air coolant sensor

the way the aircoolant sensor works is as the temperture increseas,resistance decreases and the voltage through and from the sensor increases

thermo fan switch.

the way a thermo switch works is, it will keep the circuit broken until the ecu decides it needs the fan to work, like switch.
Inside of thE switch is a open circuit untill the right tempure ture is reached then it will be a full circuot.

we did a small test on the fan swtich to check the resistance at the diffrent tempertures graph to show readings.

water temperature sensor

we boild some water up and tested the water temperature sensor.

graph to show readings.



It was a NTC type temperture sensor.

The way is works, there is a verble resistor in the sensor and when the temperture changes it can react to it and and the resistance will change,when the resistance gets higher we need more voltage to push. the raitoe changes betwwen the verble and the known resistor.

vane or flap air flow meter (ATM)

we check how this sensor operates when the vane is oprn the resistance gose up so more voltage is allowwd thorgh, and the sensor can pick that up, when its closed vise wise

map

I tested the MAP and the way it works is..
its reading the vacuum, so when the vacuum pressuer gose up the voltage gose down,there is a verable resistor in the circuit that veres when more vacuum is accring then resists more so less voltage is going to be applied.

Testing throttle body swiytch

i tested the T.B.SWITCH and the way it works is like a switch on or of..

exp.#8

we had to set up a circut and check the voltage at vce,vbe,ic,ib useing
1NPN diode bc547
1resistor 470
vary resistors2.2K,47k,220k,270,1M
then i had to record the voltage at the trasistor at vce,vbe,ic,ib for each of the resistors.

the voltage at vce was geting bigger as we put bigger resistance because we need more preusser to push through the bigger resistor.

the voltage at vbe didnt chage much as we didnt chage the resistance.

The voltage at ib keep droping as we put bigger resistors becase ther was less current.
the ic chaged because ther was less current going throgh, and we getting bigger resistance on the base

then we out our reading frm ic and vce on a graph and create a loadline.

circuit number 3

we had to build a ocean sesnior, we got given a plan sheet had to figer out the the all diffrence resistors we new the value of R6 at 10k

I stared from D2 the !st diode after V.S we needed to kno what was the V.D whichh is 0.6 we end up with 11.4v after the diode to R5 we didnt have the value use smple ohms law to figer out the value.We new the Z diode was at 9.1v there is a capacitor ther aswell to keep the voltage stayble. I worked out that R6 V.D is gota be 8.4V because we need 0.63v (I= V/R = 0.00085a). we kno from kurchofs law that in a series circuit the current shold always be the same at any point.
Then again we used ohms law to figer out R8 R= V/I = 740ohms
same for R7 we had the voltage after R8 which was 0.23v (R=V/I).

R5 13411 ohms
R6 10k ohms
R7 270ohms
R8 740ohms

then we go to the paraelle side of the circuit and had to work out R2,R3,R4.
we new 11.4 is come to the output of the op amp going to the L.E.D haveing the voltdrop of 1.8v, useing ohms vs-vd=9.6 and we kno the current is 9.5mA V/I=R 9.6V/9.5mA = 1010ohms. we did same thing for R4.
for R3 we had a diode and a l.e.d 0.6vd for the diode and 1.8vd for l.e.d which gives us 9v R=V/I 9V/9.5MA=947OHMS.
I KNO HOW THIS CIRCUIT WORKS IN THERORY BUT DIDNT BUILD IT

WORKING OUT:
R8 v/i=r 0.63/0.00085 = 740ohms
R7 " " 0.23/0.00085 = 270ohms
R2 " " vs-vd = 9.6,9.6/9.5mA = 1010ohms
R4 " " " " " " " " " "
R3 vd+vd = 2.1 vs-tvd 9 v/i=r 9v/9.5mA=947ohms

exp.5

we had to build a small circuit using these components:

3 diffrent resistors




we 1st calculated how much time it would take the capacitor to charge up. them measure the time take to reach applied voltage with a oscilloscope then make a graft we use 100 uf for each resistance at 2v.

we had to build a small circuit using these components:

3 diffrent resistors x2 1k 0.1k 0.47k








we 1st calculated how much time it would take the capacitor to charge up. them measure the time take to reach applied voltage with a oscilloscope then make a graft we use 100 uf for each resistance at 2v.

graph for exp.5

circuit #	capacitance        (uf)	resistance	calculated time	observed time  (ms)
1	100	1k	0.5	0.6
2	100	0.1k	0.5	49
3	100	0.47k	0.234	236ms
4	330	1k	1.63	175ms

exp no.6

we had to use a multimeter and check a BJT transistor  and lean how to tell between PNP and NPN also be able to fine out which of the three post are the emitter colcetor and  base.
Then we analised that the BJT will not work if you dont get the POLARITIES right, we found that by useing a multimeter.  

