O2 sensor

Use of a oxygen sensor.

Main reason for a O2 sensor is to measure and produce the perfect air/fuel ratio of 14.7:1. how it works is the oxygen level in the exhaust from there it sends a voltage reading to the car ECU. From the result it recives it compairs the result to the build in log table and sends out the correct signal to the injectors to allow X amount of fuel to be spray out in the cylinder this gives maximum power with the best fuel economy. The O2 sensors only works when it is hot enough, that's why some sensors have heaters in them so it starts working sooner than a non heated (one wire) sensor.

Tools needed:

Voltmeter on DC

O2 sensor

Car or gas torch

Run the car until the O2 sensor is hot then unplug the conection and use your volt meter  red lead to signal wire and black lead to earth. when idleing the volt should display around 0V but rev the car and it should read around 1V which is rich. If the car is staying at one number all the time or 0V all the time then the sensor is broken.

http://www.aa1car.com/library/o2sensor.htm

4842 Engine

PICTURE OF WORK, MEASUREMENTS WILL BE ON WORKBOOK


Remove rocker cover

Remove cam shaft

remove cam sprocket from cam shaft

Remove water pump driver

Turn engine around and remove sump

Further remove the sump













Remove Big end

remove jurnals and piston should fall out repete with the other 3

 

Remove crank shaft













Left with just the short block

4841 Logic Probe

A logic probe is a hand held pen circuit tester its is good for testing +/- ECU signals and injector pulses. 


How it works

The one we build can only test volts less that 24 volts.For our logic probe to work connect the red to positive and black to negative. Both green and red led should light up,this shows that the logic probe is working and is connected correctly. With the brass tip touch the positive of the battery  now the green LED should go out and the red LED gets brighter. If you touch the probe to the earth(negative) the red LED should go out and the green LED gets brighter.

4841 Electrical circuits

Voltage (volts) 

The force or pressure to move the electron in a circuit (electrical driving force)

Think of how fast the water is flowing down the stream

Available Voltage

Negative lead of voltmeter ( Black) to negative of battery or ground/earth and Positive (red) to the point in the circuit being tested.

Voltage Drop

Leads of voltmeter to both side of a component in the circuit 

Ampere (Amps) current

When  electron move along a conductor,this means there is now current flowing in the circuit ( this is measured in Amps). The larger amount and flow of electrons the larger the current.

Think of how much water is flowing down the stream.

Measure I=Current (Amps)

To measure current the circuit needs to be broken and put the merer in series

Resistance (Ohms) 

Resistance is put in a circuit to limit the flow of current and voltage

Think of it as a funnel in the stream

Wattage (P= power)

Work being done. P=VxI

Available voltage at

Power source      13.12v (red lead on positive power supply,black lead on negative power supply)

Before switch      13.11v

After the switch   13.11v

Before light Bulb  13.08v

After light bulb     0v

negative on power supply 0v  (Both lead on power supply)

The voltage decrease as it progresses through different components of the circuit because of the build up of resistance.

Voltage drop at
Power supply to in switch   3.3mv

in switch ot switch               2.2mv
ot switch to in bulb              2.1mv

in bulb to ot bulb                 13.08v

ot bulb to N- supply            0v

The bulb has the biggest voltage drop because the bulb is using a bigger load.

Wire befor bulb 0.23A

resistance of light bulb

R= V/I           

R=13.08/0.23

R=56

Watts used at light bulb
P= VxI
P=13.08x0.23
P=3W


LARGER BULB


Voltage Drop
wire before switch  23.6mV
switch                    163.5mV
wire before bulb     172.5mV
light bulb                12.66V
wire after bulb        22.4mV


current through whole circuit 1.8 Amps


The circuit with the larger bulb has a higher amp than the circuit with the smaller bulb.


resistance of larger bulb
R=V/I
R=12.66/1.8
R=7.03


As the resistance decrease the current increase. The smaller the bulb the higher the resistance.


Watts used with larger bulb
P=VxI
P=12.66x1.8
P=22W


A higher watt bulb will be brighter than a lower watt bulb in this case the bigger bulb is brighter.

voltage drop
wire before switch  2.5 mV
switch                    13.6 mV
wire before bulb1   13.6 mV
Bulb1                     5.75 V
between bulb1&2   1.2 mV
bulb2                      7.31 V
wire after bulb2       2.2 mV


Voltage available at battery  13.12 V


The current gets restricted as it moves along the different component in the circuit in result the voltage drops as well.


wire before switch  0.15A
wire before bulb1   0.15A
between bulb1&2   0.15A
wire after bulb2      0.15A


The amp in this circuit stays consistent at 0.15 amp because this is because ohm's law amount entering equals amount exiting.


