hoeij

Where does the excess energy go?

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Suppose you have a PF train, and suppose it is a cargo train that you

want to run slowly, say at 4 Volts (that's fairly close to speed setting #2).

So you want your train to run at the speed of 4 volts, however, the batteries

deliver 9V.

My question is, the energy in the remaining 9-4 = 5 volts, where does

it go? This has to be dissipated somewhere in the form of heat,

but where? Does this happen inside the motor?

With the 9V system, this excess energy (the energy that needs to be

wasted to go down from 9V to the desired voltage) this energy gets

dissipated as heat inside the 9V controller (that means that if you

choose a low speed setting for your 9V train, then the 9V controller

will get warmer than it would at a high speed setting).

But with PF, the PF receiver is small, so it can not dissipate much heat.

The only other place I can think of where this excess energy can be dissipated

is in the motor.

Question is, is this true? (if so, then a train motor powered by the PF

system should get warmer than the same train motor powered by a 9V train

controller). Has anyone compared that?

Also, if this is true, then the PF train motor should get warmer at low

speed settings than it does at high speed settings. Has anyone observed

if that is true?

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I always thought that there was some kind of PWM system to regulate the power delivered to the motor, from the 8878 battery or the ir-receiver. Hence there is no (resistive) power dissipation that needs to be cooled off. But I actually have no idea how this is implemented or if this is the case.

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Ever heard of a voltage regulator, and the pulse width modulation system I believe is the communication setup for the remote to talk to the controlling electronics similar to a TV remote system.

'hoeij' your concerns about missing voltages to a electronics tech like me sound unfounded, your not losing energy but not using it or anything of the sort....it's drawiing current that flattens batteries. Running a train motor at 4 Volts means less current being drawn, but it might not have the guts to overcome a slope or rises in the track. :wink:

Brick On ! :grin:

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Here is what I think I know on the subject:

There are 2 ways to control the speed in PF set ups. One is through the IR controller. Since the voltage from the battery is only the driving force, and the current delivered from the battery, through the IR reciever, to the motor is really what drives the speed of the motor. DC current goes up, the motor runs faster. DC current goes down, the motor runs slower. The IR reciever has something in it that I know limits the current that leaves it to a threshold (not sure what that number is). But the IR Reciever, when used, controls the current in the circuit. The second way is with the PF LiPo battery. It has a dial on top that can control the current output from the battery directly without the use of the IR reciever. You may actually be able to get a higher top end speed out of the set up by eliminating the IR Receiver, since I do not know if the battery itself has the same current limit (if any). Its not really a matter of discipating energy. The battery stores a set amount of energy. The energy is discipated from the battery as the current leaves the cell. If that energy was being "wasted" somewhere to control the speed, then the life of the battery (time it takes to drain it) would be constant regardless of the speed you operate the train. One other thing that comes into some play here is that battery life is also dependant on discharge rate. If you discharge the battery at a high rate, you get a lower total amount of energy out of it that is usable (to the train) as you would at a lower discharge rate. In this case, the "wasted" energy is lost due to heat, as a battery heats up more at higher discharge rates (every use a 15 min charger on rechargables?) I think that may be part of the rationale of putting in a current limiting device in the IR Reciever, as it may be a safety concern with heat generation in plastic parts.

In the 9V world, the controller dial controlled the current applied to the rails by using a rheostat. It is essentially a long coil of wire that acts as a variable resistor. The position of the dial controls how far the current travels in the coil before it leaves to the rest of the circuit (everything after the pick up is not used). Increasing speed makes the length of the coil shorter (less resistance), resulting in higher current for the same voltage. With a constant applied voltage (from the wall), the rheostat varied the resistance of the circuit, thus varying the current, and subsequently the speed of the train on the tracks.

Edited by Tearloch33

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I always thought that there was some kind of PWM system to regulate the power delivered to the motor, from the 8878 battery or the ir-receiver.

You sir, while being the least certain of your answer, are also the most correct.

The PF receiver controls the speed of the motor via PWM, or pulse width modulation.