NPN TYPE

vbe 0.825
veb 0.829
vbc O/L
vcb 0.822
vce O/L
vec O/L

PNP TYPE

vbe 0.825
veb O/L
vbc 0.822
vcb O/L
vceO/L
vecO/L

ex no4

We had to build a circuit with 1 diode 1 zener diode and one resistors at 1k ohms voltage supply was 10v. We then had to do some voltage drop test then change the voltage to 15v and do the same thing.from the test we found out voltage was almost the same but the current would change. for 5vs at the resistor it was 4.61 and for 10v it was 9.56. then we use ohms law to get the current
V = AxR
4.61/1000=.004

no.2
V= A x R
9.56/1000=.009

when we put the voltage higher the currant gets higher at the resistor but almost the same everywhere else in the circuit

exp no7

i had to bulid a small sircuit with one signal npn type tresistors and 2 resistors with @1kohms and another @ 10k ohms, we then had to test a transistor by doing a voltdrop test between the base and the emmiter. the resultwas 0.79v which ment that is was at the saturated point , i kew this result by looking at the region chart, when the result is over 1v it is in active region and when the result is more then it will not operate                          step 2
then we did a voltdrop test between the colector and the emmiter, the result was 0.5v at the saturated point.its fully open thats why we only need 0.5v to pust it throgh.
at the VCE of 3v the power disspated by the transisots is 45w we figured this out by useing these calcultions p/w = A x V
15 x 3 = 0.45w
 then we had to work out the beta of the transistor useing this formula
B = ic/ib

this is the kind of chart we use to find out what region the transistor is working in

 

      Components use:

• capacitor X 2
• rectifier diode X 2
• l.e.d
• Z diode
• voltage regulator
• 3 X resistor @ 1800ohms 2700ohms 8200ohms

How it works:
12v going through diode then to z.d then  to the cap (store voltage) and then input of the voltage reg with the feed back resistor for protection its regulates the voltage down to 5v. we also have a gound out put, 12v out put, l.e.d to show that it is working. at 1st it didnt work as it was ment to work because D12 was shorting

we worked out the resistor by using ohms law R = V/I

volt drops

• diode13 0.6v
• cap12.2v
• volt reg 5v
• R1 3.9v
• led 1.78
• R3 3.3v
• R2 1.14v

        components use

• transistor X 2  PNP TYPE

•         l.e.d X 2 
•         560 ohms resistors X 2
•         860 ohms resistor X 2.

  We started of making the circuit on a lock master to plan it out 1st. before that we had to work out what resistance to use. after hooking up the components supplied 12v but the clear l.e.d did not work, i looked at the bottom peace their was a lose connection on R3 fix that and the light was glowing like never has.

Formulas used
B = IC/IB
V - VD
V=R x I
VP-VDC/IB=R

how it was working.

the volts goes though the circuit and makes the current flow, the 12v is going through R14 R15 then goes to the l.e.d then to the collector. the 5volts  PWM going to R13 R16 and them going to the base of the transistor which then open the gate to C and E of the transistor to let flow through.

volt drop test.
R14 R15 v.d 9.6v
R13 R16 v.d 3.6v (vs 5v)
L.E.D v.d 1.76
L.E.D (CLEAR) v.d 2.1v
V.B.C 12.3

         

experiment no.3

we had to build a circuit on a breadboard using the following components, 2 100ohms resisters, 1  5V, 1 400mw zen er diode (ZD)

R = 100 OHMS AND RL = 100OHMS VS = 12V
We then had to work out the value of vz which is 5v

when we vary the VS from 10 to 15vs the value of the VZ is almost the same

• At 10vs the voltage drop is 4.2v and at 15vs the VD was 4.6 

• this kind of circuit is used to vloatge regulate voltage, z diode  to keep the voltage stabile.

experiment no.2

Now we look at some diodes.

A diode as two posts anode and cathode (anode more positive then negative, cathode more negative then positive)  and electrically can only flow one way inside a diode. the way to tell between A and K is by looking for a stripe on the diode and that should be the cathode and on a l.e.d type diode the longer metal leg is the anode. that's so you don't set it up the wrong way. the way to check if you've set it up the wrong way just do a  test with the multimeter and if you have set it up reverse biased direction you shouldn't get a reading something like 0.00v,(got this little diagrams form google to show things bit more clear)

then i built a little circuit with a resistor diode and/or l.e.d 1st using a       multimeter I found out witch how to tell the difference between anode and cathode one both l.e.d and/or a diode..when i measured the l.e.d in forwards biased i got a reading of 1.796V when i measured the diode  i got a reading of .544V,  when i measured both of them in reverse biased i got a reading of 0.00V. which means that a diode and/or a l.e.d will not work in reverse biased. we also  had to identify which leg is the anode and cathode on the both l.ed or diode,on the diode it was easy there was a tripe on it and that is the cathode and one the l.e.d the longer metal leg is the anode

then i built a little circuit with a resistor diode and l.e.d

 did some calculation using ohs law(A = V/R )

VS=5V,R=1KOHMS=1N4007

          5V-.6/1K= .0044A                      

 Then we calculate and measured few things to check our a answers useing this formula

V = I x R =(VS-VD) / R

these brackets are showing they must be caculated first

by looking a t the data sheet given to us we check how mucj current can flow throgh the diode at what tempurtrure and it was 1.0A @ 75degree

and for a 1 K resistor the didoe needs value of 1000v to operate in safe region.

i replaced the  diode by an l.e.d and calculate the current by useing ohms law.

VS - VD = X/R = I

5 - 1.75 = 3.3/1K = 0.32MA

Experiment 1

In this experiment i looked at resistors and we measured resistors by the colour codes and to see how much ohms each one is and how much tolerance they can have. We did this by adding up the numbers looked at the colour chat (their is a pic of one) to give us the value (and tolerance) after that we did some small experiment that showed me that when you measure the resistors in parallel you will always get a lower reading as it has to be lower then the lowest resistor in the circuit something is not right, but when you measure the resistors in series you simply add them up.     


colour chat for resistors