108 Chapter 3 Diode Rectifiers
Example 3.1 Finding the Performance Parameters of a Full-Wave Rectifier with a
Center-Tapped Transformer
If the rectifier in Figure 3.2a has a purely resistive load of R, determine (a) the efficiency, (b) the
FF, (c) the RF, (d) the TUF, (e) the PIV of diode D1, (f) the CF of the input current, and (g) the
input power factor PF.
Solution
From Eq. (3.11), the average output voltage is
Vdc =
2Vm
= 0.6366Vm
π
Idc =
Vdc
0.6366Vm
=
R
R
and the average load current is
The rms values of the output voltage and current are
Vrms = c
Irms =
2
T L0
T/2
(Vm sin ωt)2 dt d
Vrms
0.707Vm
=
R
R
1/2
=
Vm
12
= 0.707Vm
From Eq. (3.1) Pdc = (0.6366Vm)2/R, and from Eq. (3.2) Pac = (0.707Vm)2/R.
a.
b.
c.
d.
From Eq. (3.3), the efficiency η = (0.6366Vm)2/(0.707Vm)2 = 81%.
From Eq. (3.5), the form factor FF = 0.707Vm/0.6366Vm = 1.11.
From Eq. (3.7), the ripple factor RF = 21.112 - 1 = 0.482 or 48.2,.
The rms voltage of the transformer secondary Vs = Vm/12 = 0.707Vm. The rms value
of transformer secondary current Is = 0.5Vm/R. The volt-ampere rating (VA) of the
transformer, VA = 22Vs Is = 22 * 0.707Vm * 0.5Vm/R. From Eq. (3.8),
TUF =
0.63662
22 * 0.707 * 0.5
= 0.81064 = 81.06%
e. The peak reverse blocking voltage, PIV = 2Vm.
f. Is(peak) = Vm/R and Is = 0.707Vm/R. The CF of the input current is CF = Is(peak)/Is =
1/0.707 = 12.
g. The input PF for a resistive load can be found from
PF =
Pac
0.7072
=
= 1.0
VA
22 * 0.707 * 0.5
Note: 1/TUF = 1/0.81064 = 1.136 signifies that the input transformer, if present, must be
1.75 times larger than that when it is used to deliver power from a pure ac sinusoidal voltage. The
rectifier has an RF of 48.2% and a rectification efficiency of 81%.
M03_RASH9088_04_PIE_C03.indd 108
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238 Chapter 5 DC–DC Converters
Assuming a lossless converter, the input power to the converter is the same as the output power and is given by
kT
1
1
Pi =
v i dt =
T L0 0
T L0
kT 2
v0
R
dt = k
V 2s
R
(5.10)
The effective input resistance seen by the source is
Ri =
Vs
Vs
R
=
=
Ia
kVs >R
k
(5.11)
which indicates that the converter makes the input resistance Ri as a variable resistance
of R/k. The variation of the normalized input resistance against the duty cycle is shown
in Figure 5.2c. It should be noted that the switch in Figure 5.2 could be implemented by
a BJT, a MOSFET, an IGBT, or a GTO.
The duty cycle k can be varied from 0 to 1 by varying t 1, T, or f. Therefore, the output
voltage Vo can be varied from 0 to Vs by controlling k, and the power flow can be controlled.
1. Constant-frequency operation: The converter, or switching, frequency f (or chopping period T) is kept constant and the on-time t 1 is varied. The width of the pulse is
varied and this type of control is known as pulse-width-modulation (PWM) control.
2. Variable-frequency operation: The chopping, or switching, frequency f is varied.
Either on-time t 1 or off-time t 2 is kept constant. This is called frequency modulation. The frequency has to be varied over a wide range to obtain the full output
voltage range. This type of control would generate harmonics at unpredictable
frequencies and the filter design would be difficult.
Example 5.1 Finding the Performances of a Dc–Dc Converter
The dc converter in Figure 5.2a has a resistive load of R = 10 Ω and the input voltage is
Vs = 220 V. When the converter switch remains on, its voltage drop is vch = 2 V and the chopping frequency is f = 1 kHz. If the duty cycle is 50%, determine (a) the average output voltage
Va, (b) the rms output voltage Vo, (c) the converter efficiency, (d) the effective input resistance
Ri of the converter, (e) the ripple factor of the output voltage RFo, and (f) the rms value of the
fundamental component of output harmonic voltage.
