IAT Sensor Mod (Resistor Mod)

[SVOboy]

I was going to try out larry's IAT mod on my friends saturn tomorrow, I even had two 220 ohm resistors hooked up in parallel to do it:

But he wussed out last minute. So I have a few questions.

Has anyone besides larry done this?
To larry: I know you said the heat isn't different, but are you worried about running too lean or oxidation or anything? How many miles have you been running this set up? I know you have EFIE, but with this can you figure out your AFR? That's all I can really think of now, but mehbe I'll ask some more later. I'd love to try this, just wanna mehbe get some answers to convince this guy since I'm in the middle of my acetone test.

EDIT: If anyone wants a set of resistors in series that add up to 110ohms (like the ones on the bottom) I have tons of resistors around and I'd be willing to send some out to you so you don't have to buy a bunch of crap or locate a bunch of crap. That way we can also get you on testing a bit faster. Also, if you're non obd0/1 honda, I could prolly figure out what resistance to use, but it'd take me some time if possible.

[diamondlarry]

I have been running this mod since April. Up until 3 weeks ago, I drove 200 miles/week just for work. With all of the other running around, it probably ended up being 300-350 miles or more a week. At 300 miles a week this figures out to around 6,000 miles. Back when I did my sparkplug testing, I had my plugs out fairly regularly and never noticed any signs of running severely lean. As far as AFR goes, I don't have any way to check that.

[SVOboy]

Yeah, I'll show this to him and try to sway him, but I might just wait for my own testing to be done for it.

[SVOboy]

What's the best way to figure out the resistance I need for my stuff? Should I boil some water and dip the sensor in there and measure the resistance with a multimeter?

[diamondlarry]

That should work.

[SVOboy]

I don't have a multimeter handy, so I'll do it tomorrow hopefully and post up the results. I'm still hoping on acetone, but it seems like every time I start the car the needle drops randomly and won't go back up to where I started it from so it worries me, haha.

[SVOboy]

So, it's 74 degrees out now and my multimeter just read 61.5 ohms of resistance across the IAT sensor. So I'm wondering if the relationship is linear or what, cuz if it was I could just heat it up a bit and test it and then make a line to figure it out, but I think it'll be above 110 ohms.

[SVOboy]

Okay, forget what I measured just now, cuz it's no good. Here's the real numbers for the 88-91 Honda D15B2:

100 ohms = 248 degrees
20,000 ohms = -40 degrees

Seems that all obd0 and obd1 stuff uses the same values, so that should be good for all the honda stuff at least, I dunno about obd stuff outside of honda.
I'll see if I can get a b18c1 thing for flatland...

[SVOboy]

This is a picture of my IAT for Matt. This on a 91 crx dx with a d15b2 and dpfi.

[dfoxengr]

just a quick resistor ?

im not so good with electronics, but if i wanted a 12v power wire in my car to be reduced to say 3v or 6v or 9v, how would i do that?

[SVOboy]

Yep, that's right, if you want I could go shove my resistors in the sensor holes and take a picture. I'd be happy to if you want, pretty simple, shouldn't be affected by the other electrical stuff you have done on the car.

[dfoxengr]

can anyone answer my ?

[SVOboy]

Do you mean to add something to a circuit to convert 12 volt on one side to 6 volt on the other?

[dfoxengr]

like putting something inline with a wire.

[SVOboy]

The U = I * R formula hold here well.

The circuit you have is something like this:

+24V --- RESISTOR --- LED --- GROUND

In this circuit you had measured 24V on the one end of resistor (connected to 24V power source).
The other resistor end (one connected to LED) has 2V. Both measured agains circuit ground.


The circuit works in the following way:
The resistor has 22 volt voltage drop over is (24V-2V=22V). This voltage drop depends on the current going through resistor.
The LED has 2V voltage drop over it. This 2V voltage drop is typical voltage drop that stays pretty much constant on LED normal operating conditions (from 1 mA to 20 mA current nicely).

In this circuit the rest of the voltage drop (22V) then tries to be left to the resistor. When the resistor has 22V voltage drop, from the equation U = I * R we get to know that to generate 22V voltage drop on resistor the current flowing through the resistor must be around 8 milliamperes. In this application whe votlage drop over the resistor is pretty much pre-set, so the resistor limits the current flow through it to value that gives the predetermined voltage drop.

