Goodness, 'hoeji' what have you started ? (only joking )

I feel you are trying to make something out of nothing, Coulombs.....gee, I haven't heard that since trade school 1985 !

I think it is fair to say that I've spent too many messages on this topic already. Since I already went way overboard on this topic, why not do some more.

I was trying to decide whether I should run my cargo train with 9V or with PF (it is a 10 feet train, with 2 Maersk engines followed by 8 light-weight cars (some Maersk with 1 container, some MOC).

This train has two 9V motors, and when running at a constant speed, it consumes 0.42 amp (i.e. 0.21 amp per motor). This number is pretty much the same for any speed setting that I've tried. Regardless what speed it runs, as long as you keep the speed constant, then it uses 0.42 amp. This test was done with the 9V system.

I want to run it at speed setting #2, which is just above 4 volts. Now I wanted to know how long the batteries would last if I turned this into a PF train. So I computed it, in two ways:

Method 1 counts Amps: I'm using 0.42 amp, or 420 mA. If I use 1100 mAh batteries, then it is a simple computation: 1100 mAh divided by 420 mA, that's 2.62 hours. So that's how long the train should run before the batteries run out.

Method 2 counts energy: I'm using 0.42 amp at 4 volts, so that's 0.42 * 4 = 1.68 Watt. The battery pack contains 9V * 1100 mAh of energy, which is 9.9 Watt-hour, so 9.9/1.68 is 5.9 hours.

Here is the puzzle: I computed the same thing in two ways, and get a different number. If I use amps to compute the battery life, then I get 2.62 hours. But if I use Watts, and the energy content of the batteries, then I get 5.9 hours of computed running time (assuming 100% efficiency for the PF system).

I know that at least one of these computations must be wrong. But it's not so easy to understand why this is so.

The one thing I know for sure is that every second, you have to get 0.42 Coulomb through the two motors. If you do not, the train won't have enough torque to overcome friction. This matches not only observations, but also makes perfect sense if you view the motor through the laws of motion and electromagnetism.

If the PF system were 100% energy efficient then this train really has to run for 5.9 hours. Of course you never get 100%, but if the train ran for say 5 hours, then that would mean that the system has a good efficiency.

On the other hand, if the efficiency was no better than simply wasting away the unwanted voltage in a resistor, then we should get 2.62 hours of running time. Initially, that's what I thought would happen. But there is a clear consensus that the efficiency is better than that. That means that method 1 is wrong. But method 1 is a simple computation, and it's not so easy to understand why it is wrong. The people that told me that I'm wrong are probably right, but have not given a consistent explanation (a few self-contradictory explanations have been given though).

Of course, no explanation trumps observation. The only way that would really settle this is to just put in a fully charged battery pack and just let the train run until it runs out.

You need to forget everything about the motor running at a constant speed and everything about the 4V, they may be true on a 9V system but they aren't for PF.

Imagine we have a motor which runs at a speed of 2 rotations per second (for easy math purposes) when a constant 9V is applied. It also follows that, assuming no slippage, it will move at an average speed of 2c m/s where c is the circumference of the wheel.

Now if we take that same motor, still applying the same 9V so the motor is rotating at the same speed, but having the motor on for half a second, then off for half a second. We will therefore accomplish 1 rotation of the wheel and, if we stuck to that pattern we'd get an average speed of c m/s (albeit a very jerky movement). Since we've only run the motor half the time, we can reasonably assume only half the amount of energy is used.

We can get rid of the jerkiness by making the on/off pulses very,very short. We'll still only be providing 9V for half of every second, so we'll still only get an average speed of c m/s. The motor is still only on for half the time, so only half the energy is used. This is how PF controls speed.

Now all of that is in an ideal world. In truth we'd probably need slightly more than half the energy to run at half speed due to ineffiencies in the switching etc, but it's a safe bet that we will, overall, get a longer battery life by running at lower speeds.

If the PF system were 100% energy efficient then this train really has to run for 5.9 hours. Of course you never get 100%, but if the train ran for say 5 hours, then that would mean that the system has a good efficiency.

On the other hand, if the efficiency was no better than simply wasting away the unwanted voltage in a resistor, then we should get 2.62 hours of running time. Initially, that's what I thought would happen. But there is a clear consensus that the efficiency is better than that. That means that method 1 is wrong. But method 1 is a simple computation, and it's not so easy to understand why it is wrong. The people that told me that I'm wrong are probably right, but have not given a consistent explanation (a few self-contradictory explanations have been given though).

