2011 Automobile Maintenance

Advanced Course for the

Industrial Technical
Instructors

Class No. : ~wp bh~ \~bh

Name: ----------
Instructor : _ _~C.;..;;...;;ha;;..;;..;;..;;n......
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 Unit 1: Fundamental instruments and measuring experiments in

electronic component characteristics

1. Tests and instruments using for common electronic components

• Diode test
(I) Set the multimeter on QxIO.
(2) As shown in Diagram A-19(a), the current may pass through the diode as it is forward bias. The
resistance is extremely low, and the needle should deflect significantly.
(3) Exchange the red and black probes as shown in Diagram A-19(b). Due to reverse bias, the current is
unable to pass through. The resistance is therefore very large, and the needle does not deflect.
(4) If the needle ofthe multimeter shows no deflects no matter how the red and blac!{ probes are
switched, this indicates that the diode has a broken circuit.
(5) If the needle of the multimeter deflects significantly regardless ofthe switching of the red and black
probes, this indicates that the diode has a short circuit.

t:lJ Xlf)
EC3
Xl0
+ - +

Black Red tf.! Black Red ;tr

r .. T' I ~ ~
p '" N • ~ ...
(a} (Il)

Diagram A-19 Test of

• LED test
(I) Set the multimeter on QxIO.
(2) As shown in Diagram A-19(a), the LED connection is forward biased. The needle significantly
deflects, and the LED brightens. If the LED does not illuminate, it is thus a defective product.
(3) Do not connect the LED as in Diagram A-19(b). The situation is a reverse bias. The needle will not
deflect, and the LED will not illuminate.
(4) The LED has two pins, one long and one short. The long one is the P-electrode, and the short one is
the N-electrode.

-1-
• Transistor test
5-1 Determining the polarity (NPN or PNP) with a multimeter
(1) Switch the multimeter function to Rx1k or Rxl00. Let the probe have contact with two of the three
pins. If t he m ultimeter need le deflects significa ntly, t hen one of the two pins is the base (B)
electrode.
(2) Move any test rod toward the third pin (the previously unconnected pin). If t he needle of the
multimeter still significantly deflects, thus the pin which was not tested is the base B. If the needle
deflects very little when the test rod is used on the third pin, the pin where the rod was removed is
the base electrode.
(3) In the above test, if the needle deflects significantly when the red probe was used, the transistor is
thus a PNP type; on the contrary, if the needle deflects significantly when the black probe was used,
transistor is an NPN type.

5-2 Determining the transistor electrodes (E, B, C) with a multimeter (emitter, base, collector)
The appearance and pin relations of the commonly seen transistor are illustrated in Diagram 1.3-9.
It would be naturally more convenient if the relations among the electrodes are known when testing a
transistor. If it is unknown, it is not difficult to determine the electrodes with a multimeter. The method
is described below:
(1) First determine the B electrode in accordance to the method in section 5-1.
(2) As electrode B is determined. The next step is to determine which is C and which is E of the
remaining two electrodes, as shown in Diagram 1.3-10.
(3) Using the NPN transistor as an example, switch the multimeter function to R x lk. Connect the
black rod (with a positive voltage output) to the assumed C electrode, and connect the red rod to the
assumed E electrode. Next, use your finger to hold (press) electrodes Band C, but do not let the two
have direct contact. The needle will have a bias response if the assumption is correct. If the ne~dle

shows no bias response as the finger holds (presses) electrodes Band C, thus your assumption is
opposite with the actual situation. So how is this explained? This can be clearly understood from
Diagram 1.3-11.
(4) If the testing is for a PNP transistor, the situation shall be the opposite of (3). The black rod is to be
connected to electrode B, and the red rod to the assumed electrode C. (see Diagram 1.3-12)

Diagram 1.3-10

-2-
(a) If the assumption is correct. The needle shall respond with a forward bias
with the finger's resistance via the transistor. The needle indicates a low
resistance.

Red :tr
'Vi,

\
\\"
... 'E-I
RXlk
.,
.."..