The battery box provides a certain voltage (slightly different depending on type, ie 9V for 6 alkaline batteries, ~7.2 for the rechargeable etc.) and the receiver simply allows this through to the motor in different width pulses. The pulses are widest at full speed and narrower at lower speeds. The frequency of the pulses is high enough that the motor doesn't jerk but moves smoothly, though you may have noticed, there is a high pitched humming sound with PF trains due to the frequency of the pulses, the tone of which varies a bit with the speed of the train. If you power a 9V train motor with a PF receiver it will hum too, which shows it's the power delivery causing it rather than the motor.

Since the current is in effect being turned on and off again there is no excess power to be dissipated like in the 9V controller. Modern DDC controllers for scale model railroads also use this method, partly to get away from heat dissipation problems.

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'hoeij' your concerns about missing voltages to a electronics tech like me sound unfounded, your not losing energy but not using it or anything of the sort....it's drawing current that flattens batteries. Running a train motor at 4 Volts means less current being drawn,

I measured this, both for 9V trains and for 12V trains. The current that a train draws is (within just a few percent) practically independent of the speed of the train. As long as it is moving, it amount of current it draws depends only on things like the weight of the train, and whether it is in a curve or on straight track. But the amount of current is practically independent of the speed.

So the batteries will not last longer if you run the train slowly.

If that energy was being "wasted" somewhere to control the speed, then the life of the battery (time it takes to drain it) would be constant regardless of the speed you operate the train.

I think that that's exactly what happens, that the battery won't last longer at a slow speed, and my Emerald Night will take 3 hours to empty the 8878 rechargeable battery, regardless of the speed setting.

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I measured this, both for 9V trains and for 12V trains. The current that a train draws is (within just a few percent) practically independent of the speed of the train. As long as it is moving, it amount of current it draws depends only on things like the weight of the train, and whether it is in a curve or on straight track. But the amount of current is practically independent of the speed.

So the batteries will not last longer if you run the train slowly.

I think that that's exactly what happens, that the battery won't last longer at a slow speed, and my Emerald Night will take 3 hours to empty the 8878 rechargeable battery, regardless of the speed setting.

Wrong, it's like "peterab" said: speed of motor is controled with PWM. Simplest explanation is that electronics inside PF reciver (or LiPo Battery) turns the motor on and off very fast. Let's say that you want 25% power: electronics will turn motor on and off in such rate thet it will be 25% of time turned on and 75% turned of. Motor will of course not stop because all of this is happening very fast, but will run at 25% speed. This cold be noticed if you run motor at low speed and you will hear strange noise or humming (I dont know exact word). Battery usage will be depending on the speed of the train (because the current is also depending on the time the motor is on, in this case 25%)

It's the same for LiPo battery, which already has builit in electronic for charging and they just added PWM control for speed.

9V system has completely different type of regulation, with several outputs from the transformer (yes, it's not linear, you could easily measure this with voltmeter), which you change by turning the knob.

Edited by Cwetqo

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I measured this, both for 9V trains and for 12V trains. The current that a train draws is (within just a few percent) practically independent of the speed of the train. As long as it is moving, it amount of current it draws depends only on things like the weight of the train, and whether it is in a curve or on straight track. But the amount of current is practically independent of the speed.

So the batteries will not last longer if you run the train slowly.

The problem with your statement is that the 9V and 12V systems are not battery systems. Basic electrical theory tells us that in a closed DC electrical circuit, Voltage (V) = Current (I) * Resistance ®. Voltage is not controllable for a battery system without some fancy stuff, which I do not believe is part of the PF system. So the voltage in this case is constant. The circuit, either inside the IR Reciever, or that LiPo battery itself with its built in control, alter the circuits resistance, thus controlling the current. The variations you see for turns or based on weight is due to a concept where a motor actually creates a counter current (acts as a generator) when turning. The resistance to the motor turning (in this case, caused by turns or the amount of weight it pulls) varies the amount of counter current generated in the motor, which must be overcome by the battery, making the output current of the battery go up.

