Using resistors vs battery run time

[gregh]

Over in the topic http://www.gscalecentral.net/tm?high=&m=94204&mpage=1#95588

Bigjack was asking about potentiometers, resistors and diodes for reducing train speed. Del Tapparo made a comment about using these types of items to reduce speed, ?uses more power and reduces run times.? I presume he means compared to pulse width modulated type speed controllers.

I don?t means to criticize Del, but we have to be careful what we are comparing here. In actual fact, reducing the voltage using diodes or resistors, will NOT reduce your running time from a single charge very much at all. (That?s if we?re talking about reductions in voltage of only say 25% )

Using a PWM system will INCREASE your run time if you run at a lower voltage, compared to what you get at full volts.

It?s not that you LOSE runtime by using a resistor/diodes but you GAIN runtime by using a PWM controller and running slower.


WHY IS THIS SO???

Well a train motor is pretty much a ?constant current? type of load. If you take a given train on a given bit of track, it will take ABOUT the same current irrespective of voltage. Motors don?t act like resistors where doubling the voltage will double the current. The current drawn is a function of the load ? train drag, gradient and loco efficiency. The voltage just sets the speed.
That doesn?t mean the current never varies ? it will if you make your train heavier or send it up a steep grade.
If you don?t believe me do your own tests. The current may increase a little with voltage, but not much!

So lets imagine a given train taking 1A on a given piece of track powered by a 2Ah, 10V battery. It will run for 2 hours (2Ah/1A) and the battery will have supplied 20 watt-hours of energy to the train.

Now lets put a 2 ohm resistor or 3 diodes in series with the motor. It will still take 1A (maybe a tad less). So we?ll have 2V across the resistor and 8V across the motor. It will go slower (around 80% of original speed), BUT the battery is still supplying 1A and the train will still run for 2 hours. Sure energy is wasted in the resistor ? 4 watt-hours and the motor only gets 16 watt-hours. The energy used by the train is less because it doesn?t travel as far, but it still runs for 2 hours!

Now lets use a PWM controller to reduce the battery voltage to 8V. PWM (pulse width modulation - means that the supply to the motor is rapidly turned on and off, maybe 1000 times a second or more. The ratio of on to off time determines the average voltage that the motor receives.)

In this case the motor is turned on for 80% of the time and off for 20%. During the off time the momentum of the train keeps it moving and we don?t notice it?s going on-off-on-off??

So while the motor is on it takes 1A again, but the average current taken from the battery is only 0.8A, and the average voltage on the motor is 8V. It goes slower. BUT now we can run our train for 2.5 hours on a battery charge! 2Ah/0.8A=2.5 hours. It will cover the same distance as a full voltage train.

So a couple of diodes and a switch to short them out, costing a few dollars or pounds, can give you a speed reduction without reducing run times.

 

[stockers]

Thanks for that. A nicely explained situation - even I understood it.

 

[robsmorgan]

Agree with Alan! Thanks Greg,

Very useful information - as you know I am using Del's Railboss control very happily.... I know I don't run my setup very often, but easily get 2 hours out of 10 x 2800mAh Ni-Mh AA rechargeables.... Now I begin to understand why!

Keep the info coming

Regards
Rob

 

[Gizzy]

An excellent explaination Greg....

 

[whatlep]

Thanks Greg for the clear explanation on power vs. run time.

A question which may be daft, but I don't have enough know-how to judge... :thinking:
If using a PWM solution, does the motor or any of the on-board electronics have a tendency to heat up or be stressed more than with a continuous current feed?

 

[spike]

WHY IS THIS SO???

Well a train motor is pretty much a ?constant current? type of load. If you take a given train on a given bit of track, it will take ABOUT the same current irrespective of voltage. Motors don?t act like resistors where doubling the voltage will double the current. The current drawn is a function of the load ? train drag, gradient and loco efficiency. The voltage just sets the speed.
That doesn?t mean the current never varies ? it will if you make your train heavier or send it up a steep grade.
If you don?t believe me do your own tests. The current may increase a little with voltage, but not much!

So lets imagine a given train taking 1A on a given piece of track powered by a 2Ah, 10V battery. It will run for 2 hours (2Ah/1A) and the battery will have supplied 20 watt-hours of energy to the train.

Now lets put a 2 ohm resistor or 3 diodes in series with the motor. It will still take 1A (maybe a tad less). So we?ll have 2V across the resistor and 8V across the motor. It will go slower (around 80% of original speed), BUT the battery is still supplying 1A and the train will still run for 2 hours. Sure energy is wasted in the resistor ? 4 watt-hours and the motor only gets 16 watt-hours. The energy used by the train is less because it doesn?t travel as far, but it still runs for 2 hours!

The current is mainly the product of the back EMF.
The back EMF is opposite in polarity to the applied voltage and decreases the effective value of that voltage.

