On Thu, 25 Jun 1998, David VanHorn wrote:

> >I am working on an automotive timing circuit that uses a toyota reluctor
> pick
> >up. It's output is a pseudo sine wave that changes in frequency and
> amplitude
> >as the rpm changes. The ideal point to in the wave to send a position
> >interrupt is as the wave passes zero. I've tried running it through an op
> amp,
> >letting the wave swing from a 5v pedestal, and computing the midpoint
> between
> >the two peaks but this eats up too much time is an "after the event has
> come
> >and gone" solution. Anybody have a suggestion?
> >
> >Bruce
>
>
> Don't amplify it as a voltage.  Use a current to voltage converter.
> That's even a simpler circuit..   Magnets moving past coils create
> current. (evidenced by the messnier effect, since a superconductor
> has no resistance, there can be no induced voltage!)

This is not quite true, AFAIK. Faraday's law says EMF = k * d phi / dt

where k is a constant depending on what system of units you are using and
phi is the magnetic flux thru the coil. Notice that current is not
mentioned here.

Current comes into the picture as resistance approaches zero. In this
case, it is easier to use Lenz's law which states that with zero
resistance, the amount of current flowing in a coil will be such that the
field due to the coil will cancel out can flux changes due to itself or
the external environment.

So, unless I am wrong (I could be, but I just took physics E&M last
semester), even in something which has near zero resistance, the voltage
induced in the entire loop will be as given by Faraday's law. A
superconductor is a special case and standard analysis techniques usually
applied to wire won't work very well. It represents a mathmatical
singularity (because is creates a division by zero in several of the
formulas involved).

>
> This also as the effect of minimizing the effect of the pickup
> inductance since it's working into a short. It ruins the Q, making
> the transducer much more wideband, and removes almost all
> of the self-resonance problems.
>

I think that this is the REAL reason why the current to voltage converter
works best, because the loop sees a short. The easiest law to apply to
this case will be Lenz's, implying that the loop is inherently a current
producer, however, I think that if you measured the total resistance of
the circuit and multiplied by the current, the voltage you would see
would be that given by Faraday's law.

 > We made this mistake with magnetic heads for
credit cards
too.
> The voltage mode circuit is hideous, and has to be tweaked to
> a range of frequencies.
>

I know that what I am saying is REALLY splitting hairs, but I just want
to see how well what I have learned applies to real life situations :)

Sean