Engine Electronic Control Systems, Ignition circuit digram

Connect your circuit up to the function generator and fire the spark plug

                                                                                                                                  

R3 = unknown

R1 = Resistance of the primary coil 3.6 ohms 

Transistor: TIP31C

R1 coil resistance (primary) = 3.6 ohms 

Amperage = 3.0 A This is value of current that can handle form the data sheet.

We have to find the value for R2 to make the current not more 3 A in the circuit .

V = IR1

I = 12/3.6 = 3.33 A

So Rt = V/I = 12/3 = 4 ohms

The resistor we need for our circuit is:

R2 = 4 - 3.6 = 0.4 ohms

0.4 ohms resistor required

Ignition Circuit 

Ignition system has the capability of raising voltage the low voltage (12 volt) to a very high voltage (3000+volts) and distributing it to the spark plugs where spark plus ignites fuel air mixture.

when the igniting switch is on current flows through the primary winding of the ignition coil and a strong magnetic field is developed when the current flows is interrupted or switched up, the magnetic field is collapsed across the secondary winding a high voltage and this high voltage is discharge from the ignition coil through the high tension lead to the spark plugs where is ignites the fuel air mixture.

I have built an electronic ignition circuit, this includes a transistor, function generator, ignition coil. spark plug, and unknown resistor R3.

I did calculate the value for unknown resistor for my circuit which the amplifire can work and can handle the current.