Hi all with a 6V '48 converted to 12V, the heater motor is 6V. If I put in a voltage reducer inline, how muck amp draw will there be from the 2 speeds of the heater motor?
Reamer

 

I never took mine out to check actual amp draw - I just used one that handles up to 20 amps. 4th winter its doing just fine.

 

Hi mtflat
Where did you get the reducer that handles 20A? thats a 120W reducer?
Thanks
Reamer

 

Back when I converted my system to 12V some of the posts I read said to install a reostat to power the heater motor. The said as long as you don't run it on high the 6V motor will last a long time. I then took my heater apart and found the motor in my Magic Air had already been switched to 12V.

 

I found this one at Hot Rod & Custom Supply, Cape Coral, FL. Don't know if they're still around or not. Their pn was 9122 and the price 5 years ago was $30

Used to be online at www.rodncustom.com

edit: here's the link to the item in question

http://www.parts123.com/parts123/yb....rta~partsort~2

 

Howdy,


I just replaced the motor in my heater with a Siemans/VDO motor from National Autoparts for about $45. It's a permanent magnet motor so it will go CW or CCW just by reversing the wires. (the older motors are brush type motors and they only turn 1 direction...my old motor WAS a 12v motor....it just turned the WRONG way!)



According to my 54-55 manual the current requirement for the 6v motor is about 10 amps at 6.8v

To calculate the resistor needed to make it work with 13.8v you simply use OHMS law. V=IR V=voltage, I=Amps, R=resistance in ohms

The voltage across the dropping resistor will be the voltage required by the motor subtracted FROM the supply voltage(your battery).

13.8v-6.8v= 7v

Since the motor will draw 10A at 6.8v, the dropping resistor will also have 10a flowing thru it (it's a simple series circuit)

Using V=IR and R=V/I (V divided by I) , you get 7v/10A = 0.7 OHMS


Now we need to know the power dissapation required of the resistor. It will be all HEAT in WATTS.

Power in any device is calculated as Current squared times R ( IxIxR) In this case it's 10x10x0.7 = 70W

You can also calculate it as Voltage(across the resistor...7v)squared divided by the resistance in OHMS--->VxV/R = 7x7/0.7=70w


That's a pretty good size resistor. You don't want to use a 70w resistor though, you want to size it larger so it doesn't get REALLY hot! I'd use at least a 100w resistor.

Now where do you get a 0.7 OHM 100w resistor? It's not that easy. But you can get several smaller wattage resistors that are say 5 ohms or so... and wire them in parallel to get the required resistance( by the way 1 ohm would probably work just fine). I don't think you can buy them at Radio Shack though.

Here's a simple parallel resistance calculator http://www.1728.com/resistrs.htm if you want to try to buy resistors at radio shack. For example you could buy (10) 10 ohm, 10w resistors and connect them in parallel and you'd have a 1 ohm 100w combination.

Someone somewhere said that you can buy a "voltage reducer" from your local autoparts store. That might be the best bet. What ever it is it will be generating at least 70w of heat.

It's not hard to make it work. I just thought it easier to get a 12v motor.



Whew.........don't ask me what time it is.........electrical engrs always try to tell you how to build the clock!


Cheers,


Rick

 

so 2-50W resistors in parallel = 100W?
How do ya hook'em up parallel?
Reamer

 

Yup, two 50W resistors in parallel will dissipate 100W - IF they are in open air (not bundled up) and IF they are not in a confined space. I'm really leary of putting dropping resistors in high power circuits as the resistors will usually get pretty hot in service.

I'd go with a voltage regulator designed for your application - although it is more expensive. Mid-Fifty has a 15 Amp unit for $60 - part number 3223-4. That should handle your motor easily.

 

Sounds like the $45 12v motor might be a better choice and you can probably find the motor locally.

I have never liked using dropping resistors in high current applications either.

One way to ensure that you have plenty of airflow for dropping resistors if you must use them is to put them inside the heater duct itself. This way you ensure that when the motor is drawing current it also supplies the airflow to cool the resistors. You also get a little heat out of them too! Energy recovery!


The store bought voltage regulator will also dissapate heat. there's no way around it unless you use some pretty exotic electronics. Most DC voltage dropping regulators use simple "pass" transistors that get hot too. It will probably have a heat sink with fins and it will provide a constant voltage.... But at a cost of more than a new motor it's just not worth it.


If you want one that will run your heater motor, wiper motor, and 6v radio you're probably talking much more money since that would require a unit that would provide 25 or more amps running all that stuff at the same time.

This is why I'll use the 12v heater motor and I'll get a 12v wiper motor too.


Regards,

Rick

 

Here's a little blurb on how a parallel circuit goes together. If you click around in there you can see other circuits and you can drag and drop the actual components into the circuit to see what it all looks like!

http://www.ndt-ed.org/EducationResources/HighSchool/Electricity/parallelcircuit.htm

The circuit with your motor would be actually a series-parallel circuit as shown in the following link. http://www.ndt-ed.org/EducationResources/HighSchool/Electricity/seriesparallel.htm

R1 would be your motor and R2 & R3 would be the parallel dropping resistors.

 

JC Whitney carries this 12V to 6V windshield wiper control.

http://www.jcwhitney.com/autoparts/Product/showCustom-0/Pr-p_Product.CATENTRY_ID:2005027/c-10101/Nty-1/p-2005027/Ntx-mode+matchallpartial/N-10101/tf-Browse/s-10101/Ntk-AllTextSearchGroup?Ntt=wiper

 

Hmmm, If I do the voltage drop, Since It is now neg. ground, not pos.ground the fan will rotate reverse?

 

It shouldn't make a difference... That motor is probably a "brush" type motor. so it won't matter what the polarity is. Try it.

Mine was a 2 speed 12v motor and it turned the same direction regardless of how I hooked it up! IT STILL TURNED THE WRONG WAY! That's why I bought a new motor. It was pretty worn out anyway so the new motor was probably a good idea anyway. For $45 I didn't think I could go wrong.

By the way the new motor only draws about 3-4 amps at 12v.

I had the autoparts person look in their Siemans/VDO catalog for heater motors etc..... It had pictures, dimensions, and shaft length/size so it was easy to match up a motor with what I needed.