OPERATIONAL AMPLIFIERS ("OP-AMP") & MOSFETS

Vehicle Electrical and Electronics 4847 

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.

Fig. 1. Symbol of an Op-Amp

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)

Fig. 2.

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

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:

1. Depeletio type
2. 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

Mosfets are voltage controlled and based on conventional current flow theory (from positve to negative).

References:

1. http://forums.overclockers.com.au/showthread.php?t=962539
2. http://www.knowledgerush.com/kr/encyclopedia/Transistor/
3. http://www.societyofrobots.com/electronics_advanced_components_tutorial.shtml

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

Obtain 6 different resistors of different values and determine their value in two different ways:

• 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 Calculator Color Band 1 Color Band 2 Color Band 3 Results OR Resistor Color Code Chart Band Color Options Band #1 Possible Band #2 Possible Band #3 Possible Multiplier Value For Band 3 Band #4 Value Tolerance Black 0 1 1 Brown 1 1 1 10 Red 2 2 2 100 Orange 3 3 3 1,000 Yellow 4 4 4 10,000 Green 5 5 5 100,000 Blue 6 6 6 1,000,000 Violet 7 7 10,000,000 Gray 8 8 100,000,000 White 9 9 1,000,000,000 None 20% Silver 10% Gold 5% Resistor Color Code Information The resistor color code is a long standing standard in both the electronics and electrical industries, indicating the value of resistance of a resistor. Resistance is measured in ohms and there is a foundation for it called Ohm's Law. (Want to know about Ohm's Law? If so, please click here or click here!) Each color band represents a number and the order of the color band will represent a number value. The first 2 color bands indicate a number. The 3rd color band indicates the multiplier or in other words the number of zeros. The fourth band indicates the tolerance of the resistor +/- 20%, 10% or 5%. In most cases, there are 4 color bands. However, certain precision resistors have 5 bands or have the values written on them, refining the tolerance value even more. There is no standard (TANS) however, for the 5th band. From one manufacturing company to another, the 5th band may indicate 2%, 1%, 1/2% or even closer, according to their own standards. Color bands are usually found on resistors that have a wattage value of 1/8 to 2 watts; though it is rare, there are some 5 watt resistors that are banded. There are also some capacitors that are color coded. See our Capacitor Color Code Calculator.

 

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

Explain how to identify the cathode without a multimeter
Diode: Cathode side of a doide has a band marking.

LED: Short leg of the LED is the cathode.

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)

For R= 100 Ώ, RL= 100 Ώ, Vs= 12V; What is the value of Vz?
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)

capacitance 100 uF, Resistance 1kΏ

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)

capacitance 330 uF, Resistance 1kΏ

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.