3
\$\begingroup\$

I'm looking for explanation of some basic induction concepts.

It's common to see an RPM sensor for small two and four stroke engines where an open-ended wire is coiled around the spark plug's ignition coil wire at one end, and the other end of the wire is simply routed to pass in front of a conductive plate connected to some circuitry which then apparently does the pulse counting. Hooow exactly does this work?

1) From what I understand, the ignition coil's pulsing current produces a pulsing magnetic field, which is now aligned with the "sensor wire's" coil around it, thus inducing a voltage potential in the sensor wire? In a closed circuit where there's actually a current flow, I can understand that induced current, but I actually don't remember anything from my uni EE courses about induced voltage.

2) On the other end of the open-ended sensor wire, where it simply passes closely in front of a conductive plate, how exactly does that plusing voltage potential in the wire affect the conductive plate such that it can be measured? Obviously there must be inductance happening here as well, but that's still not clear to me when there's no current.

I've definitely done a lot of video watching and reading, but I've been unable to connect these dots.

enter image description here

\$\endgroup\$
4
\$\begingroup\$

In both cases, the coupling mechanism is capacitive, not inductive. It's the high voltage of the spark plug wire (relative to the chassis ground) that drives the pulse counter, not the current in it.

Wrapping one end of the wire around the spark plug wire forms one capacitor, and laying the other end across the metal plate forms another.

\$\endgroup\$
3
  • \$\begingroup\$ Wow. How much voltage is it? That's cool! \$\endgroup\$ Mar 21 '20 at 4:29
  • 1
    \$\begingroup\$ @MicroservicesOnDDD: Spark plugs typically operate in the range of 20-30 kV peak. It doesn't require much capacitance to couple a few volts of that to a logic circuit. \$\endgroup\$
    – Dave Tweed
    Mar 21 '20 at 11:04
  • \$\begingroup\$ Ohhh! Well that explains why all of my reading wasn't turning up anything familiar; I was barking up the wrong tree using the wrong terms. I will do some reading. Thanks. If anyone has a similar reference design for the signal conditioning on the sensor end, that'd be extra helpful. \$\endgroup\$
    – seth
    Mar 21 '20 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.