Cicuit board 3-O2 sensor tester

Component list.
One LM324 quad opamp.
LED red
LED yellow
LED green
3 diodes 1N4007
1 Zener diode 9.1v
Capacitor x2 0.1uF
R2=1000
R3=1000
R4=1000
R5=380
R6=10000
R7=270
R8=470

Calculations
To find R8 and R7
9.1-(.23+.63)=8.24
8.24/3=2.74
2.74+2.74=5.48
270 and 550 ohm. but i used 470 ohms

Technical explanation
This circuit tests the O2 sensor in your car which will ultimately tell you if your car is running rich or lean. The lower the voltage the leaner the car is.
When the Green LED is on:
The Green LED is on when there is under .23V coming from the signal going into the 12 input. As the input is smaller than the .23 going into the 13 input the ouput at 14 negative, which switches the Green LED on. The yellow and red LEDs are off as there is less than .23V.

When the Yellow LED is on:
The signal voltage is between .23V and .63V. The signal goes into the 9 and .23V goes into the 10. As the signal is higher than the input voltage the output is negative and acitvates the yellow LED. It does not activate the Red LED as the input voltage is higher there. And does not activate the Green as the input voltage is to high for it and so there outputs would be positive.

When the Red LED is on:
There is more than .63V from the signal going into the 6 input. As this is greater than the .63 voltage input, the output is negative which activates the Red LED. As the signal input is also at 3 and the input voltage at 2 the output is postive which flows through D3 down to the negative side of the yellow LED, this shorts the Yellow circuit causing it to not light up when the Red is lit up. The Green LED will not light up also as it is getting a positive output, but it needs a negative to work.
Voltage drop readings.

	Red is on .65V	Yellow is on .3V	Green is on .1V
Red LED	1.99	1.34	1.34
Yellow LED	1.70	1.97	1.10
Green LED	1.25	1.24	2.08
R2	8.64	0	0
R3	8.49	8.12	0
R4	0	0	8.68
R5	2.14	2.15	2.12
R6	8.50	8.51	8.49
R7	0.22	0.22	0.22
R8	0.38	0.38	0.38
D1	9.10	9.10	9.10
D2	0.69	0.70	0.69
D3	0.65	9.04	10.43
D4	0.50	0.69	0.05

Problems with the circuit:
The first problem I had with my first board was I didnt break the tracks properly, so when i first connected it to the tester I shorted out the tester and my board as there was a short directly from negative to positive. From there it was problem after problem. The Green LED circuit worked, but the Red and Yellow didnt. I found another short accross the red and yellow circuits. I think when the first short occured I think that i shorted D2 as both the Yellow and Red LED would come on together.
On my 2nd circuit, I redesigned my circuit to have a common signal and common negative so to simplify the design. The first attempt at building the 2nd board was a success.

If I were to redo this board again, I would simplify the circuit once again. I would also double check the board before connecting it to a power supply. As the first short is what destroyed my board.

Circuit board 2

Component list
Diodes x2
LED
Capacitors x2
Zener diode 22v
Resistors carbon film resistor x3 160, 270 and 820 ohms.
LM317 Regulator
Red and black wires.

To find R1
5-1.8=3.2
3.2/0.02=160 ohms.

To find R2 and R3
5=1.25(1+R3/R2)
5/1.25=1+R3/R2
4=1+R3/R2
3=R3/R2
R2x3=R3
This is a 3:1 ratio.
I used a 270 ohm for R2 and a 820 ohm for R3. It is the closest 3:1 ratio I could get.

The circuits aim is to regulate 12v into 5v.
The Zener diode and normal diode act to protect the regulator by controlling voltage spikes and smooths the input voltage. They also protect the regulator incase of shorts, giving easier paths to earth. The capacitors also work in the same way by absorbing voltage spikes. The D12 diode protects the output of the regulator. The adjustment or negative side of the regulator controls the output voltage by the resistors on the negative side and across the negative and output. The LED shows that the circuits in on.

Vd readings,
D12=6.3 This is the difference between supply voltage and output voltage.
D13=0.7 This is the knee voltage of the diode.
D15=11.5 This is the voltage supply minus the knee voltage being regulated by the zener.
Vin to output=7.56 This is how much the regulator regulates to get to 5V from Vs.
R1=2.87 This uses up the leftover 5V that the LED doesnt use.
R2=1.25 Is the how much the resistor uses to flow through it.
R3=3.84 Is 3 times the voltage of R2 to adjust the output voltage to 5V.
LED=2.22 Knee voltage for the LED.


