4.4.1 SPWM
• Natural sampling
– Amplitudes of the triangular wave
(carrier) and sine wave
(modulating) are compared to
obtain PWM waveform
Modulating Waveform
+1
M1
Carrier waveform
0
−1
Vdc
2
0
−
t0 t1 t2
t 3 t 4 t5
Vdc
2
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SPWM (2)
– Implementation example
Analog comparator chip that
compares the 2 waveforms
Generation of the carrier signal
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SPWM (3)
Generation of the modulating signal
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SPWM (4)
• Regular sampling
– Asymmetric and symmetric
T
+1
M1 sin ω mt
sample
point
3T
4
T
4
5T
4
π
4
t
−1
Vdc
2
asymmetric
sampling
t0
t1
t2
t3
t
symmetric
sampling
V
− dc
2
Generating of PWM waveform regular sampling
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SPWM (5)
MODULATION INDEX = M I :
Amplitude of the modulating waveform
MI =
Amplitude of the carrier waveform
M I is related to the fundamental (sine wave)
output voltage magnitude. If M Iis high, then
the sine wave output is high and vice versa.
If 0 < M I < 1, the linear relationship holds :
V1 = M I Vin
where V1, Vin are fundamental of the output
voltage and input (DC) voltage, respectively.
−−−−−−−−−−−−−−−−−−−−−−−−−−−−
MODULATION RATIO = M R (= p )
MR = p =
Frequency of the carrier waveform
Frequency of the modulating waveform
M R is related to the " harmonic frequency". The
harmonics are normally located at :
f = kM R ( f m )
where f m is the frequency of the modulating signal
and k is an integer (1,2,3...)
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SPWM (6)
• Bipolar switching
– Pulse width relationships
∆
δ=
∆
4
modulating
waveform
carrier
waveform
π
2π
π
2π
kth
pulse
δ 1k
δ 2k
αk
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SPWM (7)
– Characterisation of PWM pulses
for bipolar switching
∆
+ VS
2
δ0
δ0
δ1k
V
− S
2
δ0
δ0
δ 2k
αk
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SPWM (8)
– Determination of switching angles
for kth PWM pulse
AS2
v
AS1
Vmsin( θ )
+ Vdc
2
Ap2
Ap1
V
− dc
2
Equating the volt - second,
As1 = Ap1
As 2 = Ap 2
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SPWM (9)
The average voltage during each half cycle
of the PWM pulse is given as :
 Vdc  δ1k − ( 2δ o − δ1k ) 

V1k = 

2δ o
 2 

 Vdc  δ1k − δ o 
 Vs 
 = β1k  
=

 2  δ o 
2
where β1k
 δ1k − δ o 

= 
 δo 
Similarly,
 δ 2k − δ o 
 Vdc 

V2k = β 2k 
 ; where β 2 k = 
 2 
 δo 
The volt - second supplied by the sinusoid,
As1 =
αk
∫ Vm sin θdθ = Vm [cos(α k − 2δ o ) − cos α k ]
α k −2δ o
= 2Vm sin δ o sin(α k − δ o )
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SPWM (11)
Since,
sin δ o → δ o for small δ o ,
As1 = 2δ oVm sin(α k − δ o )
Similarly,
As 2 = 2δ oVm sin(α k + δ o )
The volt - seconds of the PWM waveforms,
 Vdc 
 Vdc 
Ap1 = β1k 
Ap 2 = β 21k 
2δ o
2δ o ;
 2 
 2 
To derive the modulation strategy,
Ap1 = As1; Ap 2 = As 2
Hence, for the leading edge
 Vdc 
β1k 
2δ o = 2δ oVm sin(α k − δ o )
 2 
Vm
⇒ β1k =
sin(α k − δ o )
(Vdc 2)
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SPWM (12)
The voltage ratio,
Vm
MI =
is known as modulation
(Vdc 2 )
index or depth. It varies from 0 to 1.
Thus,
β1k = M I sin(α k − δ o )
Using similar method, the trailing edge
can be derived :
β 2 k = M I sin(α k − δ o )
Substituting to solve for the pulse - width,
δ −δo
β1k = 1k
δo
⇒ δ1k = δ o [1 + M I sin(α k − δ o )]
and
δ 2 k = δ o [1 + M I sin(α k + δ o )]
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SPWM (13)
Thus the switching angles of the kth pulse is :
Leading edge : α k − δ1k
Trailing edge : α k + δ1k
The above equation is valid for Asymmetric
Modulation, i.eδ1k and δ 2k are different.
For Symmetric Modulation, δ1k = δ 2k = δ k
⇒ δ k = δ o [1 + M I sin α k ]
– Example
For the PWM waveform shown,
calculate the switching angles for all the
pulses.
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SPWM (14)
carrier
waveform
2V
1.5V
π
2π
modulating
waveform
1
t1
t2
2
3
t3 t4 t5 t6
4
5
6
7
8
9
π
t13
t15
t17
t7 t8 t9 t10 t11 t12
t14
t16 t18 2π
α1
– Harmonics of bipolar PWM
waveform
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SPWM (15)
Assuming the PWM waveform is half - wave
symmetry, harmonic content of each (kth)
PWM pulse can be computed as :
1T

