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Section 2

The document outlines the operation and characteristics of a six-step inverter, detailing the switching actions and current conduction for each switch. It provides calculations for output voltage, current, power, and switch ratings, including Peak Inverse Voltage and average current. Additionally, it discusses the Fourier expression for voltage and current harmonics, along with Total Harmonic Distortion (THD) calculations for both voltage and current.

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moaazh203
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12 views4 pages

Section 2

The document outlines the operation and characteristics of a six-step inverter, detailing the switching actions and current conduction for each switch. It provides calculations for output voltage, current, power, and switch ratings, including Peak Inverse Voltage and average current. Additionally, it discusses the Fourier expression for voltage and current harmonics, along with Total Harmonic Distortion (THD) calculations for both voltage and current.

Uploaded by

moaazh203
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Section 2

Six step inverter → switching action occurs every 60°.


180° Conduction → each switch conduct for 180°.
Three switches (not on the same leg) are “on” at the same time, so the input
3
resistance with respect to supply is always R.
2

Vdc 2Vdc
Idc = = (supply/ input current)
3R⁄2 3R
Output:

2π
1 2
▪ VL = √ ∫03 vdc
2
dωt = √ vdc
π 3

VL √2
▪ Vph = = Vdc
√3 3

Vph √2
▪ IL = Iph = = vdc (star connected load)
R 3R

2 2V2dc
▪ Po = 3 Iph R=
3R
Switches ratings:

It works the same way for the other switches: when the switch is on, the current from
the connected phase flows through it and is positive. When it is off, the voltage across
it matches the supply voltage.

▪ Peak Inverse Voltage (PIV) = Vdc

Iph Vdc
▪ ISW,(rms) = =
√2 3R

1 π vdc 2 vdc vdc 2 vdc


▪ ISW,(avg) = ( + + )=
2π 6 3R 3R 3R 9R

vdc vdc 2 vdc


▪ ISW,(max) = Idc = = 3 =
Rin R 3R
2
Fourier expression:
Voltage:
4 vdc π π
▪ Vab = ∑∞
n ( cos (n )) sin (n(wt + ))
nπ 6 6

n=1,5,7,11 …etc.

Odd non triple-n harmonics. Then the LOH is 5th.

4 vdc π
▪ Van = ∑∞
n ( cos (n )) sin(nwt)
nπ√ 3 6

√V2L,rms −V21,rms
Vharmonics
▪ THD = =
V1,rms V1,rms

2 4 Vdc π √6
∵ VL.rms = √ Vdc , VL1,rms = ( cos ( )) = Vdc
3 π √ 2 6 π

2
2
VL.rms 2
− VL1,rms (√ vdc )2
3
∴ THD𝑣 = √ 2 =√ − 1 = 𝟑𝟏%
VL1,rms √6
( π vdc )2

Current
4 vdc π
i o = ∑∞
n ( cos (n )) sin(n(wt − θn ))
n π√3 Zn 6

where: - Zn = √R2 + (nωL)2 the load impedance at harmonic (n)

nωL
θn = tan−1 ( ).
R

√ ∑∞
n (Io n,rms )
2
THDi =
Io1,rms

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