•."0 9'
~~~.-----.-/

(b) If the assumption is incorrect, the needle indicates to a high resistance.

Diagram 1.3-11 Transistor testing (NPN)

Finger's
resistanc
e

(a) If the assumption is correct. The needle shall respond with a forward bias
with the finger's resistance via the transistor. The needle indicates a low
resistance.

Red

(b) If the assumption is incorrect, the needle indicates to a high resistance.

Diagram 1.3-12 Transistor testing (PNP)

-3-
• Variable resistor
The variable resistor (VR) has a total of three terminals, as shown in Diagram D-4(a). When the wiper
rotates clockwise, the resistance between terminals :1)~]) shall increase, where the resistance between
terminals @@shall decrease. Also, the resistance between ::D::[> is a constant, which is the sum of the
resistances between terminals ~l.)(V and@®. The resistance printed on the outer case of a variable
resistor usually indicates the resistance between terminalsW@ .

Terminals

(a) Illustration (b) Electronic symbol

Diagram D-4

-4-
• Photo rcsistor

The illustration in Diagram D-9 shows a photoresistor, made from cadmium sulfide (CdS). The
photoresistor has the property of having a large resistance in the dark, but shall lower with light. With
increasing light intensity, the resistance shall become smaller. This is caused as the resistance of a
photoresistor varies to the intensity of light. Therefore, the device is commonly used in automatic control
circuits to detect the intensity of light.

(\ ~~.
.•
s::?,.'- . ji ellS

(b) Illustration
Y
(b) Electronic symbol

Diagram D-9 Photoresistor

-5-
 • Solderless plugboard (breadboard)

W B · 2N - r

Diagram 1-71 Photos of the solderless plugboard

Diagram 1-72 Application example

.........................,. , ' .'16 - -111..... . .....

.....................
~~..........,..... v~,......
-. .........
~ ...... ,. • ~: .....,.+--+
»t.+-~
•• ~
..... "............

11!!!lllllIllll!!111II111111l!!!lllllmlllllllllllll!Ill!I!!llillll!!!I
1111111111!!111!1111111111!11111!111111!~111111111111111111111111111111
_ -.~"' .~_. . . ., .. T , • . . . ... . . . .- . . . ,.~ • • > ~ -"., __ ~.-,. ...-_~...
....,. . . .~~_ ... ~ ......... -4! ,................ ~ • •~ ., ........ .. . . . - - - -• • • ~

Diagram 1-73 Internal structure

-6-
I -2 Wave test

I. Practice: Wave pattern test

II. Purpose: To observe various voltage wave patterns, and the testing of voltage amplitudes and periods
III. Instrument and materials: Multimeter, oscilloscope, function generator
IV. Basic knowledge
1. An AC wave pattern can be divided into sine and non-sine. For example, Diagram l(A) shows a
sine wave, where a non-sine wave is shown in Diagram 1 (B). Commonly seen non-sine waves
include square waves, pulse waves, triangle waves and sawtooth waves.

(A)
Sine wave

(B)
Non-sine wave

f . CJ~CJ
()v~.
I I r~
_ .LJ . LJL:J -
Square wave

Pulse wave

V '\ j1\61\1\
\,/ . \1:\7 \//
Triangle wave

Sawtooth wave

Diagram 1 Wave pattern for AC voltage

-7-
2. The number of cycles per unit time is called the frequency. The period is the duration of one cycle in a
repeating event. The unit of frequency is hertz (Hz). The two are related as follow:

Frequency (f) = l/period (T)

Period (T) = lIfrequency (f)

3. AC voltage: If the wave is sine, an oscilloscope can be used to measure the peak voltage, and the value
measured by a multimeter is the valid voltage (Vrms); for non-sine waves, the oscilloscope can detect the
voltage value between peaks. An example is shown in Diagram 2, where the peak to peak voltage for a
triangle wave is 24 V P_I'.

- 12'1

Diagram 2 Triangle wave

V. Procedure:
1. Square wave test
(1) Connect the function generator with power. Switch the function to square wave.
(2) Adjust the signal frequency to 1000 Hz
(3) Connect the oscilloscope with power. Connect the probe rod to CH-l.
(4) Connect the function generator output to the probe rod.
(5) Properly turn the VoltslDIV knob on the oscilloscope.
