Oxygen Sensor unit
Theory and operation
The oxygen sensor is located in the exhaust system and it detects the amount of oxygen in the exhaust by sending a voltage signal which ranges between 0.2V and 1.2V. This data to the ECU which then determines the amount of fuel required to keep the engine running at its optimum level depending on factors such as load, speed, etc. Note for the oxygen sensor to starts to work when the engine reaches operating temperature.
A oxygen sensor produces a voltage normal oxygen sensor signal voltage ranges between 0.2V and 1.2V. When more air less fuel (lean condition) is sensed by oxygen sensor outputs a 0.2V~0.4V and when a near perfect Air/Fuel ratio (stiochiometric range) it outputs 0.5V~0.65V and when more fuel than air (rich condition) is sensed the output voltage by the oxygen sensor is about 0.65V~1.2V
List of Components used for this project:
- 1x Op-Amp, Max supply voltage ±16 or 32V
- 1x 10kΩ, 3x480 kΩ, 1x 270Ω, 1x 470Ω, 1x 270Ω Resistors
- 3x LED (1x Green, Red and Orange)
- 2x Diodes (1N4001), Max Pd = 2.5W, Max I = 1A @ 75ºC, Max reverse voltage = 100V
- 1x Zener Diode (9V1), this a 9.1V capacity Zener diode 2x Capacitors (0.1uF), the capacitors are used to smoothen out the current in the circuit.
Source upply voltage(Vs) = 12V
LED = 1.8V
IzRm = 5.6mA (This current value was abtained from the data sheet)
We have 12v coming in from the source and as it passes D2 diode there is a voltage drop of 0.6V. So we have 11.4V going to R5. The supply voltage is further used up by R5 and the voltage drop across R5 is (11.4-9.1)=2.3V
Using Ohms Law
V = I*R
R = V/I
R5= 2.3/0.0056 = 410.7 Ohms
Now we know R6 = 10K, so we can find current along R6
the voltage drop across R6 is (9.1V-0.63V) = 8.47V
therefore current across R6 is: I = V/R = 8.47/10000 = 0.000847A (0.847mA)
Since it is a series circuit the same current flows through R6, R7,and R8
R8 = V/I = (0.63-0.23)/0.000847 = 472 Ohms.
similarly R7 = 0.23/0.000847 = 271.5 Ohms
To find R1, R2, R3, and R4 we were given current across this part of the circuit (9.5mA)
Voltage drop across R2 is 12-0.6-1.8 = 9.6V
V = I*R
R2 = 9.6/0.0095 = 1010 Ohms
voltage drop across R3 = 12-0.6-0.6-1.8 = 9V
R3 = 9/0.0095 = 947 Ohms
voltage drop across R4 = 12-0.6-1.8 = 9.6V
R4 = 9.6/0.0095 = 1010 Ohms.
Explnaton how the circut work.
1. GREEN LED on This happens when pin 13 of the opamp is getting 0.23V and pin 12 wich is conected to sensor input is getting less than 0.23V from the sensor input this allwos the out put at pin 14 ground the circut wich turns the green light on when the sensor input gose above 0.23 available voltag at pin 12 exceds that of pin13 the green LED is turn off.
2.Yellow LED
Pin 10 is getting
0.23V but when available voltages at pin 9 and pin 3 are greater than 0.23V and
at the same time pin 2 is getting 0.63V. What happens is that available voltage
at pin 9 is greater than pin 10 and at the same time pin 2 is greater than pin 3
and so there is virtually no power flowing through D3 diode hence, power coming
from the supply or source (9.6V) is grounded at pin 11 by the circuit and the
yellow LED turns on. But when when input at pin 9 drops below 0.23V the yellow
LED is turned off and the green LED turns on instead.
3.RED LED This happens when input at pin 6 is greater than 0.63V, and since pin 5 is
connected to constantly to 0.63V. During this time power at source flows
through and the circuit is grounded and the red LED is turned on. When sensor
input (pin 6) is less than 0.63V then the red LED is turned off
OPERATIONAL AMPLIFIERS ("OP-AMP") & MOSFETS
OPERATIONAL AMPLIFIERS ("OP-AMP") &
MOSFETS
What is Operational Amplifier
"OP-Amp"?
Operational Amplifiers are electronic circuits. They are ideal linear devices and thus are used for signal amplification, filtering. They awere initially used for mathematical comutations like addition, subtraction, etc.
An Op-Amp has two inputs and an output (Vout) which are connected
voltage rail (+Vss & -Vss)
+ input is called "Non inverting input"
- input is called "Inverting input"
How Op-Amps works?
Op-Amps are current controlled. Op-Apms work in many different ways. They can be used as a inverting/non-inverting voltage amplifier.