resistance or circuit
R= V/I
R= 13.12/0.15
R=87.7


Watts used at each bulb
P= VxI
W= 13.12x).15
W= 1.97w


Bulb 1 and bulb 2 watt stays the same because the watt stays the same but the current is shared throught the circuit because its in series circuit.

wire before switch  2.3 mV

switch                     9.5 mV

wire before bulb1    10.5 mV

bulb1                      3.6 V

between bulb1&2   8.6 mV

bulb2                      4.51 V

between bulb2&3    28.1 mV

bulb3                      4.9 V

wire after bulb3      1.4 mV

voltage available     13.12 V

The voltage drop less through each component with the 3 bulb in series because the the voltage is shared with more component.

amps at 

wire before switch  0.12A

wire before bulb1   0.12A

between bulb1&2  0.12A

between bulb2&3  0.12A

after bulb3             0.12A

The amp in this circuit stays the same as the amp in the circuit with 2 bulb but the amp with 3 bulb is lower because it has to share among 3 bulb. 

resistance of circuit

R= V/I

R= 13.12/0.12

R= 109.3

watts at each bulb

P= VxI

P= 13.12x0.12

P= 1.57w

The wattage with 3 bulb is less than with 2 bulb so the brightness with 3 bulb is dimmer.

Available voltage
power supply  13.12 V

in switch         13.11 V

ot switch         13.09 V

bulb1              13.08 V

ot bulb1           9.39 V

in bulb2           9.39 V

ot bulb2          4.90 V

in bulb3           4.89 V

ot bulb3          1.5 V

negative to n-  0V

Available voltage is the voltage available from the component to the end of the circuit (negative) 

voltage drop is the voltage used up by the component.

2 BULB PARALLEL

available voltage

bulb1  13.11V

bulb2  13.07V

voltage drop
bulb1  13.10V

bulb2  13.06V

The voltage in theory should stay the same  because the voltage is not shared.

amps through the circuit

bulb1  0.19A

bulb2  0.19A

total    0.38A

Current in parallel divides evenly through the other brunches where as in series the current goes only one way.

resistance of each bulb
R=  V/I

R= 13.12/0.38

R= 34.5

resistance of circuit

1/Rt = 1/R1 + 1/R2

1/Rt = 1/4.99 + 1/4.99

Rt=2.5

watt of cuircuit

P= VI

P=13.12 x 0.38

P=4.99watt


watt of each bulb
P=VI
P=13.12x0.19
P=2.49w


The watt doubles in parallel circuit compare to series circuit because the current doubles in the parallel circuit.

current at
bulb1   0.20A
bulb2   0.20A
bulb3   0.20A
total     0.57A


The current increase 0.01A when i added the 3rd light bulb. this happens because now the circuit is drawing more amp.


Ohm's law states that the current in a circuit is the same to the circuit resistance. This theory is true in both  parallel and series circuits.


available voltage
bulb1   13.11V
bulb2   13.10V
bulb3   12.98V


voltage drop
bulb1    13.10V
bulb2    12.98V
bulb3    12.96V


There is no difference in Voltage available with the 3 bulb compared with 2bulb.
because in parallel circuit the same voltage is presented across all 3 bulb.

There is also no difference in Voltage drop with the 3 bulb compared with 2bulb.

because in parallel circuit the same voltage is presented across all bulb there for they drop equal in theory.

resistance of each bulb

R=V/I

R13.12/0.57

R=23.02

resistance of circuit

1/Rt = 1/R1+1/R2.......

1/Rt =1/23+1/23+1/23

1/Rt=3/23

Rt=23/3 = 7.7

watt circuit

P=VI

P=13.12x0.57

P=7.48w

available voltage

switch                         13.11V

before parallel bulb1   13.11V

before parallel bulb2   13.10V

after parallel bulb1      11.54V

after parallel bulb2      11.54V

before series bulb       11.53V

after series bulb           1.5mV

voltage drop

parallel bulb1       1.6V

parallel bulb2       1.6V

series bulb           11.53V

current at 

the switch          0.12A

parallel bulb1     0.06A

parallel bulb2     0.06A

series bulb         0.12A

watt at each parallel bulb

P=VI

P=1.6x0.06

P=0.096w

watt at series bulb

P=VI

P=5.6x0.12

P=0.67w

The watt in this circuit is the same as the resistance in this circuit.

Because i have added 2 bulb in serise with the parallel circuit now the voltage is devided into 3 there for the 2 bulb in parallel is dim.

Voltage in a series circuit divide evenly across each bulb in the circuit. I.n parallel circuit the same amount of voltage is presented to every bulb

In series circuits the current stays the same but the voltage drops.In parallel circuits the voltage stays the same but the current.