Re: bleed resistor

Incidentally, this business of time constants is why you get a better
current rise time in a L/R drive for a stepper motor by increasing
the voltage & using a series resistor to limit the current. If you
connect a resistor and an inductor in series & apply a voltage across
them, the time taken for the current through the inductor to reach
63.2% of its maximum value (i.e., the L/R time constant) is equal to
the inductance in Henries divided by the series resistance in Ohms
(including the resistance of the inductor itself). So, increase the
series resistance (and increase the voltage to achieve the same max
current) and the current rise time in the inductor decreases
proportionately.

Regards,
Tony

--- In CAD_CAM_EDM_DRO@y..., Scot Rogers <scotr@b...> wrote:
> Thanks to you too Jon. It's times like these and people like you
that make
> me wish that I'd learned more in school.
>
>
> Scot Rogers
>
>
>
>
> ~>
> ~> Like I said 3300uf at 80V of capacitor rating.
> ~
> ~Well, the 80 V is the most important thing. So, you can put a 5 K
Ohm
> ~resistor across it, and it will draw 16 mA. You can put an LED in
> ~series with
> ~
> ~this and have a caps still charged indicator. The time constant
> ~(RC) = 49.5 seconds, so the caps will drain to 36% of 80 V in 49.5
seconds.
> ~That will be 28 V. That will be safe to touch, but not safe to
> ~accidentally
> ~short
> ~out. When the LED is out, it will be safe to short it out, too.