Left Hand Rule Applied to a Winding

[maxwells_demon]

yup. next, set the rotor as in the second position, but with shaft in line with the needle center. slowly turn the shaft while keeping this position. is there any kind of response from your compass?

 

[jake jot]

yup. next, set the rotor as in the second position, but with shaft in line with the needle center. slowly turn the shaft while keeping this position. is there any kind of response from your compass?

Yes, it rotates. I think the magnetic configuration is like this.

Thanks :)

 

[jake jot]

Yes, it rotates. I think the magnetic configuration is like this.

View attachment 273903

Thanks :)

So the DC brushless motor is a perfect example of motor that runs by pole attraction and absolutely not by force exerted by current carrying conductor, isn't it?

All drones are powered by DC brushless motor so pole attraction can be quite strong.

 

[maxwells_demon]

Yes, it rotates. I think the magnetic configuration is like this.

View attachment 273903

Thanks :)

ayt, nice one mate.

 

[jake jot]

hope somebody chimes in for how to get those 3-phase voltages from your BLDC, I'm almost at the limits of my practical knowledge.

Your last sentence above was puzzling. Isn't you just explained in the same message that:

let's address first why this is considered a 3-phase motor. if you watch the video, you will come to this point, and why this sequence of energizing the coils is helpful in the smooth operation of the BLDC. despite being a dc motor, the controller circuit that supplies the BLDC feeds it 3 distinct periodic voltage waveforms. these waveforms are similar, with the same amplitude and the same period, the only difference is WHEN each is triggered. it's like the 3-phase voltage we get from the grid (i'll assume you know what a 3-phase sinusoidal waveform is), but instead of sinusoids, BLDC uses something like a two-polarity (+V and -V) square-wave (the appropriate technical term might be a pulse-width modulated signal).

The above is how to get those 3-phase voltages from your BLDC, isn't it?

 

[maxwells_demon]

So the DC brushless motor is a perfect example of motor that runs by pole attraction and absolutely not by force exerted by current carrying conductor, isn't it?

All drones are powered by DC brushless motor so pole attraction can be quite strong.

indeed. all thanks to stronger permanent magnets (those neodymium magnets are fun to play with, just be careful not to hurt yourself), low reluctivity pole shoes, and precise control circuits, all of which are great for miniaturized/small-scale applications.

 

[maxwells_demon]

Your last sentence above was puzzling. Isn't you just explained in the same message that:
The above is how to get those 3-phase voltages from your BLDC, isn't it?

try and see how the solenoid is wound, and you'll notice you'll run into the exact same question that made you start this thread in the first place.

edit: albeit this time, you'll be dealing with voltage polarity instead of direction of force.

 

[jake jot]

try and see how the solenoid is wound, and you'll notice you'll run into the exact same question that made you start this thread in the first place.

edit: albeit this time, you'll be dealing with voltage polarity instead of direction of force.

What? your youtube example on top is the same as the amazon motor at bottom, the only difference is the amazon motor has 9 windings. What's wrong?

 

[maxwells_demon]

What? your youtube example on top is the same as the amazon motor at bottom, the only difference is the amazon motor has 9 windings. What's wrong?

for just a while, let's limit our analysis to one stator pole, as follows:

with a rotor rotation direction given by the yellow counter-clockwise arrow, and each single coil of your solenoid having its return path being subjected to the same direction of flux, at the same relative rotational speed and direction, how much voltage would you measure at the terminals of this solenoid?

 

[agrumpyoldphysicstec]

TBH I've no idea how to analyse this sort of rotor, with an iron cored rotor and shaped field poles. You probably need and electrical engineer for that. (I expect Jim Hardy would have known). I just knew the basic principle of the motor with wires in a uniform field, as taught in Physics textbooks. After that I just put my faith in Eric Laithwaite's principle that the more iron and copper, the more efficient the motor! It's actually always puzzled me that the wires in most motors don't seem to be in any magnetic field: nearly all the flux seems to go through the iron, bypassing the wires. The magnetic flux "links" the coils and I've seen analysis of transformers based on that idea, but not motors. It's probably my bad understanding of field and flux.

