The Oxygen Sensor Tester

This is my own work and represents my learning in this assessment. Any information sourced from elsewhere has been appropriately acknowledged and referenced. I have and will continue to maintain the privacy of any person I have referred to in this assessment and have sought their permission to have their information shared here. I acknowledge a copy of my work may be used for moderation purposes. I have a back up copy of the work I have presented in this assessment if it is required.

Vladimir Borzenko

References:   http://www.electronics-tutorials.ws

                       http://www.allaboutcircuits.com

                       http://www.circuit-magic.com

                       Automotive Electrics & Electronics Gregory’s First Edition 1999

The circuit diagram is on the photo:

The are two ways to calculate resistors for  this operational amplifier. R6, R7 and R8 are divider’s resistors. So, they can be calculated via the voltage drop formula.  V out/ V in = R bottom/ (Rbottom+Rtop). For our diagram (Va-Vb)/Va=R6/(R6+R7+R8). From this we have :

R7+R8=(Va*R6)/(Va-Vb)-R6=(9.1*10,000)/(9.1-0.63)-10,000=743 Ohms

Rtotal=R6+R7+R8=10,000+743=10,743 Ohms
R7=Rt*Vc/Va=10,743*0.23/9.1=271 Ohms

R8=((Vbc*Rtotal)/Va)-R7=((0.63-0.23)*10,743)/9.1V)-271Ohms=471 Ohms

I3=Va/Rtotal=9.1/10,743=847 uA

Second variant of R7 and R8 calculation:

I3=Vdab/R6=(9.1-.63)/10,000=847uA

R7=Vd/I=.23/847uA=271 Ohms

R8=(0.63-0.23)/847uA=471 Ohms

I1=I3+I2, where I2=IzRm=5.6 mA

I1=847uA+5.6mA=6.447mA

R5=(VRaw-Vdd2-Va)/I1=(12-.6-9.1)/6.447mA=356.76 Ohms

R2=(VRaw-Vdd2-Vdled)/Iled=(12-.6-1.8)/9.5mA=1010.5 Ohms

R3=(VRaw-Vdd2-Vdd4-Vdled)/Iled=(12-0.6-0.6-1.8)/9.5mA=1010.5 Ohms

R4=R2=(VRaw-Vdd2-Vled)/Iled

R4=R2=(12-.6-1.8)/9.5mA=1010.5 Ohms

Thus, results for resistors in accordance to E24(5% tolerance) row:

R2=R4=1 kOhm

R3=910 Ohms

R5=360 Ohms

R7=270 Ohms

R8=470 Ohms

layout of components was drawn with Lochmaster programme:

Voltage measurements and diagramm reading on the O2 senser tester circuit board:

Voltage input of 12.05 V is rectified by D2 and smoothed by C1. As a result, voltage at the LED’s anodes is 11.34V. From the very beginning I must mention that light difference in voltages between wire diagram and measured figures occurred due to electronic components and measuring equipment tolerance. From the figures above it’s clear that Vs is slightly higher than on the paper. For all the reason above, the zener diode D1 stabilises voltage at the point A at 9.09V. Current flows through divider circuit.  There is 0.67 V after R6 at the point which I named B and on the pins #2 and #6. There is 0.27 V after R8 at the point C. That gives the same potential on the pins #10 and #13. It’s clear from the first glance that op-amps are connected in pairs in a such specific way when incoming signal goes onto the non-inverting input of the upper op-amp and onto the inverting input of the low op-amp. Incoming signal from the PL2 contact comes onto the pins #12, #9, #6, and #3 simultaneously. If this signal level is within 0 - 0.27 V intervals there is 0.941 V on the pin#14 which is V out of the lower op-amp. Thanks to that the LED6, Green one, is forward biased. It glows. Once the incoming signal reaches 0.27 V and goes higher it overcomes voltage on the non-inverting input of op-amp. This voltage difference is amplified and as a result there is 10V on the pin#14, outcome of the low op-amp. Since this moment the LED#6 is no more forward biased. It’s off. At the same time this signal overcomes voltage on the pin#10 and comes onto the pin#9 which is inverting input for the op-amp connected to the Yellow LED5. At this moment op-amp amplifies voltage difference. But it inverts the incoming signal. For that reason the D5 is forward biased and glows. This figure remains negative but the highest op-amp “closes” LED5 forcibly after 0.67 V level of incoming signal. The #1 pin is connected through diode D3 straight away to the LED5 cathode. It gives high potential at the LED5 cathode. From this time Yellow LED is no more forward biased and it’s off. Once the input signal reaches 0.68 V and rises higher it allows to have the inverted outcome on the pin#7 what provides forward bias for Red LED1. It is on since this moment. The Yellow LED is off because of reason mentioned above. Resistors R2, R3 and R4 are limiting resistors.

