Register now to get rid of these ads!

Technical Resistor switch getting hot, wiring help needed

Discussion in 'The Hokey Ass Message Board' started by jt7282, Mar 22, 2014.

  1. jt7282
    Joined: Sep 29, 2009
    Posts: 149

    jt7282
    Member
    from Midland Tx

    I have a 1951 ford f1 and I have the original 6volt heater. The truck has been converted to 12v. I have a universal heater switch with a built in resistor. The coil is getting glowing hot. What is the fix for this? Any help would be greatly appreciated. Thank you
     

    Attached Files:

  2. jt7282
    Joined: Sep 29, 2009
    Posts: 149

    jt7282
    Member
    from Midland Tx

    Having trouble uploading pics
     

    Attached Files:

  3. jt7282
    Joined: Sep 29, 2009
    Posts: 149

    jt7282
    Member
    from Midland Tx

    Here's the switch
     

    Attached Files:

  4. George/Maine
    Joined: Jan 6, 2011
    Posts: 949

    George/Maine
    Member

    If you get a ignition resister 1.5 ohms they are white ceramic used to reduce current.
    If you were to drop 3 volts that would cool it down. 2amps x 1.5 ohms = 3 volts
    That would be placed inline with switch.
     
    Register now to get rid of these ads!

  5. 33WIRE
    Joined: Aug 8, 2009
    Posts: 83

    33WIRE
    Member

    Could you swap out the blower motor with a 12v one?
     
  6. jt7282
    Joined: Sep 29, 2009
    Posts: 149

    jt7282
    Member
    from Midland Tx

    Thank you George, I would love to 33wire. But I'd have to find one that'd fit.
     
  7. Install this in series in the power line to the switch.... this will be the cheapest/easiest solution short of a new motor. The resistor will get very hot so mount it where it won't cause problems; inside the heater in the fan airflow would be an excellent place.
     
  8. reverb2000
    Joined: Apr 17, 2005
    Posts: 440

    reverb2000
    Member
    from Houston TX

    A resistor converts voltage to heat. Resistor size is directionally proportional to heat it gives off. I know that doesn't answer your exact question but that's why it's hot.


    Posted using the Full Custom H.A.M.B. App!
     
  9. OldFord39
    Joined: Aug 23, 2011
    Posts: 64

    OldFord39
    Member
    from Monroe, Wi

    Is there a tag on your 6 volt motor giving the amperage draw ? If your 6 volt motor draws 5 amps at 6 volts it would draw about 2.5 Amps at 12 volts so now we want to use Ohm's Law to reduce your motor back to 6 volts

    So if we take 6 Volt / 5 amps we get 1.2 Ohms and 6 volt * 5 amps we get 30 Watts

    So if we take 12 volts / 2.5 amps we get 4.8 Ohms and 12 * 2.5 amps we get 30 Watts

    So if my guess of your motor being rated at 5 amps is correct your field winding resistance whould be 1.2 ohms.

    So if we subtract the 1.2 ohms from the 4.8 ohms we get 3.6 ohms; we want a
    resistor that would be around 3.6 ohms and we want a resistor rated at least 30 watts.
    If it were me I would get a resistor rated at 100 watts and mount it to a piece of aluminum to act as a heat sink (because it will get hot). Here is an e-bay link for a 10 ohm resistor but if you let them know when and if you order it you can specify a 4 ohm resistor. http://www.ebay.com/itm/1PC-100W-Wa...393?pt=LH_DefaultDomain_2&hash=item2ecab26c59

    Regards,

    Brian
     
  10. 42merc
    Joined: Dec 19, 2010
    Posts: 607

    42merc
    Member

    It gets hot because it is a resistor.
     
  11. Ebbsspeed
    Joined: Nov 11, 2005
    Posts: 4,791

    Ebbsspeed
    ALLIANCE MEMBER

    Might want to re-think the math on this one. Putting 12 volts across a 6-volt motor will result in a higher current, not lower. You've increased the voltage, but not the resistance. In order to make the motor operate like it did on a 6-volt system, you need to have the same voltage and current going through the motor windings, which in a 12-volt system means you need to drop 6-volts across a resistor, and that resistor needs to have a wattage rating to deal with the resulting heat. So going with the original assumptions of the motor requiring 5 amps of current and it being a resistive load of 1.2 ohms, you need to have a dropping resistor of 1.2 ohms. The 30 watt calculation is correct, because you've got 5 amps flowing through a resistor that has a 6-volt drop across it.