Also, please try testing your statement of the battery not lasting longer for slower speeds.

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Also, please try testing your statement of the battery not lasting longer for slower speeds.

This is a bit time consuming, it takes 3 hours for my Emerald Night to deplete the 8878, and I have to do that twice, once at high speed, and once at low speed. I was hoping that someone here might have already done this test. It takes less time to simply measure the current. I've done that at high speed and at low speed, and the result is the same.

I have a 10-feet long train with 2 Maersk engines, and a number of light-weight cars behind it. At the moment, it has two 9V motors. I was trying to decide whether to use the 9V system or the PF system to run this train, so I measured its electricity usage. It uses 0.42 amps (so 0.21 amps per motor) and this doesn't really change with speed (at low speed it takes 0.42 amps and at high speed it takes 0.43 to 0.44 amps). If I use rechargeable AAA batteries with a capacity of 1,000 mAh (which I do not have, I'd have to buy them) and I need 420 mA to run the train, then it should run 1000/420 = 2.4 hours before the battery is empty. I don't see how one can get around that. Slow or fast, either way the train pulls 420 mA according to my multi-meter. There are only so many mAh in the battery, so it seems inevitable that the batteries will run empty after 2.4 hours, regardless of the speed setting. Of course, this still needs to be tested, I may be overlooking something in this reasoning.

There is one other test that one can do, one that is not as time consuming as running the battery empty twice: we can check if (as I suspect) that a PF train motor running at low speed becomes warmer than one running at a high speed.

Incidentally, does anyone here know how many mAh there are in a fully charged 8878?

Edited by hoeij

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This is a bit time consuming, it takes 3 hours for my Emerald Night to deplete the 8878, and I have to do that twice, once at high speed, and once at low speed. I was hoping that someone here might have already done this test. It takes less time to simply measure the current. I've done that at high speed and at low speed, and the result is the same.

As far as I am concerned there is absolutely nothing wrong with your measurements, Hoeij.

[True, PWM is on/off modulation, so 50% of 9V in PWM means 9V turned on and off at 50% duty cycle. Whatever you measure with you meter depends on the electronics in the meter rather than what is going into the motor (that is 50% of 9V times amps at 100% minus losses due to phase shifts; nerdy, I know and it doesn't matter).]

The very moment you have accelerated your train to final speed, all the additional energy you need from now on is to overcome friction forces. And not to establish this speed. Friction forces originating from your train design, from wheel vs track friction, and from aerodynamic friction, and what not. The latter before "what not" is next to nothing compared to the rest. (That is very different for aircraft, but we are talking about mass monsters - trains, going at comparably low speed.)

Going LEGO slowly means that you have to overcome basically the same losses as compared to LEGO high speed. In real world, friction forces go dramatically up with speed. But LEGO trains don't ever go that fast that you will notice this (tiny) increase in frictional loss as compared to - "design flaws" created loss. Design flaws in terms of "running a plastic axle through a reasonably matching plastic hole". Or - non-matching wheel-axle width/track width. Or, even worse, poorly matched rubber bands.

There is so much energy required to keep the LEGO machines up and running at constant speed - slow or fast simply doesn't matter.

Hope that makes sense, give me Flak, I am not a physicist ...

Regards,

Thorsten

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The problem with your statement is that the 9V and 12V systems are not battery systems. Basic electrical theory tells us that in a closed DC electrical circuit, Voltage (V) = Current (I) * Resistance ®. Voltage is not controllable for a battery system without some fancy stuff, which I do not believe is part of the PF system. So the voltage in this case is constant. The circuit, either inside the IR Reciever, or that LiPo battery itself with its built in control, alter the circuits resistance, thus controlling the current.

That would be true if the PF system worked by varying resistance, it doesn't. V, I and R are all constant in a PF system. Everything is instead done through PWM, effectively varying the amount of power in the system by altering how often it is actually 'on' for over a very short period of time. As mentioned previously, the advantage of using PWM is that you don't have the problem of trying to dissipate excess amounts of heat energy as you would with a variable resistance arrangement. The downside is that you get a kind of whining/buzzing noise at lower speeds as motors are rapidly switched on/off.