The output is dependant on the voltage difference between the two
inputs. The output takes the value of the greater of the two input values. for
example, in the fig 2. below since the greater Vin is 5v, hence the output would
be whatever is on the negative (-Vss)
What is a Mosfet?
Metal Oxide Semiconductor Field Effect Transistor is one type of a semiconductor and it is mainly used in electronics as a switching device. MOSFETS's have three terminals Gate (G), Source (S), and Drain (D). There are two types of Mosfets N-channel and P-channel
References:
Operational Amplifiers are electronic circuits. They are ideal linear devices and thus are used for signal amplification, filtering. They awere initially used for mathematical comutations like addition, subtraction, etc.
Fig. 1. Symbol of an Op-Amp
+ input is called "Non inverting input"
- input is called "Inverting input"
Op-Amps are current controlled. Op-Apms work in many different ways. They can be used as a inverting/non-inverting voltage amplifier.
Fig. 2.
Metal Oxide Semiconductor Field Effect Transistor is one type of a semiconductor and it is mainly used in electronics as a switching device. MOSFETS's have three terminals Gate (G), Source (S), and Drain (D). There are two types of Mosfets N-channel and P-channel
Fig 3 Mosfet
G (Gate), D (Drain), S (Source). P-Chanel are
doped with holes and the N-Chanel are doped with electrons.
Types of Mosfets:
-
Depeletio type
-
Enhancement
Characteristics of
Mosfets:
- Mosfets are voltage controlled i.e. O2 sensor
- Mosfets have positive temperature co-efficient (conduct less current as its temperature increases)
- Mosfets must be handled with care to protect against damage by static elctrictricity.
- Mosfets can get very hot so again handle with care
VEHICAL ELECTRONIC AYOB AGHAZI 4847
Identifying, Testing and Troubleshooting Semicondutor Components
Identifying, Testing and Troubleshooting
Resistors:
Experiment No.1: Resistors
Resistors are used to control flow of current. From Ohm's law we can see that the resistance is inversely proportional to the current flow; that is as the resistance increases the current decreases.
Resistors are very important and widely used in electrical and electronic components.
Resistors are colour coded for identification purposes. There are four main colour bands.
- First two or three bands are the numbers to be written down
- Next band is the multiplier (how many zeros to add the number)
- Gold multiplier makes one decimal place smaller, silver makes two decimal places smaller
- Last band to the right is tolerance values
Colour Codes for
Resistors
-
Use the coulr code to calculate the value of the resistor
-
Include the maximum and minimum tolerance value of each resistor
-
Then measure the resistor value with a multimeter.
Value (colour
codes)
|
Measured value
(multimeter)
|
Yellow (4),
violet (7), black (0), black (0), gold
(5%)
|
467Ω
|
Resistor Color Code Chart
| |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Brown (1),
black (0), red (2)
|
0.98kΩ
|
Green (5), blue
(6), red (2), gold (5%)
5600±5% or 280=
min5320Ω and max 5880Ω
|
5.54kΩ
|
Orange(3),
orange(3), brown(1),
gold(5%)
|
327Ω
|
Brown (1),
black (0), brown (1), gold
(5%)
|
98.3Ω
|
Brown (1),
black (0), orange (3), gold (5%)
|
99.6kΩ
|
Choose two resistors and record their indivitual
Ohm resistance values and measured with a multimeter
Resistor 1: 468Ω Resistor 2:
10kΩ
1.1 Resistors in series:
calculated value 1 & 2 in series: Rt = R1+ R2
= 10468Ω
measured value 1 & 2 in series: 10480Ω
1.2 Resistors in parallel:
calculated value 1 & 2 in parallel: Rt=
R1*R2/R1+R2 = 447Ω
measured value 1 & 2 in series: 10480Ω
In experiment 1. I measured and and calculated
resistor values in two way. first from the colour code and then I measured the
resistor value with a multimeter to verify my result from the colour code
values. the result of both methods of resistor value calculation was about the
same thus proofing both methods to be right.
In addition to calculating resistor values, we
also did a small test to check the difference between when resistors are in
series and parallel.
We have also prooved that when resistors are
connected in series their total resistance is the sum of the indivitual
resistors (Rt = R1 + R2 + R3 + .... + Rn ).
Similarly, we have a shown that when resistors are
in parallel their total resistance is smaller than the smallest resistor in the
circuit and can be calculated by using this formula
(1/Rt = 1/R1 + 1/R2 + 1/R3 + ....+ 1/Rn).
Experiment 2. Diodes
Diodes are electrical device that only allows current flow in one direction (dirction of arrow). positive side is anode and negative is cathode.
A diode can behave as an insulator and a conductor depending on how it connected (formard or reverse). when current flows in the forward direction the voltage drop accross the doide is very small so it does not take much effort to bush through the doide. And in the reverse direction no current flow hence the zero voltage drop recorded.
