Oxygen sensor circuit

Today in practical session we started working on building our next project, which is an oxygen sensor. The first step was to calculate the value of the resistors to go into the circuit.

Calculations

The first resistor I calculated using ohms law was R5. To get this value I had to use the formula R = V/A which is 12V – 0.6 (voltage drop across D2) – 9.1 (Zener-diode D1) equals 2.3v. The Amperage is shown on the sheet, which is 5.6mA that I have to convert to Amps so it would be 0.0056A. Therefore R = V/A = 2.3/0.0056 = 410.71 ohms. So for R5 I would use a resistor around this value.

The next resistor I calculated was R8. So R = V/A. the voltage I will be using is the voltage drop across this resistor and to work this out I had to minus 0.63v from 0.23v which were the available voltages before and after R8. Therefore the answer is 0.40v. The current is calculated using the formula A = V/R but before we can move on I have to calculate the current flow by using the values of R6. 9.1v is the available voltage before R6 and 0.63v is the available voltage after R6 (9.1 – 0.63 = 8.47v) so the voltage drop across this resistor is 8.47v, which is used for the voltage to calculate the current. The value of R6 is already known and is 10K. Therefore the current flow is A = V/R = 8.47/ 10000 = 0.000847A. So back to working out the value of R8 we now have the current to finish of the formula R = V/A. so 0.40/ 0.000847 = 472 ohms. So for R8 I will be using a resistor around this value.

To calculate the value of R7 I used the formula R = V/A. The available voltage is 0.23 before R7 and 0v after it so the voltage drop will be 0.23v and the current will be the same as R8, which is 0.000848A. So the value of R7 will be 0.23/0.000848 = 271ohms.

Now I will be calculating the values of R2, R3 and R4. So to calculate R2 I need to use R = V/A. First I need to work out my voltage I will be using and to do this I need to minus 0.6v which is the voltage drop across D2 IN4001 diode and 1.8v which is the voltage drop across LED1. I need to minus this from 12v so (0.6 + 1.8 – 12) = 9.6v. The current is already shown on the sheet and is 9.5mA that I have to convert to Amps which is then 0.0095A. So now I can calculate the value of R7. R = V/A = 9.6/0.0095 = 1010ohms.

The next resistor is R3. I will be using R = V/A to calculate its value. First I need to workout the voltage drop across this resistor. So there’s 12v coming in and 0.6v drops across D2 and another 0.6v drops across D4 and 1.8v drops across LED 5 and added up equals (0.6 + 0.6 + 1.8 – 12) = 9v. The current remains 9.5mA (0.0095A). So the value of R3 is R = V/A = 9/0.0095 = 947ohms. I don’t know if there is a 947ohm resistor, so Il be using the nearest 1 to this value which would probably be 1K or 1000ohms.

The final resistor to calculate is R4.  Its just like above were I have to add up the voltage drops before the resistor. So 12 – 0.6 + 1.8 = 9.6v.  The current is still the same in this part of the circuit, which is 9.5mA. Therefore R = V/A = 9.6/0.0095 = 1010ohms.

Component list

After calculating the values of R5, R8, R7, R2, R3 and R4 I had to show the tutor my workings of how I got these values. Once the tutor checked it I then moved on to build the circuit on a breadboard. The components I used are

Three 1N4001 Rectified diodes for D2, D3 and D4, according to the relevant data sheet these diodes have a maximum Pd of 2.5W, max current is 1A  @ 75°C and max reverse voltage of 100V.

One 9v1 zener diode for D1

Two 0.1 uF capacitors for C1 and C2

Three LED’s, one red for LED1 (I will use this to show when I’m running lean), one yellow for LED5 (to show when I’m running in between lean and rich) and one green for LED6 (for when I’m running rich or when accelerating because green represents GO)

A 380ohm resistor for R5

A 10,000 or 1Kohm resistor for R6

A 470ohm resistor for R8

A 270ohm resistor for R7

Wires


Op Amp


Below are pictures of the circuit i built on bread board:

After building this circuit on bread board i then tested it and it worked. I will go into more detail with a video i will be recording next week because i had no time in class to do it on friday. I also finished building this circuit on the green board which I still need time to record a video and show how the circuit works. But for now I will include a topic from ''subjects to study'' from moodle to add to my blog.

Zener Diodes

A zener diode is just like a normal rectifying diode that only allows current to flow in one direction from anode to cathode (forward bias) when a voltage of 0.6v is reached to open the gate and allow electrons to flow. But with a zener diode current can also flow in reverse direction (reverse bias) when the voltage is above a certain value, for example 6volts. This is called the break down voltage known as the zener voltage.