Of course, no explanation trumps observation. The only way that would really settle this is to just put in a fully charged battery pack and just let the train run until it runs out.

I think you're getting somewhere now. Method 2 is correct.

I'll try and illustrate why with a bit of a thought experiment. I'll idealize a whole bunch of things for the purpose of making the illustration of how the PF receiver is operating clear. Later I'll try and address some of the other issues people have brought up in the thread.

A thought experiment to clarify PWM

Consider running your train motor for some fixed time period with the 9V controller output at 4.5V, say one minute, and for arguments sake lets say it makes 16 rotations around your layout.

Now consider using a PF battery box to run to motor for one minute instead. Since a battery box only outputs 9V, the train will run twice as fast so ideally it will make 32 rotations.

That's not really equivalent so we next consider using the battery box to run the motor for 30s and turning off the battery for then next 30s. Now we have the same energy usage, the same distance traveled, and over the whole minute the same average speed for both the 9V and PF case.

There is still a difference in how fast and when the train travels in each case. To make the PF case more like the 9V case, we might like to turn the battery box on and off more frequently, and thus spread its movement more evenly over time. As an example lets turn the battery on for half a second sixty times during the minute, once each second. We still have the same energy usage, and in the ideal situation where the train reaches the speed one would predict from the voltage instantly, the same distance traveled and the same average speed.

BY now you're probably thinking that I've made a bunch of idealizations and the poor motor is going to be heating up and screaming and jerking around, and you're sort of correct. It's not a problem though if we consider what the DC motor will in fact do in response to a rapidly changing square wave input.

If we think about the shape of the velocity over time graph ideally it would be a nice square wave output, but practically the motor will take a bit of time to get up to speed when the battery gets turned on, and inertia will carry it on after the battery is turned of. In reality it will look a bit more like a smoothed off square wave. In fact the higher the frequency with which we switch the battery the more smoothed off it will look. In fact it will start to look a lot like you would expect the output to look like if you averaged the input over time, ie had a constant 4.5V for the whole minute, maybe with a little noise at the frequency of switching overlaid on it (thats where the hum comes from in PWM systems, since the frequency used is in the audible range).

Hopefully you now can see how a train motor could appear to be powered from 4.5V but in actual fact be getting 9V pulses.

Duty Cycles and speed control

So far we can see how to get a train to move at half speed, but for completeness I'll illustrate a few more cases.

Zero speed is easy, just don't turn the battery on. Full speed is also easy, don't turn it off after turning it on.

How about quarter speed? If we agree that the DC motor will tend to react the same way to high frequency inputs as it would to their time average, any square wave in which the power is high for 25% of the time will result in quarter speed. Its called a 25% duty cycle because the power is on 25% of the time. It could be done by having the width of the on pulse from the 50% case, or you could keep the same width of pulse but just skip ever second pulse, ie turn the power on, off, off, off, on, off, off, off. I suspect the Lego solution does the later because the frequency of the noise the motor emits gets higher with speed as you would expect in this case.

For three quarters speed any square wave in which the power is high for 75% of the time will do. I suspect Lego do something like on,on,on,off,on,on,on,off.

Toastie's Friction Considerations

So far I've very carefully avoided any consideration of tourque or friction and also of any losses in either the PF receiver or 9V controller. What I've described is the voltages at the inputs to the train motor and its response. Toasties viewpoint at this point might be helpful.

If we consider the two cases of PF at 50% duty cycle, and 9V controller outputting 4.5V, we can see that they consume the same energy over the same time period. Thus we would expect they would give pretty much the same distance traveled from Toastie's correct observation that pretty much all the energy available at the motor will be used overcoming friction (it doesn't really matter for this argument if friction depends on speed or not since both cases are at the same speed). This doesn't account for any losses in either the 9V controller or PF receiver.

Comparison of 9V and PF losses

I think we all agree that the voltage divider in the 9V controller will lose energy (dissipated as heat). If 4.5V is reaching the motor, another 4.5V is being dissipated. This will be of the same magnitude as the energy used powering the motor, since similar current is flowing in both.