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That would be true if the PF system worked by varying resistance, it doesn't. V, I and R are all constant in a PF system. Everything is instead done through PWM, effectively varying the amount of power in the system by altering how often it is actually 'on' for over a very short period of time. As mentioned previously, the advantage of using PWM is that you don't have the problem of trying to dissipate excess amounts of heat energy as you would with a variable resistance arrangement. The downside is that you get a kind of whining/buzzing noise at lower speeds as motors are rapidly switched on/off.

If the PWM does not dissipate the excess energy, then it'll have to be dissipated somewhere else.

The whining/buzzing noise explains at least partially where that excess energy goes; obviously, you can't generate noise without spending at least some energy.

Here's one way to think about the various scenarios:

(1) Turn on 4.5 volts, constant voltage (as in the old 9V system).

(2) Rapidly switch 0 and 9V, with each on 50% of the time (as in PF).

Clearly, option (2) will produce a whining noise, and option (1) will not. There is another scenario to consider. Suppose you did this:

(3) Rapidly switch between -4.5 volts, and +4.5 volts, with each 50% of the time.

In option (3), the train would stand still, but it would make a whining noise. And clearly the motor would dissipate some energy as heat because the coils inside have resistance, so they must warm up when current runs through.

Notice now that option (2) is really the same thing as the sum of options (1) and (3).

0 and 9 is the same as 4.5 added to -4.5 and +4.5.

So the way that the PF train runs is as though you are doing (1) and (3) at the same time. The energy for (1) is spent to move the train, while the energy for (3) is dissipated as heat and a whining sound.

The conclusion is that if I have two 9V trains, of equal length/weight/speed/etc, and if one is powered by PF and the other by the old 9V train controller, then the one powered by PF has to get warmer (i.e. dissipate more energy) than the one powered by the old 9V controller.

I know this seems counter intuitive, after all, both trains are equally heavy and go equally fast. But still, the whining noise produced by the motor that is powered by the PF system is a clear indication that energy is indeed being dissipated in the motor. If no energy was lost, then why would it whine?

Although I have not yet tested it, I'm almost sure that in PF, the batteries will not last longer in low speed than they would in a high speed setting.

At speed setting #2, my cargo train moves gently over the track. At a high speed setting, it thunders over the track so fast that I fear it might derail in the curves and fall of the table. Seeing that, it seems impossible to believe the amount of current is pretty much the same in both cases. But that's exactly what my multi-meter is telling me.

Edited by hoeij

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That would be true if the PF system worked by varying resistance, it doesn't. V, I and R are all constant in a PF system. Everything is instead done through PWM, effectively varying the amount of power in the system by altering how often it is actually 'on' for over a very short period of time. As mentioned previously, the advantage of using PWM is that you don't have the problem of trying to dissipate excess amounts of heat energy as you would with a variable resistance arrangement. The downside is that you get a kind of whining/buzzing noise at lower speeds as motors are rapidly switched on/off.

I am kind of old school in my electrical theory, being a mechanical engineer with military power generation experience. PWM was not considered by me until pointed out. Makes sense based on size constraints for PF parts. My understanding of PWM is that it basically turns a DC source (ie battery) into an AC source by rapid on/off cycles. The pseudo AC signal is just periods of source voltage followed by periods of no voltage. So I get your point about it not being a variable resistance circuit in nature. Basically the speed control in the IR Reciever or the LiPo battery itself changes the effective frequency by altering the duty cycle of the on/off periods, thus controlling the speed at which the motor operates at (speed of the train). It also makes sense how LEGO was able to up the power/torque output of the PF motor while lowering its consumption (based on their words). Once the motor is at speed, energy consumption is only to overcome frictional losses.

Edited by Tearloch33

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If the PWM does not dissipate the excess energy, then it'll have to be dissipated somewhere else.

The whining/buzzing noise explains at least partially where that excess energy goes; obviously, you can't generate noise without spending at least some energy.

