Physics | Questions and Answers

QUESTION 6.
a) The arrangement of a closed loop DC motor speed control system with inner
current loop control.
b) IGBT based four-quadrant DC motor drive is built with a PM dc motor having an
armature resistance of 0.8 ohms, armature inductance of 2mH, and a torque-
constant of 1.2 Nm/A. Input DC voltage to the drive is 250 V. Friction torque can be
ignored. The drive is controlled in unipolar voltage mode at a switching frequency
of 10 kHz. Calculating,
I. Sketching the arrangement of the power circuit of the drive system.

II. Determining the switching duty-factor when the motor is running at 1200
rpm in steady state, against a shaft torque of 45 Nm.
Torque of the motor is given by the relation;
T =Kt Ia
Thus, armature current is derived as shown below.
I a= T
Kt
I a= 45 Nm
1.2 Nm / A =37.5 A
Armature voltage drop
Ea =I a Ra
Substituting the values;
Ea =37.5 A × 0.8 Ω
Ea =30 V
Switching duty factor is calculated as shown below.
Ea = ( 2 D−1 ) V s

Where V s −¿is the source voltage.
Rearranging the equation algebraically to make D the subject;
Ea =2 D V s−V s
2 DV s=Ea +V s
Therefore, switching duty factor is
D= Ea+V s
2 V s
=30+ 250V
2 ×250 V
D=0.56
III. Finding the magnitude of the peak-peak ripple in the armature current
waveform.
One switching cycle duration is given by;
T = 1
f s
Where f s−¿is the switching frequency.
T = 1
10× 103
T =1 00 μ s
Thus, closing and opening duration for G1 is calculated as;
Closing;
T close=T × D
T close=100 μs ×0.56
T close=56 μs
Opening;
T open=100−56 μs
T open=44 μs
When G1 is closed, the potential difference across the armature inductance is;
V ind=V s−Ea

V ind=250−30 V
V ind=220 V
The volt-second accumulated by the inductance during ON state is;
V ( T ) =V ind × Tclose
V ( T ) =220 V × 56 ×10−6 ms
V ( T ) =12.32× 10−3 Vs
Ripple current is calculated as;
∆ I= V ( T )
La
Where La−¿is the armature inductance.
∆ I= 12.32× 10−3 Vs
2 ×10−3 H
Thus, ripple current is;
∆ I=6.16 A
IV. Finding the efficiency of the motor drive system.
Rated output power of the motor is given by;
Prated =V s I a
Substituting the values;
Prated =250 V ×37.5 A
Prated=9375 W
Power losses in the armature resistance;
Ploss=Ra I a
2
Substituting the values;
Ploss=0.8Ω× 37.52
Ploss=1125 W
Efficiency of the machine;
η= Prated−Ploss
Prated
×100 %
Replacing with the values;