bnk = 2 ∫ f (v) sin nθdθ 
π

 0

α k −δ1k


2
 Vdc 
=  ∫ −
 sin nθdθ 
π α −2δ  2 

 k o
α +δ

2  k 2 k  Vdc 
+  ∫ 
 sin nθdθ 
π α −δ  2 

 k 1k
α + 2δ

2  k o  Vdc 
+  ∫ −
 sin nθdθ 
π α +δ  2 


k
2k
Which can be reduced to :
Vdc
{cos n(α k − 2δ o ) − cos n(α k − δ1k )
bnk = −
nπ
+ cos n(α k + δ 2 k ) − cos n(α k − δ1k )
+ cos n(α k + δ 2 k ) − cos n(α k + 2δ o )}
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SPWM (16)
Yeilding,
2V
bnk = dc [cos n(α k − δ1k ) − cos n(α k − 21k )
nπ
+ 2 cos nα k cos n 2δ o ]
This equation cannot be simplified
productively.The Fourier coefficent for the
PWM waveform isthe sum of bnk for the p
pulses over one period, i.e. :
p
bn = ∑ bnk
k =1
The slide on the next page shows the
computation of this equation.
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SPWM (17)
– Harmonics spectra
M = 0.2
Amplitude
M = 0.4
1.0
M = 0.6
0.8
0.6
M = 0.8
0.4
Depth of
Modulation
0.2
M = 1.0
0
p
2p
3p
4p
Fundamental
NORMALISED HARMONIC AMPLITUDES FOR
SINUSOIDAL PULSE-WITDH MODULATION
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SPWM (18)
– Spectra observations
Amplitude of fundamental
decreases/increases linearly in
proportion to the depth of modulation
(modulation index). Relationship given
as:
V1= MIVin
Harmonics appear in “clusters” with
main components at frequencies of :
f = kp (fm)
k=1,2,3....
where fm : frequency of the modulation
signal
“Side-bands” exist around main
harmonic frequencies
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SPWM (19)
Amplitude of the harmonics changes
with MI. Its incidence (location on
spectra) does not
When p>10, or so, the harmonics can be
normalised (as shown in Figure). For
lower values of p, the side-bands clusters
overlap, and the normalised results no
longer apply
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SPWM (20)
– Normalized Fourier coefficients
0.2
0.4
0.6
0.8
1.0
1
0.2
0.4
0.6
0.8
1.0
MR
1.242
1.15
1.006
0.818
0.601
MR +2
0.016
0.061
0.131
0.220
0.318
h
MI
MR +4
2MR +1
0.018
0.190
2MR +3
0.326
0.370
0.314
0.181
0.024
0.071
0.139
0.212
0.013
0.033
2MR +5
3MR
0.335
0.123
0.083
0.171
0.113
3MR +2
0.044
0.139
0.203
0.716
0.062
0.012
0.047
0.104
0.157
0.016
0.044
3MR +4
3MR +6
4MR +1
0.163
0.157
0.008
0.105
0.068
4MR +3
0.012
0.070
0.132
0.115
0.009
0.034
0.084
0.017
0.119
0.050
4MR+5
4MR +7
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SPWM (21)
– Example
Note : for full bridge single - phase bipolar PWM,
vo = vRR , = vRG − vR 'G
= 2vRG
The harmonics are computed from :

