starting system , LE THANH PHUC

University of Technical Education
Ho Chi Minh City

STARTING SYSTEM

Le Thanh Phuc
Faculty of Automotive Engineering
Email:
1


Introduction
The starting system is a combination of
mechanical and electrical parts that work
together to start the engine.
The starting system is designed to change
the electrical energy, which is being stored in
the battery, into mechanical energy.
To accomplish this conversion, a starter or
cranking motor is used.

2


Direct-Current Motor Principles
Magnetic lines of force flow from the
north pole to the south pole.

3

Direct-Current Motor Principles
Interaction of two magnetic fields

4


Direct-Current Motor Principles

Conductor
movement in a
magnetic field.

5


Direct-Current Motor Principles
Simple electromagnetic motor

6


Direct-Current Motor Principles
Field coil wound
around a
pole shoe.

Rotation of the
conductor is in the
direction of the
weaker field.

7


Direct-Current Motor Principles
Starter armature

Starter and
solenoid
components

8


Direct-Current Motor Principles
Lamination construction
of a typical motor
armature.

Lap winding diagram

9


Direct-Current Motor Principles
Wave-wound armature.

10


Direct-Current Motor Principles

Field coils mounted to
the inside of starter
housing

Magnetic fields in a 4-pole
starter motor.

11


Direct-Current Motor Principles

12


Direct-Current Motor Principles

13


Direct-Current Motor Principles

14


Direct-Current Motor Principles
One-way clutch

15

Direct-Current Motor Principles
Starting circuit

16


Direct-Current Motor Principles
Starting circuit

17


Characteristic of the starter motor


Speed characteristic of the starter motor: n = f (I)

e = B.l.v
B.l.π .n.D
e=
60

P.n
e = B.l.τ .
30

P.n
e = Φ.
30

ω .D
V =
2
π.D
τ=
2P

π .n
ω=
30


N
NP
Eng =
.e =
Φ. n
2a
a.60

Eng = Ce . n . Φ
n=

E ng
C e .Φ



Eng = Eo − ( ∆U ch + IRaq + IRd + IRkd )

∆U = I .rch
Eng

Eo − ∆U ch − I ∑ R
n=
=
Ceφ
Ceφ


n

Eng = Eo − ( ∆U ch + IRaq + IRd + IRkd )

Eo
I.Rd I.Ra
∆Uch

I.Rkd
Eng
Io

Inm/2

Inm


Eng

Eo − ∆U ch − I ∑ R
n=
=
Ceφ
Ceφ

n
no

φ
Eo

n
Io

Inm/2

Inm


Small load:

φ ≅ KφI

E0 − ∆U ch − I ∑ R
n≈
Ce .Kφ .I

a1
n=

I − a2

E 0 − ∆U ch
a1 =
C e .K φ
∑R
a2 =
C e .K φ