A zener diode can be used to make a simple voltage regulation circuit. The output voltage is fixed at the zener voltage of the zener diode used and so can be used to power devices requiring a fixed voltage. If you have a regulated fixed voltage - say 12 Volts from a power supply, and you want to power something requiring a lower voltage, it is possible to simply place a zener diode in series with the load device. You would choose a diode with a zener voltage equal to the supply voltage minus the voltage drop across the load.

For example, if you have a 1W 6V lightbulb to power from a 12 Watt regulated power supply, a 6.2V zener diode could be placed in series with the bulb giving the bulb 5.8V, or you could overpower the bulb a little using a 5.6V zener diode and dropping 6.4V across the bulb. Heat would be generated in the zener diode so it is essential to calculate the power lost in it so a suitably rated diode could be chosen.

Below is a picture of a zener diode I got of the internet.

http://www.reuk.co.uk/What-is-a-Zener-Diode.htm

Injector circuit.

In practical class today we got handed a sheet with an injector circuit that we had to calculate what value resistors we had to put in the circuit and what the available voltage and voltage drop would be after each

component. We also had to find out how much current flow there will be going through the diodes and transistors and check that reading with the data sheet to make sure we don't exceed the maximum values.

Component list

The components I used in creating this circuit were calculated first on a piece of paper we got handed. Ill show why I picked these particular components under my next heading Calculations. So I used four 470 ohm resistors, two 1N4001or 7 Light Emitting Diode’s and two Bipolar Junction NPN BC547 transistors. I built this circuit first on a breadboard shown in the picture below. 

After building the circuits on breadboards we used a software tool on the computer called Loch master 3.0 were we practiced making the injector circuits on PCB boards. It includes all components al l we had to do was select what component we needed then soldered it to the board. The good thing about this tool is that it allows you to practice building a circuit on the computer were you could easily change things around and figure out were your going to solder the component so that its tidy and evenly spaced so that everything fits properly. After designing the circuit on the computer we used it to build it on an actual board made with copper strips in lines to conduct electricity. The boards are about 50 x 40mm long with holes to insert components to solder. Below is a picture of my board I designed on loch master

Technical explanation of how this circuit works:

Resistors R1 and R3 have one leg connected to a 12v supply and the other leg to the anode side of LED’s D1 and D2 were a voltage drop of .6v is used to turn the led’s on. The cathode probes are connected to the collector terminals of the transistors. R2 and R4 are resistors connected to a 5v modulator, which acts as the ECU sending output pulses to the LED’s, which are meant to be the injectors. In creating this circuit I had to make cuts in between resistors R2 and R4 so voltage can be divided through the resistors to allow a current flow to the base terminals of the transistors. Both the emitter terminals are connected to earth.

This picture shows what it looks like on the other side of the board were soldering the components is done.  

I then made this circuit on to a pcb board.

Calculations

The PD across LED’s 1N4007 is calculated using the formula, pd = ID x VD and A = V/R from ohms law. This formula is used to see if we can use the type of resistor for this diode so that it doesn’t damage the LED.  LED’s have no resistance in them so it important to use the right resistor.

First the amperage flow must be calculated through the LED.

I = V/R therefore voltage is 12 – 0.6 / 470 = 11.4/470 = 0.02A

So now that I have amperage flow I can use this to calculate the power rating of the diode and check with the data sheet if it can handle this amount. Therefore Pd = ID X VD = 0.02 x 0.6 = 0.12 W

According to the data sheet for this led the total device dissipation above 25°C is 2.5W and the average current is 1.0A @ TA 75°C. 

Below is a video of my injector circuit connected to a 12v supply and a 5v modulator connected to the base of the transistor to show how the ecu will change the frequency of the pulse width of the injectors, in this case the LED's. In a car if i started to accelerate the ecu will send signals to the injectors and the injectors will spray at whatever signal they receive from the driver. In my video below i change the frequency high and low which in turn changes how the LED's flash. As i increase the frequency the led's start flashing faster and faster until i have fully increased it, and at this point the LED's stay constantly on and you cant really see them flashing only a tiny but if you look hard. 

Reflection


In making my injector circuit if i had the opportunity to make it again i will and make it better. I will spend more time on loch master designing my circuit so that it looks more professional when I make it. On the other hand I had a fair idea of how the injectors will spray by watching how the LED's flashed. 

EXPERIMENT No.8

EXPERIMENT No. 8

Summary: Vary the base resistor and measure changes in voltage and current for Vce, Vbe, Ic, and Ib. Then plot a load line.

