Circuit Board #3

Oxygen sensor tester

This circuit basically tells you how your O2 sensor is functioning. Whether or not it is receiving the right signals for the ever-changing air/fuel ratio mixture.
 It has an input from the sensor that varies, and depending on the input voltage given from the sensor, it uses operational amplifiers to give various signals. We use LEDs to display the different outcomes as a result of the varied input signal from the sensor (O2 sensor). Red LED displays a rich Fuel/Air ratio mixture, Green displays a lean mixture and yellow is Lambda1 (correct ratio mixture).

When the car is running lean the oxygen sensor puts out a very low voltage. This would make the input at pin12 lower than the 0.23v at pin13 and in turn allows pin14 to ground out through pin11. The green LED will be on.
When the car is running rich the oxygen sensor produces a higher voltage. In this case, it will be higher than 0.63v at pin6. The red LED will then be grounded through pin11, and be turned on.


To find;

R2 = [Vs - Vd (led1) - Vd (d2)] / A
      = (12v - 1.8v -0.7v) / 9.5mA (minimum current needed to flow through the LEDs)
      = 9.5v / 0.0095A
      = 1000ohm
      = 1k/ohm resistor

R3 = [Vs - Vd (led5) - Vd (d2) - Vd (d4)] / A
      = (12v - 1.8v - 0.7v - 0.7v) / 9.5mA
      = 8.8v / 0.0095A
      = 926.315ohm
      = 1k/ohm resistor

R4 = [Vs - Vd (led6) - Vd (d2)] / A
Components:
      = (12v - 1.8v -0.7v) / 9.5mA
      = 9.5v / 0.0095A
      = 1000ohm
      = 1k/ohm resistor

R5 = [Vs - Vd (d2) - Vz (zener diode)] / A
      = (12v - 0.7v - 9.1v) / 5.6mA (current needed flow through the zener diode)
      = 2.2v / 0.0056A
      = 392.857ohm
      = 390ohm resistor

R6 = 10k/ohm
      = 10000ohm (as given)

To find R7 & R8, we must firstly find the current in the circuit they are on, so;

Voltage drop across R6 = the voltage going IN (Vin) minus the voltage coming OUT (Vout);
      = 9.1v (Vin) - 0.63v (Vout)
      = 8.47v

Now we divide the voltage drop into the resistance as ohms laws requires and;
Current across R6 = 8.47v / 10000ohm
      = 0.000847A
      = 0.847mA

This determines current across R7 & R8, and from here we can calculate their resistance values by dividing their voltage drops (Vin - Vout) into the current. So;

R7 = Vd (Vin - Vout) / A
      = (0.23v - 0v) / 0.847mA
      = 0.23v / 0.000847A
      = 271.546ohm
      = 270ohm resistor

R8 = Vd (Vin - Vout) / A
      = (0.63v - 0.23v) / 0.847mA
      = 0.4v / 0.000847A
      = 472.25ohm
      = 470ohm resistor


Components needed:

1 x 14 pin out op-amp

3 x 1N4001 diodes

3 x LEDs (red, yellow and green)

1 x Zener diode (9.1v)

3 x 1 k/ohm resistors

1 x 10 k/ohm resistor

1 x 390 ohm resistor

1 x 470 ohm resistor

1 x 270 ohm resistor

2 x 0.1 uF capacitor

Below is a couple links to videos of my oxygen sensor tester in operation. You can see how, as the throttle opens it gives a rich signal and as the throttle is released a lean signal is given.

 http://www.youtube.com/watch?v=4icfs1xGLcc

http://www.youtube.com/watch?v=X8-9xLdYL5s


The oxygen sensor tester displaying a rich mixture





The oxygen sensor tester displaying a lean mixture

Under normal conditions the tester will fluctuate between the rich and lean mixture signalling LEDs. This is because of the ECU receiving these same signals and is adjusting the amount of fuel added to the air-fuel mixture constantly. This in turn keeps the emissions relativly close to Lambda1 which is good.

A fault in the circuit is giving a zero volt (0v) reading before and after each LED.
This means that the fault is somewhere before all the LEDs. Back tracking and testing components reveals 12v is available before the diode D2. Although after the diode available voltage is zero.
 At first glance, I could not visually see anything abnormal. But slighty playing with diode revealed an open circuit. Cracks between the solder and the track had stopped current from flowing in through the first component to the rest of the circuit.
 Repairing this gives us correct voltage drops across the LEDs, and the rest of the circuit again.

To do this experiment again would allow me to be a bit more delicate with my soldering to prevent shorts and/or open circuits.

Circuit Board #2

Circuit Board #2

Construct a circuit, that simulates a Voltage Regulator

This circuit represents a 5V output from an ECU used to signal the fuel injectors.