The loses in the PF receiver will be whatever is involved in running the IC's for the pulse generation typically much less energy than for running a motor. Switching transistors typically use orders of magnitude less current in the switching input than the power input. There has also been some discussion of the energy lost to the hum in the motor. Since this is a noise (in the signal sense) artifact it should be plain that it would be poor design of the electronics not to minimize it, so we can expect it also to be a fraction of the useful output. Since the PF receiver doesn't ever change the voltage of the battery pack, there is no equivalent to the 9V controllers voltage divider, so no equivalent energy loss. Their will also be some energy required in running the communication with the remote, but given the limited range it will also be a different order of magnitude than running the motor.

We can see that the expected losses are different orders of magnitude for the 9V and PF electronics, hopefully illustrating why PWM was chosen rather than a voltage divider for the PF system where battery life is limited.

Other Considerations

I think LT is confusing PWM for control of motors with switched mode power supplies which do something similar as far as I remember to control the output voltage. I want to be really clear that, if I understand the way the PF system works correctly, the only voltages being output from the PF receiver are 9V (or whatever maximum voltage the battery supplies to its input) and 0V, the speed control is via changing the duty cycle not the voltage.

There may be PWM being used to encode the signal from the remote but this is entirely separate from this discussion.

I think the observed 3-4 hour life of the rechargeable battery pack fits pretty well with the theoretical analysis and expected loses discussed above.

Since we've only run the motor half the time, we can reasonably assume only half the amount of energy is used.

I have two resistors, 10 Ohm each. One of them I subject to 4.5 Volt continuously. The other one gets 0V for 0.01 seconds, then 9V for 0.01 seconds, then 0V, 9V, 0V, 9V, .. etc.

So the first one is "on all the time" at 4.5 V.

The second one is "on half of the time" but when it is on, it gets 9V.

Question: which one will get hotter?

(note: the internal resistance of the lego train motor is roughly 10 Ohm if I remember correctly, or perhaps a little bit less. I can measure it again if you want a more precise number).

Question: which one will get hotter?

From a pure energy perspective I can't see that there would be any difference over an extended time period as long as the resistors have identical properties.

From a pure energy perspective I can't see that there would be any difference over an extended time period as long as the resistors have identical properties.

Do others here see it the same way?

I have two resistors, 10 Ohm each. One of them I subject to 4.5 Volt continuously. The other one gets 0V for 0.01 seconds, then 9V for 0.01 seconds, then 0V, 9V, 0V, 9V, .. etc.

So the first one is "on all the time" at 4.5 V.

The second one is "on half of the time" but when it is on, it gets 9V.

Question: which one will get hotter?

(note: the internal resistance of the lego train motor is roughly 10 Ohm if I remember correctly, or perhaps a little bit less. I can measure it again if you want a more precise number).

What are you talking about ?

MEH !

Applying 4.5 volts across 10 Ohms is a current draw of 450mA and said 10 Ohm load will radiate 2 Watts and then you have a 9 Volt square wave at 100Hz - then double the power and you then ask which is hotter. Check anything that is digital for heat and you'll get it.....just feel the top of a cable or FTA set top box they get pretty warm !

You can not compare an inductive load to a pure resistive load and besides with a standard electric motor the rotors current is broken constantly to maintain rotation.

You can not compare an inductive load to a pure resistive load and besides with a standard electric motor the rotors current is broken constantly to maintain rotation.

The load is not purely inductive, because the wiring inside the motor is not super-conducting. Any current that runs through any wire that is not super-conducting will produce a certain amount of heat (the square of the current, times the resistance). This is true regardless of whether the current is AC or DC, if the wire is in an inductor or not, etc.

The motor has about 7 Ohms of resistance. So the heat (in Watts) that this produces is the square of the current, times 7 Ohm.

If you double the current, the number of Watts dissipated as heat in this way goes up a factor 4. This is why, when the current is applied unevenly (but with the average staying the same) then the total number of Watts dissipated as heat in the motor will have to be more than when the current is applied evenly. If the PF system has some components to smooth out the current through the motor, then that should increase efficiency. I do not know to what extend the PF system does that, I have no oscilloscope so I can't see (the train motor itself contains of course a small capacitor, that's of course already helpful. If you put a bigger one in there, you should get less of that high-pitch sound). One way to check the PF efficiency is to see if the batteries last longer on low speed settings than on high speed settings, and if so, by how much.