An increase in voltage will increase the armature current, increasing the torque and acceleration.
As speed increases the back EMF will increase causing current and torque to decrease until torque again equals the load or the back EMF equals the applied voltage.

A decrease in voltage will decrease the armature current and torque making the motor to slow down.
The back EMF can be higher than the applied voltage, causing the motor to act as a generator.

 

[bigjack]

I'm a mechanical (now CCTV Installer) engineer by trade, and have always found the technical aspects of electrtics a bit of a mystery But since getting into garden railways a lot has been explained and things are now starting to make sence. It's great when someone takes the time to explain things The post at the top of the page needs marking as helpful, if only I knew how to do it.

 

[bigjack]

Just to let youiall know with sound and lights on the loco ran for 3 1/2hrs, with lights and sound off, it ran for about 10 mins longer

 

[gregh]

Thanks Greg for the clear explanation on power vs. run time.

A question which may be daft, but I don't have enough know-how to judge... :thinking:
If using a PWM solution, does the motor or any of the on-board electronics have a tendency to heat up or be stressed more than with a continuous current feed?

Can I take that question on notice? I'll be at the beach for a week. It's one of my pet topics, so I'll get back to you.

Thanks everyone for your comments. I expected everyone's eyes to glaze over and no replies at all.

 

[whatlep]

If using a PWM solution, does the motor or any of the on-board electronics have a tendency to heat up or be stressed more than with a continuous current feed?

Can I take that question on notice? I'll be at the beach for a week. It's one of my pet topics, so I'll get back to you.

Now that's really a bit cruel, bearing in mind the overnight temperature last night here was 2 degrees Centigrade and you haven't asked me to join you at the beach! :impatient:

Oh, go on then, I'll forgive you and try doing vigorous warm-up exercises while you sun yourself...

 

[Del Tapparo]

Thanks Greg for the clear explanation on power vs. run time.

A question which may be daft, but I don't have enough know-how to judge... :thinking:
If using a PWM solution, does the motor or any of the on-board electronics have a tendency to heat up or be stressed more than with a continuous current feed?

Older PWM controllers operating at low frequencies (100-1000 HZ) had that effect on some motors. These low frequencies also caused an audible hum in the motor, especially at low speeds. Most newer systems (including all of my controls) run around 20,000 HZ. No hum, and no ill effects. At least I've never heard of anybody losing a motor due to PWM.

 

[gregh]

A question which may be daft, but I don't have enough know-how to judge... :thinking:
If using a PWM solution, does the motor or any of the on-board electronics have a tendency to heat up or be stressed more than with a continuous current feed?

First off I have to just avoid the question of electronics & PWM. Some sound systems, DCC etc just aren?t designed to work with PWM. You have to find out from the manufacturer.
Maybe some other contributors can post a listing of what works and what doesn?t.

Confining the discussion to the motors only, it?s true that PWM will always cause more heating in a motor than pure DC. The parameters that determine how much more heating are:
the frequency of the PWM, the input voltage to the PWM controller, and the duty cycle. ie the motor volts you are running at divided by the supply volts. the inductance of the motor. (Inductance is just a sort of measure of how strong the motor?s magnetic field is.) [/UL]
When running on DC the motor heating is the same for low or high speeds (volts) (remember the current?s constant irrespective of speed). But with PWM, the heating gets worse for lower speeds.

So generally speaking, if you have a low frequency PWM, high input voltage, low inductance and are running at low speeds, you can expect problems.

But how relevant is this to G scale?
Well I?ve been building PWM controllers for 40 years (N, HO and G) and haven?t had any failures yet. Our G size motors have enough inductance coupled with the modern higher frequency PWM controllers to reduce the extra heating to manageable levels.

Track power PWM will be worse than battery locos using PWM because with track power the input voltage will be higher and you will spend more time running at higher duty cycle.
So if you?ve got a high voltage track power (say over 25V) and you run at slow speeds (say 6V on the motor) for long periods, you could have problems. But battery power PWM will never be a problem.

Del Taparo summed up some of my feelings?

Older PWM controllers operating at low frequencies (100-1000 HZ) had that effect on some motors. These low frequencies also caused an audible hum in the motor, especially at low speeds. Most newer systems (including all of my controls) run around 20,000 HZ. No hum, and no ill effects. At least I've never heard of anybody losing a motor due to PWM.

(I have to add that I have built and am using PWM controllers operating as low as 25 Hz with no problems in G scale. But I don?t do continuous running for hours at a time.) ( I?m not sure if Del really means 24000 Hz or 2400 Hz. The Hobbyking RC motor controls I use operate at around 2400Hz and you can hear them ?whistle?.)

PWM probably got its bad reputation in the old days with HO and N scales and low frequency.

Some day I will write a more detailed (ie complicated) explanation in a separate topic, with some test results.