Exercise: Identifying the anode and cathode of doides using a multimeter.
Voltage drop in forward biased
direction
|
Voltage drop in reverse biased
direction
| |
LED
|
1.8V
|
0
|
DIODE
|
0.57V
|
0
|
Diode: Cathode side of a doide has a band marking.
Calculate the current flowing through diode and compare it against a measured result usung meter.
Calculated : Measured using meter:
known: V=5v, R=1kΏ. (1000 Ώ.) I = 0.005mA
V=I*R
I=V/R= 5/1000 = 0.005A (5mA)
Is the reading as you
expected; explain why or why not?
Yes, the reading was as expected.
From Ohm's law we learnt that resistance and current are inversely proportional.
In this case we were working with a relatively big resistor 1kΏ. And that is the reason for the
tiny current flow. On the other hand if we had low resistance our current would
have increased proportionally.
Calculate the voltage drop across the diode:
Since we have all ready know the
current flowing through the diode to find the voltage drop across diode we use
Ohm's law again:
V=5v, I= 0.005A
V = I*R = 0.005 x 5 =
0.66V
Using the table of data above, the maximum value of the current that can flow through ther given diode is 1A.
For R = 1kΏ, the maximum value of Vs so that the diode operates in a safe region is 1000V.
Using same circuit we only replace diode with an LED, calculate the current, then measure and check answer.
Calculated : Measured using meter:
known: V=5v, R=1kΏ. (1000Ώ.) I = 0.003mA
V=I*R
I=V/R= 5/1000 = 0.005A (5mA)
The result shows that there is no change to current flow in the circuit whether we use an LED or a doide. The reason is that it is the high resistance that is restricting the flow of current in the circuit.
Experiment 3:
Zener Diode: Components: 2x resistors, 1x 5V1 400mW Zener diode (Zd)
Answer: Vz= 4.97V
Vary Vs from 10V to 15V
what is the value of Vz?
when Vs= 10V Vz= 4.72V
Vs= 15V Vz= 5.1V
Zener doide behaves just like any other doide when forward based until a certain reference voltage is exceeded it becomes reverse biased and maitains the 5V reference voltage regardless of how much higher Vs gets above reference voltage of 5V.
The zener voltage stays relatively constant regardless of the Vs. This the characteristics of a zener diode. It acts as a voltage regulator and are used to maitain constant voltage. Experiment 4:
Components: 1x resistor, 1x 5V1 400mW Zener diode, 1xdiode 1N4007
Exercise: Obtain a breadboard, suitable components and build the circuit.
Vs= 10 and
15V, R= 1kΏ
Voltage drops
|
10
Volts
|
15
Volts
|
Volt drop V1:
|
4.64V
|
4.82V
|
Volt drop V2:
|
0.67V
|
0.70V
|
Volt drop V3:
|
5.31V
|
5.50V
|
Volt drop V4:
|
4.74V
|
9.50V
|
Calculated current
(A)
|
I=10/1000= 0.01A
|
I= 15/1000=
0.015A
|
After connecting the various components according to the circuit above we have tested the voltage drops across V1, V2, V3, and V4.
When Vs is varied between 10 and 15
volts, the voltage drop at V1 is maintained at about 5V just about the reference
voltage of the Zener diode. While the voltage drop across the normal diode
remains at about 0.57V. The current flow in the circuit is very small (10-15mA),
hence the seires connection of the circuit.
The Zener diode blocks off any voltage below 5V and bleeds off excess voltage that is why voltage drop at V4 is that high in both cases. The diode regulates the amount of current in the circuit and keeps it within the circuit proper operating conditions and protects it from exposure to excessive voltage that may damage the components.
Capacitors:
Capacitors are used as a charge storing devices in electrical circuits. It consists of two metal plates separated by an insulator. The unit standard of a capacitor is Farad (F).
Experiment 5;
Exercise: First calculate how much time it would take to charge up a capacitor. Then connect the circuit as shown above. Measure the time taken by the capacitor to reach the applied voltage on an oscilloscope. Fill in the chart below and draw the observed waveforms in the graphs below.
1uF= 0.000001 F
Circuit Number
|
Capacitance (uF)
|
Resistance (kΏ)
|
Calculated Time (ms)
|
Observed Time
(ms)
|
1
|
100
|
1
|
500
|
400
|
2
|
100
|
0.1
|
50
|
100
|
3
|
100
|
0.47
|
235
|
210
|
4
|
330
|
1
|
165
|
1600
|
Calculation procedure in here I will
only show calculation of circuit number 1;
To convert 100uF to Farad (F) multiply 100 by 0.000001 = 0.0001F
T= R*C*5
T= 1000*(100*0.000001)*5= 0.5s (500ms)
Graphs:
Circuit 1: (10V/Div and 500ms/Div)
Circuit 2: (10V/Div and 100ms/Div)
capacitance 100 uF, Resistance 100Ώ
Circuit 3: (10V/Div and
100ms/Div)
Circuit 4: (10V/Div and 1s/Div)
from the result of both the calculated and graph we can see that as the resistance is directly proportional to charge time and also the higher the capacitance the longer the charge time.
