CIRCUIT BOARD 3

COMPONENTS: Resistors x7 carbon film, R2,R3,R4= 1kohms, R5= 380ohms, R6= 10kohms, R7= 260ohms, R8= 470ohms. Quad op-amp. L.E.D x3- LED1 red, LED5 yellow, LED6 green. all 1.8volts. Rectifing diodes x3, D2,D3,D4 1N4001. Zener diode- D1= 9V1. Capacitors x2 C2,C1= 0.1uF. 12v supply. sensor input= 0-1volt

CALCULATIONS: I=V/R= 8.47/10000= 0.000847amps. R=V/I. R8= .4/.000847= 472.2ohms. R7= .23/.000847= 271.5ohms. R6= 10kohms. R5= 2.2/.0056= 392.8ohms. R4= 9.5/.0095= 1000ohms. R3= 8.8/.0095= 926.3ohms. R2= 9.5/.0095= 1000ohms

HOW THE CIRCUIT WORKS: As the supply voltage comes through it powers the op-amp and goes to the zener where you could say is regulated down to 9.1volts and the other 2.9v charge and store in the capacitor. The 9.1volts then go down through R6,R8,R7 and voltage is divided because of the different size resistors. The voltage at points between these resistors go to the op-amp at there specific points. This is so when the sensor input changes between 0-1volt the op-amp will non-invert and invert amplify causing different LEDs to light up and turn off.

TESTING: We know by the diagram what voltage should be at certain points. I used available voltage to test for these known voltages and also to find out if I had any open circuits. Also I did some voltage drop testing to make sure components weren't high in resistance or open or shorted circuits.
Test 1 my board didnt work. All LEDs were on when only the red LED should of.
Test1: RESULTS: VD- R2, R3, R4 9.5v, 8.8v, 9.5v. LED1, 5, 6 2.03v, 3.66v, 1.29v.
Test2 my board still didnt work but this time the zener was earthed to the common ground.
Test3 my board finally worked.
Test 3: RESULTS: When the sensor input was .91v LED1 Vd= 2.05volts, LED5, 6 Vd= 0volts. sensor input .41v LED5 Vd= 3.91volts, LED1, 6 Vd= 0volts. sensor input .12v LED6 Vd= 2.62volts, LED1, 5 Vd= 0volts.

PROBLEMS: I had many problems with my board. My first problem I come across was I never earthed the zener to common earth. I solved this problem when I tested for available voltage and found 12.3v before the zener and 0v after, but i should have got around 3v. My second problem was only the red LED was lighting up, and changing the input signal still didnt change a thing. I tested available voltage with 12Vs and no input signal. I found I had the correct voltage readings at the specific points, this meant that the problem was in the op-amp. But I had diode tested it at the start and it was fine. I ended diode testing it again and found circuits were shorted together. I then scraped the board so the circuits werent shorting any more. Then I set it up for testing again and it finally worked.

If I had the chance to do the circuit board again I would triple check everything. The soldering, the earth (all of them), the power to the op-amp and earth from the op-amp, and diode test close wires and make sure they're not shorting to each other.

CIRCUIT BOARD 2

COMPONENTS: Rectifing diodes x2, Resistors x3 R1=1kohms R2=260ohms R3= 750ohms, Capacitors x2 C15, C16= 330uF, Zener diode, Voltage regulator LM317, L.E.D white 1.8v 20mA, 12v supply.

CALCULATIONS: Vout= Vref(1+R3/R2) or Vout/Vref-1= R3/R2. 5/1.25-1=R3/R2. 3=R3/R2. R1= V/I= 5/20mA= 1000ohms.

This circuit regulates the 12v supply voltage down to 5v to send. This circuit has a zener diode to protect it from back EMF. As the 12v come through the first rectifier then into the voltage regulator, it is regulated to 5volts to the flow on into the L.E.D and out to processor via ribbon. It has a rectifier parallel to the regulator so when the capacitor is full excess voltage will flow back around the regulator to the input side. A wire is linked before the first rectifier so a 12v signal is also sent to processor via ribbon. Capacitors in parallel in the circuit collect excess voltage and charge and store voltage.

To test to check if the circuit is working voltage drop testing can be done and also available voltage. You also need to remember voltage before the regulator is 12v and after it is 5v.
My RESULTS: 1. 0.74v 2. 11.65v 3. 11.65v 4. 3.92v 5. 1.26v 6. 5.17v 7. 3.26v 8. 1.91v 9. 6.47v vd across in/out= 6.48v, in/adj= 7.73v, out/adj= 1.25v. All these results tell me that the circuit is fine. Because the circuit is mostly parallel you would expect to have close to supply voltage across the components before the regulator and after the regulator 5volts across the components or adding up to 5volts if in series. eg. R1 and LED1. R1 was 3.26v and LED1= 1.91v. Both added equaled around 5v. The voltage drop across 1 meant the rectifing diode is working well. If it was high means there is high resistance in the diode.

I had no problems with my circuit board 2. The tests come up fine, there were no bad soldering connections. It worked good with output readings of 12.08v and 5.17v.

If I had a chance to make the board again I would compact it more because it is abit to spaced out. Maybe design it in a different shape because it is to long and rectangle. Maybe design it into a small square shape would be a good design if it doesn't get to hot being so compact.