 

[mike]

worrieingly.. this is starting to make sensce

 

[whatlep]

First off I have to just avoid the question of electronics & PWM. Some sound systems, DCC etc just aren?t designed to work with PWM. You have to find out from the manufacturer.
(huge snip)
Track power PWM will be worse than battery locos using PWM because with track power the input voltage will be higher and you will spend more time running at higher duty cycle.
So if you?ve got a high voltage track power (say over 25V) and you run at slow speeds (say 6V on the motor) for long periods, you could have problems. But battery power PWM will never be a problem.
(huge snip)
PWM probably got its bad reputation in the old days with HO and N scales and low frequency.

Greg

Thanks for such a comprehensive (and comprehensible) answer! You've confirmed several things which I had at the back of my mind, quite a few from many years ago, as you suggest in the last sentence. About 1984 I managed to utterly destroy an N gauge loco by leaving it running slowly for about 30 minutes on a PWM controller: the bodywork melted as the motor overheated.... :crying:

 

[gregh]

Going back to the resistor/battery power topic, I thought you might be interested in some test results.
I tethered an Aristo motor block on my test bench so its wheels spun on the flanges. This provided some ?drag? to simulate running on track with a light train.

Then I applied different DC voltages and measured the current. And repeated using a Hobbyking 2.4kHz PWM controller with a 15V input.

Here?s a graph of the results.

The black line is on pure DC. At 3V (motor just turning) it took 0.35A. By 12V this had increased to 0.48A and it never increased beyond that. Even at the lower speeds a 400% increase in volts caused only a 160% in current.

The pink line is with the PWM showing the MOTOR current. The current is constant above 9V.

The blue line shows the current coming from the SUPPLY (or battery). Note that it is almost linear - at 7V the current is 0.25A and at 14V the current is doubled. This is exactly as I stated before ? if you run at a lower voltage with PWM, the current from the battery will reduce and you will run for longer.

You might wonder why are the 2 currents different? Well the PWM controller acts like a transformer with quite low losses (mine used 60mA at all voltages).
The power from the battery is battery volts x battery amps.
The power in the motor is motor Volts x motor current.
And these are equal, so
Battery amps = motor V x motor amps / battery volts.
If motor amps are constant as explained above, the battery amps are proportional to motor V / battery Volts. Lower the motor Volts and the battery amps must be lower.

 

[gregh]

Just as a follow uo to this 2 month old topic, I've added a lot of this discussion to my webpages, along with some more info on batteries and charging.
Have a look at:
http://www.members.optusnet.com.au/satr/battery.htm

 

[spike]

The power from the battery is battery volts x battery amps.
The power in the motor is motor Volts x motor current.
And these are equal, so
Battery amps = motor V x motor amps / battery volts.
If motor amps are constant as explained above, the battery amps are proportional to motor V / battery Volts. Lower the motor Volts and the battery amps must be lower.

Hi Greg.
Not wishing to be pedantic but the bit in red..
I think it better if this read power supplied to motor.
If you meant power developed by the motor then this is not correct.

If
V = applied voltage
Eb = back e.m.f.
Ra = armature resistance
Ia= armature current

then the power equation for the motor is
VIa = EbIa +I2aRa

Power supplied to motor = VIa
Power developed by motor = EbIa
Copper loss = I2aRa

the I2aRa losses could amount to approx 5%

So to get the motor power you need to know the armature resistance and backemf.

Cheers Mike

 

[gregh]

Hi Mike,
It's nice to know someone reads my stuff, especially someone with some electrical knowledge.

While your equations are all correct for the power into a motor and out of the shaft, that's not what I was meaning. I was comparing the current out of the battery and current INTO the motor. Or another way of saying this could be " the current into and out of the PWM SPEED CONTROLLER". I wasn't getting into the internals of the motor itself.

Concerning the motor power and armature resistance, I try to measure resistance by measuring the stalled rotor current with a 12V battery supply. I'm not sure it's an accurate way, but it's all I can think of. Using an ohmmeter never gives the same results twice. Maybe you have another method? I've only got Bachmann and Aristo motors to measure.

The results are that my G scale motors armature resistances are in the 3.5 - 4.5 ohm range. (HO and N are more like 10-20 ohms) Now that means that the losses in our motors could be up to 50%, not the 5% you expect in big motors.
I get that 50% figure by assuming a well loaded motor taking 1A at around 12V in. That means the armature volts are only 8V, shaft power 8W and the armature losses are 4W = 33% losses.

(I tried to add an Excel table of my measured loco data here but it didn't work)

 

[spike]

Hi Greg.

Your doing a good job of explaining things.

Re armature resistance.
I would say you got the method ok, thats a good way to measure it.
Using the ohmeter is a bit dodgy as you will rely on good brush contact and
this can vary quite a bit.

I'm surprised at the losses......I will look into this.
I work in electric motor repairs but most dc motors we get are quite large and
we would expect the losses somewhere in the region I quoted.

Now small model motors may be a different kettle of fish.
As you know the construction is normally 3 or five pole, assuming iron cored so
the losses may be quite large and normal.

HNY spike