    And as Brian pointed out in his post, the resistor needs to be a higher wattage rating so it can dissipate that heat. The one the fellow has now is a hell of an ignition source.
     
  12. nukeman
    Joined: Mar 17, 2007
    Posts: 131

    nukeman
    Member
    from Michigan

    Actually, with motor loads, there is no linear formula that works because the resistance is controlled by the counter-EMF of the motor and that changes with speed. Ohms law can only be applied if you know the resistance at running speed and that does not change in proportion to voltage.

    Fortunately, in this case it's not critical and experimentation is the easiest way to determine the right resistor.
    You want to use a resistor with enough ohms, in series with the motor, that allows the motor to run at about the same speed it did on 6v. It also needs a high enough rating in watts so it doesn't burn itself out.
    Putting the switch and resistor in the airflow of the fan would be ideal.
     
  13. George/Maine
    Joined: Jan 6, 2011
    Posts: 949

    George/Maine
    Member

    I think those ign resisters will take 3 amps, but if you are still running to hot put 2 resisters in parallel and need more put one in series. That should bring it down to about 6 volts.
    Pick a speed you like and adjust.
     
  14. nukeman
    Joined: Mar 17, 2007
    Posts: 131

    nukeman
    Member
    from Michigan

    No.
    Putting two resistors in parallel will make his switch hotter not cooler (but not as hot as no resistor). However, if his resistors are getting too hot then he can try two resistors in parallel.

    http://www.calculator.net/resistor-calculator.html
     
  15. Again, any resistors installed should be put into the circuit ahead of the switch, not between the switch and the motor. After the switch will increase the load on the switch, making the problem worse. Before the switch will reduce the load on the switch.

    You'll need a wire-wound, ceramic-type resistor for a load this large and most ignition resistors should be good for 7-8 amps. If not, find out what the full-speed current draw at 6 volts is for the motor, then calculate the watts (volts times amps) and use a resistor with a watt rating at least as large, preferably a bit larger; a higher watt rating will translate into a cooler operating temp. To semi-accurately determine the ohms size of the resistor needed, use the formula R = VD/I where R is ohms, VD is the voltage reduction you want (i.e. from 13 to 7 is 6), and I is the motor amps.

    If you have difficulty finding a resistor with a high enough watt rating, you can parallel two but you'll need to increase the ohms on each. The formula for this is R1+ R2/ R1 x R2 = R total. Two the same size in parallel equals 1/2 their single rating.
     
  16. Ebbsspeed
    Joined: Nov 11, 2005
    Posts: 4,791

    Ebbsspeed
    ALLIANCE MEMBER

    Yeah, I know, just didn't want to get into the inductive reactance formulas here. There's already enough confusion on this board about things electrical:D

    Crazy Steve, the resistor is in series, so doesn't matter where it is.
     
  17. Actually, it does. Nukeman called me out on this in a PM too...:)

    Yes, it's a series circuit, but there is a logic to what I said.

    First, introducing an additional resistance into the circuit will increase the total load, and that load in amps will be the same everywhere (series circuit, right?). Because the motor load amps isn't linear by voltage but will go down with the voltage, the 'final' total would have to be measured. Each circuit resistance will have it's own voltage drop across it, correct? Add all the voltage drops together, that equals the circuit voltage. And each drop will consume X amount of watts, as watts is volts times amps. So if the dropping resistor is before the switch, the load the switch sees in watts will be reduced by however many watts the resistor uses, although current will be the same everywhere. Make sense now?
     
    Last edited: Mar 23, 2014
  18. Mr48chev
    Joined: Dec 28, 2007
    Posts: 25,898

    Mr48chev
    ALLIANCE MEMBER

    Yep, a resistor switch is basically a little heater just like the electric heater with the coil of wire element. If it doesn't get hot it isn't working.

    There is the off chance that you may have one with the wrong rating for what you are doing but every resistor like that I have ever used will brand you in short order if you touch it when it is on.
     
  19. One other thing needs looking at; the switch. Installing a dropping resistor to reduce voltage to the motor will introduce additional load into the circuit across the resistor. I can't say that reducing the motor voltage will reduce the motor amps enough to offset the load from the resistor, so the switch may be undersized. A higher-amp-rated switch may be needed once the motor speed is finalized.
     
