| Electronics > Beginners |
| Coil charging |
| << < (4/4) |
| injb:
--- Quote from: magic on May 13, 2019, 09:26:36 pm ---Okay, this thread is getting a bit weird so maybe I'll just answer some of the original questions ;) --- Quote from: injb on May 12, 2019, 06:27:44 pm ---I found this calculator (https://daycounter.com/Calculators/Inductor-Current-Power-Calculator.phtml) which uses the formula: Ton_max= I_sat * L/V and it gave me approx 3.8ms for the values above, which is close to the charge time I'm seeing on the scope. But I don't know if I really understand what this formula is saying. Is it "this is how long it takes to reach I_sat, given L and V" ? Is the resistance value implied, because I've specified I and V? --- End quote --- It assumes zero (or small/irrelevant) resistance. In such case current rises at the rate of V/L and so you get that particular formula relating time to current. If your measurement is accurate, it looks like this is indeed happening. The current rises almost linearly and then somehow becomes limited at 9A. --- Quote from: injb on May 12, 2019, 06:27:44 pm ---I thought the above formula might just be a rearrangement of: T = L / R But this formula is supposed to tell me the time it takes to reach 63.2% of maximum current, right? I know that the ECU of the car limits the current to 9A. Does this mean that the current I'm seeing after approx 3.8ms is 63.2% of the maximum that would flow if it wasn't limited by the ECU? --- End quote --- Nope, this one is for an inductor with significant series resistance. As current increases, increasing amount of voltage is wasted on overcoming resistance and the voltage that actually "accelerates" current through the inductor decreases and therefore, by V/L, the rate of current rise slows down. You get an exponential decay curve like one of those but upside down. After each L/R time, current gets 63% closer to its final value, which is V/R. This doesn't look like what you observe. --- End quote --- Thanks, that answers my questions perfectly. I was thinking of a constant voltage, but I didn't think about how the voltage drop would increase as current increases. |
| langwadt:
--- Quote from: injb on May 13, 2019, 01:11:44 pm --- --- Quote from: Jwillis on May 13, 2019, 05:59:49 am ---How "old school are you talking" . ... --- End quote --- Bosch Motronic 3.1 from an 89 Porsche 944 Turbo --- Quote ---Even though the coil current limits for some time before the spark, the ECM is still likely to be calculating an optimum turn on time before the spark is needed to minimize dissipation. BTW: a constant voltage applied to an inductor will cause a mostly linear current ramp up(or down) so long as any resistive voltage drop is a small fraction of the applied voltage. --- End quote --- Yes it uses a map for dwell time which is rpm vs battery voltage. --- End quote --- dwell time should be constant with rpm, dwell angle changes |
| Navigation |
| Message Index |
| Previous page |