Solution
Vs = 220 V, k = 0.5, R = 10 Ω, and vch = 2 V.
a. From Eq. (5.8), Va = 0.5 * 1220 - 22 = 109 V.
b. From Eq. (5.9), Vo = 10.5 * 1220 - 22 = 154.15 V.
c. The output power can be found from
Po =
1
T L0
kT 2
v0
= 0.5 *
M05_RASH9088_04_PIE_C05.indd 238
R
dt =
1
T L0
1220 - 22 2
10
kT
1Vs - vch 2 2
R
dt = k
1Vs - vch 2 2
R
(5.12)
= 2376.2 W
25/07/13 3:41 PM
5.3 Principle of Step-Down Operation 239
The input power to the converter can be found from
kT
Pi =
1
1
Vsi dt =
T L0
T L0
kT
Vs 1Vs - vch 2
R
dt = k
Vs 1Vs - vch 2
R
220 - 2
= 0.5 * 220 *
= 2398 W
10
(5.13)
The converter efficiency is
Po
2376.2
=
= 99.09,
Pi
2398
d. From Eq. (5.11),
Ri = Vs/Ia = Vs 1Va/R 2 = 220 * 1109/102 = 20.18 Ω
e. Substituting Va from Eq. (5.8) and Vo from Eq. (5.9) into Eq. (5.6) gives the ripple
­factor as
RFo =
f.
Vr
1
=
- 1
Va
Ck
(5.14)
= 31/0.5 - 1 = 100 ,
The output voltage as shown in Figure 5.2b can be expressed in a Fourier series as
vo 1t2 = kVs + a
∞
Vs
sin 2nπk cos 2nπft
nπ
n=1
+
Vs ∞
11 - cos 2nπk2 sin 2nπft
nπ na
=1
(5.15)
The fundamental component (for n = 1) of output voltage harmonic can be determined from
Eq. (5.15) as
1Vs - vch2
[sin 2π k cos 2πft + 11 - cos 2πk2sin 2πft]
π
1220 - 22 * 2
=
sin12π * 1000t2 = 138.78 sin16283.2t2
π
v1 1t2 =
(5.16)
and its root-mean-square (rms) value is V1 = 138.78/12 = 98.13 V.
Note: The efficiency calculation, which includes the conduction loss of the converter, does not take into account the switching loss due to turn-on and turn-off of
practical converters. The efficiency of a practical converter varies between 92 and 99%.
Key Points of Section 5.3
• A step-down chopper, or dc converter, that acts as a variable resistance load can
produce an output voltage from 0 to VS.
• Although a dc converter can be operated either at a fixed or variable frequency,
it is usually operated at a fixed frequency with a variable duty cycle.
• The output voltage contains harmonics and a dc filter is needed to smooth out
the ripples.
M05_RASH9088_04_PIE_C05.indd 239
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244 Chapter 5 DC–DC Converters
Condition for continuous current: For I1 Ú 0, Eq. (5.25) gives
a
e kz - 1
E
b Ú 0
z
e - 1
Vs
which gives the value of the load electromotive force (emf) ratio x = E/Vs as
x =
E
e kz - 1
… z
Vs
e - 1
(5.31)
Example 5.2 Finding the Currents of a Dc Converter with an RL Load
A converter is feeding an RL load as shown in Figure 5.4 with Vs = 220 V, R = 5 Ω, L = 7.5 mH,
f = 1 kHz, k = 0.5, and E = 0 V. Calculate (a) the minimum instantaneous load current I1,
(b) the peak instantaneous load current I2, (c) the maximum peak-to-peak load ripple current,
(d) the average value of load current Ia, (e) the rms load current Io, (f) the effective input resistance Ri seen by the source, (g) the rms chopper current IR, and (h) the critical value of the load
inductance for continuous load current. Use PSpice to plot the load current, the supply current,
and the freewheeling diode current.
