0195136047.pdf

764 BASIC CONTROL SYSTEMS

Ia=

( 2 Vm/π )cosα−Kmωm

Ra

(16.1.38)

Substituting in Equation (16.1.32), we get the following equation after rearranging terms:

ωm=

2 Vm

πKm

cosα−

Ra

Km^2

Ta (16.1.39)

which is the relationship between speed and torque under steady state. The ideal no-load speed

ωm 0 is obtained whenIais equal to zero,

ωm 0 =

Vm

Km

, 0 ≤α≤

π

2

(16.1.40)

ωm 0 =

Vm sinα

Km

,

π

2

≤α≤π (16.1.41)

For torques less than the rated value, a low-power drive operates predominantly in thediscontinu-

ous conductionmode, for which a zero armature-current interval exists besides the duty interval.

With continuous conduction, as seen from Equation (16.1.39), the speed–torque characteristics

are parallel straight lines whose slope depends onRa, the armature circuit resistance. A filter

inductor is sometimes included to reduce the discontinuous conduction zone, although such an

addition will lead to an increase in losses, armature circuit time constant, noise, cost, weight, and

volume of the drive.

EXAMPLE 16.1.2

Consider a 3-hp, 220-V, 1800-r/min separately excited dc motor controlled by a single-phase fully

controlled rectifier with an ac source voltage of 230 V at 60 Hz. Assume that the full-load efficiency

of the motor is 88%, and enough filter inductance is added to ensure continuous conduction for

any torque greater than 25% of rated torque. The armature circuit resistance is 1.5.

(a) Determine the value of the firing angle to obtain rated torque at 1200 r/min.

(b) Compute the firing angle for the rated braking torque at−1800 r/min.

(c) With an armature circuit inductance of 30 mH, calculate the motor torque forα=60°

and 500 r/min, assuming that the motor operates in the continuous conduction mode.

(d) Find the firing angle corresponding to a torque of 35 N·m and a speed of 480 r/min,

assuming continuous conduction.

Solution

(a)Vm=

√

2 × 230 = 325 .27 V

Ia=

3 × 746

0. 88 × 220

= 11 .56 A

E= 220 −( 11. 56 × 1. 5 )= 220 .66 V at rated speed of 1800 r/min

ωm=

1800 × 2 π

60

= 188 .57 rad/s