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Re: Inductive Kick Effects - was- Re: cap firing voltage sc
> Original Poster: "The Flavored Coffee Guy" <elgersmad-at-email.msn-dot-com>
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> The ramp voltage doesn't indicate a ramp current. When you supply a voltage
> to and equal amount of voltage there is no current flow. An inductor can
> only follow it's time constant, and current can only increase at a rate that
> equal to it's time constant. Alright if you calculated what the inductive
> reactance would be in the first nano second as if it were a frequency, then
> you see the kind of inductive reactance that is limiting current. Once
> current has been established through the coil, and this must be a greater
> time period than a quick shut off can produce.
A time constant is always associated with more than a single reactive
component. The rate of current rise in a pure inductance is related
both to the inductance and the voltage applied across it. Reading
that another way, the time taken for the current to rise to a
nominated value depends on the inductance and the voltage applied
across it. In an LR circuit the time component is dependent on both L
and R. In an LC circuit, both L and C play a part.
> A ramp is produced by a capacitor as it charges. When you move the
> calculation to 2 nano seconds as if it were a frequency the inductive
> reactance drops. A coil is self inductive, and if you suddenly try to halt
> current flow though it, it will try to compensate. The higher the frequency
> of cut off, and the higher the voltage of the self inductive pulse, and this
> can be felt through the secondary. Use a single generator and test that one
> out with a true ramp current. You will see a difference. Then maybe use a
> MOSFET to amplify the current in the situation.
A voltage ramp is produced by a capacitor *which has a constant
current forced through it*. This is a common method of generating
ramp, triangular and sawtooth waveforms.
Attempting to interrupting the current flow in an inductor causes
a voltage rise *which depends on the load applied as well as the
inductance*. If the load is purely capacitive, the voltage is easily
defined by assuming the energy stored in the inductor is transferred
in its entirety to the capacitor. The time taken is a simple function
of the resonant frequency of inductor/capacitor. If no external
capacitor is connected, the self-capacitance of the windings dictates
the magnitude of the voltage hike.
Malcolm
<snip>