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I wanted to use an outside signal other than the vehicle sensor, to use on sensor circuit. I don't have a problem, my question is generic. I figured I could run a wire from positive battery posts with a resister in series, to cut back the voltage. I don't know how to pick the correct resister? Depending on the circuit, the computor might expect to see a signal of less than one volt. I know a flashlite battery is about 1.5 volts, still a little high for some circuits. I guess I could hook wire and resister to battery and check voltage drop across resister, then subtract that value from the battery value? I'm not literate when it comes to electronics, ok. How do I pick the correct resister? Any thoughts? thanks
A resistor will not reduce the voltage unless there is current flowing through it. What you want is a resistor divider. If you want to create a node that is 1V and proportional to the battery voltage, you will need at least 2 resistors. Connect both in series from the pos to the neg terminal. Seclect them so that the total resistance is quite high so that the current is low and you don't just kill the battery. Select the ratio between the two resistors so that you get the voltage you want. <p>Many combinations of resistors will give you any given voltage. You now need to think about the circuit that will use this voltage. If its input is high impedance and requires little or no current, use a very high resistance for your divider to be efficient. If the ckt needs some current, adjust the value of the larger resistor so that at least that much current can flow through it.<p>Finally select the wattage of the two resistors acording to the current flowing through them and their value. For a low current high resistance combination, even 1/8W resistors would be enough.<p>For a 12V battery and 100uA load you would need a total of 120k ohms. To get 1V at the junction the top resistor needs to be 110k and the bottom one needs to be 10K. Scale this example to your needs.
You could use a pot and the above principal. Since It looks like you're trying to fool a sensor input and you probably don't know the exact voltage you want you could adjust the pot until you get the desired effect.
yes, with a voltage divider, you tap into the junction between the two resistors. IE: where one end of one resistors hooks into the other end of the other resistor.<p>I guess you could use this formula for the voltage:<p>Vout = R1/R2 * Vin<p>make sure you listen to haklesup and use an overall high resistance so you don't draw too much current from the source.<p>A 100k potentiometer (variable resistor) would be a good choice. This would allow you to change the voltage to whatever gives the desired result, as Joesmith suggests.<p>There would be three terminals on a generic pot. The two outer ones go to the positive and ground connections, and the center one is the output of the voltage divider.<p>Of course, you may be wanting a different solution than this for your application. Can you tell us more about what exactly you're trying to do so we can tailor advice more towards what exactly you need?<p>Cheers,
Thanks for all the info. In the past, I've seen info where an auto technician, when checking for a computor reaction to a signal on the 02 sensor circuit, unplugged the connector to the sensor with engine running, stuck a nail in the connector end that went to the computor, held in one hand, touched the positive bat. terminal with the other hand, I guess the body acted as a resister, I'm not suggesting anyone try this. The computor saw .8 volts--rich signal--the computor leaned the engine, because the engine wasn't rich to begin with, the engine died or stumbled, letting the technician know, that the computor was reacting to the signal. I just thought using a wire and proper resister would be a better way? I never heard of a voltage divider until you folks mentioned it. I did find a voltage divider calculator on the net. With the sourceV-- outV and current, it gives the proper resistors to use. But now I'm concerned about the current? I thought 1/2 amp would be safe, but now I'm not sure? I guess I could put my dvom in series with the circuit and read current. I don't know, I need to get myself an electronics beginners book or something. thanks
A half amp (500mA) may be an acceptable current if you just plan to use it intermittantly. I was thinking you wanted a perminant connection to the battery and therefore a low current would prevent premature discharge of the battery while you were parked.<p>I don't know about the car tech's procedure but it is common to test the o2 sensor by disconnecting it and ensuring that the engine reacts to the change. This is done to confirm the emmissions controls are working properly (part of the standard CA smog check) and the exact procedure varies by make and model. Essentially, you fool the car into thinking there is no oxygen in the exhaust and the emission system compensates by adjusting the fuel air mixture. Naturally with a sensor not responding, the mixture will go to its maximum limit and the engine will run crappy. The check engine light should also come on. Replacing the sensor with a resistor will fool the engine into thinking the o2 content is stable at whatever level the resistor sets it to. This may be what the tech was trying to do in a sloppy way. Was this test in the context of a smog inspection or was it a diagnostic procedure used for a repair.<p>Perhaps the repair manual for your car (with wiring diagrams and procedures) will shed more light on your application than a basic electronics text.
The test was a diagnostic procedure. I've seen it listed in a couple of different places, never tried it, though. Anything I tried would just be for diagnostic purposes. I've heard mechanics say, unplugging sensors for diagnostic purposes was poor diagnostics. It just seemed to me, if you could imitate the proper sensor signal on a suspect circuit, it might tell you something? thanks
If I understand this question correctly.<p>You would like to rig a signal that replaces the signal typical of an automotive O2 sensor.<p>Could this be accomplished from a potentiometer ?<p>How is this processed in the cars computer ?
"I've heard mechanics say, unplugging sensors for diagnostic purposes was poor diagnostics."<p>It depends on what you are doing. For example if you have a very poorly running engine presumably due to a faulty sensor, then unplugging the bad one may have no effect because it is already bad, unplugging a good sensor may also tell you little because the engine is already running so poorly.<p>If on the other hand, you have an engine which fails emissions for CO but otherwise runs well, removing a few choice sensors like the o2 may tell an experienced tech if that sensor is working properly. (I suppose some intuition comes into play here).<p>Many engine sensors act simply like potentiometers. They act as half of a voltage divider (the other resistor is on the engine computer's board). The resistance changes wrt. the thing it is sensing (o2 in your case) Look in the specs for the part or measure one with a ohmmeter to know the range of operation. Then you can more wisely select a resistor to imitate the sensor. Expect the value of the sensor to change as the engine warms up.<p>To measure the R on a running engine you would have to briefly remove the connector from the sensor and make the measurement but you can probably derive a better feeling for how it changes by measuring the voltage across it while the engine warmes up. If it has 2 wires measure between, if it is one wire, measure from connector to sensor body (engine block or neg terminal on bat)<p>In any case, shorting the sensor connector to the battery (+ terminal)through a resistor might be unwise if you don't understand what and why you would do so. That wire probably leads directly back to the ECM (engine control module) and that would cost a couple hunderd $ to replace. Connecting it to ground (or chassis) through a resistor should be OK because that is what happens in the sensor. (sensor body is a connection to ground itself)
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