I got a standard ground and a pin on the ECU. I need to put 2.5 volts on that pin in the ECU. What size resistor do I use? Or is there some other device I can use to put that much voltage on that pin? I've taken an electrical course in Physics but those equations require amps and I don't know what level of amps would be working here. Basically I need a 9 volt drop... on a 12 volt circuit.
Technical electrical question
Silvertune
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read the page.
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Bitter is that a decent source to order from online? Those things are cheap.
Perhaps a stupid question but wouldn't I need the negative regulator since it'd be taking a negative input or is the ECU power source a positive?
Perhaps a stupid question but wouldn't I need the negative regulator since it'd be taking a negative input or is the ECU power source a positive?
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Bitter is that a decent source to order from online? Those things are cheap.
Perhaps a stupid question but wouldn't I need the negative regulator since it'd be taking a negative input or is the ECU power source a positive?
Perhaps a stupid question but wouldn't I need the negative regulator since it'd be taking a negative input or is the ECU power source a positive?
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Hmm. This is so much more complicated then it should be. I wish I understood more about this. In a single resistor circuit, the voltage drop across the resistor equals the voltage on the battery. So how is there partial voltage drops across pins to the ecu? There must be a resistance level in the ECU in order to make that happen.
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the ECU has some of those V-Regs inside of it. they're magic. Usually heatsinked to the side of the case to dump heat or the case has some fins on it to help out. Really the ECU only needs one to drop 10-15v to 5v for things, not too many things need many amps of 5V so it's not a big one.
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Hm. But what creates the voltage going into the pin on the ECU? I mean, what creates a drop that isn't to 0, there must be resistance involved with the pin to get a voltage there, otherwise if it were like a ground it would be at zero volts and the drop would be 100% across the sensor. I'm definitely not up to par on my electronics knowledge though, it makes sense somehow.
I'm going to go to radioshack to see if they have these.
I'm going to go to radioshack to see if they have these.
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I had an idea. There is a resistance sheet for all the sensors in the car. Of course it doesn't have the FTS but it has the other temp sensors. Based in their resistance ratings and resulting voltage I can get away with wiring up a simple 2k resistor. I'm going to try this. Truth is I don't really care if I kill this ECU anymore.
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12-15 volts goes into the ECU, the ECU has ground. +5 comes out for certain sensors, it uses a voltage regulator inside the magic box. Beyond that, I don't know and never really needed to know since with what I do, if the ECU isn't making its 5 volts for sensors then it's defective and it gets replaced or rebuilt by an outside company.
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I'm going to jump the 5v sensor supply witha 2.5k resistor back to the fuel temp pin. It should work.
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If you take 5V, put a resistor in line, you still get 5V out the other end. Can you map out what you are going to do or do you have a link?
Or are you just figuring that the resistor is acting like a temp sensor and just going with the 5V back to the ECU instead of 2.5? Will that act like super hot (or cold) fuel?
If there's any chance you're headed to the MNSC meet tomorrow (Noon, near 494 and 100) I will be there. I don't have any magic answers but maybe we could bounce ideas around.
Or are you just figuring that the resistor is acting like a temp sensor and just going with the 5V back to the ECU instead of 2.5? Will that act like super hot (or cold) fuel?
If there's any chance you're headed to the MNSC meet tomorrow (Noon, near 494 and 100) I will be there. I don't have any magic answers but maybe we could bounce ideas around.
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What meet is that? I'm right by 494 and 100, maybe 2 miles away. Toss me an address and I'll see if I can head over there.
Resistance causes a voltage drop, that's how all the temp sensors in the car work. They get a 5v input, the temp varies the resistance the sensor puts on the circuit which results in a voltage drop. The ECU reads that voltage which translates into a temperature readout that it uses to calculate its needs.
I'm tapping the 5v output and running a simple circuit through a 2.5k resistor then to the fuel temp pin on the ECU. That will trick the computer into thinking the fuel is around 70 degrees Fahrenheit, which isn't all that far off. And since it won't be driven in winter it really won't be an issue. I'm also throwing in a set of stock injectors. After messing with the latencies and stuff and getting worse and worse results I'm going back to high z, screw peak and hold.
Random question I'll post here: have you ever dealt with a Walbro pump that didn't actually whine? Every pump I've ever dealt with (except mine) whines like crazy, but mine is silent.