Experiment 6
Transistors are semiconductors that are uses a small current to turn on a much lager one. They are used to control current flow in a circuit.
There are two type of transistors NPN and PNP
Meter check of a Transistor:
First put meter on diode test mode
Emitter base junction has slightly higher voltage drop that the collector base junction. From here it is easy to name all three junctions of the transistor.
After naming the three junctions, we go on determine whether a transistor is NPN or PNP. To do this we connect (+) lead of multimeter to base of transistor and connect (-ve) lead of meter to either of the other two junctions if we get a small voltage reading on meter then it is a NPN transistor but if we get OL reading and then we reverse the leads (-ve) on base and (+ve) on either one of the other two and we get a small voltage reading then our transistor is PNP.
Identifying the legs of transistor with a multimeter.
Transistor No.
|
Vbe
|
Veb
|
Vbc
|
Vcb
|
Vce
|
Vec
|
NPN
|
0.738V
|
OL
|
0.73V
|
OL
|
Ol
|
OL
|
PNP
|
0.739V
|
OL
|
0.74V
|
0.735V
|
OL
|
OL
|
Experiment 7:
Transistor as a switch
Components: 1x small signal NPN transistor, 2 resistors.
Exercise: connect the circuit and switch on the power supply.
Connect the multimeter between base and emitter and note voltage reading.
Voltage between base and emitter is measured to be 0.8V. We then connected between collector and base the voltage reading was very small about 0.05V.
This indicates a very small voltage going through Vbe. It is the base that controls the collector. This is sufficient voltage at base and current to flow between collector and base and hence the transistor is on but if there is no enough voltage going through base then that means transistor is off and no flow between Vbe and Vce. The small current flowing through base-emitter is controlling the flow between collector-base flow.
The base-emitter has reached the
high enough voltage for current to flow between collector-base (0.05V). since
base-emitter acts like a diode. It only creates a connection between
collector-base only when the voltage at base-emitter reaches the 0-7V
threshold.
Normally a transistor has three regions:
Saturated Region is the region where both emitter and collector are forward biased Vce (Zero). When at the saturated region and sufficient current is available then current at Vbe does not raise current at Vce. Used when a transistor is required as a switching device.
Active Region: this is the region in the middle of the chart and it is where power dissipated is highest. This region is used when amplification of small signals is required.
Cut-Off Region: In this region both emitter and collector are reverse biased. Base current is zero, hence no collector flow. Used when a transistor is required as a switching device.
Power dissipated by transistor at Vce of 3V:
P= Vce*Ic= 0.015x3= 0.045W (45mW)
The Gain is ratio between Ic and Ib. It is the multiplying factor that and the reason why transistor is able to turn a small current to control a much larger current.
β=Ic/Ib
Vce @ 2V: β= Ic/Ib= 20mA/0.8mA= 25
Vce @ 3V:
β= Ic/Ib= 15mA/0.5mA= 30
Vce @ 4V:
β= Ic/Ib= 7mA/0.2mA=
35
Gain increases with increase in Vce.
Experiment
8
set up the
circuit below on a board. Use a 470R for Rc and BC547 NPN transistor.
Vary base resistor and measure
changes in voltage and current for Vce and Vbe, Ic and Ib. Then plot a load
line.
Result:
Rb 47k
|
Vbe 0.72V
|
Vce 0.09V
|
Ib 0.10mA
|
Ic 0.14uA
|
Rb 220k
|
Vbe 0.70V
|
Vce 0.60V
|
Ib 0.8mA
|
Ic 0.08mA
|
Rb 270k
|
Vbe 0.69V
|
Vce 1.05V
|
Ib 0.6mA
|
Ic 0.12mA
|
Rb 330k
|
Vbe 0.69V
|
Vce 1.42V
|
Ib 0.2mA
|
Ic 0.12mA
|
Rb 1M
|
Vbe 0.85V
|
Vce 2.58V
|
Ib 0.08uA
|
Ic 0.15mA
|
From the test it evident that Vce
increases as Rb is increased but Vbe does not show change and stays about the
same. Vce changes as the transistor changes depending on which region the
transistor is in. low for saturated and high as it reaches the active
region.
Ib is inversely proportional to Ic . Ib reaches its highest when the resistance is at its lowest (47k) and Highest at 1M. Ic is at its maximum when
This is as expected since the more restriction to flow of current the is the lower the current is going to be.