(6) Properly turn the TIMEIDIV knob on the oscilloscope.
(7) Record the displayed wave pattern on the monitor.

Square Sinusoidal Triangle

wave wave wave

Function Oscilloscope
generator

Diagram 3 Square wave test

-8-
(8) Draw the square wave
I
TlME/D IV",

T
F

2. Sine wave test

The procedure is same as the square wave test.

-9-
(8) Draw the square wave
~~.----~""'~'~~~---

TIMElDIV""
VOLT/Dr\,,,,, - --,-, ,
T -
F

2. Sine wave test

The procedure is same as the square wave test.

-9-
 Unit 2: Clipping circuit and its application experiments
Clipping circuit is a rectifier circuit waveform ofa diode, its function is to clip off the signal which lies
above or below some specific voltage levels, and is known as clipper. The basic cl ipping circuit, according to
series or parallel output of the d iode, and whether or not it is added with bias suppl y voltage, is divided into
four types:

(1) Series clipping circuit with no bias supply

(2) Parallel clipping circuit with no bias supply
(3) Series clipping circuit with bias supply
(4) Parallel clipping circuit with bias supply

- !

fa)

~.6T-:
~.
.

Diagram 2-18 Parallel clipping circuit with no bias supply

(1) Series clipping circuit with no bias supply

Diagram 2-18 shows a basic series clipping circuit with no bias supply. In Diagram 2-18(a), when input Vl
signal is at positive half-cycle, the diode is ON to become short circuit. As the entire
Vl signal is added on to resistor RL, so the output voltage Vo is equal to input voltage. If the input signal is at
negative half-cycle, then the diode will reverse bias to become open circuit. The circuit passes through RL is
closed to zero, and so the output terminal 's voltage is also closed to zero. In Diagram 2-18(b), the diode
direction is opposite to diode direction in Diagram 2-18(a), and its working principle is similar to the former
one but only that its output voltage is opposite.

(II) Parallel clipping circuit with no bias supply

Diagram 2-19 shows a basic parallel clipping circuit with no bias supply. In Diagram 2-19(a), when input
signal V l is at positive half-cycle, the diode will forward bias to become ON, the output voltage is equal to the
potential barrier of diode. At such, the signal voltage will cross over to resistor Rs. When signal Vl is at
negative half-cycle, the diode will reverse bias to become open circuit, and all the input voltage will fall on the
diode. Therefore, the output voltage Vo is equal to V1. In diagram 2-19(b), its working principle is similar to (a),
but as both diodes are in opposite direction, so the output signal is also opposite.
R,

v,

Diagram 2-19 Parallel clipping circuit with no bias supply

-10-
(Ill) Series clipping circuit with bias supply
Diagram 2-20 shows a basic series clipping circuit with bias supply. In Diagram 2-20(a), when input
signal V I is smaller than VE, the diode will reverse bias to become open circuit. As the current flowing through
RL is zero, so the output voltage Vo is equal to V E. When the input signal V I has increased to exceed VE, the
diode will forward bias to become short circuit, the output voltage is equal to input voltage. In this circuit, V E
works as a lower limit to restrict output.

In diagram 2-20(b), when input signal VI is smaller than bias voltage VE, the diode will forward bias to
become short circuit, and so the output voltage Vo is equal to Vl. When VI has increased to exceed VE, the
diode will reverse bias to become open circuit. Circuit I is equal to zero, and so output voltage Vo is equal to VE.
In this circuit, VE works as the upper limit to restrict output.

In Diagrams 2-20(c) and 2-20(d), their principles are similar to that in Diagrams 2-20(a) and 2-20(b),
respectively, the only difference is that the bias voltage Vo is negative, and so their restricted levels are precisely
opposite.

-.<
,=,,...----l!'~: ---'-l-"::
. tJ
r. t "''i

';:" ", ... ,o-,!o+-- -..,...----_

F.~.~ ~~
l
.:
,
~