The table below  supplements voltage reading. It helps to check voltage for falts finding also.

Pin#	Green LED on	Yellow LED on	Red LED on
1	0.006	0.003	10
2	0.7	0.7	0.7
3	0.3	0.3	0.3
4	11.56	11.57	11.54
5	0.7	0.7	0.7
6	0.3	0.3	0.3
7	10.3	10.27	0.9
8	10.3	0.9	0.9
9	0.3	0.3	0.3
10	0.3	0.3	0.3
11	0.0001	0.0001	0.0001
12	0.3	0.3	0.3
13	0.3	0.3	0.3
14	0.95	10.27	10.27

These photos illustrate that diodes change one another from the Green one to the Red one with increasing input signal. This proves that all the calculation and the diagram reading are correct. This tester can be used to check the O2 sensor output and conclude if fuel mixture is reach or lean.

Board#2 5V Voltage Regulator

This board is designed to build a voltage regulator with V out = 5V. The main component is regulator LM 317T. Values of R3 and R2 must be calculated. All calculations are made on the basis of voltage deviders theory and formula Vout= Vref *(1+R3/R2), where Vref=1.25 V

 

All calculations are on the photo below:

LIST OF COMPONENTS

Diodes2x 1N4001;Zener diode; 2 Electrolytic capacitors 33uF, 25V; LM317T Voltage Regulator; Yellow LED, 20 mA; R1=160 Ohms; R2=270 Ohms; R3=810 Ohms.

N.B. Initially, calculations were done for the Red Led with voltage drop which differ from the Yellow LED. But I was supplied with Yellow LED to build the circuit board. As a result, voltage drop measured across the LED gives higher value than it must be.

Voltage drops at the bottom of the last photo describe how this voltage regulator works: 12v input signal comes through the half wave rectifier (D13) with voltage drop 0.699 what tells us that it works at the 'knee voltage'. Rectified signal is stabilized by Zener diode with Vd =10.93V and smoothed by C15. This 10.93V signal comes to the input of Regulator LM317T. The output of the Regulator is 5.13V. Voltage for the adjustment electrode of the Regulator is preset by R2 and R3 devider.D12 diode protects the LM317T regulator from the storage charge in the capacitors at the moments of switch- on or switch-off modes. The outcome signal is once again smoothed by C16 and provides voltage for the LED via limiting resistor R1. V drop across R1 is 1.217V.Voltage drop across Yellow LED is 3.9V.In practice, all these fluctuations are due to preliminary calculations for the RED LED as I mentioned above and also because of elements tolerance.

LM317 is one of the most common regulators which can be used to provide stable 5V signal for electronics devices. Calculation and implementation are simple and affordable.

Board1 Injector Simulator.

 Image. Wiring diagram.

Red LED and  Green LED were chosen for this circuit. According to the LED datasheet forward current for RED LED 10ma<I<30mA
GREEN LED 10mA<I<25mA
Let'assume that working current for both LED's is 20mA.

Vdrled- voltage drop for Red LED is 1.785 V
 Vdgrled-voltage drop for Green LED is 1.755 V
 Transistor BC547A is chosen for this circuit. From this transistor datasheet:
  Vbe=700 mV
  Vce=90 mV
  Thus: R14=(Vcc-Vdrled-Vce)/Ic=(12V-1.785V-0.09V)/0.02A=506 Ohms
From resistor value table choose the resistor with 5% tolerance. This is E24 series.
Therefore, R14=510 Ohms
R15=(Vcc-Vdgrled-Vce)/Ic=(12V-1.755V-0.09V)/0.02A=507.75 Ohms
For the same reason as it is mentioned above in R14 calculation, choose resistor 510 Ohms > R15=510 Ohms
At the next stage let's calculate Ib=Ic/b, where b=Hfe- current gain
b varies from 110 to 800 for this type of transistors. To be on the safe part of the road choose b=110. That means transistor will work under minimum conditions. If b is higher the transistor will work for sure.
Ib=Tc/b=0.02A/110=0.18 mA
However, returning back to the datasheet, we can see Ib=0.18 mA is not enough to get the transistor hard saturated. According to the datasheet for BC547A Ib must be 5mA.
Thus, R13=R16=(Vpwm-Vbe)/Ib=(5V-0.7V)/0.005 A= 860 Ohms
Choose R13=R16=910 Ohms from the resistor value table
Limiting resistors which were calculated will provide saturation mode.
The results are :  R13=R16=910 Ohms + - 5%
                          R14=R15=510 Ohms + - 5%
                          T1=T2= BC547A