One way to check the PF efficiency is to see if the batteries last longer on low speed settings than on high speed settings, and if so, by how much.

Hoeij,

you are still thinking in terms of a perfect electrical world, right? The very moment you increase speed, friction forces may go everywhere (curves vs straights and what not). I am pretty sure that the amount of electrical energy you need to compensate these varying mechanical "losses" or better sinks may actually dominate.

So, here is a proposal: Why don't we run a nice, conclusive experiment (all the theory is just that, anyway: Theory ... who wants to hear that? I have a chemical reaction dynamics class tomorrow and it just prevents me from going to sleep - they hate it. Theory ...)

All we need is proper measurements. So lets figure out what you want to prove and design the experiment. I am ready (sort of lost the initial idea). What do you want to show? I have a 9V power pick-up motor, straight 9V motors and PF motors, receivers, LiPos, and DC (0-25V) power supplies along with the LEGO 9V train power "regulator".

If we run the experiment appropriately, we can probably satisfactorily come to a conclusion with everyone being happy.

All the best,

Thorsten

So lets figure out what you want to prove and design the experiment. I am ready (sort of lost the initial idea).

Thorsten, what I want to know is if the battery or the Lipo lasts longer at low speeds than it does at high speeds, or if the battery life is the same regardless of the speed setting.

My dual-motored train uses 420 mA (210 mA per motor). I measured the current, and found that it uses 420 mA at any speed setting (I tried speed settings 1,2,3,4, which is 3.0, 4.2, 5.4 and 6.6 Volts). The friction force is pretty much the same at any speed setting, and so is the current. Those are measured facts.

The question is: can one compute from that how long the train will run on the batteries? The LiPo has 1100 mAh, so divide 1100 by 420 and you get an estimate for how long this train can run. If I use 6 rechargeable AAA batteries, you get a similar number.

The problem is that this number is independent of the speed setting (which is a bit unexpected). Question is: at a low speed setting, will the train run longer than the computed number (1100 divided by 420)? Letting the train run for hours on end is not really an option here (it's temporarily in the TV room, there would be complaints).

In some sense the question is moot now, I've decided I'll run this train with the 9V system at the train show. But this means quite a bit of wiring because the track is too long to supply power to just one place on the track. To make sure that the power is evenly divided over the track, I've cut three 9V track connectors and added wiring to make them 10, 10, and 16 feet long.

I did this because I thought (dividing 1100 by 420) that the train wouldn't run long enough in the PF system and I'd have to replace batteries too often. But I don't actually know how long the batteries would have lasted.

So, here is a proposal: Why don't we run a nice, conclusive experiment

ha ha ha... it took over a week to get to the easiest point and solution that takes just about 5 hours: 2 ovals, let's say 16 curves and 16 straights and two pf engines.

if there is going to be a poll, my vote is going to "low speed, longer run"

anyway, even if I'm not that much into electronics & c., it's been a really interesting "teaching topic"!

have a nice train lego experiment day

mrBlue

hoeij, when you say you have measured the current, what kind of meter were you using to make the measurement?

With the PWM system, the voltage (and thus current) is switching on and off at a frequency of tens if not hundreds of thousands of times per second -- depending on the circuit design, the PWM switching frequency is some fixed value likely between 20 kHz and 100 kHz. At these high frequencies, most hand-held consumer multimeters will not have the measurement bandwidth to accurately report a value, and the value that it does display may be misleading or meaningless.

While you're still driving a DC motor, the output voltage from the controller is very much an AC signal, although not a sinusoidal one.

Edited June 1, 201113 yr by blargh

ha ha ha... it took over a week to get to the easiest point and solution that takes just about 5 hours: 2 ovals, let's say 16 curves and 16 straights and two pf engines.

It'd be easier to take friction (and space) out of the equation by just running the motors raised up and seeing how long they last.

It'd be easier to take friction (and space) out of the equation by just running the motors raised up and seeing how long they last.

yes, it would be easier, but it will not be a true test... how many people runs their trains upside down and not on tracks? for sure few have just an oval tracks too (with 50% curves and 50% straights), but at least it would represent a bit more a "real" common scenery

have a nice train lego day

mrBlue

hoeij, when you say you have measured the current, what kind of meter were you using to make the measurement?

