 ### Network # Figuring spark time values for needed circuit - inductive ignition type

#### russwr

Dec 20, 2019
31
How to manipulate formula for values as necessary for spark time , so as can get back EMF from inductor greater than 100v , when spark time is to be 34% duty cycle of 360 degrees flywheel turning, using 6 amps DC, inductor time extension, 20 initial volts DC, Spark as only from inductive type ignition. The AI computer recently gave sample formula that said 100v output as from 11.11 milliseconds fast discharge, - but probably won't match the required 34% on time of spark. The sample inductor 1.5Millihenry charges , fires forward , then sends back EMF to output also using side diode guide, after polarity reversal. to the same output. V = L * (di/dt) * t. Some values need to be changed so as circuit matches for TDC ignition firing to 34% on before spark cut off.

#### Delta Prime

Jul 29, 2020
1,219
Interesting! That AI did not ask if there was any ferromagnetic material involved?
Since the EMF (voltage) = d/dt(magnetic flux), and the magnetic flux is proportional to the current, then the voltage is proportional to d/dt(current). The proportionality constant is the inductance and or inductor L. Tell that to AI.

I think you're looking for the dwell- time here's an example.
In a typical four-cycle petrol engine, the spark repetition rate can be calculated with this formula- t=120/ (N*RPM), where “t” is seconds, and “N” denotes the number of cylinders. Thus, for V8 engine running at 6000 RPM, we can calculate that 120 seconds divided by 48 000 (the number of cylinders multiplied by the RPM), yields a time of .0025, or 2.5 milliseconds between sparks.

Moreover, at 6000 RPM, the crankshaft rotates at a rate of 36 000 degrees per second (one rotation=360 degrees multiplied by the RPM, and dived by 60 to arrive at a rotation rate/second). Put in another way, the crankshaft rotates at a rate of 28 milliseconds per degree at 6 000 RPM, which we can calculate with this formula- 1/ (.006*RPM).

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