I had some problems with this circuit.
First of all I had a 220 ohm resistor in, instead of a 270 ohm.
Secondly I misunderstood how a regulator works. I thought that the output was negative and adj was the 5v. Because I misunderstood this, I then wired in the regulator the wrong way, but thinking it was the right way I spent all day trying to figure out the problem with this board. Once I realised that the Output is the signal and not the negative I fixed the problem in 5min.

If I were to redo this board i would compact the board, and I find out exactly how the components work that I'm not sure about.

Circuit board 1

Component list
NPN Transistors x2 C547
LED x2 yellow, white/red.
470 ohm resistor x2
820 ohm resistor x2
Red and white wires.

12-1.8=10.2
10.2/0.02=510 There are no 510 ohm resistors so I used 470 ohm resistors for R3 and R4.

5-0.7=4.3
Using the data sheet to find the current needed to saturate the transistor.
4.3/0.005=840 There are no 840 ohm resistors so I used 820 ohm resistors for R1 and R2.

The LEDs have a constant 12v power supply, but are not earthed. When the 5V power supply is pulsed to the base of the transistors which earths the LEDs power supply causing them to flash which acts like an injector turning off and on.

VD Readings.
VCE T1 0.05 Using this reading on the load graph I can find the Ic and B.
VCE T2 0.05 Using this reading on the load graph I can find the Ic and B.
Vbe T1 0.82 This is the knee voltage drop.
Vbe T2 0.83 This is the knee voltage drop.
R1=9.75 This resistor reduces Voltage input at the collector by 9.75.
R2=10.02 This resistor reduces Voltage input at the collector by 10.02.
R3=4.14 This shows how much the voltage is being reduced before the base.
R4=4.14 This shows how much the voltage is being reduced before the base.
All the readings were correct this time.
This board worked first time, if i was to redo this board again i would compact the components into a smaller portion of the board.

Experiment 8-4824

	Rb	Vbe	Vce	Ib	Ic
R1	4.7K	0.77	0.03	0.9ma	6.7ma
R2	22K	0.74	0.06	0.2ma	6.5ma
R3	27K	0.73	0.07	164µ	6.45ma
R4	330K	0.68	1.49	14µ	3.67ma
R5	1M	0.65	2.7	5µ	1.24ma

The Vce increased as the resistance increased. This caused the transistor to go from saturation to active, as there was less current to use.

There was close to no change at Vbe as the transistor always needs 0.7-0.8 to operate.

Ib decreased as the resistance increased. There became less current for the transistor to work with.

Ic also decreased as the resistance increased. This provided less current to be amplified.

My Graph of the results.

Using this formula; B=Ic/Ib I found the Beta at each Vce.

The Beta of this transistor at:

1. Vce of 0.03 6.7ma / 0.9ma B= 7
   
2. Vce of 0.06 6.5ma / 0.2ma B=32
3. Vce of 0.07 6.45ma / 0.164ma B=39
4. Vce of 1.49 3.67ma / 0.014ma B=262
5. Vce of 2.7 1.24ma / 0.005ma B=248
   

The graph can show us the areas of the activeness of the transistor. At the highest Vce the transistor is in the cut off area as it is almost off. At 1.5 Vce the transistor is active, but is still not fully active. The three Vce readings under 0.2 show that the transistor is saturated and is now fully active.

Experiment 7-4824

By using a voltmete accross the B to the E, I found a 0.8 Vd, this is indicating that the transistor uses 0.8 V to function. The 0.8 is its knee voltage.

The Vd accross the C to E is 0.05V. This V difference is very small. The more the transistor is on, the less the difference in Vd. The less its on the greater Vd accross the C to E.

The lower the amps at the base, the lower amps will be at the collector, therefore a high voltage drop. The more amps at C and B the more the transistor is on, so less Vd down to the 0.8 needed for it to operate.

So 'A' in the graph is in the region of Saturation and means it is fully working with minimal Vd, whereas 'B' is in the 'off' region and the transistor will not function.

The power dissipated by the transistor at Vce of 3Vd is:
Ic=13ma Vd=3
13x3=39W

The Beta of the transistor at Vce 2,3 and 4 is:

At 2Vd Ic=20ma ÷ Ib=0.8ma Beta=25

At 3Vd Ic=13ma ÷ Ib=0.5ma Beta=26

At 4Vd Ic=5ma ÷ Ib=0.2ma Beta=25

Experiment 6-4824

Transistors can be used as amplifiers or switches. They come in NPN type and PNP type, because of this there is often to way of telling which wire is B,C or E.To find the B, C and E of a transistor, I can use the Diode test function on my multimeter to find which is which wire.