Solution
Vs = 220 V, R = 5 Ω, L = 7.5 mH, E = 0 V, k = 0.5, and f = 1000 Hz. From Eq. (5.23),
I2 = 0.7165I1 + 12.473 and from Eq. (5.24), I1 = 0.7165I2 + 0.
a. Solving these two equations yields I1 = 18.37 A.
b. I2 = 25.63 A.
c. ∆I = I2 - I1 = 25.63 - 18.37 = 7.26 A. From Eq. (5.29), ∆I max = 7.26 A and
Eq. (5.30) gives the approximate value, ∆I max = 7.33 A.
d. The average load current is, approximately,
Ia =
I2 + I1
25.63 + 18.37
=
= 22 A
2
2
e. Assuming that the load current rises linearly from I1 to I2, the instantaneous load current can be expressed as
i1 = I1 +
∆It
kT
for 0 6 t 6 kT
The rms value of load current can be found from
Io = a
kT
1/2
1/2
1I2 - I1 2 2
1
i21 dtb = c I 21 +
+ I1 1I2 - I1 2 d kT L0
3
(5.32)
= 22.1 A
f.
The average source current
Is = kIa = 0.5 * 22 = 11 A
and the effective input resistance Ri = Vs/Is = 220/11 = 20 Ω.
M05_RASH9088_04_PIE_C05.indd 244
25/07/13 3:41 PM
5.4 Step-Down Converter with RL Load 245
30 A
Probe Cursor
A1 9.509m,
A2 9.0000m,
dif 508.929u,
SEL
0A
I(R)
25.455
17.960
7.4948
30 A
0A
- I(Vs)
30 A
0A
0s
5 ms
I(Dm)
10 ms
Time
Figure 5.6
SPICE plots of load, input, and diode currents for Example 5.2.
g. The rms converter current can be found from
IR = a
kT
1/2
1/2
1I2 - I1 2 2
1
+ I1 1I2 - I1 2 d i21 dtb = 1k c I 21 +
T L0
3
(5.33)
= 1kIo = 10.5 * 22.1 = 15.63 A
h. We can rewrite Eq. (5.31) as
VS a
e kz - 1
b = E
ez - 1
which, after iteration, gives, z = TR/L = 52.5 and L = 1 ms * 5/52.5 = 0.096 mH.
The SPICE simulation results [32] are shown in Figure 5.6, which shows the load current I(R), the supply current - I 1Vs 2 , and the diode current I(Dm). We get I1 = 17.96 A
and I2 = 25.46 A.
Example 5.3 Finding the Load Inductance to Limit the Load Ripple Current
The converter in Figure 5.4 has a load resistance R = 0.25 Ω, input voltage Vs = 550 V,
and ­battery voltage E = 0 V. The average load current Ia = 200 A and chopping frequency
f = 250 Hz. Use the average output voltage to calculate the load inductance L, which would
limit the maximum load ripple current to 10% of Ia.
M05_RASH9088_04_PIE_C05.indd 245
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260 Chapter 5 DC–DC Converters
Condition for continuous inductor current and capacitor voltage.
average inductor current, the inductor ripple current ∆I = 2IL.
Using Eqs. (5.56) and (5.60), we get
If IL is the
VS 1 1 - k 2k
2kVs
= 2IL = 2Ia =
fL
R
which gives the critical value of the inductor Lc as
Lc = L =
11 - k 2R
2f
(5.64)
If Vc is the average capacitor voltage, the capacitor ripple voltage ∆Vc = 2Va. Using
Eqs. (5.56) and (5.63), we get
Vs 11 - k 2 k
8LCf 2
= 2Va = 2kVs
which gives the critical value of the capacitor Cc as
Cc = C =
1 - k
16Lf 2
(5.65)
The buck regulator requires only one transistor, is simple, and has high efficiency
greater than 90%. The di/dt of the load current is limited by inductor L. However, the
input current is discontinuous and a smoothing input filter is normally required. It
provides one polarity of output voltage and unidirectional output current. It requires a
protection circuit in case of possible short circuit across the diode path.
Example 5.5 Finding the Values of LC Filter for the Buck Regulator
The buck regulator in Figure 5.17a has an input voltage of Vs = 12 V. The required average output voltage is Va = 5 V at R = 500 Ω and the peak-to-peak output ripple voltage is 20 mV. The
switching frequency is 25 kHz. If the peak-to-peak ripple current of inductor is limited to 0.8 A,
determine (a) the duty cycle k, (b) the filter inductance L, (c) the filter capacitor C, and (d) the
critical values of L and C.