  20. nukeman
    Joined: Mar 17, 2007
    Posts: 131

    nukeman
    Member
    from Michigan


    Sending you a PM so you can correct your mistake before it's pointed out in public is the opposite calling you out. But whatever.

    Adding resistance in a series circuit will always decrease total load because load is current times volts. It's Ohms law in its simplest form.

    Your logic is wrong from the first sentence on because it's the opposite of Ohms law.

    Think of an on/off switch as adding near infinite resistance when it's off. Will the load increase or decrease? Hint: adding resistance means less current, and off is about the least current possible, right? And load is current times volts, right? So more resistance also means less watts or load.

    Make sense now?
     
  21. George/Maine
    Joined: Jan 6, 2011
    Posts: 949

    George/Maine
    Member

    I think if you get a 1ohm resister with about 10 watt you will be ok.
    6 volts and guessing 6 amps, ohms law would be 1 ohm resister.
    now add 1 ohm to circuit making that 2 total.
    with 12 volts now with 2 ohms=6 volts.
    power would be about 6 watts put it up 10 on safe side.
    You maybe able to adjust with the switch or get one with a reastate.
    If you run on low be happy.
     
  22. If we were talking a simple resistance-only circuit, you'd be right. But we're not; this is a motor circuit. Motors (any motor) need a certain amount of power to operate any given load. That is measured in HP. HP can be converted into watts; 746 watts = 1 HP. So however much power is needed to turn that fan for a given RPM, the watts needed will be the same. Lower the voltage, the amps go up. Raise the amps, the volts needed go down. Now, it's not quite that simple because of the before-mentioned reactances involved so it won't be completely linear, but you still need the same power for a given load. This is why motors heat up when put under a heavy load; voltage remains the same, so the motor draws more current (= more heat) in order to perform the added work.

    So if this motor draws 10 amps at it's rated 6 volt input (or 60 watts), putting a resistor in front of it to reduce voltage to 6 volts will not reduce the motor current; it will still draw 10 amps, and you've added load because the resistor will consume power by 'converting' the extra 6 volts to heat. Will this represent an 'added' load compared to running the motor on 12 volts? Hard to say; too many unknown factors, but I suspect you'll either find little difference in the measured load either way or a small increase. This will not reduce current flow to the level that an equivalent 12V motor would have.
     
    Last edited: Mar 23, 2014
  23. nukeman
    Joined: Mar 17, 2007
    Posts: 131

    nukeman
    Member
    from Michigan

    No. The first problem is with your belief in DC motors making a constant power.

    Lets say we have a 6vdc 1/10th HP blower motor that turns at 800 rpm and draws 12.4 amps at full load. That's 74.4 watts.
    From experience we know it will turn faster if we double the voltage, not twice as fast but lets say 1200 rpm.
    Where do you suppose it gets the power to turn 50% faster with the same load?
    Wouldn't that imply 50% more power?
    How many amps do you think it draws now?
    You suggested half which would be 6.2.
    Maybe it's around 9.3 amps?
    How many watts? 111.6?

    Obviously my numbers are purely hypothetical but hopefully you can see that the power does not stay the same when you change voltage and the rest of your discussion was based on this misunderstanding.

    As I stated in my PM I have read your other thread and think you are doing a great job there.
     
  24. Ebbsspeed
    Joined: Nov 11, 2005
    Posts: 4,791

    Ebbsspeed
    ALLIANCE MEMBER

    While you (Crazy Steve) are correct in that the total power consumed (measured in watts) and heat generated will be more, because in addition to what the motor uses/generates, the resistor is basically turning those additional wasted "6 volts X whatever current the motor needs" into heat.

    BUT, your following quote is absolutely wrong about the placement of the resistor affecting the load, depending on where is is in the circuit. It IS a series circuit, and Ohms law cannot be broken.

    I swore to myself a long time ago that I wouldn't get involved in any of these "Ohms law" threads, because they normally spiral downward and turn into arguments. But I do try to stop bad information from being propagated to folks who think that automotive wiring is a black art.
     
    Last edited: Mar 23, 2014
  25. nukeman
    Joined: Mar 17, 2007
    Posts: 131

    nukeman
    Member
    from Michigan

    I agree with you 100% (again).