True, a very insignificant amount of energy is dissipated inside the power supply. Switched DC converters operate by turning on and off very quickly, on the order of hundredths to thousandths of a second and beyond. The motor only sees the average value of these pulses. There is no excess power dissipated as in a voltage divider circuit because the batteries are constantly being switched on and off. Your multi-meter will measure average values, the only way to see the actual pulses would be to hook up an oscilloscope.

The batteries will absolutely last longer running on lower power vs. higher power, that is the point of switched power supplies. If this were not true, no one would spend the money designing and building them. A voltage divider circuit with a fixed resistor or rheostat is way cheaper to design and build.

Edited by Brick Tamland

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The batteries will absolutely last longer running on lower power vs. higher power, that is the point of switched power supplies.

If you are powering a light, then yes, but if you're powering a DC motor, then I do not believe that that is true.

If this were not true, no one would spend the money designing and building them. A voltage divider circuit with a fixed resistor or rheostat is way cheaper to design and build.

The quick on/off switching is definitely the most practical thing to do, because that way you can make a PF receiver that doesn't generate heat (in other words: a PF receiver that can be made small enough to be practical for lego).

The reason for building it this way is because: it works, it is cheap, practical, and small. But if the battery pack produces 9V and you want to run a train at 4.5V DC, then I do not think that this system is any more energy efficient than simply dissipated half of the voltage away in a resistor (note: this applies only to the motor. If you're powering a light, then it is more efficient).

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If you are powering a light, then yes, but if you're powering a DC motor, then I do not believe that that is true.

The quick on/off switching is definitely the most practical thing to do, because that way you can make a PF receiver that doesn't generate heat (in other words: a PF receiver that can be made small enough to be practical for lego).

The reason for building it this way is because: it works, it is cheap, practical, and small. But if the battery pack produces 9V and you want to run a train at 4.5V DC, then I do not think that this system is any more energy efficient than simply dissipated half of the voltage away in a resistor (note: this applies only to the motor. If you're powering a light, then it is more efficient).

This is untrue. The amount of power delivered to the load(and hence the energy consumed by the load) is independent of the type of load.

Switched DC power converters are immensely more complex and much more expensive than the alternative, which is a voltage divider circuit consisting of a single resistor in series with the load. For a switched converter, you need at minimum, an inductor, a capacitor, and two transistors just for the converter. Then you need a microcontroller to generate the switching commands to implement PWM. To reiterate what I said before, batteries will last longer on low power with a switched converter than they would on high power. I don't mean to seem argumentative, but this is a fact.

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This is untrue. The amount of power delivered to the load(and hence the energy consumed by the load) is independent of the type of load.

A motor behaves very differently than for example an incandescent lamp. With a motor, the current depends mostly on the torque that the motor is producing. With an incandescent lamp, the current depends on voltage, so that's a completely different behavior.

Switched DC power converters are immensely more complex and much more expensive than the alternative, which is a voltage divider circuit consisting of a single resistor in series with the load. For a switched converter, you need at minimum, an inductor, a capacitor, and two transistors just for the converter.

I agree with that. But does the lego PF receiver have all those things?

I've read that the only thing it does is rapidly switching the power on/off/on/off... But if you want to be efficient from an energy point of view, then you have to do more than just rapidly switching on/off, you also need the other components you mentioned.

To turn 9V DC into 4.5V DC without wasting half of the energy, that's a complicated thing to do. Because it means that you have to turn for example 9V with 100mA into 4.5V with almost 200mA. It is possible to do that, but it is complicated, and I question if the lego PF receiver actually does that (turn 9V, 100mA into 4.5V with more than 100mA).

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A motor behaves very differently than for example an incandescent lamp. With a motor, the current depends mostly on the torque that the motor is producing. With an incandescent lamp, the current depends on voltage, so that's a completely different behavior.

I agree with that. But does the lego PF receiver have all those things?

I've read that the only thing it does is rapidly switching the power on/off/on/off... But if you want to be efficient from an energy point of view, then you have to do more than just rapidly switching on/off, you also need the other components you mentioned.

