IAT Sensor Mod (Resistor Mod)

[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.