
66
CHAPTER 7. DC DRIVES
1. the torque developed by the motor τd ,
2. the speed ω,
√
Solution: Vs = 440V , Vm =
2 × 440 = 622.25V , Ra = 0.25Ω, Rf = 175Ω, αa = 60◦ , and Kv = 1.4V /A − rad/s.
1. The maximum field voltage (and current) would be obtained for a delay angle of αf = 0 and
Vf =
2 × 622.25
2Vm
=
= 396.14V
π
π
The field current is
If =
Vf
396.14
= 2.26A
=
Rf
175
The developed torque is
τd = τL = Kv If Ia = 1.4 × 2.26 × 45 = 142.4N · m
2. The armature voltage is
Va =
2Vm
2 × 622.25
cos 60◦ =
cos 60◦ = 198.07V
π
π
The back emf is
Ea = Va − Ia Ra = 198.07 − 45 × 0.25 = 186.82V
The speed is
ω=
186.82
Ea
= 59.05rad/s or 564rpm
=
Kv If
1.4 × 2.26
Example
A 220V, 3hp, 1800rpm separately excited DC motor is controlled by a one-phase full-wave converter with an AC
source of 230V at 60Hz. Assume that full load efficiency of the motor is 88% and enough field inductance is added to
ensure continuous current for any torque greater than 25% of rated torque. Ra = 1.5Ω.
1. Determine the firing angle α to obtain the rated torque at 1200rpm.
2. Compute the firing angle for the rated breaking torque at -1800rpm.
3. Find the firing angle corresponding to a torque of 35N ·m and speed of 480rpm, assuming continuous conduction.
Solution
1. Since η = Pout /Pin and Pin = Va Ia , we have
Ia =
Pin
Pout η
746 × 3 × 0.88
= 11.56A
=
=
Va
Va
220V
It follows from Va = Ea + Ia Ra that the back emf at rated speed of 1800rpm is given by
(Ea )1800 = Va − Ia Ra = 220 − 11.56 × 1.5 = 202.66V
Note that (Ea )1800 = Kv ω1800 If and (Ea )1200 = Kv ω1200 If . Then, we have
(Ea )1800
ω1800
=
(Ea )1200
ω1200
As a result, the back emf at the speed of 1200rpm is
(Ea )1200 =
ω1200
1200
× 202.66 = 135.11V
(Ea )1800 =
ω1800
1800
In order to obtain rated torque at 1200rpm, the armature voltage should be
Va = (Ea )1200 + Ia Ra = 135.11 + 11.56 × 1.5 = 152.36V
which is the output voltage of the full-wave converter, that is,
√
2 2Vs cos α
2Vm cos α
Va =
=
π
π
which implies that
Va π
152.36π
cos α = √
= √
= 0.736
2 2Vs
2 2 × 230
So, the firing angle is α = 42.6◦ .
7.2. SINGLE-PHASE FULL-WAVE CONVERTER DRIVE
67
2. Note that (Ea )−1800 = −202.66V , which means that the armature voltage for the rated braking torque at
-1800rpm is
Va = (Ea )−1800 + Ia Ra = −202.66 + 11.56 × 1.5 = −185.32
To get -185.32V, reverse the armature connections. Then, we get
185.32π
Va π
= 0.8953
= √
cos α = √
2 2Vs
2 2 × 230
So, the firing angle is α = 26.45◦.
3. The speed in radian per second for n = 1800rpm is
ω=
2π × 1800
2πn
=
= 188.57rad/s
60
60
Note Ea = Kv ωIf . Then the motor voltage constant Kv satisfies
Kv If =
202.66
Ea
=
= 1.075
ω
188.57
The armature current corresponding to the torque of 35N · m is
Ia =
τd
35N · m
= 32.56A
=
Kv If
1.075
The back emf is given by
(Ea )480 = Kv If ω480 = 1.075
2π × 480
= 54V
60
The armature voltage is calculated as
Va = (Ea )480 + Ia Ra = 54 + 32.56 × 1.5 = 102.84V
Then, we get
Va π
102.84π
= 0.4967
cos α = √
= √
2 2Vs
2 2 × 230
So, the firing angle is α = 60.2◦ .
68
CHAPTER 7. DC DRIVES
Chapter 8
AC Drives
8.1
Induction Motor
A set of three-phase AC voltages is needed to drive a three-phase induction motor. The per-phase equivalent circuit
of an induction motor is shown in Fig. 8.1.
