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Old 05-12-2009, 01:09 PM   #1
tbox56
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Member#: 199784
Join Date: Jan 2009
Default Faking sensors, how much resistance?

I have been working on a coversion usign a subaru 2.5lNA, and i recently got it back on the road. The only problem is that i have a handfull of sensors that I need to "fake". Basically i need to send 2-5v to the ecu on several different wires. I have the list of the ecu IO signals, so i know the voltage I need to send, I just need to figure out roughly how many ohms of resistance i need to put in the line to reduce that voltage from 12-14 to the 2-5. I am planning on using precision 10 turn potentiometers to allow me to dial this in, but i need to know what ohm range i need to buy.

Thanks in advance,
Thomas
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Old 05-12-2009, 01:21 PM   #2
UkNuck
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this sounds like a bad idea but fwiw:

- you don't "put ohms of resistance in line" to reduce voltage

- you make a voltage divider (i.e. put your 14.8V across the 'outer' contacts of the potentiometer and take 0-14.8 volts from the 'wiper')

- what absolute pot values you need will depend on the current that is required

- OR you use a proper solid state zener / voltage source / emitter follower depending on the required accuracy / stability / current requirement
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Old 05-12-2009, 09:12 PM   #3
Spore
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Quote:
Originally Posted by tbox56 View Post
I have been working on a coversion usign a subaru 2.5lNA, and i recently got it back on the road. The only problem is that i have a handfull of sensors that I need to "fake". Basically i need to send 2-5v to the ecu on several different wires. I have the list of the ecu IO signals, so i know the voltage I need to send, I just need to figure out roughly how many ohms of resistance i need to put in the line to reduce that voltage from 12-14 to the 2-5. I am planning on using precision 10 turn potentiometers to allow me to dial this in, but i need to know what ohm range i need to buy.

Thanks in advance,
Thomas
These things can be calculated pretty easily given you have electrical experience/background. Remember there will always be voltage loss when voltage goes through a resistor (variable or not.) You measure the amount of current the circuit is using. Then you find out how much voltage you need to lower it by. You then divide the voltage with the current and you will find out what resistance value you will need for your signal.

But I would highly suggest you do not "fake" signals if your not familiar with electronics. A simple mistake can fry your ECU (or melt some wires.) So make sure you are comfortable to tackle this before doing so.

Quote:
Originally Posted by UkNuck View Post
this sounds like a bad idea but fwiw:

- you don't "put ohms of resistance in line" to reduce voltage

- you make a voltage divider (i.e. put your 14.8V across the 'outer' contacts of the potentiometer and take 0-14.8 volts from the 'wiper')

- what absolute pot values you need will depend on the current that is required

- OR you use a proper solid state zener / voltage source / emitter follower depending on the required accuracy / stability / current requirement
Either I have misread your post or there is some information that are passed down incorrectly. A potentiometer (aka variable resistor) is a voltage divider. When you turn the knob (or screw) you vary the resistance and therefore alter the voltage passing through the device. And you do put resistors inline with the circuit to lower its voltage. Putting resistors in parallel of the circuit will just increase its current. You are correct on the amount of resistance is based on the current of the circuit.
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Old 05-12-2009, 10:43 PM   #4
tbox56
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I understand that these can be calculated easily with 2 of the three pieces of information, meaning that i need to know the current to claculate the resistance necessary. Unfortunetly i don't have any way to figure out what the current on the circut is. I just need a general idea of how many ohms will be required to make this reduction of voltage possible. (50ohms, 100, 200, 1k, 2k, ect)

Or can someone supply the information necessary to complete the formula so that i can calculate the restance necessary?

I do have electrical background and i feel confident that i can fake these without any problem, I have already reduced the wiring harness down from every wire in the car to just what i need without any problems. I am not going to just throw one of these resistors in the line without checking my voltage either, i will tune it in with a voltmeter and then plug it into the ecu. I am aware of the risks, however this is the only way that i know of to get rid of those darn fuel tank codes...

Thanks
Thomas
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Old 05-12-2009, 11:23 PM   #5
UkNuck
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Join Date: Jul 2002
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02 GroceryGettAr
Nonemore Black

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Quote:
Originally Posted by tbox56 View Post
I understand that these can be calculated easily with 2 of the three pieces of information, meaning that i need to know the current to claculate the resistance necessary. Unfortunetly i don't have any way to figure out what the current on the circut is. I just need a general idea of how many ohms will be required to make this reduction of voltage possible. (50ohms, 100, 200, 1k, 2k, ect)

Or can someone supply the information necessary to complete the formula so that i can calculate the restance necessary?

I do have electrical background and i feel confident that i can fake these without any problem, I have already reduced the wiring harness down from every wire in the car to just what i need without any problems. I am not going to just throw one of these resistors in the line without checking my voltage either, i will tune it in with a voltmeter and then plug it into the ecu. I am aware of the risks, however this is the only way that i know of to get rid of those darn fuel tank codes...

Thanks
Thomas
See right there is the problem with using a SERIES resistor as a voltage dropper. You are relying on the resistance of the load to get the 'right' drop. You CAN'T tune it before plugging it in because the voltage drop depends on it it being plugged in.

The proper way to do it is what is called a voltage source as I tried to explain above - that is, something that supplies the correct voltage regardless of the ECU current draw. The simplest way is with a voltage divider in which the resistors in the divider carry a current that is large (e.g. 10x for a 10% error) compared with the load current.

These may help

http://en.wikipedia.org/wiki/Voltage_divider

http://en.wikipedia.org/wiki/Buffer_amplifier
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