Half-Wave Rectifier Feeding a DC Motor

7/30/2019 Half-Wave Rectifier Feeding a DC Motor

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PE424UCR/HWM Dr. Guru Feb. 06, 2006

Half-wave rectifier feeding a DC Motor

This section, let us determine the operation of a dc motor fed by a half-wave rectifier. The

motor has been operating for a long time so that it has obtained its stead-state. Let us assumethat in its steady-state operation, the motor's speed is nearly constant. When the flux createdby the field winding is held constant or it is a permanent-magnet motor, the induced emf (orback emf) in the motor is constant. The equivalent circuit of the motor operating from a 120-V(rms), 60 Hz supply is given below. The diode is assumed as ideal.

Motor's parameters: R 4 Ω. L 10 mH. Ea 50 V.

AC Source: v s t( ) Vm sin ωt( ). V.

where Vm 170 V. and ω 377 rad/s

Motor's impedance: z R j ω. L.

Z z Z 5.497 Ω=

φ arg z( ) φ 0.756 rad= or φ 43.304 deg=

The diode begins conduction only when the the input voltage goes above the back emf ofthe motor. The conduction begins when

α asinEa

Vmα 0.299 rad= or α 17.105 deg=

When D is on, the differential equation is

Ldi t( )

dt. R i t( ). Ea Vm sin ωt( ).

1




Expected solution: i t( )Vm

Zsin ωt φ( ). Ea

RK e

ωt

tanφ.

where tanφ ωL

R. tanφ 0.943=

Vm

Z30.928 A=

Ea

R12.5 A=

The constant of integration K is determined by applying the initial condition:

i α( ) 0 => K e

α

tanφ Ea

R

Vm

Zsin α φ( ).. K 35.902 A=

The extinction angle is obtained by setting the current to zero. To do so, let us first guess itsvalue as

β π

Using the root function, we obtain the extinction angle as

β rootVm

Zsin β φ( ).

Ea

RK e

β

tanφ. β,

β 3.512 rad= β 201.206 deg=

Let us sketch the waveforms: ωt 0 0.01, 2 π...

Input (source) voltage: v s ωt( ) Vm sin ωt( ). V.

Output voltage: v o ωt( ) if α ωt β v s ωt( ), Ea,

Current in the motor: i ωt( )Vm

Zsin ωt φ( ). Ea

RK e

ωt

tanφ.

2




Current through the diode: i D ωt( ) if α ωt β i ωt( ), 0,( )

v o ωt( )

v s ωt( )

ωt180

π

.

0 20 40 60 80 100120140160180200220240260280300320340360200

150

100

50

0

50

100

150200

i D ωt( )

ωt180

π

.

0 20 40 60 80 100120140160 180200220240260280300320340 3600

369

12151821242730

Theoretical analysis: α 17.105 deg= β 201.206 deg=

α 0.299 rad= β 3.512 rad=

DC current through motor: I odc1

2 π. α

β

ai a( ) d. I odc 6.379 A=

DC Power supplied to the motor: P odc I odc Ea. P odc 318.928 W=

3




Effective current through motor: I orms1

2 π.

α

β

ai a( )2

d. I orms 10.339 A=

Average power dissipated by R: P R I orms2

R. P R 427.599 W=

Total power supplied to the motor: P oT P R P odc P oT 746.527 W=

Apparent power input:

S input

Vm

2I orms

.S input 1.243 kVA=

Power factor: pf P oT

S input

pf 0.601=

rad 1 W 1 VA 1 A 1 V 1 Ω 1 mH 0.001 kVA 1000

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Half-Wave Rectifier Feeding a DC Motor