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Stock blower motor draw lots of current

S

summitlt

NAXJA Forum User
Location
Maine
I was messing around in my truck and founf that the blower motor for my heater draws tons of curent. It will move my voltmeter more than my 100W each pair of lights. And more than my electric fan for the motor!

Is that normal?
 
I pulled the ignition switch out of three different XJs (2 87s and an 89) and all were melted around the wire that feeds the blower motor and radio. Since none of them have anything special for radios I have to conclude that the blower motors are the cause. I would hope this is not an issue with a new blower motor but rather a result of a worn out one (in one of them it was squeeling loudly). Only problem is a new blower motor is a bit expensive IIRC. I read somewhere here about drilling a hole in the center of the back to allow some lubrication to be inserted, might help.
 
They all seem to do that, at least on the older body styles (pre-'97).

You could always use a relay and wire the blower motor directly to the battery. I did this on my YJ and the current draw is FAR less.
 
iroc86 said:
They all seem to do that, at least on the older body styles (pre-'97).

You could always use a relay and wire the blower motor directly to the battery. I did this on my YJ and the current draw is FAR less.
Click to expand...
Not really you just don't see it.
 
Yup, those old motors defineately draw some current! When mine went in the MJ last winter I yanked one out of a parts rig and took it to the shop to test it out before putting it in. threw it on the alternator/starter test bench and hooked up some (small, maybe 16-18ga) jumpers to it. Hit the switch and was treated to a wonderful smoke show as the jumpers instantly vaporized. I figured the motor had a short, but after getting some heavier wire it turned out to be fine.
 
langer1 said:
Not really you just don't see it.
Click to expand...

The draw CAN be reduced if you use a heavier gage of wire than the original or if the original was being overloaded by other loads connected to it.

If the wire is heavier gage then you get more effective voltage in the conection and the load (amps) are going to be less as the motor draws power from the system and:

motor power draw = AMP load x effective line voltage

SACEM
 
sacem said:
The draw CAN be reduced if you use a heavier gage of wire than the original or if the original was being overloaded by other loads connected to it.

If the wire is heavier gage then you get more effective voltage in the conection and the load (amps) are going to be less as the motor draws power from the system and:

motor power draw = AMP load x effective line voltage

SACEM
Click to expand...
So your saying a blower motor set on low speed, (High resistance) will draw more power from the battery than when it on high?
I think your thinking about an AC motor not a DC one. DC motor speed can be controlled with resistance but not a AC motor.
 
For a mathematical explanation, wire resistance can be calculated using the formula R = rho * (L / A), where R is resistance, rho is the resistivity of the material (a constant), L is wire length, and A is wire area. You can see how increasing the length increases resistance, while increasing the area (i.e thicker gauge) reduces it.

langer1 said:
So your saying a blower motor set on low speed, (High resistance) will draw more power from the battery than when it on high?
Click to expand...

I want to say no. As sacem alluded, think of Ohm's Law (I = V / R) and Watt's Law (P = V * I), where I is current, V is voltage, R is resistance, and P is power. As the resistance increases, current draw decreases. According to Watt's Law, power and current are proportional -- so less power would be used with a higher resistance. Keep in mind that even without a resistor inline (i.e. high fan speed), the wire still acts as a resistor of sorts.

Hmm, that sounds acceptable... but I never did pay much attention in my electrical circuits class :). Somebody correct me if my thinking is off.
 
langer1 said:
That reminds me, I got to get mine fixed soon. It squeels some and sure pulls the voltage down.
Click to expand...

Mine was fine in Sept, but after sitting when we went to drive it across the country to here, it was squeeling something awful, and in December going through the rockies, no heat was not an option. Brand new blower motor only cost me $45US so really not too expensive at all.
 
seymouj said:
I pulled the ignition switch out of three different XJs (2 87s and an 89) and all were melted around the wire that feeds the blower motor and radio. Since none of them have anything special for radios I have to conclude that the blower motors are the cause. I would hope this is not an issue with a new blower motor but rather a result of a worn out one (in one of them it was squeeling loudly). Only problem is a new blower motor is a bit expensive IIRC. I read somewhere here about drilling a hole in the center of the back to allow some lubrication to be inserted, might help.
Click to expand...

The blower is easy to take out, at least in a 96, and drill a hole on the closed end. Then your able to oil both ends of the armature shaft and all is well. I did it over a year ago and it's still working fine. I used a thin oil I have for my engine lathe's headstock bearings.
 
Before you go too far, check the voltage at the battery and don't rely on the built-in voltmeter. The blower seems to share some circuitry with the voltmeter, and will always show a big voltage drop even when it's healthy.
 
iroc86 said:
For a mathematical explanation, wire resistance can be calculated using the formula R = rho * (L / A), where R is resistance, rho is the resistivity of the material (a constant), L is wire length, and A is wire area. You can see how increasing the length increases resistance, while increasing the area (i.e thicker gauge) reduces it.



I want to say no. As sacem alluded, think of Ohm's Law (I = V / R) and Watt's Law (P = V * I), where I is current, V is voltage, R is resistance, and P is power. As the resistance increases, current draw decreases. According to Watt's Law, power and current are proportional -- so less power would be used with a higher resistance. Keep in mind that even without a resistor inline (i.e. high fan speed), the wire still acts as a resistor of sorts.

Hmm, that sounds acceptable... but I never did pay much attention in my electrical circuits class :). Somebody correct me if my thinking is off.
Click to expand...

Thaks iroc86 for the explanation you have answered the question much more clearly than I would have been able. Really I deal with this things daily and know how it works but have to look up the laws that explain them.

SACEM
 
Well, I took matts suggestion and foudn that the blower motor draws about as much as my blinkers. I had all of the things I could think of running (two 100w lights, two headlights, two 55 watt lights, blower mtoor, electric fan, brake lights, radio) and my voltmeter on the dash was bordering the red zone, wasnt quite on it, but pretty close. My portable meter said 13.3 volts. I assume the red zone is 12 volts. So its all good anyway.
 
The volt meter running near 13 V that only means that your alt was working at maximum output to cope with all the load that you had on at that moment but it does not show you anything else.

I suppoe that it was cheaper for AM/DC to install a voltmeter than an Ampmeter because you can not use it for much as a charge situation indicator, but it comes just that way.

SACEM
 
Yes, but that also told me different than my voltmeter on the dash. The voltmeter on the dash was signifying that I was close to draining off the battery.
 
summitlt said:
Yes, but that also told me different than my voltmeter on the dash. The voltmeter on the dash was signifying that I was close to draining off the battery.
Click to expand...

You are right, the only problem is that not many can interpret a voltmeter like you do.

Cheers

SACEM
 
iroc86 said:
For a mathematical explanation, wire resistance can be calculated using the formula R = rho * (L / A), where R is resistance, rho is the resistivity of the material (a constant), L is wire length, and A is wire area. You can see how increasing the length increases resistance, while increasing the area (i.e thicker gauge) reduces it.



I want to say no. As sacem alluded, think of Ohm's Law (I = V / R) and Watt's Law (P = V * I), where I is current, V is voltage, R is resistance, and P is power. As the resistance increases, current draw decreases. According to Watt's Law, power and current are proportional -- so less power would be used with a higher resistance. Keep in mind that even without a resistor inline (i.e. high fan speed), the wire still acts as a resistor of sorts.

Hmm, that sounds acceptable... but I never did pay much attention in my electrical circuits class :). Somebody correct me if my thinking is off.
Click to expand...
Absolutely true accept one small thing, your formula is for a resistive load, not an inductive load.
 
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