Myboleiytr

4.4.1 SPWM
• Natural sampling
– Amplitudes of the triangular wave (carrier) and sine wave (modulating) are compared to obtain PWM waveform
+1 M1

Modulating Waveform

Carrier waveform

0

−1

Vdc 2
0
t0 t1 t2

t 3 t 4 t5

−

Vdc 2

Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002

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SPWM (2)
– Implementation example
Analog comparator chip thatcompares the 2 waveforms

Generation of the carrier signal

Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002

2

SPWM (3)
Generation of the modulating signal

Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002

3

SPWM (4)
• Regular sampling
– Asymmetric and symmetric
T +1
sample point

M1 sin ω mt

T 4

3T 4

5T 4

π 4

t

−1

Vdc 2asymmetric sampling

t0

t1

t2

t3

t
symmetric sampling

V − dc 2
Generating of PWM waveform regular sampling

Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002

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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". Theharmonics 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...)
Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002 5

SPWM (6)
• Bipolar switching
– Pulse width relationships
∆

δ=

∆ 4

modulating waveform

carrier waveform

π

2π

kth pulse

δ 1k

π
δ 2k

2π

αk

Power Electronics andDrives (Version 2): Dr. Zainal Salam, 2002

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SPWM (7)
– Characterisation of PWM pulses for bipolar switching
∆

+ VS 2

δ0

δ0

δ0 δ 2k

δ0

δ1k

V − S 2

αk

Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002

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SPWM (8)
– Determination of switching angles for kth PWM pulse
AS2 AS1
v Vmsin( θ )

+ Vdc 2

Ap1

Ap2

V − dc 2

Equating thevolt - second, As1 = Ap1 As 2 = Ap 2
Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002 8

SPWM (9)
The average voltage during each half cycle of the PWM pulse is given as :  V  δ − ( 2δ o − δ1k )   V1k =  dc  1k   2δ o  2    Vdc  δ1k − δ o   Vs   = β1k   =   δ   2  2  o where β1k Similarly,  δ1k − δ o    = δo   

 δ 2k − δ o   Vdc  V2k = β 2k   ; where β 2 k =   δ   2    o The volt - second supplied by the sinusoid, As1 =
αk α k −2δ o

∫ Vm sin θdθ = Vm [cos(α k − 2δ o ) − cos α k ]

= 2Vm sin δ o sin(α k − δ o )
Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002 9

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 V β1k  dc 2δ o = 2δ oVm sin(α k − δ o )    2  Vm ⇒ β1k = sin(α k − δ o ) (Vdc 2)

Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002

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SPWM (12)
The voltage ratio, Vm MI = is knownas 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, δ1k − δ o β1k = 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 )]
Power Electronics and Drives (Version 2): Dr. Zainal Salam, 2002...