Single Phase Inverter With AC Motor As Load
In this section, Single Phase Inverter With AC Motor As Load is discussed. At first, the output waveform is drawn and the output current relation is established by the superposition theorem seperately combining the current due to D.C source, internal e.m.f of motor and transient current due to R and L.
Summary
In this section, Single Phase Inverter With AC Motor As Load is discussed. At first, the output waveform is drawn and the output current relation is established by the superposition theorem seperately combining the current due to D.C source, internal e.m.f of motor and transient current due to R and L.
Things to Remember
1) Diagram of it is same as that of single phase inverter but the components of a.c motor is installed instead of that resistive load as shown in single phase inverter.
2) The output current relation is established by the superposition theorem seperately combining the current due to D.C source, internal e.m.f of motor and transient current due to R and L.
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Single Phase Inverter With AC Motor As Load
Single Phase Inverter With AC Motor As Load


Let us assume that the sinusoidal back emf of ac motor is in synchronism with inverter switching frequency and is purely sine wave without any harmonics.
Let inverter frequency fp = 1/tp Hz, eo = Eo sin(wt + εo) = internal e.m.f of motor where Фo = Angle of lead with respect to Vo .
For positive cycle of Vo ,
Vs2−eo−Roio−Lodidt=0
Solution of equation(i) gives equation for io for positive half cycle,
These equation can be easily be solved by superposition theorem.
Io= Io1 + Io2 + Io3
Where Io1 = Current due to D.C voltage
Io2 = Current due to eo
Io3 = Transient Current
1)A steady state part associated with driving function Vs/2 and is given by
Io1 = Vdc/(2 Ro)
2) Io2 = A steady part associated wth driving function Eo sin(wt + ε o) and is given by
Io2=Eo√RO+(ωL)2sin(ωt+εo−δ0)whereδ0isgivenbyδ0=tan−1ωLRO
3) Io3 = Transient part given by solution of following equationRoio+Lodidt=0Solutionisi03=I3e−t./TOwhereTo=LO/RO(timeconstant)anditdependsontheinitialconditionofthecircuiti03=I3e−θ/tanδo
The complete time domain equation is
IO = IO1 + I02 + I03
= I1 - I2 sin(wt + ε o- δO) + I3 e-t/T0 --------(i)
The value of I3 can be calculated at time t = 0 by assuming IO = Ia and substituting in the above equation
I3 = Ia - I1 + I2 sin( ε o- δO)
The value of I3 can be substituted on equation (i) and we get the value of Ib.
Solution for negative half cycle
IO = -I1 - I2 sin(wt + ε o- δO) + I4 e-t/T0
To calculate I4 at pi radian, Io = Ib
I4=Io+I1+I2Sin(π+εo−δo)e−θ/tanδo
Lesson
Inverter
Subject
Electrical Engineering
Grade
Engineering
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