Sunday, 15 May 2011

testing ignition coils

the first thing i did in this task is get two coil and get there specifications.Getting secification for the coils is easy you just have to find the part number that would be on the coil then eneter this numer in google and it will give you a website to get all the secifications of.I did this and got:
coil 1:

no:  cic32r
voltage:  12v
primary:  1.8 ohm
secondary: 8 to 9 ohm

coil 2:

no:  cic31
voltage:  12v
primary:  3to4 ohm
secondary  7to8 ohm

After finding the specifications of these coils i tested them and found:
coil1:
primary: 1.8 ohm
secondary: 0 ohm
earth leakage test:no

coil2:
primary: 3.3 ohm
secondary: 9.75 ohm
earth leakage test: no

coil one is not serviceable because at secondary there was no resistance. this shows that this coil is faulty.
coil two is servicable because it meets specifications or is close enough to the specifications.

Next test was testing a wasted spark coil pack
testing the three coils:
coil 1:
secondary: 12.67 kohm
primary: 1.2 ohm
coil2:
secondary:12.73 kohm
primary: 1.2 ohm
coil3:
secondary:12.75 kohm
primary: 1.1 ohm

Next thing to test was ballast resistotrs. first i had to get the specifications
ballast resistor 1:
no:BR2
ohm spec: 1.1ohm
ballast resistor 2:
no:BR3
ohm spec: 1.7ohm

measured resistance:
ballast resistor no1: 0.9ohm
ballast resistor no2: 1.7ohm
both of these ballast resistors are serviceable
Testing the ballast restors was done with a mulltimeter set on ohms the black terminal coming out of com goes onto the -of the ballast and the red terminal coming out of v/ohm goes on the + of the ballast.

Next task was measuring current draw and voltage drop.To do this i had to wire up a ballast resistor in series with a coil primary windings.To test this i had to connect an ameter in series and get the current draw.I did this and got a result of 5.4A. Next i calculated the voltage drop across the coil by using the formula AxR=V from ohms law:
4x1.3=7.02 the 4 is the amprege in the circuit the 1.3 is the resistance in there circuit therefor:
calculated voltage drop across the coil is 7.02v
measured voltage drop across the coil is 4.7v
I calculated the voltage drop across the ballast resistor:
5.4x1.7=9.18 same formula as before 5.4=amprege and 1.7=resistance therefor:
calculate voltage drop across the ballast resistor is 9.18v
measured voltage drop across the ballast resistor is 6.83v
my calculated voltage did not equal my measured voltage the reason for this was my coil was on for to long and if coil are left on for to long they lose power.

1 comment:

  1. Coils have the properties that they have a high impedance when they first turn on (Faraday's law) but its impedance lowers as the coil charges up which means it has a higher current draw and lower volt drop the longer it is left on until it is saturated

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