Day 4 - Fuel injection circuit and fault finding

Theory
Kirchhoff's voltage law:

    In any given circuit, the algebraic sum of the applied EMFs is equal to the algebraic 

    sum of the voltage drop. (Etotal = V1 + V2 + V3 + ... +Vn)

Kirchhoff's current law:
    The sum of the currents entering a junction equals the sum of the currents leaving
    that junction. (Itotal = I1 + I2 + I3 + ... + In)
npn Transistor

Between base and emitter, the small current flows.
Between collector and emitter, the large current flows.






Current passing between the base and the emitter permits a much larger current to pass between the collector and the emitter. The greater the voltage difference between the base and the emitter, the more current will pass, and a proportionally higher current can pass between collector and emitter.
In an NPN transistor, the base must be positive with respect to the emitter for collector-to-emitter current to pass.

This is the part of the fuel injection circuit diagram.
Before reparing automotive electronic parts, it is very important to diagonise firstly.
There are some kinds of faults as follow:
1. short to ground(unwanted connection)
2. short to power
3. open
4. high resistance










Unitec auto repair reference center:
http://arrc.epnet.com.libproxy.unitec.ac.nz/autoasp/index.asp?sid=101118592&uid=unitec.main.autorefctr

Practical

Experiment No.7
Exercise:  Connect the circuit as shown in Fig 12 and switch on the power supply

Connect the multimeter between base and emitter.

Note the voltage reading and explain what this reading is indicating
VBE = 0.749V

Connect the multimeter between collector and emitter.

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

VCE = 0.056V

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

A: Saturated

B: Cut-off

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

The saturation and cutoff regions are useful as switching circuits.

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

PD = Vce x IC

      = 3 x 12.5mA

      = 37.5mW

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

β = Ic / Ib

at Vce = 2V     β = 20/0.8 = 25
at Vce = 4V     β = 5/0.2 = 25
at Vce = 3V     β = 12.5/0.5 = 25

Measure the available voltage of the circuit below.

From the experiment, there was a voltage drop(0.3V) after diode because of tiny current flowing through the diode.











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