Oxygen Sensor Complete

Oxygen Sensor Write Up Parts Used

LED 35000 MCD Emitted Colour: Red Vf (V) Max: 2.6 If mA: 75 Vf (V) Type: 2.2

LED 35000MCD Emitted Colour: Yellow Vf (VMax:2.6 If mA: 75 (Vf) Type:2.2

LED 35000MCD Emitted Colour: Green Vf (V) Max: 2.6 If mA: 75 (Vf) Type 2.2

Electrolyte Capcitors x2

9V1 Zener Diode

LM324AN 14pin Intergrated Circuit

R1 390R +/- 5% 0.5 Watts

R2 470R +/- 5% 0.5 Watts

R3 10K +/- 5% 0.5 Watts

R4 270R +/- 5% 0.5 Watts

R5/R6/R7 1K +/- 5% 0.5 Watts

D1/D2 1N4001 50V Silicone Diode 1 Amp

Jumper Wires

Additional 1V power supply

How it Works

Connected with a 12V supply, current flows through the first diode. The voltage is then lowered to 11.3V due to the operating voltage of the diode. Current then flows through the capacitor and towards the first LED, dropping the voltage by a further 1.8V, (9.5V)which then goes towards a 1K Resistor, and then to the 7th pin of the IC. With the Yellow LED, current flows through another 1N4001 Diode, which drops the voltage by 0.7V (10.6V), and then through the LED itself dropping the voltage by 1.8V (8.8V), through another 1K resistor and then to the 8th pin of the IC. For the last LED, there are no additional diodes, so voltage drop only occours across the LED itself (9.5V), which goes through a 1K resistor and then to the 14th pin. When the board is connected to a 12v supply on the positive and negative rails, only the Red LED should turn on. There is a lead on the 12th pin, (called the sensor input) we may now alter which LED's will come on, depending on the varying 1V voltage. As voltage is increased to 0.3V, the red LED will turn off and the yellow LED will turn on. When voltage is increased to 0.7 V, the yellow LED then turns off, and then the green LED will turn on.

Fault Finding

My first board I constructed had faults due to bad soldering. My LED's would blow and components would not operate, so I completely redesigned my board to make it much more user friendly and less complex. When my second board was constructed, My LED's would not turn on when additional voltage was added. After hours and hours of trying to solve the problem, I realised my IC was wired up wrong, and had to change my design once more. With only 1 day left to complete this circuit, I constructed one more, only to find out that my new board shorted out when I connected it to a power supply. With no other alternatives I had to scrap my board and start my write-up.

Power Circuit Complete

Parts Used

LED RED 35000MCD
Emitted Colour: Red
Vf (V) Max: 2.6
if mA: 75
Vf (V) Type 2.2

Electrolyte Capacitor x 2
Radial 25V 33uF

D1 1N4001
50V Silicone Diode
1 Amp
D2 iN4001
50V Silicone Diode
1 Amp
D3 Zener Diode
5v1 1 Watt

R1

100R +/- 5% 0.5 Watts

R2

330R +/- 5% 0.5 Watts

R3

1K +/- 5% 0.5 Watts

Voltage Regulator LM317T
Output Current in Excess 1.5A
Output Adj 1.2V - 37V

Jumper Wires
2cm x 3

How it works:
This Power Circuit is very similar to a Voltage Regulator circuit we constructed, apart from a few new components such as the Zener Diode and a different type of regulator. 12V is supplied to the positive rail and enters the input pin of the regulator.There is a 0.7 VD across D1 from being in forward bias which drops the voltage to 11.3. D2 in reverse bias prevents any current flowing through it which will prevent the components from being destroyed if they are connected the wrong way. The zener diode and capacitors are unneccesary and don't play any role into regulating the voltage but as filters for the regulator. The Resistors in the circuit determine how much voltage will be regulated from the output by using Vout = Vref (1+R3/R2). If Vout has to equal 5v, I could use any combination of resistors, as long as one is 3x higher then the other. In this case I used a 1k and a 330R. With this resistance I should now get an output of 5v, which can now pass through my 110R and LED, lighting it up. This circuit has 3 wires at the end of the Positve 12v supply, the output of 5V and Earth. Meaning I can power anything with my board with 5V, when connected to a 12V supply or more.