-_.... :" ..,~,~ """'- -.

Diagram 2-20 Series clipping circuit with bias supply

-11-
(IV) Parallel clipping circuit with bias supply
Diagram 2-21 shows a basic parallel clipping circuit with bias supply. In Diagram 2-21 (a), when VI is
smaller than VE, the diode will reverse bias to become open circuit. Current I is equal to zero, and so VO=VI.
When VI has increased to exceed VE, the diode will forward bias to become short circuit, and so VO=VE. In this
circuit, VE works as an upper limit to restrict output.

In Diagram 2-21 (b), when VI is smaller than VE, the diode will forward bias to become short circuit, and
so output VO=VE. When VI is greater than VE, the diode will reverse bias to become open circuit. Current 1=0,
and so Vo=VI. In this circuit, VE works as the lower limit to restrict output.

R"
~-A,/V\,..'·"_·__ ',,_~

--", I
I ,. ~[:r:
--~
I (!~

(0;

p. r
~'-'~---8'
-'\
)

_ =___)
i -;
:rs~-
i-

Diagram 2-21 Parallel clipping circuit with bias supply

v.

J
"

Diagram 2-21 (continued)

In Diagrams 2-21 (c) and 2-21 (d), their principles are exactly similar to that in Diagrams 2-21 (a) and
2-21 (b), the only difference is that the bias voltage VE is negative, and so their restricted levels are in opposite
direction.

-12-
 The diode clipping circuit is able to clip the original sine wave input signal VI to become output square
wave as shown in Diagrams 2-22 and 2-23. In Diagram 2-22, VE2 is greater than VEl When VI is smaller than
VEl, the diode DI will forward bias to become short circuit, the diode D2 will reverse bias to become open
circuit. Therefore, output Vo is equal to VE L

p~
~·/'Y...-------(-

!
:; 1:
!
,') ,:!:

Diagram 2-22 Two level clipping circuit

.. ,
V·
·~"""''--''''''''''''''''''.r---r-i
• --~,---<.>
r.>
! I
... .%.
~i ;t.
"
I' \
,!
l ~\
\.
/
:.,<~ -=-- {.=:..!""---""~--.--- -::;-
-I
f

Diagram 2-23 Sine wave being clipped to become symmetrical square wave

When VI is greater than VEl and smaller than VE2, the diodes 01 and 02 will reverse bias to become open
circuit, and so VO=VI. When VI is greater than VE2, the diode 01 will reverse bias to become open circuit, the
diode D2 will forward bias to become short circuit, and so VO=VE2, the output waveform of which is shown in
Diagram 2-22. Ifwe adjust VEl and VE2 in equal value but in opposite pole, then we would get a symmetrical
square wave output voltage as shown in Diagram 2-23 .

Besides capable of allowing ordinary diode to generate square wave clipping circuit, we can also use Zener
diode to achieve this clipping circuit as shown in Diagram 2-24. In Diagram 2-24(a), when V I is at positive
half-cycle, and VI is smaller than the collapse Zener diode voltage Vz, the diode will reverse bias to become
open circuit, and so Vo=VI. When VI is greater Vz, the diode will reach the collapse point, and the voltage at
both terminals of the diode will maintain at voltage Vz. When input VI is at negative half-cycle, the diode will
forward and reverse bias to become short circuit. Therefore, Vo=O.

",f
i
!

':- .._...._-_... _,- --

Diagram 2-24 Clipping circuitformed by Zener diode

-13-
 _.,
~

-~-"'7"'·'·'''---'T~-'='''''';<

..
}~t
!
>':,~ ~ - ~ - --
'L_____'~___---"T""l:
,-_-}~_\.~- -___-,..-
- .~/:-,--
! ---~-
\ i ----
\- - /-
Diagram 2-24 (continued)

In diagram 2-24(b), the two Zener diodes are connected together. When input signal V1 is at positive
half-cycle and is small than VZ2, the diode 02 will reverse bias and be cut-off, and so Vo is equal to V1. When
V1 is greater than VZ2, the diode 02 will reach the collapse point, 01 will forward bias to become ON, and so Vo
is equal to VZ2. When V1 is at negative half-cycle and is smaller than VZ1, the diode will reverse bias and be
cut-off, and so Vo is equal to V1. When V1 is greater than VZ1, the diode 01 will reach the collapse point, 02
will forward bias to become ON, and so Vo is equal to VZ1.

[ Experiments]
I. Series clipping circuit with no bias supply

Step:
(1) Connect the circuit according to Diagram E2-3.1.
(2) Set the signal generator in 10Vp-p, 1kHz sine wave to assume an input signal of AC 1Ov.
(Note: Tfyou don't have a signal generator, you can also use a 110/1 0 transformer to get the sine wave)

CHl

CHZ
2() K

-Grounding clip

CHI
CH2

AC
Ii}V --- ZO K Oscil scope

Grounding clip

Diagram E2-3.1

-14-
 (3) Connect the oscilloscope according to the diagram, then input waveform (CH I) and output waveform
(CH2) so that they can all be clearly seen on the oscilloscopes. Watch out for the input waveform
section that has been clipped off.
(4) Draw out the output waveform.

2. Two level clipping circuit

Step:
(1) Connect the circuit according to Diagram E2-3.5.

2:OK CHl
/\e oscilloscope
lOV V;;, CHZ
+
Grounding clip
{b}

Diagram E2-3.5

(2) Set up VE1=VEF3Y.

(3) Observe and draw out its output waveform (Vo).

-15-
 Unit 3: Full-wave rectifier circuit and its application experiments
The differences between full-wave and half-wave circuits: Full-wave rectifier falls in the entire output
cycle, and allows unidirectional circuit to pass through the load, but half-wave rectifier only allows the output
cycle to fall at the positive half-cycle. The result of full-wave rectifier is that every output of half-cycle is
presented in DC output voltage pulse as shown in Diagram 2-7. The average output voltage value of full-wave
rectifier is twice that of half-wave as shown in the following equation:

f\Lj \ 1\ \,
o~"V~' ''·

Diagram 2-7 Full-wave rectifier

V. : peak voltage

(a) The input at positive half-cycle, DI and 02 will forward bias;

and 0 3 and 0 4 will reverse.bias

'-", JJ ~.~.
u~~ ;~TL: : :~l::::;:::::=:;:;~~~
(b) The input at negative half-cycle, 0 3 and 04 will forward bias;
and 01 and 02 will reverse bias

Diagram 2-10 Operations of bridge recti fier

Diagram 2-10 shows a full-wave bridge rectifier circuit that uses four diodes. In Diagram (a), when input
cycle is positive, the diodes 01 and 0 2 will forward bias, and 03 and 04 will reverse bias. Therefore, the circuit
will flow towards the direction of resistor RL as shown in the diagram . Its waveform is similar to positive
half-cycle of input. In Diagram (b), when the input is at negative half-cycle, the diodes 03 and D4 will forward
bias, and 01 and 02 will reverse bias. At such, the current that passes through RL direction is similar to the
circuit that passes through the RL direction at positive half-cycle. Therefore, it will form an output waveform
similar to positive half-cycle on both terminals of RL, and generate a full-wave rectifier output voltage on both
terminals of RL, the peak output voltage is similar to the voltage peak value of the secondary coil.

-16-
Rectifier filter circuit

The function of filter circuit is to filter the rectified DC pulse voltage into a more stab le DC voltage as
shown in diagram 2-11. It means that to connect the rectified output circuit to the output terminal of filter
circui t to obtai n a smooth and flat DC outp ut current. The fi lter circuit features a capacitor filter, R-C filter, L-C
filter and 7[ -type fi lter, etc. Here, we onl y discuss capacitor fil ter and R-C fil ter.

Diagram 2-11 Filtered wave supplied by power source

[ Experiments]
1. Connect the circuit according to Diagram 3-1

lit
t
V bet

,________l_k_a...: . ~

Diagram 3-1

2. Input signal VAC is 6V sine wave of frequency 1kHz.

3. Please use an oscilloscope to determine the waveform on both terminals of RL, and plot it in Diagram 3-26.

Oscilloscope
Vertical= V/cm
Horizontal=- - - - - ms/cm
Diagram 3-26

-17-
4. Again, connect the circuit according to Diagram 3-2.

Diagram 3-2

5. Please use an oscilloscope to determine the waveform on both terminals of RL, and plot it in Diagram 3-28.

I
I

I"

I
Oscilloscope
Vertical= V/cm
Horizontal=- -- - - ms/cm
Diagram 3-28

-18-
 Unit 4: Voltage division circuit and its application experiments

[Practice 4-1 ]