Set up the following circuit on a bread board. Use a 470R for Rc and a BC547 NPN transistor.

Pick five resistors between 2K2 and 1M for Rb. You want a range of resistors that allow you to see Vce when the transistor is the saturated switch region and when it is in the active amplifier region. I used 47K, 220K, 270K, 330K and 1M, but this can vary depending on your transistor. Some may need to use 2K2. Put one resistor in place, and measure and record voltage drop across Vce and Vbe. Also measure and record the current for Ic and Ib. Then change the Rb resistor and do all the measurements and record the new readings. Do this for each of the resistor values above.

  Record here:

            Rb 47K            Vbe 0.69V       Vce 0.08V       Ib 0.02mA        Ic 6.41mA

Rb 220K          Vbe: 0.67V      Vce: 0.15V      Ib: 0.03mA       Ic: 6.24mA

Rb 270K          Vbe: 0.69V      Vce: 0.20V      Ib: 0.03mA       Ic: 6.21mA

Rb 330K          Vbe: 0.67V      Vce: 0.30V      Ib: 0.03mA       Ic: 6.01mA

Rb _____         Vbe: _____     Vce: _____     Ib: _____         Ic: _____

Your voltage drop measurements across Vce should vary from below 0.3 v (showing the transistor is in the saturated switch region) to above 2.0 v (showing the transistor is in the active amplifier region) If this is not the case, you may have to try a smaller or bigger resistor at Rb. Talk to your teacher to get a different size resistor, and redo your measurements.

 Discuss what happened for Vce during this experiment. What change took place, and what caused the change? 


During this experiment the different resistors caused different voltage drop readings at collector to emitter (Vce). The base resistors take up most of the voltage, which is 4.3 volts and the supply voltage is 5 volts. So the base resistors (Rb) take up 4.3 volts, so the voltage are low at 'Vce'...

Discuss what happened for Vbe during this experiment. What change took place if any, and what caused the change?

During this experiment the voltage at base to emitter (Vbe) got around .7 of a volt like a diode in its forward bias of 0.7 volt drop across it. So the resistors took up 4.3 volts and 0.7 volt remaining is used to push current from the base to emitter. So the resistors voltage and the voltage - base to emitter added up to the supply voltage which was 5 volts.

Discuss what happened for Ib during this experiment. What change took place, and what caused the change?

 Discuss what happened for Ic during this experiment. What change took place, and what caused the change? 

Plot the points for Ic and Vce on the graph below to create a load line. Plan the values for so you use up the graph space.  Use Ic as your vertical value, and Vce as your horizontal value.

Using Vbe on the Vce scale, plot the values of Ib so the finished graph looks similar to fig 13

Calculate the Beta (Hfe) of this transitor using the above graph.

  

 Explain what the load line graph is telling you. Discuss the regions of the graph where the transistor is Saturated, Cut-off, or in the Active area. 

EXPERIMENT No.7

EXPERIMENT No. 7

Transistor as a switch

Components: 1 x Small Signal NPN transistor, 2 resistors.

Exercise:  Connect the circuit as shown in Fig 12 and switch on the power supply.

Fig 12

Connect the multimeter between base and emitter.

Note the voltage reading and explain what this reading is indicating.

The voltage between base and emitter is 0.8v which is the voltage required to push electric charge through the insulating layer from the base to emitter. A small current appears.

Connect the multimeter between collector and emitter.

Note the voltage reading and explain what this reading is indicating.

The voltage between collector and emitter is 0.6v which means the base has free holes for the electrons to move along and are closer so you don’t need a higher voltage like .8v to move electrons. A higher current appears (20mA) because electrons can move/flow easier.

In the plot given below what are the regions indicated by the arrows A & B?

Fig 13

How does a transistor work in these regions? Explain in detail:

A – Cutoff = Transistors that are fully off

B - Saturated = Transistors that are fully on

What is the power dissipated by the transistor at Vce of 3 volts?

P= I x V

   =15 x 2.7

   =15mA x 2.7(Vce)

   =0.4micro W / 400 mW

   =0.0405 Watts

What is the Beta of this transistor at Vce 2,3 & 4 volts?

2Vce - B=Ic/Ib = 20mA/0.8mA = 25

3Vce - B=Ic/Ib = 15mA/0.6mA = 25

4Vce - B=Ic/Ib = 5mA/0.2mA = 25

EXPERIMENT No.6 Meter check of a transistor

EXPERIMENT No. 6

Meter check of a transistor

Bipolar transistors are constructed of a three-layer semiconductor “sandwich,” either PNP or

NPN. As such, transistors register as two diodes connected back-to-back when tested with a

multimeter’s  “diode check” function as illustrated in Figure 9.1. Low voltage readings on the base with the black negative (-) leads correspond to an N-type base in a PNP transistor. On the symbol, the N-type material corresponds to the “non-pointing” end of the base-emitter junction, the base. The P-type emitter corresponds to “pointing” end of the base emitter junction the emitter.