The relationship between torque and speed for an induction motor is given by
τd =
sωsync
h
3Rr Vs2
i
2
Rs + Rsr + (Xs + Xr )2
where Rr is the per-phase resistance of the rotor winding referred to the stator winding, Xr is the per-phase reactance
of the rotor winding referred to the stator winding, Rs is the per-phase resistance of the stator winding, Xs is the
per-phase leakage reactance of the stator winding, Rm represents the per-phase resistance for core losses, Xm is the
m
per-phase magnetizing reactance, s = ωs −ω
is the slip of the motor, ωs = 4πf
ωs
P is the synchronous speed of the motor,
ωm is the speed of the motor, f is the supply frequency, P is the number of poles in the motor, Vs is the rms phase
voltage applied to the stator winding.
Starting Torque is the torque when s = 1, that is, ωm = 0, which is given by
τs =
3Rr Vs2
ωsync [(Rs + Rr )2 + (Xs + Xr )2 ]
The maximum developed torque during motoring is called pullout torque or breakdown torque which is given by
τmm =
3V 2
p s
2ωsync [Rs + Rs2 + (Xs + Xr )2 ]
which is reached at
Rr
sm = p
2
Rs + (Xs + Xr )2
The maximum regenerative torque is the developed torque when s = −sm , which is given by
τmr = −
3V 2
ps
2ωsync [−Rs + Rs2 + (Xs + Xr )2 ]
If the motor is supplied from a fixed voltage at a constant frequency, the developed torque is a function of the slip
and the torque-speed characteristics can be determined by the equation above. A typical plot of the developed torque
as a function of slip or speed is shown in Fig. 8.2.
There are three operation regions:
1. Motoring or powering, 0 ≤ s ≤ 1:
In this region, the motor rotates in the same direction as the rotating magnetic field. As the slip increases from
zero, the torque also increases (almost linearly). Once the torque reaches its maximum Tmm at s = sm , the
torque begins to decrease with the increase in slip.
69
70
CHAPTER 8. AC DRIVES
Figure 8.1: Three-phase AC induction motor equivalent circuit
Regenerating
ωm>ωs>0
Motoring
0<ωm<ωs
τ
Plugging
ωm<0,ωs>0
mm
τs
−1
2ω
s
−s
m
0 sm
ω
s
1
0
2
−ωs
τ
mr
Figure 8.2: Three-phase AC induction motor torque-speed curve
8.1. INDUCTION MOTOR
71
2. Regeneration, −1 ≤ s ≤ 0:
In the regeneration region, the speed ωm is greater than the synchronous speed ωs with ωm and ωs being in the
same direction and the slip is negative. As a result, Rsr is negative, so is the torque τd . This means that power
is being fed back from the shaft into the rotor circuit and the motor operates as a generator. The motor returns
power to the supply.
3. Plugging, 1 ≤ s ≤ 2:
In plugging, the speed is opposite to the direction of the rotating magnetic field and the slip is greater than
one. This may happen if the sequence of the supply source is reversed while motoring, so that the direction of
the rotating magnetic field is also reversed. The developed torque, which is in the same direction as the field,
opposes the motion and acts as braking torque. Since s > 1, the motor currents will be high, but the developed
torque will be low. The energy due to a plugging brake must be dissipated within the motor and this may cause
excessive heating of the motor. This type of braking is not normally recommended.
It can be noticed from the torque equation that the speed and torque of the induction motor can be varied by one
of the following methods:
1. Stator voltage control
2. Rotor voltage control for wound rotor motors
3. Frequency control
4. Stator voltage and frequency control
5. Stator current control
6. Voltage, current and frequency control
Example: A three-phase 460V 60Hz four-pole Y-connected induction motor has the following per-phase equivalent
circuit parameters: Rs = 0.42Ω, Rr = 0.23Ω, Xs = Xr = 0.82Ω. The rotor speed is 1750rpm. Determine ωs , s, τd , τs ,
sm , τmm , τmr .