To turn 9V DC into 4.5V DC without wasting half of the energy, that's a complicated thing to do. Because it means that you have to turn for example 9V with 100mA into 4.5V with almost 200mA. It is possible to do that, but it is complicated, and I question if the lego PF receiver actually does that (turn 9V, 100mA into 4.5V with more than 100mA).

Again, you are incorrect. Torque is a product of flux, which is a product of current and inductance. It doesn't work the other way around. You are also wrong about the light bulb. Take your multimeter and try to measure the voltage across it when its not plugged in. Then measure the resistance. For a fixed current, the voltage drop across the bulb is a product of resistance. For a fixed voltage, the current is a product of resistance.

The switching converter will generate a constant current regardless of what is connected on the other end(it will actually oscillate slightly as a function of circuit components, but its close to DC). The other components I mentioned do not dissipate energy. Transistors will have a slight leakage, but its so small its negligible. In periodic steady state, the average voltage across an inductor is zero. The average current across a capacitor is zero. Power is equal to current times voltage, so its obvious that neither of these can dissipate power. All of these components are necessary to create a switched converter, so yes, in some form or another, the PF receiver has them. The transistors do the actual switching, the capacitor and inductor filter out high frequency ripple from the current, ensuring you don't destroy your motor. They have no effect on energy efficiency.

The point of the converter is not to turn 9V 100ma into 4.5V 200ma, as the power consumed would be equal. The point of the converter is to turn it into 4.5V 100ma and half the power consumption. Its not really all that complicated if you understand switched converters, which you don't(no offense intended). I can recommend some books if you want to learn about them, but it will require a solid Calculus background(Calc IV level) and a good understanding of electromagnetic theory to really get anything out of it.

Edited by Brick Tamland

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To turn 9V DC into 4.5V DC without wasting half of the energy, that's a complicated thing to do. Because it means that you have to turn for example 9V with 100mA into 4.5V with almost 200mA

I think this is the key to your misunderstanding. The PF receiver doesn't do this, it simply controls the duty cycle, or percentage of the cycle the power is flowing. When current is flowing, the voltage is 9V, otherwise it's 0V.

While the current is flowing it it being used to drive the motor, overcome friction etc, just like the 9V system.

Consider a 9V motor running at 4.5V at 100mA, to achieve the same speed the PF receiver would need to power it at 9V at 100mA at a 50% duty cycle (ie for half the time). With the 9V controller the other 4.5V at 100mA is being dissipated as heat. There is no need for this with PF since there is no other large resistive load. Obviously this is an idealized description and there will be small losses in the receiver, some loss to the humming sound etc, but none of them are significant enough to be anywhere near the loss to heat in the 9V controller.

The other components Brick Tamland mentions are required just to switch the 9V (or in fact whatever voltage the battery provides, ie 7.2V for rechargables etc) on and off, though as he correctly states the circuit will be designed to minimize losses.

BTW I've studied physics, electromagnetic theory, calculus and scientific instrumentation and worked as a repair tech on an IBM motherboard assembly line.

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Again, you are incorrect. Torque is a product of flux, which is a product of current and inductance.

I think I wrote the same thing but the other way around. In any case, current and torque are almost linearly related to each other, and to produce the torque needed to move the train, you have to run at least a certain amount of current through it.

All of these components are necessary to create a switched converter, so yes, in some form or another, the PF receiver has them. The transistors do the actual switching, the capacitor and inductor filter out high frequency ripple from the current, ensuring you don't destroy your motor. They have no effect on energy efficiency.

The capacitor and inductor must have a major impact on the energy efficiency. If they weren't there, the high-pitch humming sound that the PF train motor makes would be much louder, and you can't make sound without spending energy.

I've been drawing diagrams as to how this might work, and I think that you need a diode too.

