The ignition system includes the components as shown below -
Battery/Alternator
Ignition switch
Ballast resistor
Coils(Primary winding/Secondary winding)
Igniter(Ignition module)
Distributor
HT leads
Spark plugs
The primary winding has 150 turns of heavy wire and the secondary winding has 20,000 turns of fine wire. The secondary winding has about 100 times the number of turns of the primary winding.(Turns ratio 1:100)
Formular
Np: number of turns in primary winding, Ns: number of turns in secondary winding
Vp: voltage in primary winding, Vs: voltage in secondary winding
Ip: current in primary winding, Is: current in secondary winding
Example:
If Vp=30V, Np=10 and Ns=1000 then calculate Vs?10 / 1000 = 80 / Vs
Vs = 80000 / 10 = 8000V
If Ip=6A then calculate Is?
10 /1000 = Is / 6
Is = 10/1000 x 6 = 60mA
Practical
Testing ignition coils
Coil#1 No CIT-118
Coil#1 Voltage 12V
Coil#1 Primary 1.2Ω (Test result: 1.2Ω)
Coil#1 Secondary 8.5k~9.5kΩ (Test result: 9.29kΩ)
<- Secondary winding testing
<- Primary winding testing
Earth leakage testing; 0L
Coil#2 No Bosch SU12R
Coil#2 Voltage 12V
Coil#2 Primary 1.5Ω (Test result: 1.5Ω)
Coil#2 15kΩ (Test result: 14.76kΩ)
<- Primary winding testing
<- Secondary winding testing
Earth Leakage testing: 0L
Wasted Spark Coil Pack
Coil#1 Secondary 7.24kΩ
Coil#2 Secondary 7.05kΩ
pin#1 - pin#2 : 1.2Ω
pin#3 - pin#4 : 0.1Ω
pin#2 - pin#4 : 0.6Ω
pin#1 - pin#4 : 0.6Ω
How can we recognise which one is which?
1. Measure the resistance between each two pins.
pin#1-pin#2 : 1.2Ω
pin#2-pin#3 : 0.6Ω
pin#3-pin#4 : 0.2Ω
pin#1-pin#3 : 0.6Ω
pin#1-pin#4 : 0.6Ω
As shown above measurments, we can see that there are two coils and they are connected.
Testing Ballast resistors
Specifications
Ballast resistor spec 1.6Ω
(Test result: 0.2Ω)
Ballast resistor No2 BR1
Ballast resistor spec 1.0Ω
(Test result: 1.2Ω)
Measuring Currnet draw and Voltage drop
Current draw: 6.02A
Coil calculated Voltage drop 10.59V
Coil measured Voltage drop 10.19V
Ballast resistor calculated voltage drop 1.41V
Ballast resistor measured voltage drop 0.04V
Wiring up ignition systems
1. Wire up an ignition module using a function generator to trigger the module
3. Wire up the wasted spark ignition system using the function generator to trigger the module
Waste-spark system uses one ignition coil to fire the spark plugs for two cylinders at the same time.
Waste-spark ignition is another name for the distributorless ignition system or electronic ignition. A 4-cylinder engine uses two ignition coils and a 6-cylinder engine uses three ignition coils.
Both spark plugs fire at the same time. When one cylinder is on the compression stroke, the other cylinder is on the exhaust stroke. The spark that occurs on the exhaust stroke is called the waste spark, because it does no useful work and is only used as a ground path for the secondary winding of the ignition coil. The cylinder on the compression stroke uses the remaining coil energy.
4. Wire up the coil over ignition system using the function generator to trigger the module
#1 - Battery, #2 - Earth, #3 - Trigger
Coil-on-plug system uses a single ignition coil for each cylinder with the coil placed above or near the spark plug. The advantage of the Coil-on-plug systems is that it doesn't need the spark plug wires thatoften cause electronagnetic interference.
5. Build the ignition circuit on a breadboard using two 2N2222 transistors
Ic = 800mA
Measure the resistance of the ballast resistor and primary winding of the coil.
Resistance of the ballast resistor: 2Ω
Resistance of the primary winding of the coil: 3.2Ω
To calculate Rc, firstly refer to the datasheet of the transistor which is TIP31C. According to the datasheet, maximum Ic is 3A. For the safety factor, divide the maximum current by two. RT=12 / 1.5
RT=8 Rc=8-(2+3.2) Therefore Rc = 2.8Ω
Rb=(5-0.7) / 1.5 = 2.9Ω
Second experiment
Measure the resistance of the ballast resistor and primary winding of the coil.
Resistance of the ballast resistor: 2Ω
Resistance of the primary winding of the coil: 3.2Ω
To calculate Rc, firstly refer to the datasheet of the transistor which is TIP31C. According to the datasheet, maximum Ic is 3A. For the safety factor, divide the maximum current by two. RT=(12-0.7-0.05) / 1.5
RT=7.5 Rc=7.5-(2+3.2) Therefore Rc = 2.3Ω
Rb=(5-0.7-0.7) / 400mA = 9Ω (why Ib=400mA? - from the datasheet, the maximum current is 800mA, for the purpose of safety factor, divide the maximum current by two. Therefore I = 400mA)
Third experiment
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