(VˆRG )n  VDC

 2 
as a function of M I
Example :
In the full - bridge single phase PWM inverter,
VDC = 100V, M I = 0.8, M R = 39. The
fundamentalfrequency is 47Hz. Calculate
the values of the fundamental - frequency
voltage and some of the dominant harmonics.
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SPWM (22)
– Three-phase inverters
Effect of odd triplens
For three-phase inverters, there is
significant advantage if p is chosen to
be:
odd and multiple of three (triplens)
(e.g. 3,9,15,21, 27..)
With odd p, the line voltage shape looks
more “sinusoidal”
Even harmonics are absent in the phase
voltage (pole switching waveform) for p
odd
Spectra observations
The absence of harmonics no. 21 & 63
in the inverter line voltage due to p as a
multiple of three
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SPWM (23)
Overall, spectra of the line voltage is
more “clean” (lower THD, line voltage
is more sinusoidal)
More concern with the line voltage
Phase voltage amplitude is 0.8
(normalised) for modulation index =0.8
Line voltage amplitude is square root
three of phase voltage due to the threephase relationship
Waveform
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SPWM (24)
π
Vdc
2
−
−
2π
V RG
Vdc
2
Vdc
2
VYG
Vdc
2
Vdc
VRY
− Vdc
p = 8, M = 0.6
Vdc
2
−
−
V RG
Vdc
2
Vdc
2
VYG
Vdc
2
Vdc
VRY
− Vdc
p = 9, M = 0.6
ILLUSTRATION OF BENEFITS OF USING A FREQUENCY RATIO
THAT IS A MULTIPLE OF THREE IN A THREE PHASE INVERTER
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SPWM (25)
Harmonics
Amplitude
1.8
0.8 3 (Line to line voltage)
1.6
1.4
1.2
1.0
0.8
0.6
B
0.4
19
37
23
41
43
47
59
61
65
67
79
83
0.2
85
89
0
21
19
A
63
23
37
39
41
43
45
47 57
59
61
81
65
79
67
69 77
83
85
87
89
91 Harmonic Order
Fundamental
COMPARISON OF INVERTER PHASE VOLTAGE (A) & INVERTER LINE VOLTAGE
(B) HARMONIC (P=21, M=0.8)
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SPWM (26)
− Overview
It is desirable to push p to as large as
possible. When p is high, the harmonics
will be at higher frequencies based on : f
= kp(fm), where fm is the frequency of the
modulating signal
Although the voltage THD improvement
is not significant, but the current THD
will improve greatly because the load
normally has some current filtering effect
If a low pass filter is to be fitted at the
inverter output to improve voltage THD,
higher harmonic frequencies is desirable
because it makes smaller filter
component.
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SPWM (27)
− Example
The amplitudes os the pole switching
waveform harmonics of the red phase of a
three-phase inverter is shown in the
following Table. The inverter uses a
symmetric regular sampling PWM scheme.
The carrier frequency is 1050Hz and the
modulating frequency is 50Hz. The
modulation index is 0.8. Calculate the
harmonic amplitudes of the line-to-voltage
(i.e. red to blue phase) and complete the
Table.
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SPWM (28)
Harmonic
number
1
Amplitude (pole switching
waveform)
1
19
0.3
21
0.8
23
0.3
37
0.1
39
0.2
41
0.25
43
0.25
45
0.2
47
0.1
57
0.05
59
0.1
61
0.15
63
0.2
65
0.15
67
0.1
69
0.05
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Amplitude (line-to
line voltage)
27
SPWM (29)
• Unipolar switching
– 2 pair of switches operating at carrier
frequency
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SPWM (30)
– Frequency spectrum, MI = 1
– Normalized Fourier coefficients
(Vn/VDC)
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SPWM (31)
– 2 pair of switches operating at carrier
frequency, other pair at reference
frequency
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