Hi Guys! Heater blower question

[reporter000001]

Hello,

Long time viewer,first time poster.

Can anyone tell me if true, the cabin heater/ac blower gets 12v on high?? is this confirmed? does any one have the voltages for low med high/ "off" low med high?

I just need to know if high bypasses resistor.

THANKS!! much usefull info on here..

p.s. searched found my answer but just making sure, no way to test my haness is on my floor hahaha

[greg64]

Howdy. Yes, when on high, the resistor network is bypassed, so full alternator voltage goes to the fan blower motor. Medium and low cut in one of two resistors, which are cleverly mounted in the airstream to keep them from overheating.

[reporter000001]

perfect!!!! thank you very much!!!

[wilkin250r]

I've never actually tested the voltage, but you might be surprised how little the voltage actually drops across those resistors. Based on my experience with electronics in general, I'd guess that "medium" is only about a 1V drop, and "Low" is probably only a 2V drop. Maybe even less.

[greg64]

Ahhhh, but fan output goes down with the square of the voltage. Say the fan motor is 1 ohm resistance. So with 12v to the fan, the current will be 12 amps (12/1). The fan power is 144 Watts (12x12). Now if we add a 1 ohm resistor in series with our motor, then the current will now be 6 amps (12/2). The motor has 6V and the resistor has 6V. So now the motor power is 6x6 or 36 Watts, a heck of a lot less. Actually 1/4.

The point being that the voltage drop doesn't have to be much to cause a large drop in motor output.

[reporter000001]

Originally Posted by greg64

Ahhhh, but fan output goes down with the square of the voltage. Say the fan motor is 1 ohm resistance. So with 12v to the fan, the current will be 12 amps (12/1). The fan power is 144 Watts (12x12). Now if we add a 1 ohm resistor in series with our motor, then the current will now be 6 amps (12/2). The motor has 6V and the resistor has 6V. So now the motor power is 6x6 or 36 Watts, a heck of a lot less. Actually 1/4.

The point being that the voltage drop doesn't have to be much to cause a large drop in motor output.