Solution
Vs = 12 V, ∆Vc = 20 mV, ∆I - 0.8 A, f = 25 kHz, and Va = 5 V.
a. From Eq. (5.56), Va = kVs and k = Va/Vs = 5/12 = 0.4167 = 41.67,.
b. From Eq. (5.59),
L =
5112 - 52
0.8 * 25,000 * 12
= 145.83 μH
c. From Eq. (5.61),
C =
M05_RASH9088_04_PIE_C05.indd 260
0.8
= 200 μF
8 * 20 * 10-3 * 25,000
25/07/13 3:42 PM
5.9 Switching-Mode Regulators 261
d. From Eq. (5.64), we get Lc =
From Eq. (5.65), we get Cc =
5.9.2
11 - k2R
2f
=
11 - 0.41672 * 500
2 * 25 * 103
= 5.83 mH
1 - k
1 - 0.4167
=
= 0.4 μF
16Lf 2
16 * 145.83 * 10-6 * 125 * 103 2 2
Boost Regulators
In a boost regulator [8, 9] the output voltage is greater than the input voltage—hence
the name “boost.” A boost regulator using a power MOSFET is shown in Figure 5.18a.
Transistor M1 acts as a controlled switch and diode Dm is an uncontrolled switch. The
circuit in Figure 5.18a is often represented by two switches as shown in Figure 5.18b. The circuit operation can be divided into two modes. Mode 1 begins when transistor.
M1 is switched on at t = 0. The input current, which rises, flows through inductor L and
transistor Q1. Mode 2 begins when transistor M1 is switched off at t = t 1. The current that
was flowing through the transistor would now flow through L, C, load, and diode Dm. The
inductor current falls until transistor M1 is turned on again in the next cycle. The energy
stored in inductor L is transferred to the load. The equivalent circuits for the modes of
­operation are shown in Figure 5.18c. The waveforms for voltages and currents are shown
in Figure 5.18d for continuous load current, assuming that the current rises or falls linearly.
Assuming that the inductor current rises linearly from I1 to I2 in time t1,
Vs = L
I2 - I1
∆I
= L
t1
t1
(5.66)
∆IL
Vs
(5.67)
or
t1 =
and the inductor current falls linearly from I2 to I1 in time t2,
Vs - Va = -L
∆I
t2
(5.68)
or
t2 =
∆IL
Va - Vs
(5.69)
where ∆I = I2 - I1 is the peak-to-peak ripple current of inductor L. From Eqs. (5.66)
and (5.68),
∆I =
1Va - Vs 2t 2
Vst 1
=
L
L
Substituting t 1 = kT and t 2 = 11 - k 2T yields the average output voltage,
Va = Vs
M05_RASH9088_04_PIE_C05.indd 261
Vs
T
=
t2
1 - k
(5.70)
25/07/13 3:42 PM
268 Chapter 5 DC–DC Converters
Condition for continuous inductor current and capacitor voltage. If IL is the average inductor current, at the critical condition for continuous conduction the inductor
ripple current ∆I = 2IL. Using Eqs. (5.86) and (5.92), we get
kVs
2kVs
= 2IL = 2Ia =
fL
11 - k 2R
which gives the critical value of the inductor Lc as
Lc = L =
11 - k 2R
2f
(5.96)
If Vc is the average capacitor voltage, at the critical condition for continuous conduction the capacitor ripple voltage ∆Vc = -2Va. Using Eq. (5.95), we get
-
Iak
= -2Va = -2IaR
Cf
which gives the critical value of the capacitor Cc as
Cc = C =
k
2fR
(5.97)
A buck–boost regulator provides output voltage polarity reversal without a
transformer. It has high efficiency. Under a fault condition of the transistor, the di/dt
of the fault current is limited by the inductor L and will be Vs/L. Output short-circuit
protection would be easy to implement. However, the input current is discontinuous
and a high peak current flows through transistor Q1.