Both an inductor, and a capacitor, can store a small amount of energy. I think that's the answer to the "where does the excess energy go" puzzle. During on "on" cycle, when 9V is coming from the batteries even though we want to run the motor at 4.5V, some of this excess energy is moved into the capacitor and the inductor. Then, during the "off" cycle, the current to the motor does not drop to 0. Instead, the motor still receives current (and hence, it still produces some torque). This current comes out of the capacitor and inductor. Because they are small, the capacitor/inductor quickly run out of energy, so to prevent the current from dropping too far, the on/off switching needs to be done quickly (hence the high pitch).

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Oh, dear is this still going on ! :laugh:

Great, now I have to dive back into Electronics Networks I which for me was back in 1985. :blush:

This talk of inductors and capacitors, oh dear.....1.) inductors oppose changes in current by the use of back-emf or back-voltage.......2.) capacitors oppose changes in voltage by discharging current back into the circuit. But we are dealing in small DC motors and low power control devices, capacitance and inductance are only minor players...we are using unless you have the rechargable pack, AA batteries which have a Ah rating of only 75mA/hr of course that depends on type of AA cell being used too.

In simple terms the power function system is just like the chopper power supply found in PC's and DVD players where it's output level is set by a PWM signal controlled from an external source (in the case of an actual chopper supply optocoupler feedback of selected voltage level to give the right output voltage to it's load.

It is more efficient than old school chunky mains transfomer types which of course waste energy.

Now this talk of flux and current, eh......it's voltage which creates magnetic flux but the current gives it strength. Stronger the current flow, greater the strength of the flux resulting in greater mechanical force (linked to torque of course). But you do need that voltage to create flux and get the current flowing in the first place.

Dam it, where's 'Mark Bellis' around when you need him.....he is the expert on PF electronics. :blush:

Ah, well......Brick On everyone ! :wink:

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I think I wrote the same thing but the other way around. In any case, current and torque are almost linearly related to each other, and to produce the torque needed to move the train, you have to run at least a certain amount of current through it.

The capacitor and inductor must have a major impact on the energy efficiency. If they weren't there, the high-pitch humming sound that the PF train motor makes would be much louder, and you can't make sound without spending energy.

I've been drawing diagrams as to how this might work, and I think that you need a diode too.

Both an inductor, and a capacitor, can store a small amount of energy. I think that's the answer to the "where does the excess energy go" puzzle. During on "on" cycle, when 9V is coming from the batteries even though we want to run the motor at 4.5V, some of this excess energy is moved into the capacitor and the inductor. Then, during the "off" cycle, the current to the motor does not drop to 0. Instead, the motor still receives current (and hence, it still produces some torque). This current comes out of the capacitor and inductor. Because they are small, the capacitor/inductor quickly run out of energy, so to prevent the current from dropping too far, the on/off switching needs to be done quickly (hence the high pitch).

The capacitor and inductor have no impact on energy efficiency. They simply eliminate ripple in the output current. The energy they store from cycle to cycle is not the same as the energy dissipated by the load, which you seem to be implying is "lost". This is why you get longer battery life at low power. However, I'm done arguing about this. I know what I've stated can be backed up with math, if you choose to believe it its your own decision. My qualifications? Masters degree in Electrical engineering, currently working in the power transmission industry.

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To run a typical lego train, you need about 200 mA, or 0.2 A,

or 0.2 Coulomb per second. So for each Coulomb, it can run about 5 seconds.

At the top speed setting, the motor is basically connected straight to the

batteries, so for every Coulomb the motor uses, one Coulomb is taken out of

the battery pack.

If the batteries are 1000 mAh, that's 1000 mC/s * h = 1 C/s * h = 3600 C.

So the train should run for about 5 * 3600 seconds, i.e. 5 hours. I think

that for a light-weight train, this is pretty accurate.

If you run the train now at a low speed, say speed setting #2, then it will

still use 0.2 A so you get 5 seconds of running for each Coulomb.

When the motor used one Coulomb at a slow speed setting, say equivalent to 4 volts,

then it consumed 4 Joule's of energy, during 5 seconds of time.

