Parts list
Operational amplifier: LM324AN
Resistors; R2, R3, R4 = 1K, R5 =390R, R6 = 10K, R7 = 270R, R8 = 470R
Capacitors; C1 and C2; Monolythic Dip C5K, 1 uF
LEDs, 1 Red, 5 yellow, 6 green 1.8V, 9.mA
Diodes D2, D3 and D4 1N4001 forward bias voltage 0.7
Zener diode D1; 9V1, IzRm 5.6mA
Veroboard; 15x25 holes, copper lines
What it Does
This tester is used to test oxygen sensors
How it works
A twelve volt supply is connected to the 12V red wire, current flows through a protection diode(D2) and a smoothing capacitor (C1) in case AC is used (not necessary as this circuit uses DC not AC). The zener protection resistor (R5) limits the current flowing to the zener diode (D1), which regulates the voltage down to 9.1 volts. Current for the various inverting inputs of LM324AN is then further limited by R6, R7 and R8. Also 0V is connected to the black lead which is connected to the earth at R7 and the IC’s pin 11 .
The sensor input (PL2) is connected to a supply of 0 to 1V which represents a real sensor output if this input is low: voltage of 0.01V the red LED will illuminate if higher voltage (0.3V) is supplied the green Led will illuminate and when even higher voltage (0.4 - 0.5V) is detected the yellow Led will light up and when a high voltage of 0.6 to 0.9 is detected the red led lights up. This is because we are using an op-amp as a voltage comparator. The output of the op-amp will be saturated fully positive if the non inverting input is more positive than the inverting input, and saturated fully negative if the non inverting input is less positive than the inverting input. So, an op-amp’s extremely high voltage gain makes it useful as a device to compare two voltages and change output voltage so that when one input exceeds the other a led will display this change.
Test points
Use a Voltmeter to check available voltage at; the 12V supply, 9.3V out from the zener, 0.64V at pin 2 and 5 and 0.23V at pin 13. Pin 4 has a supply voltage of 11.13V. Pin 11 has 0.01mV so its the earth. The pin 9 is the output from the yellow LED when its off = 9.86V, when on the voltage drops to 0.875V this shows that it is earthed then as voltage continues to rise and the red LED turns on the voltage at pin 9 = 0.9 V.
Using an ohmmeter check for connectivity across the drilled holes and copper lines should be OL if not bridged.
Voltmeter across an LED to check for correct voltage range of 1.8V. The operation of the sensor tester and the led colours and how they indicate the circuit is working are shown above in ‘How it works’ section
Calculations
R2 Led 1= (12V - 0.7V- 1.8V) / 9.5mA=1K
R3 LED 5 = (12V-0.7V -0.7V- 1.8 ) / 9.5mA= 1K
R4 LED 6 = (12V - 0.7V- 1.8V) / 9.5mA = 1K
R5 zener protection = (12V - 9.1V) / 0.0095A = 305.26ohms so use the preferred value of 390R
R6 = 9.1V / 0.0095A
And since R6= 10K use this to find the current that is the same through all the series resistors 9.1V / 10K = 0.00091A
R8 = (0.63 - .23V) / 0.00091 A = 439.56ohms so use a 470R
R7 = 0.23V / 0.00091A = 252ohms so use 270R
Faults and problems
Had trouble figuring out the resistor sizes went back and looked up the data sheets, wiki, and previous lessons to finally find the resistors by the above calculations.
One time consuming process was planning how to fit the components on a 15x25 board while minimising use of jumper wires this was solved by placing most components vertically while extending components over the IC instead of using many jumper wires and drilling holes.
The last problem was when the 12V supply was connected all the leads turned on and the circuit could not be used as a sensor tester. This had to be fixed by re soldering and drilling larger holes re testing with an ohmmeter for conductivity and re placing any faulty components. Next the available voltage test were taken as shown in test points this revealed several problems such as no earth at one point and a short at another extra earth wires where added and holes drilled to solve this problem. The circuit now works so well that another copy was created with even more compact arrangement of components and improved soldering.
Reflection
When this circuit was made again it took less time, planning and creation as the experience from the first time helped to insure that none of the above faults happened. It also looked neater because of the improved component layout.
Injector Circuit
Parts list
Two yellow LED1, LED2; 1.8V, 5mm, 30mA
Two switching NPN Transistors BC547
Four Resistors R13 R16: 470R 0.5W, R14 R15: 330R 0.5W
Wire- jump wires and coloured leads
Veroboard Grid 2.54 mm, strips, board connector
What it does
This small circuit is used in a fuel injected engine to switch the injectors (LED’s) on and off at extreme speeds and would be controlled by the ECU (5V).
How it Works
The main 12V supply passes a current flow through the collector resistors R14 & R15 and the LED’s when the 5V (ECU) is connected to the yellow wires, current flows to the NPN transistors (BC547) via R13 and R16 this switches the BJT’s to saturation mode “on” where Vce is close to zero measured is 65mV and Vbe is 850mV.
When the 5V (yellow wires) are disconnected the BJT’s go into cut off where Ib is close to zero mode turning the LED’s off also the measured vales are collector emitter Voltage 10.8 and base to emitter voltage is 1.7V .In both the on and off positions there is minimum power dissipation. These 2 LED’s, which represent injectors, can be switched separately as they are separate circuits and the measurements were taken separately but are close enough to state only one value. The OV is connected to the black ground lead which is connected to the BJT’s emitters.
Test points
An ohmmeter is used to test for continuance over the drilled holes.
Problems and faults
First problem was to find the size of the resistors. To start the values of the LED’s resistors:R14 R15 is calculated
12VS - 1.8VLed / 0.03A(Led current) = 340ohm.
Then the BJT’s resistors: R13 & R16 have to be calculated the information was found on the transistor Fairchild pdf data sheet I sourced from the Internet .
We need to find base resistance but first we must find base current
Ib = Ic/hfe so
Ib = 100mA/110
Ib= 0.909mA
Now the Base Resistance(Rb) can be found by ohms law
V = IR
R = V/IRb = 5Vs – 0.58Vbe / 0.909mA
Rb = 486ohms
So use the preferred value resistor of 470R
Conclusion
With a few tweaks this circuit was a success everything operated smoothly.
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