CIRCUIT BOARD 1

COMPONENTS: R14 & R15= 560ohm resistors R13 & R16= 1000ohm resistors L.E.D x2 1.8volts/ 20mA Transistors x2 c547w64 (NPN) Vs=12volts PWM=5volts.

Calculations: R14 & R15 = V/A= 12-1.8/ 20mA= 10.8/.020= 510ohms.

R13 & R16= V/A= 5-.7/5mA= 4.3/.oo5= 860ohms

Beta for transistor found on data sheet.

This circuit has high current flowing from the 12v supply to the collector of the transistors. When the 5v pwm signals come which is low current, it flows through the base to emitter, it opens a gate (so to speak) and allows the high current circuit to flow through the collector to emitter to ground. (or to injectors) or whatever other component you wish to switch on.

To test whether the circuit is working, simple volt drop test can be done across components. Looking at a data sheet for the transistor I found Vce= .05volts and Vbe= .08volts. I also know the L.E.D is 1.8volts. I checked these points with my multimeter and found Vce= 0.04volts for both transistors. Vbe= 0.83 from LED1 and 0.78volts from LED2. The voltage drop across LED1 was 1.86volts LED2 was 2.18volts. I found that all my results were good readings and all within specifications. I also checked the circuit by adding all my voltage drop readings up for each circuit and checking if it match Vs= 12.20volts=Vr14+ LED1+ Vce= 10.28+ 1.86+ .04= 12.18volts= Vr15+ LED2+ Vce= 9.97+ 2.18+ .04= 12.19volts. Vs=5.32volts= Vr13+Vbe= 4.50+.83= 5.33. Vs= 5.32volts= 4.49+ .78= 5.27volts.

When I done my circuit board I had no problems with design, and soldering it all together. It worked properly and first time.

If I had the chance to make it again I think I could make it abit more compact but not to compact that it could heat up fast and fail to work and maybe run earth straight to emitter instead of using jumper wires.

EXPERIMENT 8

Using different resistors at Rb: At Vce during the experiment the voltage didnt vary indicating low resistance high amperage, good switch. At Vbe voltage also didnt vary much from .7volts indicating good saturation. The current through the base had little change when Rb change due to cheap multimeter, but the concept is, more current flowing through the base will open up the gate wider letting the high current flow through the collector to emitter. (the red dotted line is the load line in the pic) Beta is the ratio between Ic:Ib. Beta = Ic/Ib. My RESULTS: 1- 3.1/5= .62beta 2- 5/4.37= 1.14 3- 5/1= 5 4- 104/21.7= 4.79 5- 5.1/4.5= 1.13

EXPERIMENT 7

TRANSISTORS: Using a multimeter and checking voltage drop across the base and emitter. My reading was .8volts indicating that there is low voltage being used and high current is still being generated. The voltage drop across the collector and emitter was .o6volts indicating little resistance in the boundary meaning its a good switch and well saturated. In fig13 the area coloured black is where you want your readings to be. low voltage high amps. (it is called the saturated region) The region coloured red is called the cut off region. The cut off region is where the transistor is not working properly. It is using to much voltage and also is not creating good amperage flow. (fully closed). The middle region is the active region where the transistor is working also. I learnt that using a transistor you can switch high current using low current through the base to emitter which then allows high current through from the collector to emitter to flow through.

EXPERIMENT 6

TRANSISTORS: There are two types of transistors positive (NPN) and negative (PNP) Using a multimeter in diode test mode you can determine the base emitter and collector also if its NPN or PNP. A NPN or PNP transistor is found when you get two readings on your meter with it having one leg in common. That one leg in common will decide what charge it has. If the common leg was when the red lead was touching it, it will be a NPN transistor. If it is the black lead it will be a PNP transistor. (the common leg will always be the base) To find the collector and emitter normally the highest reading of the two will indicate the emitter to base and the lower one will be collector to base. Transistors are used as a switch to use low current to switch high current. eg low current is passed through base to emitter and opens up the gate(so to speak) and lets the high current from the collector flow passed through the (gate) out to the emitter. (NPN transistor)

EXPERIMENT 5

CAPACITORS: Capacitors are voltage storage devices. They store charge until a switch is activated to discharge the capacitor. To calculated how long it takes to charge a capacitor is expressed- RxCx5=T eg 1000000ohms x .000001uF x 5= 5secs. Using an ocilliscope you can also calculate the charging time for a capacitor. Setting the ocilliscope to the correct settings 2volt per division and 250ms per division I was able to get good graph readings to calculate the charging time. I found in my results that using different sized resistors the charging time is affected. eg. it will take longer 2 charge a 330uF capacitor if you was to use a 100kohm resistor than compared to a 100ohm resistor the time will be shorter. Also changing the size of capacitor will affect the charging time because a bigger capacitor stores more voltage so it would take longer to reach full capacity. My RESULTS: #1: .ooo1Mohms x 100uF x 5 = 0.05secs #2: .00001 x 100 x 5 = .005secs #3: .00047 x 100 x 5 = .235secs #4: .0001 x 330 x 5 = .165secs. OBSERVED TIME: 1: 515ms 2: 67ms 3: 325ms 4: 1750ms