I . According to Diagram 2-3, connect the circuit on the PC board and supply l2V power to terminals A
and B.
Determine the voltages on terminals R I and R2 respectively, lower VI and V2 voltages and record the
results.
2. Check whether or not V1 +V2 is equivalent to Vs (12V)?
3. Use voltage division principle to calculate VI and V2 values, and compare them with actual detected
results.

V:R
-If I -.' ~'''~-'~='''~"~"~l] W;'' ----l
, •
,I a
Detected r
I \1,O\f ((023 Z ?Aj
I (L.O t
result I'
!
Calculated ,I?•
result \L I \0 i 2. ~
~ \ 2.-

Diagram 2-3 Voltage division

principle

[Practice 4-2]
I. Please determine the resistance of throttle position sensor (TPS):
(a) During idling speed (throttle door is closed), the TPS resistance is _ _ _ _ _ _ 0;
(b) When the throttle door is fully opened, the TPS resistance is ______ 0.
2. Please design and draw out a throttle detecting circuit.

3. Please determine your self-designed TPS circuit and determine the following agendas:
(a) During idling speed (throttle door is closed), the TPS output voltage is _ _ _ _ _ _V;
(b) When the throttle door is fully opened, the TPS output voltage is V_

-19-
 Unit 5: Automatic headlight circuit experiments

5-1 Darlington circuit

1. Darlington circuit
When a circuit requires a very high input resistance or a large circuit gain, a single transistor is difficult
to satisfy the requirement. At such, we can combine two transistors to form a Darlington circuit (known as
Darlington pair or Darling combination) in order to satisfY the circuit requirement.

1-1 The circuit gain of Darlington circuit

Darlington circuit is available in four combination types as shown in Diagram 10-1. In a circuit, it is
more appropriate to use silicon transistor. If we use a wrong transistor, it may cause big electric leakage and
abnormal temperature characteristic.

The input current IE of Darlington circuit is magnified first after passing through TRI and is again
magnified after passing through TRz (please refer to Diagram 10-2). Therefore, the total current gain is
roughly equal to the multiplication of the current gain of two transistors, i.e.

(10-1 )

I I~
,_. - ..- "~ ~4r:S 1" RJ'N 1~ I' !> ~

~ '.~. ~
--t:b( .. 'nt., eq~'") ' " ,.
. .. !
v-y
.... ,• . ::;, x
, ."
~.
., • .
~i

i
!t •i'•

L-.._ _ _ _ _ __ _ _ _ _ _ _ __ _ ." - . -- -- '

Diagram 10-1 Four combination types of Darlington circuit

-20-
 !
J.
1< '" 1.jOOf)

l<: 100(;0
p=- = ,~ -. _
. = l{HIO(
I. 1
=~ : )(f. ,

r.. !i:: 1•• + I.e.

= l00+l0 OflO
= 101G<O !~ 1000fr

Diagram 10-2

Diagram 10-3

The ~ value of a single transistor is normally ~ 300, but the /3 value of Darlington circuit can be easily
reached up to 1000 times.

For example: A Darlington circuit composed of transistors /3 =1 00, its total /3 =1 00 x 100 = 10000 times.

5-2 Transistor-driven relay

When the load of a control circuit is a high voltage or a big current, it is not easy to control it with a
transistor. So we would like to get help with a relay.

When a transistor is used in conjunction with a relay, as the loop circuit is simple and the cost is cheap, i
is claimed to be the best combination in automation control. In such application, it utilizes a small output
capacity from transistor to control the relay, and then uses the relay contact to open or close the load circuit.

5-3 Understanding relay

Relay is composed of a coil, armature, spring, silver contact, bracket and other components as shown in
Diagram 10-4. Its contact can be designed into various group numbers depending on different applications.

The common relay contact is indicated as COM (common). When the relay coil is in OFF state, its
conduction with COM is known as normally close (N.C.) contact or b contact. Normally, when it is not in

-21-
conduction with COM, it is known as normally open (N.O.) contact or a contact.

Diagram 10-4 shows a normal (the coil is OFF) state. When N.C. is in conduction with COM, then N.C.