Transistor number
----Vbe Veb Vbc Vcb Vcb Vce
NPN 745 OL 742 OL OL OL
PNP OL 748 OL 742 OL OL

Experiment 5-4824

Circuit number	Capacitance (uF)	Resistance (K)	Calculated Time (s)	Observed Time (s)
1	100	1	0.5s	0.5s
2	100	0.1	0.05s	0.05s
3	100	0.47	0.235s	0.235s
4	300	1	1.65s	1.65s

R x Capacitance in F x 5= total time to charge capacitor.

Eg.100-6 x 13 = 0.1

0.1x5 = 0.5s

Circuit 1

Circuit 2

Circuit 3

Circuit 4

Having a smaller resistor will result in having a faster charging time, provided the Vs and Capacitance stays the same.

Having a larger Capacitance will increase the charge time, just like as you would a bath to a pool at the same flow of water.

Experiment 4-4824

At 10V At 15V
V1=4.64 V1=4.81
V2=0.66 V2=0.70
V3=5.31 V3=5.51
V4=4.88 V4=9.86
I=0.00505 I=0.01024

The zener diode regulates both circuits to roughly 5V in reverse bias, but as there is more voltage in the 15V circuit there is more amps.

V1 shows the zener diode regulating 5v.
V2 shows the knee voltage of the regular diode.
V3 shows the combined Vd of the two diodes.
V4 shows where the resistor uses up the remaining voltage.

Experiment 3-4824

After building experiment 3's circuit, i had to find the value of Vz, for 3 different Vs. 10V,12V and 15V. After finding Vz for each Vs I found that no matter what the Vs the Zener diode would regualte the voltage down to 5V and then the resistors would use the rest of the voltage.

This type of circuit would be used to regulate voltage to other components.

When you set the zener diode in foward bias, it fails to regulate the voltage so to much voltage gets through.

Experiment 2-4824

Diodes work by using two types of materials, 'P' type material and 'N' type material. P type is more positive and N type is more negative. When in foward bias the materials are repelled from the power supply as they will not be drawn towards like atoms. This pushes the materials together until the gap is small enough to jump accross and carry on the flow of electricity.

A yellow LED has a voltage drop of 2.5 volts, a diode has a voltage drop of 0.6-0.7 in foward bias. In reverse bias they do not work and becomes an open circuit. To find the cathode on either of these components is easy to distinguish as the cathode is the shorter leg of the component. Also the strip side of the diode and the flat side of the LED is another way of disinguishing the difference.

To find the current flowing through the diode i used this formula. Vs-Vd=V V/R=I

With a 5Vs and a 1k ohms resistor the current flowing through the diode was 4.3ma. When i measured the circuit, I got 4.8ma.

Using Ohms law I calculated a right answer, but I didnt use the correct voltage, I measured the Vs after I had done these tests and found the Vs was 5.3 V. So i need to do the OCV test before every experiment.

I then calculated the Vd of the diode by reversing ohms law so that IxR=V Vs-V=Vd.
I calculated the correct Vd of the diode to be 0.7 V.

Using the data sheet i found that the maximum amperage that can flow through the diode is 1amp @ 75degrees celcius. The maximum of Vs is 1000 @ 75degrees celcius.

I replaced the diode with a yellow LED then calculated the current flowing through it to be 2.94ma and when measured I was right.

I observed that there is less amperage in a LED circuit than a diode ciruit as the LED uses more volts.

Experiment 1-4824

Picture from:
http://www.michaels-electronics-lessons.com/images/resistor-color-code-all.gif


My Estimate
Maximum/Minimum

1. 714000/646000 ohms Blue, Grey, Yellow, Gold
2. 10.5/9.5 ohms Brown, Black, Black, Gold
   
3. 10100/9900 ohms Brown, Black, Black, Red, Gold
   
4. 1575/1425 ohms Brown, Green, Red, Gold
5. 465000/313500 ohms Orange, Orange, Yellow, Gold
6. 1050/950 ohms Brown, Black, Red, Gold
   

Actual Reading

1. 691000 ohms
2. 10.5 ohms
3. 9960 ohms
4. 1488 ohms
5. 329000 ohms
6. 994 ohms

I calculated the resistance of the resistors, I then measured their resistance with my multimeter to confirm my calculations. All of the actual readings were within the tolerances.

R1= 1500 ohms R2= 1000 ohms
In seires I calculated 2500 ohms by adding 1500+1000 together, and when measured it was 2500 ohms.
In parallel I calculated 600 ohms by using the formula 1/RT= 1/R1+1/R2.

When measure it was 600 ohms.

I have demonstrated that parallel circuits have less resistance that seires circuits.