    Motor theory and power formulas and Ohms law trip up a lot of people when you try to put them all together in the same pile.
    I've been an industrial electrician for 30 years and many of my coworkers struggle with these same concepts.
    Throw in lagging power factor corrections and I get confused too.
     
  26. JEEZ, you guys just aren't getting it...

    Ok, Nukeman, you're up first.

    Where did I say that? What I did say is a motor will consume a specific amount of power for a given load. This is boilerplate electrical theory, beyond dispute. Will different design/size motors consume different amounts of power into the same load? Yes. Will changing the voltage change the power output? Yes. Will changing the voltage change the speed? Yes.

    But what I'm talking about here is a specific motor, the 6V motor the OP is trying to make work. This motor will draw a specific amount of current at it's rated voltage while turning the fan (the mechanical load) at a specific speed. Remove the fan (load), the amps will go down. Apply pressure to the shaft (adding load), the amps will go up. So right now the inputted voltage is too high, causing issues with the resistors in the switch, plus the fan is running much faster and drawing more current. So if you reduce the voltage, current and power will go down and fan speed will drop. But we haven't changed the mechanical load at all; once fan speed is back to where it was at rated voltage, you'll find that the motor current and voltage will be the same as it would be if the motor was connected to just a 6V supply. Unless you drop the fan speed to slower than 'normal' (under 6V), motor current will be the same. And you still have the 'additional' load of the resistor in the circuit. Don't believe me, ask someone familiar with motor circuits and connecting them; they'll tell you exactly the same thing.

    Ebbsspeed, you're next...

    But I'm not breaking Ohms law, I specifically stated that current will be equal everywhere in the circuit. But what I did say is that the load in watts presented to the switch would be lower, and that's absolutely true.

    Normally, this wouldn't matter. But it does in this case, and I'll explain why. Now, we do agree that each series resistance represents a voltage drop, and that watts are consumed at each drop, right? Voltage drops added together equals the circuit total voltage, adding the watts used at each drop equals the total watts in the circuit. Again, basic theory. Some of the watts are converted to mechanical energy in the motor, the rest are converted to heat.

    Let's say that the motor and the dropping resistor both present 30 watt loads for a total of 60 watts. With the switch after the resistor, you've already 'used' 30 watts, so the watt load on the switch will just be the motor. Why this matters is because the switch has resistors in it for speed control, and the difference in having 30 watts vs 60 watts going through those will have an impact on how hot they get, which was the original problem. Because you've already dropped voltage in the dropping resistor, the resistors in the switch will see a lower voltage. Volts times amps equals watts, fewer volts = less watts = less heat....

    Guys, I taught electrical theory for a number of years, this is basic stuff to me. While Ohms law will cover most of what you need for 'basics', it doesn't always cover or explain everything....
     
    Last edited: Mar 23, 2014
  27. nukeman
    Joined: Mar 17, 2007
    Posts: 131

    nukeman
    Member
    from Michigan

    Dude,

    I taught electrical theory for 30 years, not just a number of years.
    As for asking someone familiar with motor circuits and connecting them, again, that would be me.

    Your understanding of DC series circuits is wrong.
     
  28. Ebbsspeed
    Joined: Nov 11, 2005
    Posts: 4,791

    Ebbsspeed
    ALLIANCE MEMBER

    I'll continue to believe what I know is true. No use arguing this one. I'll hang a picture over my EE degree hanging here on the wall.

    I'm out.
     
  29. nukeman
    Joined: Mar 17, 2007
    Posts: 131

    nukeman
    Member
    from Michigan

    I'm out too.


    You can be proud of that degree around here, Ebb.
     
  30. V8 Bob
    Joined: Feb 6, 2007
    Posts: 2,518

    V8 Bob
    ALLIANCE MEMBER


    If you spent some time going through a NAPA/Balkamp listing, I'm sure you will find a 12 volt replacement motor, as I found one for a '41 Ford heater. You need to know the dimensions/mounting of your 6 V motor. :)
     

Share This Page

Register now to get rid of these ads!

Archive

Copyright © 1995-2020 The Jalopy Journal: Steal our stuff, we'll kick your teeth in. Terms of Service. Privacy Policy.

Atomic Industry
Forum software by XenForo™ ©2010-2014 XenForo Ltd.