and COM will become an open circuit. When current passes through the coi l, the coil wi ll generate a magnetic
force to attract the armature to allow N.O. and COM to conduct each other, and N.C. and COM wi ll become an
open circuit. At such, by contro lling only the conduction and non-conducti on of the coil, we are able to use the
relay contact to control circuit ON or OFF.

Bracket Coil
Diagram 10-4 Structural diagram of relay

~·~"C
~n~ .'.J.
~?
r " (.,

(a) Drawing method 1

d.
i _
~

(b) Drawing method 2

Diagram 10-5 Circuit symbols of relay

The circuit symbols of relay are shown in Diagram 10-5. Both drawings (a) and (b) are acceptable. The
diagrams show the examples of one contact, but a circuit can incorporate a relay with appropriate contact
groups depending on requirement.

The relay that is commonly used in transistor circuit comes in such rated voltages as DC6V, DCI2Y,
DC24V and DC48V, but DCI2V is widely used. Rated voltage refers to the actual action when the coil is added
up with this voltage and that it will not cause any damages to the coil after long period of conduction. Normally,
the relay is able to function well if its coil is added with 85%~ 110% of rated voltage.

Relay plays an extremely vital role in automatic control segment. The advantages and disadvantages are
described at below:

( Advantages J :
I. A small current input is capable of activating the contact to act, such as to open or close a large current (or
high voltage), and utilizing several contacts to open or close several circuits simultaneously.
2. The opening and closing of contact is precise. An opened contact has an infinite resistance; and a closed
contact has an extremely low, or zero resistance.
3. The contact is capable of resisting over-current or over-voltage; whereas, electronic components have no
such endurances.
4. The loop circuit is simple, and the cost is cheap.
5. While in operation, the action can be visually inspected to ease maintenance.

( Disadvantages J :
I. In places with severe vibration, the contact may come loose.
2. In the course of opening and closing the circuit, the spark generated will tend to wear off the contact, and

-22-
 there is a certain life expectancy due to mechanical abrasion. Generally, the opening/closing life of the
contact is arou nd 100,000 cycles.

' ~'~-l

-"-© I
I lRelay .:. t'",

(a) When VlO is small, 110 is small

U~' I
(b) When V 10 is big, 110 is big

I
i
--;;;- ~"~e

Iv,. ' Reljy

Jr,

L,~__.,.--.
-....+--........----1
~------=-~~~

(c) When VlO is small, IlOis extremely small (d) When V 10 is big, 110 is extremely big

Diagram 10-6 Methods of using a transistor to drive the relay

Circuit Applications
A control circuit that responds in accordance with the light strength is known as optoelectronic control.

Among all the optoelectronic detecting components, the most inexpensive one is photo resistor. The
commonly used photo resistors are made of cadmium sulfide, and so photo resistors are commonly called CdS.
When the light is strong (brighter), the resistance of CdS is small; and when the light is weak (darker), the
resistance value of CdS is big.

Diagram 10-8 shows the circuit of an optoelectronic switch. The CdS resistance is small when it is
exposed to light, and so VlO is small and the relay will not act. When the light on CdS is blocked, the CdS
resistance value will become bigger, and so VlO will rise to ease electricity conduction through the transistors
TRI and TR2, causing the relay to be attracted. The relay contact is widely used in controlling load (for example:
counter, automatic door circuit, and security alarm).

In actual applications, in order to avoid the surrounding light to cause interferences, CdS should be
covered up with a sleeve as shown in Diagram 10-9(a). Ifwe require a long distance control, then we should
add a glass lens as shown in Diagram 10-9(b).

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 CdS

i-' Diagram 10-8 Optoelectronic switch