Solution:
Vs
=
ωm
=
ωs
=
s =
τd
=
460
√ = 265.58V
3
2π × 1750
2πnm
=
= 183.27rad/s
60
60
4π × 60
4πf
=
= 188.5rad/s
P
4
ωs − ωm
188.5 − 183.27
= 0.028
=
ωs
188.5
3Rr Vs2
h
i
2
2
sωs Rs + Rsr + (Xs + Xr )
=
3 × 0.23 × 265.582
i = 119.38N · m
h
2
0.23 2
0.028 × 188.5 0.42 + 0.028
+ (0.82 + 0.82)
=
3Rr Vs2
h
i
ωs (Rs + Rr )2 + (Xs + Xr )2
=
3 × 0.23 × 265.582
i = 82.96N · m
h
2
2
188.5 (0.42 + 0.23) + (0.82 + 0.82)
sm
=
Rr
0.23
p
=p
= 0.1359
2
2
2
Rs + (Xs + Xr )
0.42 + (0.82 + 0.82)2
τmm
=
3V 2
p s
2ωsync [Rs + Rs2 + (Xs + Xr )2 ]
=
3 × 265.582
p
= 265.64N · m
2 × 188.5[0.42 + 0.422 + (0.82 + 0.82)2 ]
τs
72
CHAPTER 8. AC DRIVES
τmr
8.2
= −
3V 2
p s
2ωsync [−Rs + Rs2 + (Xs + Xr )2 ]
= −
3 × 265.582
p
= −440.94N · m
2 × 188.5[−0.42 + 0.422 + (0.82 + 0.82)2 ]
Frequency Control
In the induction motor, the rms air-gap flux can be expressed as
φ=
Vs
km ω
where km is a constant.
At the rated voltage and rated frequency, the flux will be the rated flux.
Case 1: Decrease frequency If the voltage Vs is kept constant at its rated value while the frequency is reduced
below its rated value, the flux will increase, which would cause saturation of the air-gap flux and the motor parameters
would not be valid in determining the torque-speed characteristics. At low frequency, the reactances will decrease and
the motor current may be too high. This type of frequency control is not normally used. In this case, we need to
reduce the voltage Vs to keep φ constant.
Case 2: Increase frequency If the frequency is increased above its rated value, the flux and torque would
decrease. If the synchronous speed corresponding to the rated frequency is called the base speed ωb , the synchronous
speed at any other frequency f = βfrated becomes
ωs =
4πβfrated
4πf
=
= βωb
P
P
and
s=
βωb − ωm
ωm
=1−
βωs
βωb
The reactances become
X̄s
=
2πLs f = 2πLs βfrated = β(2πLs frated ) = βXs
X̄r
=
2πLr f = 2πLr βfrated = β(2πLr frated ) = βXr
Therefore, the torque equation becomes
τd
τmm
sm
sωs
h
2ωs
h
=
=
=
3Rr Vs2
i
2
Rs + Rsr + (βXs + βXr )2
3Vs2
i
p
Rs + Rs2 + (βXs + βXr )2
Rr
p
2
Rs + (βXs + βXr )2
If Rs is negligible, then the maximum torque at the base speed is
τmb =
3Vs2
2ωb (Xs + Xr )
which is constant. The maximum torque at f = βfrated is
τmm =
3Vs2
3Vs2
3Vs2
τmb
1
=
=
= 2
2βωb (βXs + βXr )
2ωb (Xs + Xr ) β 2
β
2ωs (X̄s + X̄r )
which implies that the maximum torque is inversely proportional to frequency squared and τmm β 2 remains constant.
The slip for the maximum torque is
Rr
sm =
βXs + βXr
8.3. VOLTAGE AND FREQUENCY CONTROL
73
Example: A three-phase 1750rpm 460V 60Hz 4-pole, Y-connected induction motor has the following parameters:
Rr = 0.38Ω, Xs = 1.14Ω, Xr = 1.71Ω, and Xm = 33.2Ω. The motor is controlled by varying the supply frequency.
Calculate the speed at the maximum torque for f = 1.5frated and for Rs = 0Ω and Rs = 0.66Ω.
Solution:
Vrated
ωb
β
ωs
460
√ = 265.58V
3
4π × 60
4πfrated
=
= 188.5rad/s
=
P
4
= 1.5
= βωb = 1.5 × 188.5 = 282.75rad/s
=
The case of Rs = 0Ω
sm
=
ωm
=
nm
=
τmm
=
Rr
0.38
= 0.089
=
βXs + βXr
1.5 × 1.14 + 1.5 × 1.71
(1 − sm )ωs = (1 − 0.089) × 282.75 = 257.61rad/s
60 × 257.61
60ωm
=
= 2460rpm
2π
2π
2
3Vs
3 × 265.582
h
i =
i = 87.5N · m
h
p
p
2ωs Rs + Rs2 + (βXs + βXr )2
2 × 282.75 0 + 02 + (1.5 × 1.14 + 1.5 × 1.71)2
The case of Rs = 0.66Ω
sm
=
ωm
=
nm
=
τmm
=
8.3
Rr
0.38
p
=p
0.088
2
2
2
Rs + (βXs + βXr )
0.66 + (1.5 × 1.14 + 1.5 × 1.71)2
(1 − sm )ωs = (1 − 0.088) × 282.75 = 257.9rad/s
60 × 257.61
60ωm
=
= 2463rpm
2π
2π
3Vs2
3 × 265.582
h
i=
i = 75N · m
h
p
p
2ωs Rs + Rs2 + (βXs + βXr )2
2 × 282.75 0.66 + 0.662 + (1.5 × 1.14 + 1.5 × 1.71)2
Voltage and Frequency Control
Voltage and frequency control is to adjust the frequency and voltage at the same time so that the ratio of voltage and
frequency is kept constant. This type of control is known as volts/hertz control.