9.1: PNP transistor meter check: (a) forward B-E, B-C, voltage is low; (b) reverse

B-E, B-C, voltage is OL.

BIPOLAR JUNCTION TRANSISTORS

Here I’m assuming the use of a multimeter has a diode test function to check the PN junctions.

If your meter has a designated “diode check” function, and the meter will display the actual forward voltage of the PN junction and not just whether or not it conducts current.

Meter readings will be exactly opposite, of course, for an NPN transistor, with both PN

junctions facing the other way.

 Low voltage readings with the red (+) lead on the base is the “opposite” condition for the NPN transistor. If a multimeter with a “diode check” function is used in this test, it will be found that

the emitter-base junction possesses a slightly HIGHER forward voltage drop than the collector

base junction. This forward voltage difference is due to the disparity in doping concentration

between the emitter and collector regions of the transistor; the emitter is a much more heavily

doped piece of semiconductor material than the collector, causing its junction with the base to

produce a higher forward voltage drop.

Knowing this, it becomes possible to determine which terminal is which on an unmarked transistor.

This is important because transistor packaging, unfortunately, is not standardised. All

bipolar transistors have three terminals, of course, but the positions of the three terminals on the actual physical package are not arranged in any universal, standardised order.

Suppose a technician finds a bipolar transistor and proceeds to measure voltage drop with a

multimeter set in the “diode check” mode. Measuring between pairs of terminals and recording the

values displayed by the meter, the technician obtains the data in Figure 9.2.

Figure 9.2: Unknown bipolar transistor. Which terminals are emitter, base, and collector?

-meter readings between terminals.

The only combinations of test points giving conducting meter readings are terminals 1 and 3

(red test lead on 1 and black test lead on 3), and terminals 2 and 3 (red test lead on 2 and black test lead on 3). These two readings must indicate forward biasing of the emitter-to-base junction

(0.655 volts) and the collector-to-base junction (0.621 volts).

Now we look for the one terminal common to both sets of conductive readings. It must be the

base connection of the transistor, because the base is the only layer of the three-layer device

common to both sets of PN junctions (emitter-base and collector-base). In this example, that

terminal is number 3, being common to both the 1-3 and the 2-3 test point combinations.

METER CHECK OF A TRANSISTOR

Those sets of meter readings, the black (-) meter test lead was touching terminal 3, which tells us

that the base of this transistor is made of N-type semiconductor material (black = negative).

Thus, the transistor is a PNP with base on terminal 3, emitter on terminal 1 and collector on terminal 2 as described in Figure 9.3.

·       E and C reverse: 1(+) and 2(-): “OL”

·       E and C reverse: 1(-) and 2(+): “OL”

·       E and B forward: 1(+) and 3(-): 0.655 V

·       E and B forward: 1(-) and 3(+): “OL”

·       C and B forward: 2(+) and 3(-): 0.621 V

·       C and B forward: 2(-) and 3(+): “OL”

Figure 9.3: BJT terminals identified by meter.

Please note that the base terminal in this example is not the middle lead of the transistor, as one

might expect from the three-layer “sandwich” model of a bipolar transistor. This is quite often

the case, and tends to confuse new students of electronics. The only way to be sure which lead

is which is by a meter check, or by referencing the manufacturer’s “data sheet” documentation

on that particular part number of transistor.

Knowing that a bipolar transistor behaves as two back-to-back diodes when tested with a

Diode test function is helpful for identifying an unknown transistor purely by meter readings.

It is also helpful for a quick functional check of the transistor. If the technician were to measure

Using the Diode test function in any more than two or any less than two of the six test lead combinations, he or she would immediately know that the transistor was defective (or else that it wasn’t a bipolar transistor but rather something else – a distinct possibility if no part numbers can be referenced for sure identification!). However, the “two diode” model of the transistor fails to

explain how or why it acts as an amplifying device.

                                                                   Fig 10

Identify the legs of your transistor with a multimeter. For identifying and testing purposes, refer to the representation shown above.

Diode test (V) meter readings
Transistor number	VBE	VEB	VBC	VCB	VCE	VEC
NPN	.841	0	.836	0	0	0
PNP	0	.892	0	.886	0	0

EXPERIMENT No.5 The Capacitor

The Capacitor

The capacitor stores electric charge.