Example 5.7 Finding the Currents and Voltage in the Buck–Boost Regulator
The buck–boost regulator in Figure 5.19a has an input voltage of Vs = 12 V. The duty cycle
k = 0.25 and the switching frequency is 25 kHz. The inductance L = 150 μH and filter capacitance C = 220 μF. The average load current Ia = 1.25 A. Determine (a) the average output
voltage, Va; (b) the peak-to-peak output voltage ripple, ∆Vc; (c) the peak-to-peak ripple current
of inductor, ∆I; (d) the peak current of the transistor, Ip; and (e) the critical values of L and C.
Solution
Vs = 12 V, k = 0.25, Ia = 1.25 A, f = 25 kHz, L = 150 μH, and C = 220 μF.
a. From Eq. (5.86), Va = - 12 * 0.25/11 - 0.252 = -4 V.
b. From Eq. (5.95), the peak-to-peak output ripple voltage is
∆Vc =
1.25 * 0.25
25,000 * 220 * 10-6
= 56.8 mV
c. From Eq. (5.92), the peak-to-peak inductor ripple is
∆I =
M05_RASH9088_04_PIE_C05.indd 268
12 * 0.25
25,000 * 150 * 10-6
= 0.8 A
25/07/13 3:42 PM
5.9 Switching-Mode Regulators 269
d. From Eq. (5.89), Is = 1.25 * 0.25/11 - 0.252 = 0.4167 A. Because Is is the average
of duration kT, the peak-to-peak current of the transistor,
Ip =
e. R =
Is
∆I
0.4167
0.8
+
=
+
= 2.067 A
k
2
0.25
2
-Va
4
=
= 3.2 Ω
Ia
1.25
11 - k2R
11 - 0.252 * 3.2
= 450 μH.
2 * 25 * 103
k
0.25
From Eq. (5.97), we get Cc =
=
= 1.56 μF.
2fR
2 * 25 * 103 * 3.2
From Eq. (5.96), we get Lc =
5.9.4
2f
=
Cúk Regulators
The circuit arrangement of the Cúk regulator [10] using a power bipolar junction transistor is shown in Figure 5.20a. Similar to the buck–boost regulator, the Cúk regulator
provides an output voltage that is less than or greater than the input voltage, but the
output voltage polarity is opposite to that of the input voltage. It is named after its
­inventor [1]. When the input voltage is turned on and transistor Q1 is switched off, diode
Dm is forward biased and capacitor C1 is charged through L1, Dm, and the input supply
Vs. Transistor Q1 acts a controlled switch and diode Dm is an uncontrolled switch. They
operate as two SPST current-bidirectional switches. The circuit in Figure 5.20a is often
represented by two switches as shown in Figure 5.20b.
The circuit operation can be divided into two modes. Mode 1 begins when transistor Q1 is turned on at t = 0. The current through inductor L1 rises. At the same
time, the voltage of capacitor C1 reverse biases diode Dm and turns it off. The capacitor C1 discharges its energy to the circuit formed by C1, C2, the load, and L2. Mode 2
begins when transistor Q1 is turned off at t = t 1. The capacitor C1 is charged from the
input supply and the energy stored in the inductor L2 is transferred to the load. The
diode Dm and transistor Q1 provide a synchronous switching action. The capacitor
C1 is the medium for transferring energy from the source to the load. The equivalent
circuits for the modes are shown in Figure 5.20c and the waveforms for steady-state
voltages and currents are shown in Figure 5.20d for a continuous load current.
Assuming that the current of inductor L1 rises linearly from IL11 to IL12 in time t1,
Vs = L1
IL12 - IL11
∆I1
= L1
t1
t1
(5.98)
∆I1L1
Vs
(5.99)
or
t1 =
and due to the charged capacitor C1, the current of inductor L1 falls linearly from IL12
to IL11 in time t2,
Vs - Vc1 = -L1
M05_RASH9088_04_PIE_C05.indd 269
∆I1
t2
(5.100)
25/07/13 3:42 PM
582 Chapter 11 AC Voltage Controllers
Example 11.1 Finding the Performance Parameters of a Single-Phase Full-Wave
Controller
A single-phase full-wave ac voltage controller in Figure 11.2a has a resistive load of R = 10 Ω
and the input voltage is Vs = 120 V (rms), 60 Hz. The delay angles of thyristors T1 and T2 are
equal: α1 = α2 = α = π/2. Determine (a) the rms output voltage Vo, (b) the input PF, (c) the
average current of thyristors IA, and (d) the rms current of thyristors IR.