With the PWM system, the voltage (and thus current) is switching on and off at a frequency of tens if not hundreds of thousands of times per second -- depending on the circuit design, the PWM switching frequency is some fixed value likely between 20 kHz and 100 kHz. At these high frequencies, most hand-held consumer multimeters will not have the measurement bandwidth to accurately report a value, and the value that it does display may be misleading or meaningless.

While you're still driving a DC motor, the output voltage from the controller is very much an AC signal, although not a sinusoidal one.

I'm pretty sure hoeij has measured the 9V motor running with a 9V controller, still room for error due to the motor not being purely an ideal resistor but better than the situation above.

I'm pretty sure hoeij has measured the 9V motor running with a 9V controller, still room for error due to the motor not being purely an ideal resistor but better than the situation above.

That's right. Note that to do this measurement, it was necessary to cut a 9V connection cable so that I can insert the multimeter in the setup (this was OK, I had to cut some connection cables anyway to make them longer with additional wiring. Lego 9V wires are 22AWG, but when I make them longer I insert 20AWG wire to minimize losses).

You can see on: http://www.philohome.com/motors/motorcomp.htm that if you change the voltage while keeping the torque constant, then the current stays also almost constant (within 10% or so). So the current is nearly independent of the voltage. I observed the same thing in my measurements as well. That's why I focus on current, not on volts or watts.

Another thing I found in my measurements is that if you add cars to the train, cars with equal weight, then each additional car increases the current-consumption by the same amount. So if you've measured the current for "engine", "engine + 2 cars", "engine + 4 cars" then you can predict (with good precision) what the average current will be for "engine + 6 cars", or "engine + 8 cars", etc. Again, this behavior is in good agreement with how DC motors are supposed to behave.

yes, it would be easier, but it will not be a true test... how many people runs their trains upside down and not on tracks? for sure few have just an oval tracks too (with 50% curves and 50% straights), but at least it would represent a bit more a "real" common scenery

have a nice train lego day

mrBlue

I actually think this test would be just fine to make a comparison. The point is to determine (by experiment) if a PF setup will run longer at a slow speed than at a high speed. The conditions of the experiment really do not matter as long as the two tests are done under the same conditions. Having the train elevated or even upside will be fine to determine the battery life vs. train speed setting. Now if you want to figure out how long a particular train on a particular track setup will last (like for a train show), this will not tell you much, unless you actually set up the tracks and run the train on it. As long as the controllable factors of the experiment for both cases are the same (load on the motor ie. weight being pulled, and friction from track setup), it will do to answer this paricular question.

In terms of electrical theory, the one thing you have to consider is that a 9V system essentially has an unlimited energy supply. For the case of PF, whether its the standard battery pack (with new, non-rechargable batteries or rechargable batteries) or the LiPO battery pack, the battery itself has a LIMITED supply of total energy in it. The potential energy of the battery (stored energy that will be converted to run the train) is a simple calculation of voltage times ampere hours (To get it to a known unit, like Joules, you need to convert the hours to seconds, but its trivial for this case). The LiPo battery is 7.2V with 1100mAH. That is a total stored energy of 28.5 kJ.

For the motor, it consumes the bulk of the energy in the system. Its power is simply Voltage times current. Since power is simply energy per unit time, the you can consider 1 W as 1 J/sec. Using hoeij's current and voltage measurements (knowing they may not be accurate based on his equipment and PWM issues), on speed setting 1, you get 3.0V x 210 mA = 0.63 W (or J/sec). That means if the battery was allowed to completely drain (which the PF LiPO will not allow by design), you get a theoretical max run time of 12.5 hours. At speed setting 4, you get 6.6V x 210 mA = 1.39 W (J/sec), and a max theoretical run time of 5.7 hours.

Now of course the LiPO battery will not allow a full discharge, as it will cause the LiPO battery to explode (from what I have read). You also have energy consumed in other ways, such as heat generation, noise as pointed out, and of course frictional forces. So the max theoretical run time is probably about twice as long as you will really get out of your PF setup, depending on the overall size of your train, the amount of curves in your track set up, etc. But based on an energy perspective, the speed of the train most definitely will affect the run time of the system. It all comes down to understanding that a battery provides a fixed amount of stored energy, and that is the real factor in this. Comparisons to 9V systems are completely irrelavent as teh 9V system has an unlimited supply of energy.