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.

Experiment 5 complete

For this experiment i had to build a circuit with 2 different capacitors rated at 100 uF and 330 uF, and 3 different resistors at 100 ohms, 470 ohms and 1K. Then I calculated the time the capacitor would reach full charge by using the formula T (Time) = R (Resistance) x C (Capacitance) x 5 (Number of Steps).



My first test was with the 100uF capacitor and 1K resistor. I calculated the amount of time for the capacitor to fully charge by using the formula T=RxCx5. (1 x 100 x 5) = 500ms to fully charge. Then using the oscilloscope I measured the actual time it took the capacitor to fully charge using the 200ms per division setting.



My next test was using the same 100 uF capacitor, but replacing the 1K for a 100R. 0.1 x 100 x 50ms to fully charge. Using the oscilloscope at the slowest pace possible (200ms per diversion,) I could only estimate the amount of time it took, because the capacitor fully charged before I had time to measure anything, so I just recorded my calculation of 50ms.

Next was the 100uF and 470R. 0.47 x 100 x 5 = 235ms. Measuring with the oscilloscope I recorded 200ms.

Now using a bigger capacitor (330uF) and replacing the 470R with a 1K I calculated the time this capacitor would fully charge. 1 x 330 x 5 = 1650ms. Then using the oscilloscope I meausered a time of 1800ms to fully charge.

Changes with the resistor size will effect the charging time by restricting more current or not restricting as much current. The bigger the resistor, the longer time it will take for the capacitor to charge because of the increased resistance. The smaller the resistor, the less time it will take to charge the capacitor from the decreased resistance.

Changes in the capacitor size will effect the charging time. If a bigger capacitor is used, the longer the time it will take to fully charge would increase. A smaller capacitor will take less time to fully charge.

Self Test #1 Complete

Sketch the symbals and label the leads for the two and three terminal devices

NPN.

MOSFET.

LED

ZENER DIODE
Q2. Fill in the blanks

Q3.

Q4: If V=IR and P=IV, what is P in terms of V and R

P=V2R

Vd = 0.7V. Diodes only require 0.7V to operate.

Q6, Vs=50V R1= 10Ohms Diode = 1W Max

Current Flow = (50-0.7)/10= 4.93 Amps

Q7, Pd=IdxVd Id=4.93, Vd=0.7 4.93 x 0.7 = 3.45W

Q8, Yes the diode will be destroyed. The maximum watt dissipation for the Diode is 1 watt and the total amount of watts this circuit will create on the diode will be too great for the diode to control.

Q9, Sketch a simple voltage regulator using discreet components

The zener diode acts as a 5V regulator, so any component placed on the same track as the zener diode will recieve 5V from the Zener Diode opposed from the 12v power supply.

Q10, Is the zener diode in the foward or reverse bias arrangement. Reverse

Experiment 7 complete

With experiment 7 I was asked to construct a circuit with a 15V power supply, a 1K Ohms resistor at Rc and a 10K Ohms resistor at Rb and a BC547 NPN type transistor.

This is a Common Emitter, or open collector circuit because the Emittor has a common connection to both the input and output.

With the circuit connected to a power supply, I connected my voltmeter across the Base and Emitter and recorded .801V. This is because the Vbe only requires 0.7 volts to operate.

I then connected my multimeter across the collector and emitter and recorded 53.7mV. This is because the transistor is saturated and fully open, letting current flow through Vce.

In the saturated state, the transistor is fully on, meaning the switch is closed and the collector and emitter junctions are in forward bias.The Cut Off region is when the transistor is fully off, or the switch is open meaning the transistor is fully off. From this graph I can see that If the Vce was at 3V, The power dissipated by using Pd=Vce x Ic. If the Vce is 3v, then the current at Ic is 14mA. Then Pd will be 3 x 14 = 42mW

Using the graph I can also get the current gain (Beta) of this transistor. If the Vce was 2 volts, Ib = 0.8mA and Ic = 20mA. Beta = Ic/Ib, so 20/0.8 25Beta 3v=14/0.5=28Beta 4v = 5/0.2 = 25 Beta

Experiment 6 complete

After obtaining 2 different transistors, one a BC547 and the other a TIP32C, I was asked to identify the legs on each transistor and state whether it was a PNP or NPN type. Because i've already explained how to identify a NPN or PNP type and identify which legs the base, collector and emitter were I could do it in a minimum amount of time.

With my multimeter I tested the BC547 transistor and found out that it was a NPN type, and then tested TIP32C and found out it was a PNP Type. Now that i knew which type both transistors were i could then measure the voltage drop across their 3 pins and record the data on this sheet.

Here is a photo of me measuring the Vbc by using the Diode test function on my TIP32C PNP type transistor. My negative lead is on the left leg, which is the base on this PNP type transistor, and my positive lead is on the middle leg which is the collector pin.

Note that my positive lead is on the mounting tab of the transistor. This is because the top terminal is also connected to middle pin, or leg.