~~~
Bulb =® ::.-::-..:----=.. Q=== cds
--::::::=:---
---.
(a) CdS added with a sleeve to prevent interferences from surr~)Unding light

Light projector I~ More than----j Light receiver

-fH;=}~ fQ-------;-------
.J

(b) Using a lens to extend the distance

Diagram 10-9

[ Experiments]
1. Install a circuit according to Diagram 10-8.
2. After the installation, adjust VR to screen off the light on CdS and allow the relay to be attracted. Expose
CdS to light and allow the relay to be released .
( Note J If the CdS is covered up with a sleeve, it would increase its sensitivity.
3. Is the circuit working normally? Answer: _ _ _ _ __
4. If the positions of CdS and VR are being interchanged, what will be the operations of the circuit? Try to
install the circuit, and adjust VR until it is able to work normally.

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 Unit 6: Interior lighting delay OFF circuit experiments
Diagram 10-11 shows a lighting delay OFF circuit suitable to use in controlling the conducting time of
heater circuit in a plastic bag sealing machine. When SW is closed, Vc=O, and so Vin=Vcc-Vc=Vcc, and the
relay will act (be attracted) immediately. After passing through VR and lk O resistor, the capacitor will be
charged to increase Vc gradually. As Vin=Vcc-Vc, therefore, the gradual increasing ofVc will gradually
decrease Yin until the relay is being released. When the circuit is closed, the relay will be attracted for a while
(the time duration will depend on the magnitude ofVR) and then it will release itself.

va
, small

Relay OFF

t-.;--:-.::::....-~-=,,~--~ Time (second)

long

Diagram 10-11 Delay OFF relay

[ Experiments]
I. Install the circuit according to Diagram 10-11
2. When SW is closed, see whether or not the relay will be attracted immediately? Answer: _ _ _ __
3. After a while, see whether or not the relay will release itself automatically? Answer: _ _ _ __
4. After adjusting VR, determine the time that the relay is being attracted - shortest time (seconds);
longest time (seconds).

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 Unit 7: Electronic turn signal light circuit experiments

In automatic control circuit, flicker relay is used in engine to activate the buzzer to sound a beeping sound
or the red pilot lamp to flash during a malfunction in order to alert people. While in vehicles, the flicker relay is
used as a turn signal light.

Diagram 10-12 shows a flicker relay, of which is activated by charging and discharging of capacitor to
cause the relay to be attracted and released repeatedly. When SW is closed, the capacitor C will be charged after
passing through RI and 1k (2 resistor. When Vin has increased, the relay will be attracted, and then C will be
discharged after passing through 1k (2 resistor and R2. When Vin has dropped, the relay will be released once again.
The repeating charging of C after passing through RI and discharging after passing through R2 will cause it to be
attracted and released repeatedly. The relay used in the diagram has two contacts; one of the groups is used in
controlling the load.

_ __ ~~~~_ Time

Diagram 10-12 Flicker relay

[ Experiments]
1. Install the circuit according to Diagram 10-16. Use a 10k(2variable resistor on RI , and lOOk (2 variable resistor
on R2.
2. Switch the power ON, and after closing SW, see whether or not the relay is being attracted and released
repeatedly? Answer_ _ __
3. See whether or not you are able to change the relay ON or OFF time by rotating RI? Answer_ _ __
4. See whether or not you are able to change the relay ON or OFF time by rotating R2? Answer_ _ __

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