Using 6v motor on 12 v

[engerth399]

I have a kit which uses a 6v motor and resistor to enable it to run on 12v.  My question is: Should I wire the resistor in parallel across the motor or in series with it?  It's not made clear in the instructions.

Your advice or experience would be welcome.

Thanks

Peter

[B.Meldon]

It should be in series.
Sadly no situation is ideal!
You will find that adding the resistor will make the motor far less powerfull.
Also if this is in the Koppel A Tsugawa chassis you will find the gearing is an issue that will prevent the loco starting at a slow speed.
It will run slower once it gets going, but to get it to move off the controller will need to be turned up higher the loco will then start at a higher than ideal speed.
The chassis was designed to go round and round at a non relistic higher speed, it is not so good if you want to run it slow, for stop start or shuntting.
There is a replacement motor that can be used and it improves the running, but it does not solve the high gearing starting problem and it is a taller motor so it will not fit in some kits, the KS Wren kit for example.

Brian Meldon

[JulienW]

Putting the resistor in series with the motor will have the desired effect of protecting the motor from the additional voltage & current, at least it does with the half-dozen loco kits (including three of the SRB Pecketts),  Tsugawa chassis and Kato Portram power bogies that I've done over the years.

I'm not sure what would happen if you wired the resistor in parallel- I suspect that the current would see that as a lower resistance path compared to the motor and then the motor will not run (My A level in physics is about 30 years old and it's information I rarely so use my brain has forgotten where it's stored!).

Out of interest which kit is it?

[engerth399]

Many thanks to both for your help.  My A level in physics is far older than Julien's but I remember 1/R + 1/R so it seems it would be a lower effective resistance - probably not good.  (Actually that level of physics was O level)!!
The kit is a Marcel Jolly kit for a Verney railcar.

Thanks again

Peter

[B.Meldon]

The kit is a Marcel Jolly kit for a Verney railcar.

Thanks again

Peter

As far as I know that kit uses the Tenshodo motor bogie gears, so ignore the info about the Koppel A Tsugawa chassis.

I recomend you modify the kit to use a slightly larger 12v motor.

Brian Meldon

[engerth399]

Again thanks for the info.  I have a tramfabriek motor as an alternative.

Peter

[JulienW]

I wonder if the original motor is one of those that Nigel Lawton?

Anyhow my feeling is that if you have one of Sven's /Tramfabriek's 12 volt motors that will fit then it will probably run better on 12 volts than a 6 volt motor combined with a resistor (and if you can fit in one of his flywheels on a twin shaft motor then that will not do any harm either).

[B.Meldon]

I agree with Julien. The 12 volt Tramfabriek motor will be far better than the 6 volt motor with a resistor.

Brian Meldon

[Stephen Rainsbury]

Many thanks to both for your help.  My A level in physics is far older than Julien's but I remember 1/R + 1/R so it seems it would be a lower effective resistance - probably not good.  (Actually that level of physics was O level)!!

Luckily my A-Level Physics is more recent.I was teaching it at 11:15 this morning :-)
The resistor in parallel is not good, as current will flow through it, doing nothing apart from getting a bit warm, but your poor motor will get the full 12V, potentially killing it.  It will go very fast, but then probably burn out, a happy life, but a short one.
I can send you a circuit that will limit the voltage to 6V, without restricting performance too much. Its a bit bulky and all the action will happen over the first half of your controller knob, it is stops your motor from getting damaged.
.

[engerth399]

Thanks for this but the model was finished some time ago and appeared in September Continental Modeller.  As with most of my models it has been test run a couple of times and I've moved on to another project.  I suppose A level physics has changed since my day.

Peter

[Retiarius]

Resisters in Series

R = R1 + R2  etc.

Resisters in Parallel

1/R = 1/R1 + 1/R2  etc.

So for 2 resisters in parallel you can use

R  = (R1 x R2) / (R1 + R2)

Resistors in series are equivalent to one resistor whose resistance is the sum of each individual resistor. Resistors in parallel, on the other hand, result in an equivalent resistance that is always lower than every individual resistor.

[Stefan Krebs]

 :D :D :D
All correct if you run your engine powered by batteries, but not if you use most of the mains-power controllers, even more so with pulse width modulation (PWM) or phase chopper controllers (PCC).
The later do not supply DC in the strictest sense, but have an AC element with a frequency.
The motor resistance, which is a coil (solenoid), is frequency dependant. And that statement applies to the frequency change caused by change of rpm too.  ;D
The static motor has very little resistance, but as the speed increases the resistance increases too! That's why a motor that is stuck burns out very fast; - the power supply drives a very high current which overheats the motor sharpish.