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Tracer I'll be there around noon maybe a tad after. I'll be pulling up a gold Honda Odyssey.
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http://www.autoshop101.com/forms/h32.pdf
We may both be right (from our conversation yesterday afternoon)... A single resistor all by itself won't cause a voltage drop, or at least not an amount like your'e thinking. A voltage divider is needed somewhere along the line.
However... a single resistor might work, if the voltage divider is built directly into the ECU itself! If you look at the bottom of page 1 (and also similarly on page 2) that's exactly what they are doing (the extra resistor shown inside of the ECU itself). As the resistance in the sensor itself changes, then because of the voltage divider built into the ECU, the voltage *where the ECU is picking it up* is changing.
The pdf I linked to is for Toyota, but I'm willing to bet it's a common way of doing things.
We were thinking you had to have the voltage divider outboard, but it looks like half of it's in the ECU itself. So your single resistor may very well work (though for a totally different reason than you thought, LOL).
Only question is - what value of resistor? If you can find out what's in the ECU then just duplicate it, your 5V supplied will be 2.5V as a result. Or if you can find an actual sensor you might be able to measure the resistance across it and do it that way as well. They are probably both similar values at room temperature.
We may both be right (from our conversation yesterday afternoon)... A single resistor all by itself won't cause a voltage drop, or at least not an amount like your'e thinking. A voltage divider is needed somewhere along the line.
However... a single resistor might work, if the voltage divider is built directly into the ECU itself! If you look at the bottom of page 1 (and also similarly on page 2) that's exactly what they are doing (the extra resistor shown inside of the ECU itself). As the resistance in the sensor itself changes, then because of the voltage divider built into the ECU, the voltage *where the ECU is picking it up* is changing.
The pdf I linked to is for Toyota, but I'm willing to bet it's a common way of doing things.
We were thinking you had to have the voltage divider outboard, but it looks like half of it's in the ECU itself. So your single resistor may very well work (though for a totally different reason than you thought, LOL).
Only question is - what value of resistor? If you can find out what's in the ECU then just duplicate it, your 5V supplied will be 2.5V as a result. Or if you can find an actual sensor you might be able to measure the resistance across it and do it that way as well. They are probably both similar values at room temperature.
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Where are you getting this from ...... of course a resister in line in a circuit will produce a voltage drop if there is current flowing?
A resister just buy itself sitting on a table with nothing attached won't create a voltage drop of course but all of you seem to be floundering around some pretty basic electronics and when good advice is ignored over bogus pull it out of your ass answers something is seriously wrong.
A resister just buy itself sitting on a table with nothing attached won't create a voltage drop of course but all of you seem to be floundering around some pretty basic electronics and when good advice is ignored over bogus pull it out of your ass answers something is seriously wrong.
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Where did that come from?
A *single* resistor attached to a 5V source is not going to spit 2.5V out the other side. Multiple resistors will have amounts different than 5V across them, most definitely. For example 2 identical ones can be used and tapped in the appropriate spot to get 2.5V from that 5V source. Hence the voltage divider discussion. Look at the ECU description I linked and you'll see that a voltage divider is exactly what is used.
But "A resistor" as you put it, by itself, doesn't do anything for him.
A *single* resistor attached to a 5V source is not going to spit 2.5V out the other side. Multiple resistors will have amounts different than 5V across them, most definitely. For example 2 identical ones can be used and tapped in the appropriate spot to get 2.5V from that 5V source. Hence the voltage divider discussion. Look at the ECU description I linked and you'll see that a voltage divider is exactly what is used.
But "A resistor" as you put it, by itself, doesn't do anything for him.
OK sure :yesway:
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Sure it will. E=IR. Given the current, I, stays the same and the resistance, R, changes, then voltage, E, will also change. Simple math.
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I dunno what to tell you Greg. I put the resistor in and it stopped throwing the code. Voltage input was supposed to be anywhere from .1 to 3V to keep it off the code and it didn't throw one for 20 minutes of driving.
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Yeah, perfect. It's a voltage divider, as show in the pdf I sent. The ECU has the other resistor. You are running 2 resistors.
So if I take a multimeter and go out to my car battery, I touch the terminals with the probes, I read 12V (give or take but for this discussion).
I lift off one of the probes. I touch it to a resistor and then touch the other end of the resistor to the battery, I'm going read 6V?
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