Solution
R = 10 Ω, Vs = 120 V, α = π/2, and Vm = 12 * 120 = 169.7 V.
a. From Eq. (11.1), the rms output voltage
120
= 84.85 V
12
Vo =
b. The rms value of load current is Io = Vo/R = 84.85/10 = 8.485 A and the load power
is Po = I 2oR = 8.4852 * 10 = 719.95 W. Because the input current is the same as the
load current, the input VA rating is
VA = Vs Is = Vs Io = 120 * 8.485 = 1018.2 W
The input PF is
Po
Vo
1
sin 2α 1/2
=
= c aπ - α +
bd
π
VA
Vs
2
1
719.95
=
=
= 0.707 1lagging2
1018.2
12
PF =
(11.2)
c. The average thyristor current
π
IA =
=
1
12 Vs sin ωt d1ωt2
2πR Lα
12Vs
1cos α + 12
2πR
= 12 *
(11.3)
120
= 2.7 A
2π * 10
d. The rms value of the thyristor current
π
IR = c
= c
=
=
M11_RASH9088_04_PIE_C11.indd 582
1/2
1
2
2
2V
sin
ωt
d1ωt2
d
s
2πR2 Lα
2V 2s
4πR Lα
2
π
11 - cos 2ωt2 d1ωt2 d
1
sin 2α 1/2
aπ - α +
bd
2
12R π
Vs
c
120
= 6A
2 * 10
1/2
(11.4)
31/07/13 4:19 PM
11.5 Three-Phase Full-Wave Controllers 587
Example 11.2 Finding the Performance Parameters of a Single-Phase Full-Wave
Controller with an RL Load
The single-phase full-wave controller in Figure 11.5a supplies an RL load. The input rms voltage is Vs = 120 V, 60 Hz. The load is such that L = 6.5 mH and R = 2.5 Ω. The delay angles of
thyristors are equal: α1 = α2 = π/2. Determine (a) the conduction angle of thyristor T1, δ; (b)
the rms output voltage Vo, (c) the rms thyristor current IR; (d) the rms output current Io; (e) the
average current of a thyristor IA; and (f) the input PF.
Solution
R = 2.5 Ω, L = 6.5 mH, f = 60 Hz, ω = 2π * 60 = 377 rad/s, Vs = 120 V, α = 90°, and
θ = tan-1 1ωL/R 2 = 44.43°.
a. The extinction angle can be determined from the solution of Eq. (11.9) and an iterative solution yields β = 220.35°. The conduction angle is δ = β - α = 220.35 90 = 130.35°.
b. From Eq. (11.11), the rms output voltage is Vo = 68.09 V.
c. Numerical integration of Eq. (11.12) between the limits ωt = α to β gives the rms
thyristor current as IR = 15.07 A.
d. From Eq. (11.13), Io = 12 * 15.07 = 21.3 A.
e. Numerical integration of Eq. (11.14) yields the average thyristor current as
IA = 8.23 A.
f. The output power Po = 21.32 * 2.5 = 1134.2 W, and the input VA rating is VA =
120 * 21.3 = 2556 W; therefore,
PF =
Po
1134.200
=
= 0.444 1lagging2
VA
2556
Note: The switching action of thyristors makes the equations for currents nonlinear. A numerical method of solution for the thyristor conduction angle and currents
is more efficient than classical techniques. A computer program is used to solve this
example. Students are encouraged to verify the results of this example and to appreciate the usefulness of a numerical solution, especially in solving nonlinear equations of
thyristor circuits.
Key Points of Section 11.4
• An inductive load extends the load current beyond π. The load current can be
continuous if the delay angle α is less than the impedance angle θ.
• For α 7 θ, which is usually the case, the load current is discontinuous. Thus, the
control range is θ … α … π.
11.5Three-Phase Full-Wave Controllers
The unidirectional controllers, which contain dc input current and higher harmonic
content due to the asymmetric nature of the output voltage waveform, are not normally used in ac motor drives; a three-phase bidirectional control is commonly used.
M11_RASH9088_04_PIE_C11.indd 587
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