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Battery ignition system and spark timing calculation

Short answer

Battery ignition system: primary current builds magnetic field; when points open, field collapse induces high voltage in secondary, distributed to spark plugs. Condenser prevents arcing. For a 6-cylinder engine at 4000 rpm: 200 sparks/s (5 ms interval); spark advance 24° gives 1.0 ms before TDC at 4000 rpm, 1.33 ms at 3000 rpm; stored energy 64 mJ; secondary voltage 33.6 kV.

(a) With a neat circuit diagram, explain the working of a battery (coil) ignition system. (b) A six-cylinder, four-stroke petrol engine runs at 4000 rpm. Find the number of sparks per second and the time between sparks. If the spark advance is 24 degrees of crank angle, find the time by which the spark occurs before TDC at 4000 rpm and at 3000 rpm. The ignition coil has a primary inductance of 8 mH, breaking a primary current of 4 A, and a secondary to primary turns ratio of 120. Find the energy stored and the secondary voltage if the primary induced voltage is 280 V.

(a) Battery (coil) ignition system

 +-------[Ignition switch]--[Ballast R]--+
 |                                       |
 Battery 12V                    Primary winding (N1)
 |                                       |--+-- Contact breaker
 |                              Secondary (N2) |  + condenser
 |                                       |   Cam
 +------ ground --------------------------+
                  Secondary --> Distributor --> Spark plugs

Working

  1. With the breaker points closed, current flows from the battery through the primary winding and builds up a magnetic field in the iron core.
  2. The cam on the distributor shaft opens the points, and the primary current collapses suddenly.
  3. The collapsing field induces about 250 to 300 V in the primary and, because of the turns ratio, 20 to 30 kV in the secondary.
  4. The high voltage goes through the distributor to the correct spark plug, where a spark jumps across the gap.
  5. The condenser across the points absorbs the primary induced voltage, prevents arcing at the points and makes the current collapse faster.

In an electronic system, the points are replaced by a transistor switch triggered by a Hall-effect sensor or pickup coil, so there is no contact wear.

(b) Numerical

Sparks per second: a 4-stroke engine fires each cylinder once every 2 revolutions.

sparks/s=6×40002×60=200\text{sparks/s} = \frac{6 \times 4000}{2 \times 60} = 200

Interval =1/200=5= 1/200 = 5 ms.

Time for 24 degrees advance: crank speed at 4000 rpm =4000×360/60=24 000= 4000 \times 360/60 = 24\,000 deg/s, and at 3000 rpm =18 000= 18\,000 deg/s.

t4000=2424 000=1.0 ms,t3000=2418 000=1.33 mst_{4000} = \frac{24}{24\,000} = 1.0\ \text{ms}, \qquad t_{3000} = \frac{24}{18\,000} = 1.33\ \text{ms}

For a fixed advance angle, the time available for combustion before TDC becomes shorter as speed rises. Since combustion takes roughly the same time, a centrifugal advance mechanism increases the advance angle at higher speed.

Coil energy:

E=12LI2=12(8×10−3)(4)2=0.064 J=64 mJE = \tfrac{1}{2} L I^2 = \tfrac{1}{2}(8\times10^{-3})(4)^2 = 0.064\ \text{J} = 64\ \text{mJ}

Secondary voltage:

V2=120×280=33 600 V=33.6 kVV_2 = 120 \times 280 = 33\,600\ \text{V} = 33.6\ \text{kV}

Answer: 200 sparks/s (one every 5 ms); advance time 1.0 ms at 4000 rpm and 1.33 ms at 3000 rpm; E=64E = 64 mJ; V2=33.6V_2 = 33.6 kV.

✓ Answer checked by Claude Opus 5.5 (high effort)

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