>From: "Eisermann, Phil [Ridg/CO]" <
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>Reply-To: pic microcontroller discussion list <
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>To:
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>Subject: Re: [OT]:Current Sensor
>Date: Thu, 5 Apr 2001 16:10:10 -0500
>
> The trick is that the motor is inductive. I don't know what your
>background is, mathematical or otherwise, so i'll try to offer a simple
>explanation without any math. I hope this makes sense. Its been a while :)
>If I omitted something, or goofed it up, I hope someone will point it out
>and correct me.
>
> An ideal inductor behaves according to v=L(di/dt). In non-math
>speak, that means the voltage (v) is proportional (via the inductance, L)
>to
>the time rate of change of the current through the inductor. In other
>words,
>the faster the current is changing, the higher the voltage across the
>inductor. Similarly, if the current isn't changing (e.g a constant current
>through the inductor), there is no voltage across the inductor. Of course,
>that's the ideal. A real inductor has to be made of wire, and wire has
>resistance. But you get the idea.
>
> Now, consider what would happen if the current were to somehow
>change from 2A to 0. Never mind why. What's the voltage across the
>inductor?
>It depends on how fast the current changes, as stated above. The faster it
>goes to zero, the higher the voltage that develops across the inductor has
>to be. So what would happen if it were to change instantaneously? The
>voltage across it would have to be infinite! Can this happen? Only in a
>textbook. (This is the principle behind spark coils, btw).
>
> In the motor cicuit, once you open the switch, the current has got
>to go somewhere. It cannot instantaneously dissappear (because that would
>require the voltage to be infinite). Where does it go? back to ground,
>through your resistor. It decays exponentially, not linearly. The faster
>your PWM frequency, the less the motor current decays in the off period.
>
> So, does that make any sense at all? Also, your 10kHz sounds like
>a
>good place to start.
>