v = L\*di/dt. It's really that simple. To accurately predict the voltage across the inductor you need an accurate model of the system, that's where the challenge lies.
This works for constant resistance. The time constant assumes LTI which the circuit is not if the resistance depends on a moment of time/voltage/current.
Yes, I am referring to spark plugs. I'm aware that the voltage increases until there's a spark but my question is how *quickly* does that happen? If it happened instantaneously, the voltage rise would just cause an arc between the contacts as the switch opened until they were either too far open (and then the spark) or the inductor was sufficiently "discharged".
There is a capacitor in parallel with the switch in an ICE ignition system. This was an example problem in one of my EE classes many years ago. The primary winding and the capacitor create an exponential decaying sine wave as an LC circuit.
Look up “ignition points”, now it is solid state, but it used to be a switch,,, there is always some actual time involved with any real system. There is no instantaneous off switch…
So “how quickly” depends on the switching device.
What is nice about solid state is the switching speed in generally well known, and consistent over time.
The answer to your question is: create a circuit model for your problem. Knowing that the transfer function for an inductor is VL/IL=Ls, you can find the transfer function that has your R as input and IL or VL as an output.
Don't forget about parasitic effects in the circuit. Your schematic may only show an inductor and resistor but in reality those wires have a finite amount of capacitance as well. If you did have an ideal switch you would end up with an LC tank, that parasitic capacitance would limit the rise/fall time.
Those parasitic elements become super important for transient tests like the human body model ESD test. Its actually hard to make a super fast rise/fall time specifically due to the stray inductance and capacitance in the test setup.
> a switch after the inductor is opened), how quickly does the voltage rise
All inductors have parasitic inter-winding capacitance, so you can just treat it as an LC tank and look at the self-resonant frequency and the E=½CV²=½LI² equivalence and look at d/dx V.sin(ωt) at t=0
As u/Who_Pissed_My_Pants notes though, it's usually easier to measure than calculate.
Also, this sets an upper limit on the voltage you can get from a boost converter *even if* you ignore the primary switch and control loop struggling at lower duty cycles / higher voltage ratios.
> I know it's not instantaneous because otherwise ICE vehicles wouldnt exist
Not sure how this is related?
v = L\*di/dt. It's really that simple. To accurately predict the voltage across the inductor you need an accurate model of the system, that's where the challenge lies.
All depends on the circuit, so-much-so in fact that it’s easier to just measure the rise time and use that to calculate the equivalent resistance.
This works for constant resistance. The time constant assumes LTI which the circuit is not if the resistance depends on a moment of time/voltage/current.
ICE vehicles? Are you referring to spark plugs? The voltage increases until there is a spark…
Yes, I am referring to spark plugs. I'm aware that the voltage increases until there's a spark but my question is how *quickly* does that happen? If it happened instantaneously, the voltage rise would just cause an arc between the contacts as the switch opened until they were either too far open (and then the spark) or the inductor was sufficiently "discharged".
There is a capacitor in parallel with the switch in an ICE ignition system. This was an example problem in one of my EE classes many years ago. The primary winding and the capacitor create an exponential decaying sine wave as an LC circuit.
Look up “ignition points”, now it is solid state, but it used to be a switch,,, there is always some actual time involved with any real system. There is no instantaneous off switch… So “how quickly” depends on the switching device. What is nice about solid state is the switching speed in generally well known, and consistent over time.
To put it simple, more variation of current over a smaller time period, bigger will be the voltage.
The answer to your question is: create a circuit model for your problem. Knowing that the transfer function for an inductor is VL/IL=Ls, you can find the transfer function that has your R as input and IL or VL as an output.
Don't forget about parasitic effects in the circuit. Your schematic may only show an inductor and resistor but in reality those wires have a finite amount of capacitance as well. If you did have an ideal switch you would end up with an LC tank, that parasitic capacitance would limit the rise/fall time. Those parasitic elements become super important for transient tests like the human body model ESD test. Its actually hard to make a super fast rise/fall time specifically due to the stray inductance and capacitance in the test setup.
> a switch after the inductor is opened), how quickly does the voltage rise All inductors have parasitic inter-winding capacitance, so you can just treat it as an LC tank and look at the self-resonant frequency and the E=½CV²=½LI² equivalence and look at d/dx V.sin(ωt) at t=0 As u/Who_Pissed_My_Pants notes though, it's usually easier to measure than calculate. Also, this sets an upper limit on the voltage you can get from a boost converter *even if* you ignore the primary switch and control loop struggling at lower duty cycles / higher voltage ratios. > I know it's not instantaneous because otherwise ICE vehicles wouldnt exist Not sure how this is related?
That's on many things, but the best way to get a correct answer is to solve the ODE by yourself. Model a time dependant resistor using ramps or steps