The motor's torque is proportional to the voltage, so yes, at 12V you have roughly twice a much torque; which is what gets you going.
The motor's power is proportionate to the current, so yes, 12V drive double the current (U=RxI) driven by 6V, which provides the motor with twice the power.
Both effects make the same motor at 12V much more powerful than at 6V.
However, no matter how small the Ohm-resistance of the wire in the motor, it will turn current into heat, and if that is not dissipated, the motor will overheat. The smaller the motor the harder to get rid of the heat. It's that heat dissipation (a design feature) that defines what voltage the motor can cope with. Hence operating a 6V motor at 12V is not recommended as the motor can't dissipate the heat and will die of heat stroke.
The cheap fix is the resistor, a PWM could maintain the 12V (and torque) at 6V eff, i.e. solve the starting issue.

Hope this helps
Waldschrat

[Graeme Dunbar]

There is a bit of mis-typing or something similar here unfortunately.

Torque is proportional to current. Speed is proportional to voltage. The motor resistance does not change, but motors are also generators and generate a voltage (Back Electromotive Force - BEMF or Counter Electromotive Force - CEMF) when the shaft spins. The BEMF opposes the supply voltage which in turn reduces the current through the motor.

Mechanical power is torque multiplied by rotational shaft speed which is the result of, or is transduced from electrical power which is current multiplied by voltage. A motor develops most torque when it is starting (or otherwise stalled) and zero torque when there is no load and it is running at its maximum speed (for the given supply voltage). Power is the rate of doing work so is at a maximum between these two extremes (a lot of force, but going nowhere or running fast but unable to carry anything), at half the no load speed and half the stall torque.

When a train is going from level track to travelling uphill at constant supply voltage the train slows down as speed is traded for torque and vice versa going downhill. There will be a braking effect as gravity tries to turn the motor into a generator, pumping power back into the supply (regenerative braking) but in a model it is hardly noticeable.

[Stefan Krebs]

Dear Greame,

No mistyping, I merely tried to keep things simple,  and not a contradiction either.

If you go back to U=R x I (Ohm's law; in units: V = Ohm x A), and you consider what you wrote, you realise that the inductive component of the motor's resistance must be near zero with a motor not rotating, but higher when running (hold the motor still long enough and it will burn out at a voltage it runs all day long if allowed to spin, i.e. at a constant voltage the current is highest at standstill, and drops as speed increases).

The unit for force is N = kg m/ s E2; and Volt expressed in SI-units is: V = N m/ A s; or transposed: N = V x A s/ m. As the torque of the motor is the force on the rotor times radius, i.e. N x m, it is obviously proportionate to both voltage and current.
However, all other things being equal you don't get current without voltage, but you can get voltage without current, i.e. the physical root cause for torque is the voltage. And if the voltage increases it drives a higher current, i.e. the impact of increased voltage on the force is exponential and increases torque and power, while a change of current has only a direct proportionate effect.
More importantly, the link between voltage and speed is not a direct one, it is the power that is proportionate to voltage. The unit for power is Watt: W = V A = N m/s, i.e. torque and power are NOT the same. As discussed, an increase in voltage drives an increase of current, i.e. exponential.

So far, so easy, it all applies to both, AC and DC.

What we need to consider now, is that the commutator of a DC motor is a mechanical chopper, that switches current on and off to the various coils in the rotor of DC motor; it creates AC as otherwise the motor won't work. Obviously, the frequency is dictated by the rpm. As discussed above, the internal resistance is frequency-dependant, i.e. if the motor is under load, the rpm drop and with it the resistance.
Going along with your assumption for a moment, we use a very powerful power supply that doesn't cave in under load (not true for most model railway transformers!), i.e. the voltage remains constant, the reduction in speed on a ramp will increase current. As formula above shows, power will go up, but not enough to compensate fully for the increased power demand. Thus, most of us will notice what you describe, i.e. the engine slows down on a ramp, which is also caused by the power supply caving in, i.e. voltage dropping due to reduced resistance.

The BEFM you mention is not changing anything on above, it is the inductive resistance in the equation. It is an 'energy-loss' that applies in all situations, and is proportionate to the power-consumption, i.e. higher under load. That's why it may be used to achieve near constant speed (actually power) as it is a tell-tale as to how hard the motor is working. And any BEFM-controller will compensate by increasing voltage, not current (that's just the result), to supply more power to meet the increased demand.

If this simplified, i.e. not 100% accurate explanation isn't good enough I suggest you get yourself a good physics book and read it up. I'm sure, we've lost most readers by now.


Cheetrs
Waldschrat

[Graeme Dunbar]

Yes, I agree. Not the place to discuss the finer points of electromagnetism.