To calculate resistance needed;

[ Vs - LEDv ] / I

= [ 5 - 1.7 ] / 0.02

= 3.3 / 0.02

= 165 ohm

Thus establishing the a 180 ohm resistor used as R1. If R2 = 240 ohm (data sheet specification) then;

Vout = Vref (1 + R3 / R2)

= 5 = 1.25 (1 + R3 / R2)

= 5 / 1.25 = 1.25 / 1.25 (1 + R3 / R2)

= 4 - 1 = R3 / R2

= 3 = R3 / R2

Meaning R3 is 3 x R2;

R2 x 3 = R3

= 240 x 3 = 720

So R1 = 180 ohm, R2 = 270 ohm and R3 = 820 ohm

Components:

1x LM317 Voltage Regulator

1x LED

2x 1N4007 diodes

1x Zener diode

2x 1uF capacitors

1x 180 ohm resistor

1x 270 ohm resistor

1x 820 ohm resistor

Again using Lochmaster to create a circuit board design. Gives a great range of components.

If u hadn't noticed (and I didn't) the underneath of D4 (the black diode above) has not been cut, therefor it is being shorted, and defeats the purpose of even having one. At first my voltage regulator was not working, but Scott was quick to help and point this out.

After a long fluff around I got the board to produce a just less than average result;

Regulated Voltage = 5.9V

Voltage drop readings;

Regulator:

Input to Output = 6V

Input to Adjust = 7.3V

Adjust to Output = 1.25V

D13 = 0V

D12 = 6.05V

D15 (Zener) = 12V

LED = 4.3V

R1 = 2V

R2 = 1.25V

R3 = 4.66V

C16 = 5.9V

C15 = 12V

The reason for less than perfect result i suspect, is because of a resistor being too high. Either R2 or R3. Never the less, the concept is present.

Circuit Board #1 (Injector Circuit)

Circuit Board #1

Construct a circuit, simulating an injector circuit

This board simulates two 5V signals sent to two individual fuel injectors (LEDs are used for simulation purposes) in a circuit. We use a function generator to produce the ECU simulating pulse of 5 volts through the base of the transistors.



Injector Circuit wiring diagram



To Calculate resistance needed;

Datasheet info:

BC547:
max Ic = 100mA (0.1A) & Ib = 5mA (.005A)
min Ic = 10mA (0.01A) & Ib = 0.5mA (.0005A)

This also gives us Beta (B). Ic / Ib = B:
0.1(100mA) / 0.005(5mA) = 20
0.01(10mA) / 0.0005(0.5mA) = 20

 Datasheets are nessesary to calculate the values of the resistors (r13, r14, r15 & r16). They indicate that the LEDs needs a minimum of 20mA (0.02A) in its circuit to operate. The Collector side of the Transistor has a minimum of 10mA and a maximum constant current rating of 100mA (0.1A).

 So iv decided the current needs to be safely between 20mA (to enable the LED to operate) and 100mA. 60mA (0.060A) is good.

Therefor;

B (beta) = 20
Ic = 60mA

And if, Ib = Ic / B
= 0.060A / 20
= 0.003A

Then that would mean,
Ib = 3mA

Components:

2x NPN type transistors

2x 1 k/ohm resistors

2x 470 ohm resistors

2x LEDs

Lochmaster design (reverse)

Lochmaster design (top)

Using Lochmaster demo allows us to create a base design to work with. Its gives us a range of components to choose from. It also allows you to add labels and measurements to each of your components and/or symbols.
Using this base design we construct a breadboard sample. This will be our prototype.

 





Injector Circuit complete (top)

Injector Circuit complete (reverse)

 After construction of the board, all looks visually sound and works as it should. But, voltage drop results across components is a very important part of Autotronics and is a great skill to know and understand. It reveals operating conditions and/or malfunctions beyond visual inspection.

Voltage drop across;

LED #1 (red) = 2.2V - functioning normally

LED #2 (white) = 1.9V - functioning normally

R14 (red) = 9.69V
R15 (white) = 10.05V

R13 (red) = 4.67V

R16 (white) = 4.67V

 The measurements are taken with a seperate power source of 5V, connected up to both the base inputs of the two transistors. Also having their own seperate earth hooked up back through both emitters.

 These voltage drop readings indicate the LEDs are all functioning normally,

and there are no signs of a short, an open circuit or any other malfunction.

Experiment #7

Experiment #7

Transistor as a switch

Constructing a circuit using a 15v power supply, an NPN type transistor, a 1k/ohm resistor and 10k/ohm resistor.

A voltage drop reading across Vbe (Base to Emitter) of only 0.7v is produced. This transistor only needs this slight amount of voltage to operate. A reading across Vce (Controller to Emitter) is 0.05v, thus confirming that our transistor is in the 'saturation zone' on a transistor behaviour graph.

Ic = (Vs - Vce) / R
    = (15v - 0.05v) / 1000 ohm
    = 14.95 / 1000
    = 0.01495
    = 14.95mA

Ib = (Vs - Vbe) / R
    = (15v - 0.7v) / 10000 ohm
    = 14.3 /10000
    = 0.00143
    = 1.43mA

 Saturation zone refers to the transistor being fully open or 'maxed out'. The Active zone refers to the transistor working to the desired specifications or in its 'prime'. The Cut-off zone refers to the transistor not working or not flowing at all.