The battery pack is at 9 volts, so 4 Joule's of energy corresponds to 4/9 Coulombs

of electricity at 9V. This 4/9 Coulomb is roughly 0.5 C, minus some losses.

Now during those 5 seconds, how many Coulombs were taken out of the battery pack?

(A) If it is 1 Coulomb, then that's 9 Joules, meaning that 5 Joules must have been

wasted somewhere.

(B) If it is 0.5 Coulomb, then (except for minor losses) no energy was wasted.

Intially I thought option (A). But the consensus here is pretty solidly against

option (A). I'll have to assume then that option (A) is wrong (if option (A) were

true, then the motor should get much hotter than what we saw in the 9V system. But

we have not observed that).

The only other option is that, at a low speed setting, for every Coulomb that the

motor consumes, less than one Coulomb is taken out of the battery pack.

I don't really see a way around that. Whether the speed is high or low, either way

you use 1 Coulomb for every 5 seconds. There are only 3600 C in the battery pack.

The only way you could avoid wasting lots of energy at a low speed setting, is to

take less than 1 C out of the battery pack for every 1 C that flows through the motor.

Conversion AC --> AC is much simpler to understand. With AC -> AC it is clear that

every Coulomb at 4.5V costs only 1/2 Coulomb at 9V (with some minor losses). With DC -> DC

this is much more complicated to understand, but if little or no energy is wasted

(which is the clear consensus) then something similar must be going on.

My qualifications? Masters degree in Electrical engineering, currently working in the power transmission industry.

Look, I'm just trying to understand what's happening. I value your explanations and those of others, regardless of qualifications.

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If the batteries are 1000 mAh, that's 1000 mC/s * h = 1 C/s * h = 3600 C.

So the train should run for about 5 * 3600 seconds, i.e. 5 hours. I think

that for a light-weight train, this is pretty accurate.

And that is true. But, I'd like to come back to my "reasoning" posted in this thread before.

All you guys are doing is living in a perfect world. Your energy is flowing into motion via perfect conversion. If that would be the case, you don't have to use any energy at all after reaching final speed.

Real world means that in addition to all the calculations presented here, we have serious, if not dominating losses. Philo has shown us the electrical vs. mechanical power conversion efficiency of the motors at a certain, ideal load (he is hauling a mass basically frictionless up in his measurements). From that data set we can at least estimate the internal losses; and they are quite significant. But these would be virtually constant and not show up in any electrical power consumption vs run time comparisons.

We are hauling trains, translating to a physically serious mass at constant potential energy with respect to height (we are not going up/down in this discussion, we are moving horizontally (ideally)). So, in this ideal electrical world discussed so far, once we are up to speed, we solely have to overcome internal losses AND friction losses. The faster we go, the less mechanical friction losses become. They are greatest close to not moving at all anymore (that is the reason ABS systems are installed in cars). I have recognized that in my little movie on my RCX driven trains: You have to go to RCX power level four to get the train going. Once it moves you can go down to power level two.

I have no idea, but I guess friction is usually dominating in the battery life determining process (at least when hauling some 10 cars). So we may get an ideal picture of battery life asking the electrical gurus (I agree LT: Where on earth is Marc Bellis???) but where are the mechanical experts? Friction must have an impact. Otherwise people would not ask about lubrication, grinding, and what not.

Once you have accelerated the space shuttles to final speed, the motors are turned off; well they fall back to earth. Friction is modest up there ...

Regards,

Thorsten

Edited by Toastie

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

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

'Toastie' is correct, very correct.....nothing is perfect and your calculations don't take in friction or even reluctance....yes, that is a real thing....an electrical engineer will tell you it's real (it's resistance in magnetic fields and acts like electrical reistance opposes current flow.....in this case to opposes magnetic field strength).

That question you originally asked 'Where does the energy go ?' has two answers.

1.) Be converted in mechanical energy to move an object via electromagnetic means.

2.) Losses caused by an in-perfect world of friction, resistance, reluctance, etc.,.

Dam, it now I feel need a dam physics teacher to come on board. :laugh:

Brick On anyway ! :wink:

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