DC Motors

Power Engineering Foundation DC Motors

By Fuad Latip 2006 1

DC Motors DC motors have a DC output only because a mechanism exists that converts the internal AC voltages to DC voltages at their terminals. The mechanism is known as commutator. The Voltage in a Rotating Loop If the rotor of this machine is rotated, a voltage will be induced in the wire loop. The total induced voltage on the loop eind is given by final form of the voltage equation below

φω

π

ω22

=

=

ind

ind

e

rlBe

The Induced Torque in the Rotating Loop Supposed battery is now connected to the DC machine as shown in figure 1.

Figure 1

After several have done on it, the resulting total induced torque on the loop is given by

i

thusrlBand

rilB

ind

ind

φπ

τ

πφτ

2

2

=

==



Therefore, the torque produced in the machine is the product of the flux in the machine and the current in the machine, times some quantity representing the mechanical construction of the machine (the percentage of the rotor covered by pole faces). In general, the torque in any real machine will depend on the same three factors:

1. The flux in the machine 2. The current in the machine 3. A constant representing the construction of the machine

The formula for circuit which connected to the motor is

IReV indB += Example 1 Figure 1 shows as simple rotating loop between curved pole faces connected to a battery and a resistor through a switch. The resistor shown models the total resistance of the battery and the wire in the machine. The physical dimensions and characteristics of this machine are

r = 0.5 m l = 1.0 m R = 0.3 Ω B = 0.25T VB = 120 V

1. What happens when the switch is closed? 2. What is the machine’s maximum starting current? What is its steady-state angular

velocity at no load? 3. Suppose a load is attached to the loop, and the resulting load torque is 10 N-m. What

would the new steady state speed be? How much power is supplied to the shaft of the machine? How much power is being supplied by the battery? Is this machine a motor or a generator?

4. Suppose the machine is again unloaded, and a torque of 7.5 N-m is applied to the shaft in the direction of rotation. What is the new steady state speed? Is this machine now a motor or generator?

5. Suppose of the machine is running unloaded. What would the final steady state speed of the rotor be if the flux density were reduced to 0.20T?

(Answer will be discussed in class)



POWER FLOW AND LOSSES IN DC MACHINES There is always some loss associated in DC machines. The efficiency of DC machine in percent is defined by the equation

%100xPP

in

out=η

The difference between the input power and the output power of a machine is the losses that occur inside it. Therefore

%100xP

PP

in

lossin −=η

a. The Losses in DC Machines The losses that occur in DC machines can be divided into five basic categories: 1. Electrical or copper losses (Armature loss IA

2RA + Field loss IF2RF).

2. Brush losses, Pbrush drop losses = (Vbrush drop)*(IA) 3. Core losses 4. Mechanical losses 5. Stray load losses b. The Power Flow Diagram

Pin = VTIL

IA2RA + IF

2RF losses

Pconv = EAIA = τindωm

Core losses Mechanical losses

Stray losses

Pout = τappωm



Types of DC Motors There are five major types of DC motors in general use: 1. The separately excited DC motor 2. The shunt DC motor 3. The permanent-magnet DC motor 4. The series DC motor 5. The compounded DC motor THE EQUIVALENT CIRCUIT OF A DC MOTOR Separately Excited and Shunt DC Motors



1. A separately excited DC motor is a motor whose field circuit is supplied from a separate constant-voltage power supply.

2. The Shunt DC motor is a motor whose field circuit gets its power directly across the armature terminals of the motor.

3. The equation for the armature circuit of these motors is

AAAT RIEV += The Terminal Characteristic of a Shunt DC Motor The terminal characteristic of shunt DC motor is a plot of its torque versus speed. How does a shunt DC motor respond to a load? Suppose that the load on the shaft of a shunt motor is increased.

1. When load ↓, then the τload > τind the motor start to slow down, ω↓. 2. When ω↓, the internal generated voltages drops (EA=kφω↓) the armature

current IA = (VT-EA↓)/RA increases. 3. When IA↑ induced torque τind=(KφIA)↑ 4. Finally the incduced torque will equal to the load torque at a lower mechanical

speed of rotation, ω. Derivation of output characteristic of a shunt DC motor can be done as follows: The KVL equation for a shunt motor is

AAAAT RIIEV += The induced voltage EA=Kφω, so

VT = Kφω + IARA Since, τind = KφIA, current IA can be expressed as

φτK

I indA =

Therefore

Aind

T RK

Vφ

τφω += K

( ) indAT

KRV

τφφ

ω 2K−=



This equation is just a straight line with a negative slope. The resulting torque-speed characteristic of a shunt DC motor is shown in figure 2.

Figure 2

Example 2 A 50-hp, 250V, 1200 r/min, DC shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06Ω. Its field circuit has a total resistance Radj + RF of 50Ω, which produces a no load speed of 1200 r/min. there are 1200 turns per pole on the shunt field winding as figure 3.

Figure 3

1. Find the speed of this motor when its input current is 100A. 2. Find the speed of this motor when its input current is 200A. 3. Find the speed of this motor when its input current is 300A. 4. Plot the torque-speed characteristic of this motor. (Answer will be discuss in class)

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DC Motors