Board for this circuit is on the photo :
This is a working board examle. Soldering has revealed that power supply wire was covered by oxide . Ideally, all these wire must have been treated with flux in order to remove oxidation before soldering.
Other components were soldered with common bar solder.
The measurements on the diagramm show volatge drop accross each device.
It is obvious that voltage drop accross current limiting resistors provides accord to the calculation above with reference to the components tolerance. Both transistors are saturated. That's why for the transistor T1 Vbe=0.817V and Vce=0.041V. For the transistor T2 Vbe=0.818V and Vce=0.042V. Voltage drop accross LED1=2.129V and LED2=2.136V what allows them to glow brightly.

 

Reflection: In this particular case the elements layout is a bit irrational. If I could rebuild it I would do it in more compact way. There is no high demand for heat exchange because the transistors work in the saturation mode. 

Practical Workbook Experiment #6 BJP TRANSISTORS

It's clear from all measurements above how to identify the transistor's electrodes.Vbe is always higher than Vbc. Once you pay attention what is the polarity of multimeter lead you attach to the base and collector or emitter you can easily find what type is the transistor. For NPN type if positive lead is attached to the base and negative to collector or emitter you have Vbe and Vce reading. For PNP transistor picture will be opposite.Negative lead is attached to the base and positive to collector or emitter. Thus, you can check if the transistor is working and also identify its type and mark the electrodes.

 Vbe shows that transistor base to emitter junction voltage drop slightly exceeds so-called 'knee voltage'.
Vce is that small because the transistor is fully opened.

In the A-region transistor is saturated. That means it's fully opened and Ic reaches its maximum while Vce is in minimum.
In the B-region, which is called cut off, transistor is closed.
To calculate power dissipation we can draw  lines to the load line on the graph. Then, we can find the Pd by multiplying Vce and Ic.
Beta that refer as Hfe as well, can be find from this graph also.
Hfe=Ic/Ib. The same technique of drawing lines to the vertical and horizontal axes is used to determine the Ic and Ib.

In Experiment No 8
we changed the Rb and checked what were changes for Vbe, Vce, Ib, Ic.

Rb is the limiting resistor in this case. By increasing Rb, we limit Ib from 90uA to 10uA. The lower Ib the closer transistor to the cut off region. That's why Vce rises during this experiment. We know that Ic is beta times Ib. So that, it goes down. To simplify, we experimented how transistor can be controlled by Ib.

The graph that is drawn above on the basis of the measurements from this experiment allow us to see how all these transistor's characteristics are interrelated. Another advantage of this graph is that it makes easier to understand how transistor change its working mode.The load line also allows to model amplifier and to see input and output signals relation.

Practical Workbook. Experiments on resistors, diodes and capacitors.

Experiment1
 Resistor Colour Code helps to identify resistors; and several simple circuits in series and parallel connection refreshed the Ohm's Law application. I've found that even funny proverb helps to keep in mind row of numbers and write down tha table of resistors value . It's easy to remember and can be usful.

From the table above which is based on the resistor value calculations and experimental measurements I concluded that only one resistor failed because of poor quality.


Experiment2-Experiment4
All our experiments on diodes from the practical workbook showed that voltage drop across diodes is essential feature of this electronics components.Applying this for the calculation from the data sheet or measuring this in particular circuit and studying diagram we can read the whole picture about what is happening with electrical signal at the moment.
Another fundamental rule which is known as  Kirchhoff's current law we applied in Experiment 3.
Along with Kirchhoff's voltage law and voltage drop these are three the most offten applicable rules for circuit reading in electrical and electronics engeenering.








R= (Vs-Vz)/Is; Is=Iz+Il  where Iz from the data sheet. Il=Vz/Rl.




V1=Vz; V2=Vknee(0.7v or 0.6 v depending on the diode properties) and V4=Vs-Vz-V2.


Experiment5 The experiment on the capacitors not only reminded us that this device is used for timing but also demonstrated how the charging time can be changed. The smaller resistor rating the faster capacitor will be charged. Once the capacitor was changed onto the bigger one the charging time increased. The patterns on the oscilloscope screen demonstrate how the charging process goes. It's easier to see 5 stages as well as percentages of charge. The hihger charge the longer it takes because it's harder to move electrons.
Depending on the capasitors construction charge can be stored either for a long time or for a while due to leakage.
R*C*5=T where R is resistance in Ohms, C is capacitance in Farads and T is time in seconds.

experiment1