The ratio of the rated phase voltage Vrated to the base speed ωb is defined by
d=
Vrated
ωb
Now the synchronous speed is ωs = βωb . In order to keep d constant, the phase voltage Vs must be changed so that
Vs
=d
ωs
which means that
Vs = dωs = dβωb = βVrated
So if the frequency is decreased, the voltage Vs must be reduced to keep d constant.
At the frequency f = βfrated , the torque equations become
τd
=
sωs
τmm
=
h
h
3Rr Vs2
i
2
Rs + Rsr + (βXs + βXr )2
2ωs Rs +
sm
=
3Vs2
i
p
Rs2 + (βXs + βXr )2
Rr
p
2
Rs + (βXs + βXr )2
74
CHAPTER 8. AC DRIVES
As the frequency is reduced, β decreases and the slip for maximum torque increases. By varying both the voltage
and frequency, the torque and speed can be controlled.
The voltage at variable frequency can be obtained from three-phase inverters.
Example: A three-phase 1750rpm 460V 60Hz 4-pole, Y-connected induction motor has the following parameters:
Rs = 0.66, Rr = 0.38Ω, Xs = 1.14Ω, Xr = 1.71Ω, and Xm = 33.2Ω. The motor is controlled by varying both
the voltage and frequency. The volts/hertz ratio, which corresponds to the rated voltage and rated frequency, is
maintained constant. Calculate the maximum torque and the corresponding speed for 60Hz and 30Hz.
Solution:
Vrated
=
ωb
=
d =
460
√ = 265.58V
3
4π × 60
4πfrated
=
= 188.5rad/s
P
4
Vs
265.58
= 1.409
=
ωb
188.5
For f = 60Hz,
ωs
=
β
=
Vs
=
sm
=
ωm
=
nm
=
τmm
=
=
4π × 60
4πfrated
=
= 188.5rad/s
P
4
ωs
188.5
=1
=
ωb
188.5
βVrated = 265.58V
Rr
0.38
p
=p
= 0.1299
2
2
2
Rs + (βXs + βXr )
0.66 + (1 × 1.14 + 1 × 1.71)2
(1 − s)ωs = (1 − 0.1299) × 188.5 = 164.01rad/s
60 × 164.01
60ωm
=
= 1566rpm
2π
2π
3Vs2
h
i
p
2ωs Rs + Rs2 + (βXs + βXr )2
3 × 265.582
i = 156.55N · m
h
p
2 × 188.5 0.66 + 0.662 + (1 × 1.14 + 1 × 1.71)2
For f = 30Hz,
ωs
=
β
=
Vs
=
sm
=
ωm
=
nm
=
τmm
=
=
4π × 30
4πfrated
=
= 94.25rad/s
P
4
ωs
94.25
= 0.5
=
ωb
188.5
βVrated = 0.5 × 265.58 = 132.79V
Rr
0.38
p
=p
= 0.242
2
2
2
Rs + (βXs + βXr )
0.66 + (0.5 × 1.14 + 0.5 × 1.71)2
(1 − s)ωs = (1 − 0.242) × 94.25 = 71.44rad/s
60 × 71.44
60ωm
=
= 682rpm
2π
2π
3Vs2
h
i
p
2ωs Rs + Rs2 + (βXs + βXr )2
3 × 132.792
i = 128.82N · m
h
p
2 × 94.25 0.66 + 0.662 + (0.5 × 1.14 + 0.5 × 1.71)2
8.3. VOLTAGE AND FREQUENCY CONTROL
75
200
β=.4
β=.6
β=.8
β=1
β=1.2
β=1.4
100
β=.2
0
torque τd (N.m)
−100
−200
−300
−400
−500
−600
0
500
1000
1500
slip s
2000
2500
Figure 8.3: Three-phase AC induction motor torque-speed curve
3000