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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

Single Phase Inverter With AC Motor As Load

Single Phase Inverter With AC Motor as Load
Single Phase Inverter With AC Motor as Load
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 ,

Vs2eoRoioLodidt=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=EoRO+(ωL)2sin(ωt+εoδ0)whereδ0isgivenbyδ0=tan1ωLRO

3) Io3 = Transient part given by solution of following equationRoio+Lodidt=0Solutionisi03=I3et./TOwhereTo=LO/RO(timeconstant)anditdependsontheinitialconditionofthecircuiti03=I3eθ/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+εoδo)eθ/tanδo

Lesson

Inverter

Subject

Electrical Engineering

Grade

Engineering

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