Ms-II, V3, 2.34 code Hall sensor with pull up resistor
I'm in the middle of trying to set up my ignition and I don't know what to use as my ignition input capture. I have read the configuration article and tried running the engine both ways. Normally, a 'positive going' pulse would usually mean a pulse seen by pin #24 on db-37 that is +12v or +5v? And since it is inverted by the opto-isolator, that means falling edge would be the input capture?
Since I'm using a hall sensor that pulls a voltage low , the pull up resistor actually pulls it to +12v or +5v. This would be considered 'positive going' and again, the opto isolator inverts the signal. So I would assume I would choose falling edge for my input capture? Am I correct so far? This is what confuses me and I'm still learning the basics of this stuff. Correct me if I'm wrong. thanks.
Ignition input capture: How to know what to use?
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Read the manual to see if your question is answered there before posting. If you have questions about MS1/Extra or MS2/Extra or other non-B&G code configuration or tuning, please post them at http://www.msextra.com The full forum rules are here: Forum Rules, be sure to read them all regularly.
Read the manual to see if your question is answered there before posting. If you have questions about MS1/Extra or MS2/Extra or other non-B&G code configuration or tuning, please post them at http://www.msextra.com The full forum rules are here: Forum Rules, be sure to read them all regularly.
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Bernard Fife
- Super Squirter
- Posts: 1009
- Joined: Mon Feb 16, 2004 3:15 pm
usagoal,
When a sensor pulls a signal low, it pulls it to ground. The pull-up resistor determines the 'high' voltage, not the 'low' voltage, if it is done properly.
If you are using an extenral signal that hgoes low to siggnal the engine position, then it is invertewd by the optoisolator, and you want to trigger on the 'rising edge', as you say.
Lance.
When a sensor pulls a signal low, it pulls it to ground. The pull-up resistor determines the 'high' voltage, not the 'low' voltage, if it is done properly.
If you are using an extenral signal that hgoes low to siggnal the engine position, then it is invertewd by the optoisolator, and you want to trigger on the 'rising edge', as you say.
Lance.