does anyone know how much resistance i need to make .5 volts from 12 volts. i am hooking up a little relay system that will send the ecu a .5 voltage (14.7 a/f) when i get into boost, thus giving me full control of my fuel.
u would need a resistance of 11.5 volts
ok sorry that was kinda mean, I really don't have a clue about this stuff, and was just tryin to be funny and put a lil check in the email me replies to this thread box.
how would making the computer think its running 14.7 give you full control of the fuel? make it thinks its running 16:1 or so so it adds fuel and makes it richer
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a step down transformer would work, we use to mess around with them in electronics class
A step down transformer won't work without an AC or a pulsing DC power supply.
The simplest way to create a voltage in the range that you want is to wire up some resistors in series and take the voltage signal from there.
Here's one way:
135 Ohm 5 Ohm
+12V----/\/\/\/\-----------/\/\/\/\--------- Gnd-
|
|
| .5V for signal
This isn't the best way by any means, but it's one way. This assumes that you have 13.5V charging voltage while the car is running, and that the relay doesn't have a large amount of internal resistance. In truth, the signal will probably be above or below .5V, but it should be close and it will allow you to test your idea.
-->Slow
I know I'm gonna sound like an a$$ here but, with a name like coldfusion, dropping voltage sounds like it would be below your expertise. lol
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it is. but thanks slowolej, i will try that out and see what i get.
hypsy - if i make the ecu think it is lean, it just dumps fuel with no pattern or control. i have an e-manage so i can tune the fuel to whatever i want as long as the ecu thinks everything is stoich.
You should also cosider running a cheap 7812 voltage regulator on the circuit.
Your car never runs 12v. A fully charged battery is 12.7v and running, system voltage is anywhere from 14.7v cold to 13.7v hot.
using the circuit above
12v= .4285v
12.7v= .4535v
13.7v= .4892v
14.7v= .525v
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well i will only be using this circuit when the car is running, so it should put me very close to .5
well the pcm will kick out codes for the o2 if you do that , cause the comp needs to see it bouncing up and down the scale
the pcm automaticly reads around .4-.5 volts if the o2 is disconnected
i think your gonna have alot of problems trying to tune it , if you try it
but might as well try it and see if it works
[quote=97cavie24ls(JDM cav sedan™)]
the pcm automaticly reads around .4-.5 volts if the o2 is disconnected
?? if you disconnect the o2 it starts dumping fuel like mad.
and for the short amount of time that the ecu will see a constant .5 volts, it will not trip a code.
wouldn't he need a voltage regulator
i would say your best bet
Any voltage regulator, either 12, 10 or 5 even, broken down even use capacitors to smooth it out to get a good clean voltage. paired with a proper pair of resistors wired in series like above, depends what size regulator you decide on and make the wires as short as possible for least amount of resistance.
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coldfusion wrote:[quote=97cavie24ls(JDM cav sedan™)]
the pcm automaticly reads around .4-.5 volts if the o2 is disconnected
?? if you disconnect the o2 it starts dumping fuel like mad.
how do you know?
i'm *almost* sure it is the exact opposite. it stops pulling fuel like mad.
it should run on an internal map without reading the o2 sensor... but that thinks it is with 252cc injectors...
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how do i know?? i have done it.
and when the computer sees the voltage stay in that range it will do the same as as if its disconnected
If you disconnect the O2 sensor, it won't dump gas in the engine more than when you start it up. When the O2 is disconnect, the ECM will stay like when it just start. it will use the preset map in the E-prom.
Gilles
2.3 Ho
[quote=97cavie24ls(JDM cav sedan™)]and when the computer sees the voltage stay in that range it will do the same as as if its disconnected
no it won't, when it is disconnected, it sees a 0 voltage, so it adds fuel to get out of the lean condition. When it sees a .5 volt, it thinks that the a/f mixture is stoich and will continue at that constant amount of fuel.