radiator cooling fan resistor values?

[JayG]

Originally Posted by DennisAN

Be careful to "zero" your meter when dealing with such tiny ohm values. It is possible the measured values in the above post are in error by 0.2 ohms each, when the real values would be 0.0 ohms (i.e. a wire) and 0.6 ohms.

What YEAR is the OP's car? The resistor value appears to vary between model years.

The FSM for 2000+ cars shows only a single resistor per fan motor. Both are shown as 0.55 ohms on the schematic diagram.

To calculate the wattage you have to account for the resistance of both the resistor and the fan motor as they are operating in series to ground. Once you have the overall amperage, you can then backtrack and apply ohm's law to the resistor itself to determine its minimum wattage rating.

That makes sense, that the measured values are off by ~.2 ohm. I believe that on HIGH, the fan get the full 12v and on low it is lowered by the resistor in series

Now the fuse for each fan is 40 amp, if it was a full draw of 40 amps, it would be ~440 watts

12 volts through a .55 ohm resistor is ~260 watts or ~21 amps.

My guess is that the low speed resistor is ~ 300 watts rating if .55 ohm and ~250 watts if .8 ohm

What gets interesting is that the voltage drop on a .55 ohm resistor with a 12 v source is ~11.5 volts. That leaves ~.5 volt to the fan

If we go with .8 ohm as measured, then the current is 15 amps and the voltage drop would be 8.25 volts leaving 3.75 for the fan. This sounds more realistic than .5 volts to the fan

if the actual voltage is higher, say 13 volts than the voltage to the fan will increase slightly and the fan speed would be higher

Maybe a measurement of the voltage after the resistor would shed some light on this

Of course if my calculations are wrong, then fo-ged-about-it

[danci1973]

Originally Posted by JayG

Now the fuse for each fan is 40 amp, if it was a full draw of 40 amps, it would be ~440 watts

12 volts through a .55 ohm resistor is ~260 watts or ~21 amps.

My guess is that the low speed resistor is ~ 300 watts rating if .55 ohm and ~250 watts if .8 ohm

What gets interesting is that the voltage drop on a .55 ohm resistor with a 12 v source is ~11.5 volts. That leaves ~.5 volt to the fan

Sorry to revive an old thread, but the above is fundamentally wrong as it neglects the fact, that the electric motor has an impedance as well.

If we assume that the motor draws full 40A at 12V, it's impedance would be 0.3 Ohm. But that would be unreasonable, so if the motor drew 20A, the impedance it would be 0.6 Ohm. I would 'guess' that the actual impedance is somewhere in the middle - in fact, I'd say it's equal to the resistor, i.e. 0.55 Ohm.

Since the resistor for low speed operation is connected in series with the motor, the total impedance is 1.1 Ohm, meaning that the amperage draw is 11A and the resulting voltage drop is 6V, leaving another 6V for the fan, cutting the fan speed in half.

The power dissipation on the resistor is 66W, so a 100W 0.55 Ohm resistor should be just fine.