A capacitor consists of two metal plates very close together, separated by an insulator. When connected to a battery or power source electrons flow into the negative plates and charge up the capacitor. The charge remains there when the battery is removed. The charge stored depends on the “size” or capacitance of the capacitor, which is measured on Farads (F).

Types of capacitor:

Non-electrolytic capacitor
	·          Fairly small capacitance - normally about10pF to 1mF ·          No polarity requirements - they can be inserted either way into a circuit. ·          Can take a fairly high voltage.
Variable capacitor
	·          Adjustable capacitor by turning a knob - similar to variable resistors. ·          The maximum capacitance available is about 200pF. ·          Used in radios.
Electrolytic capacitor
	·          Large capacitances - 1mF to 50000mF ·          Warning: These must be corrected the right way round (polarity) or they can explode - the white terminal on the diagram above signifies positive. ·          Black stripe with “-“ shows which terminal is the negative (usually the short one) ·          Low voltage rating – from 25 ~ 100V DC ·          They have a significant leakage current - this means that they will lose the charge stored over time.
Tantalum capacitor
	·          These have the same properties as the Electrolytic capacitor, but they are physically smaller & have lower leakage. As a result, though, they are more expensive.

Identifying Capacitor “Size”

If the Farad “size” is not printed on the capacitor, you may find an EIA code listed. Use the table below to figure out the capacitance

μF	nF	pF	EIA Code
0.00001*	0.01	10	100
0.0001*	0.1	100	101
0.001	1.0 (1n0)	1,000	102
0.01	10	10,000*	103
0.1	100	100,000*	104
1.0	1000*	1,000,00*	105
10.0	10,000*	10,000,00*	106

* Values with asterisk are not usually expressed in this form

RC Time Delay or “Charging Time”

Capacitors take time to charge. It doesn’t happen instantly. The charge time is dependent on the resistor in the circuit and the size of the capacitor. And it is expressed in the equation: R x C x 5 = T.  This is the time it takes to charge up to the applied voltage.

For example, 1,000,000 Ω x 0.000001 F x 5 = 5 seconds to charge to applied voltage. This can also be expressed as 1 MΩ x 1 μF x 5 = 5 seconds.

Capacitors are often used for timing when events take place. And often the voltage only has to get up to about 2/3 the applied voltage, and this happens at about 1/5 the time of their charging. So this is why the 5 is built into the equation. The concept of “time constants” is used here, where whatever the time it takes for a capacitor to build up to the full charge, it takes about 1/5 of that time to build up close to 2/3 of the charge. So you can divide the charge time into 5 segments, and the first time segment is often the time you are interested in.

Practice watching the capacitors charge up in the exercise below.

 

Fig 8-Capacitor Charging Circuit

Components: 1 x resistor, 1 x capacitor. 1 x pushbutton N/O switch.

Exercise: First, calculate how much time it would take to charge up the capacitor. Then, connect the circuit as shown above. Measure the time taken by the capacitor to reach the applied voltage on an oscilloscope. Fill in the chart below. Also draw the observed waveforms in the graphs below, filling the details on each one.

Note: you will need to adjust the time base to enable you to observe the pattern.

Circuit number	Capacitance (uF)	Resistance (KΩ)	Calculated Time (ms)	Observed Time (ms)
1	100	1	500ms	500ms
2	100	0.1	50ms	50ms
3	100	0.47	235ms	235ms
4	330	1	1650ms	2000ms

Label the axis of each graph:

Circuit 1:

Capacitance 100mF                     Resistance 1000

Circuit 2:

Capacitance 100mF                      Resistance 100

Circuit 3:

Capacitance 100mF                     Resistance 470

Circuit 4:

Capacitance 330mF                      Resistance 1000

How does changes in the resistor affect the charging time?

The charge time of the capacitor depends on its size and the size of the resistor in the circuit. The lower the resistance of the resistor, for example 100W the faster it would take to charge the capacitor. A resistor with a higher resistance like 1000W will take a longer time to charge then a 100W resistor using the same capacitor because the resistor limits the current flow which charges the capacitor at a slower rate.

How does changes in the capacitor affect the charging time?

Capacitors take time to charge up depending on the value of the resistor and size of the capacitor. A 330mf capacitor with a 1kW resistor will take a longer time to charge than a 100mf capacitor and same size resistor because the capacitor is determined by the size of the plates, the distance between them and the type of dielectric material used.

   

http://www.techitoutuk.com/knowledge/electronics/components/capacitors/capac.html (reference for the above picture)