Just breifly, if the voltage at Vce is at 3v and we are trying to find the power dissipated (Pd) by the transistor, then;

Pd = Ic x Vce

If Vce = 3v, then we find using the transistor behaviour graph that Ic will = 13mA (0.0013A), so;

13 x 3 = 39W

Finding the Beta (B) gain at different Vce voltages;

B @ 2V = Ic/Ib

= 0.02/0.0008

= 25

B @ 3V = Ic/Ib

= 0.0125/0.0005

= 25

B @ 4V = Ic/Ib

= 0.005/0.0002

= 25

Experiment #6

Experiment #6

Meter check of a Transistor

Bipolar Junction Transistors are a three-layered Semiconductor. They are set up as NPN (negative-positive-negative), or PNP (positive-negative-positive). NPN transistors are most commonly used. Transistors are theoretically two diodes back to back or face to face.

NPN type transistor readings:

Vbe - 0.838
Veb - O.L.
Vbc - 0.834
Vcb - O.L.
Vce - O.L.
Vec - O.L.

PNP type transistor readings:
Vbe - O.L.
Veb - 0.841
Vbc - O.L.
Vcb - 0.834
Vce - O.L.
Vec - O.L.

Transistors will give overload readings between the pins that simulate diodes in a reverse bias.

Experiment #4

Experiment #4

Diode and Zener diode

Mixing things up little now when we're asked to construct a circuit using firstly a 10v supply, a resistor (1k/ohm), a Diode (1N4007) and a Zener diode (5V1 400mW). Run in series taking readings of each component as follows:

Volt drop across:

 V1 (Zener diode) - 4.65v

 V2 (diode) - 0.66v

 V3 (diode & Zener diode) - 5.29v

 V4 (resistor) - 4.72v

Calculated current A: 10v/1000ohm = 10mA

Now taking readings with a voltage supply of 15v:

Volt drop across:

 V1 - 4.8v

 V2 - 0.68v

 V3 - 5.48v

 V4 - 9.5v

Calculated current A: 15v/1000ohm = 15mA

Given this information, we can see that the smallest difference in readings between the two tests is from the standard diode (just 0.02v difference). The highest difference is from the resistor (4.78v difference). The Zener stays almost the same (0.15v difference). Thus teaching that access voltage is consumed by the resistor.

Experiment #3

Experiment #3

Zener diode

 

Zener diodes polarity can easily be determined like a standard diode. Visually, there is a stripe on the cathode side and the symbol is similar to the standard diode apart from the right angled legs.

Constructing a electrical circuit on a breadboard, using 2 x resistors (both 100ohm), a Zener Diode (Zd) and a 12v supply voltage (Vs). Run one resistor (Rl) in parallel with the Zener Diode (Zd) and the second resistor (R) in series with both the Zener diode and the other resistor (Rl).

The Zener Voltage (Vz) is equal to the voltage drop across the Zener Diode. The voltage drop across the Zener is 4.99v, therefor the Zener Voltage is 4.99v.

If you were to lower the supply voltage (Vs) to 10v, the voltage drop decreases to 4.78v, therefor the Zener Voltage decreases to 4.78v.

Also if the voltage were to increase to 15v, the voltage drop across the zener increases to 5.13v, therefor Zener Voltage increases to 5.13v. Zener diodes are used to regulate voltage and supply a steady current.

Reversing the polarity of a Zener Diode changes its function. Zener voltage becomes 0.8v and it functions as a normal diode would.

Experiment #2

 Experiment #2

Diodes

A diode has the characteristics of a conductor when current flows one way. They also have insulator characteristics when current tries to flow in the opposite direction. Basically diodes allow current (amps) to flow in only one direction, Anode to Cathode.

Identifying diode's and/or light emitting diode's (LED's) polarity without using a multimeter is simple. Diodes display cathode with a white stripe around the end. LEDs display this by using a smaller terminal for cathode and a slightly longer terminal for anode.

We were given a data sheet to identify the characteristics of a 1N4007 type diode. Data sheets are very handy to have at your disposal when you need them. An example is given using a 5v supply, a 1k/ohm resistor and a 1N4007 type diode. When calculating the current (I) across the diode (d) we use the formula:
Id = (Vs - Vd)/R
= (5 - 0.6)/1000
= 4.4/1000
=0.0044A (4.4mA)                                                           Measured: 4.6mA


So the answers are not identical, but they're not far off and the concept is there. To calculate the voltage drop across the diode we use the following formula:
Vd = Vs - (I x R)
= 5 - (0.0044 x 1000)
= 5 - 4.4
= 0.6v                                                                                 Measured: 0.6v


Bang on. Now the maximum value of current able to flow through the given diode is 1A and the maximum supply voltage used to keep the diode working in a safe region is 1000v.


Replacing the standard diode with a light emitting diode (LED) and calculating the current is done as follows:
I = (Vs - Vd)/R
= (5 - 1.7)/1000
= 3.3/1000
= 0.0033A (3.3mA)                                                          Measured: 5mA

This observation shows us that there is low current being used by the LED.