Fault Finding:
Originally I tried copying my circuit diagram as close to the schamatic as possible in order to minimize confusion as to where parts would go and how much board I would need. With an almost replica circuit diagram I could feel convinced my circuit would work. However when it came to constructing the board, some problems occoured when my board was short circuiting due to bad soldering. After cleaning the solder up, I wanted to check the condition of the Voltage Regulator. As I tryed heating up the board up in order for me to remove it, I accidently broke the Output pin on the regulator, and stripped all the copper around it, creating an open circuit. I could of relocated my regulator by use of jumper leads, but decided to go back to the drawing board and create a more spaced circuit, to prevent components shorting out. With my new, more efficient drawing, I then began constructing and did not run into any problems. Now that my board was complete I could test that the output of my Regulator would match the required output on the Circuit Diagram. With my board connected to a 12Vs, I measured the output with my Voltmeter and recorded 5.10V, Meaning the experiment was a success.

Injector Circuit Complete

Parts Used
LED 5MM YLW 35000MCD
Emitted Colour: Yellow
Vf (V) Max : 2.6
if mA : 75
if mA (peak) 100
Vf (V) Typ: 2.2

LED 5MM WTE 35000MCD
Emitted Colour: White
Vf (V) Max: 2.6
if mA : 75
if ma (peak) 100
Vf (V) Typ: 2.2

R1 x2

470R +/- 5% 0.5 Watts

R2 x2

47R +/- 5% 0.5 Watts

BC547 NPN Type Transistors x2

Jumper Leads 2cm x2

How it works:
Connected to a Vs of 12 across the Positive lead and Negative Lead, It appears that this circuit does not work. That is because an additional power supply of +0.7 V needs to be connected to Vbe . The additional power supply is needed because with no voltage across Vbe, the transistor switch will not close. With the additional supply added, The user can switch what light to turn on by alternating the positive lead on the inputs.

Fault Finding:
In order to make sure my components would not be destroyed, i needed to calculate a suitable Resistor to restrict the current flow. With a Vs of 12 and 25mA flowing through the LED's, I could then use Ohms Law to calculate the Resistance needed to prevent the LED's blowing up.

But I also had to take into acount the Voltage Drop across Vce, which is around .5V. So 11.5/0.025= 460R. The closest resistor I could get to 460 was 470, so I decided to use a 470R.

Then I had to calculate the resistor I would need for the base of the transistor. As there is a 0.7 voltage drop across Vbe, I had to subtract this from 5V. My first mistake was when I divided 4.3 by the amount of current flowing through Vbc, which was 25mA, when infact I needed to divide 4.3 from the current gain of the transistor. I needed to obtain the Beta by loooking online for the datasheet. The current gain of the BC547 transistor had a range of 110-800. I used the minimum of 110 and divided this from the 4.3V supply and got 39. I had the option of using either a 33 or 47 Ohms resistor and I chose to use the 47 Ohms. After constructing my circuit I noticed my white LED would turn on when connected to the positive and negative supply, which it shouldn't because the transistor is preventing current to flow through it until a second power supply has been added to the circuit. After doing some fault finding I realisised my transistors were faulty and needed replacing. Then I constructed on a new board due to solder being everywhere from trying to pull the transistors out, but this time I only constructed half of the circuit. This way I could test my circuit quicker, and identify faults with less components. When I connected my power supply to the positive rail and to earth and noticed the LED were not on I knew the transistor was working. Then, connecting the 5V supply to the circuit I could make the transistor turn on and let current flow from Ic-Ie. This would turn the LED on, and it did. I then constructed the other half of the circuit and got the circuit running properly.