Voltage drop is a pain in the ***, I did this rc circuit stuff last year, and remember it pretty well, but didn't know how to say it. So, if you add a resistor you will experience a voltage drop across that resistor, as per the equation Voltage = Current * Resistance. The thing about this however, is it all depends on however much current is running through your circuit. If you can figure out your current, you should be able to figure out what resistance to use to create a certain amount of voltage drop.

If you have 12 volts, and 1 amp, then if you put it in the equation you get:

12v = 1a * xohms
And therefore you'd have 12 ohms of resistance.

If you then go:

xv = 1a * 6ohms, you'd find that your voltage is now only at a value of 6.

So that's how the stuff goes.

[dfoxengr]

how many amps do cars put out then? i guess ill have to figure that out by what my alternator puts out huh?

im sure its like 140 or so.

[SVOboy]

Yeah, i have no idea how much current the car is running around on, I think you'd just have to measure.

[Flatland2D]

Why are you trying to measure voltage drop? Even though the ECU is really reading a voltage, it's proportional to resistence. Therefore the temp it reads is proportional to resistence.

As for current in the sensor, if you stick a 110 Ohm resistor in there, it'd draw 109mA (12V / 110 Ohms).

[diamondlarry]

I think the voltage in the IAT circuit may only be 5 volts. I know it's 5 for the MAP sensor signal but I'm not positive on the IAT.

[SVOboy]

5 volts sounds about right, but you'd just have to take a multimeter to it, I'm also putting up top that if anyone wants I will whip up some 110 ohm resistors for them, if people who aren't reading the top anymore are interested.

[nrgrebel]

Some time ago I stumbled upon a variable iat control generally used for performance tuning. It was a dash mount unit. Anybody else seen this? I've also seen mention at theoldone.com forum about tuning mixture using such a device.


http://forum.mpgresearch.com/index.php

[dfoxengr]

im trying to find a way to drop voltage so i can run say something that needs 6v off of my 12v in the car. this doesnt have anything to do with iat

[Flatland2D]

Larry, I believe you're right. The sensors do use 5V. The fuel injectors and accessories use 12V.

Originally Posted by SVOboy

Voltage drop is a pain in the ***

Welcome to the life of a EE.

[SVOboy]

Haha, yes, well, too bad for me I spose. You haven't been around recently, but I have a proposition for you sometime (I'm about to go out to do the Crop Walk though).

[dfoxengr]

so can anyone answer my ?

[diamondlarry]

Okay, now I think I understand what you're asking. It sometimes takes us old guys awhile to catch on. :-) You would have to know in advance what the load(Amperage) of the device is that you're going to power. Then it would be a matter of coming up with the resistor values. A simple way to do 12 down to 6 volts would be to use two resistors of equal value in seies. Then you would tap the junction netween the two resistors and place your item between there and ground. As I said above, you need to know the current draw because if you use to high a value of resistors you won't have enough current available; too low and things get hot.
Larry

[dfoxengr]

gotcha. why do you use 2 in series, and put the power to the item between? what formula get me the proper resitor values?

[Flatland2D]

Ah, I see now. The resistor method that Larry described is called a voltage divider. It'll only work if everything in the circuit is a fixed constant and the load of the circuit you're connecting is very small. This is generally not the best way to drop voltage. A DC motor for example will consume varying current throughout its RPM band. If the equivalent impedence of the circuit you're connecting to the divider is much higher than the resistors in the divider, then the divider won't work at all as expected. Most of the drop will be across the load. Resistor values too high and you may not have enough current. If you're trying to run an accessory of the 12V the voltage divider will only work for very simple circuits like LED lights. If you were to use this method the formula would be:

Voltage_out = Voltage_source * (R1 / (R1 + R2))

where R1 and R2 are in series and Voltage_out is the drop across R1.

A better way of doing this would be to use a zener diode with specs for the voltage drop you want. A 6V zener diode will always (well almost always) have a 6V drop across it. Better yet would be to just get a voltage regulator (looks like a large transistor) that will put out a constant voltage. Radio Shack carries them for the standard values (5V and 12V). Anything different and you've have to order it online or find an electronics store. As long as you don't try to draw too much current from the voltage regulator the output will remain constant.

[diamondlarry]

It's been too long since I've been in colege:-) I knew whe I was writing that that wasn't the best way to drop voltage.