You will not be the first to notice that electrical engineers design motors so that the conductors lie in regions of little flux. Also, the force on the conductors is small. If you think that you have a bad understanding it is because the subject is so badly taught. I have school physics textbooks and university electrical engineering textbooks and they do not agree. The university textbooks agree with observation whereas the school ones do not and also disagree with the established laws of physics.

It is fairly easy to analyze motors and generators as well as transformers in terms of flux "linking". Fleming's rules of thumb deal only with flux "cutting" which is really just a special case.

Is this a motor, or a generator, or a transformer?

 

[jake jot]

for just a while, let's limit our analysis to one stator pole, as follows:
View attachment 273906

with a rotor rotation direction given by the yellow counter-clockwise arrow, and each single coil of your solenoid having its return path being subjected to the same direction of flux, at the same relative rotational speed and direction, how much voltage would you measure at the terminals of this solenoid?

I don't get what you mean. Hope others can chime in or answer.

How does it differ to the amazon motor anyway? Did you mean it can't produce AC?

Also I noticed a great symmetry, in which motors can be converted to generator and vice versa. If one can (not practically but conceptually) rotate every motor by hand, one can get induced voltage at the plug? What motors are exception to this, perhaps the AC induction motors?

I wonder if MacGyver has episodes where he was trapped in a cellar and there was an old motor and he used it to produce electricity?

 

[Averagesupernova]

I happened to have a small Coleman generator apart today. The stupid thing is made so everything is impossible to get at unless a significant part of it is disassembled. That aside, I took a few pix. Thought I'd post them here. Generator is 120 volts 1850 watts max. Probably 1200 to 1500 continuous. Notice what I assume is a pair of diodes on the rotor. Also what I believe to be MOVs. This unit also has a DC output for automotive battery charging. Good for about 15 amps I believe. Two different sized windings are visible on the stator. One being the 120 VAC and the other the battery charging winding I have to assume. I would have looked it over closer but I'm in a hurry to use it.

 

[jake jot]

It will. but do not connect the battery directly to the generator terminals. You might burn those windings. At least put a resistor in the circuit. Start with large value resistance and wattage, then try next lower resistance values if the setup doesn't start.

If you could determine the size of the magnetic wire they used for winding that generator and find out its ampacity, it would help lots with determining the resistance to put in place, given standard battery voltages.

Here is the amazon generator. It's much smaller than it looks.

Amazon.com: EUDAX School DIY Dynamo Lantern Educational STEM Building ,Labs Demonstration Motor Activity Teaching Model Hand Cranked Power Electricity DC Electric Generator Physical Science Experiment Education: Toys & Games

The length from front to back is about 6.5 inches. The magnetic wire diameter using the caliper is about 0.20mm (0.0078 inches).

The size of the magnet (solid red and blue either side center) is about 2 inches long, the white winding plastic is also 2 inches long, and 1.5 inches wide.

Using a multimeter, the current is 10 milliampere at the fastest i can rotate it using the hand crank. The load is one LED for either direction of the hand crank directions. I don't know how many numbers of windings in the rotor. But the resistance between terminals or the winding is 26 ohms.

Using the above data. Any idea what kind of battery (3 volts? 6 volts? ) and resistor to turn it into a motor? I'd remove the hand crank to avoid more resistance in the rotor. Thanks!

(If others know. Kindly share too as maxwell_demons logs in rarely now)

 

[jake jot]

Yeah, my bad. Sophie and Merlin are right on this one. That little amazon motor is still unidirectional, contrary to what I said earlier.

You might want to chech the rotor. I think I see something like a metal strip on its end, this might be the outer end of our "iron core" block.

Such a metal block would be unnecessary for a "force on a current carrying conductor" motor. I should know. I built a simple two-pole dc motor for a friend's child's high school project using permanent magnets for the stator, with the rotor frame of just barbecue sticks. And some magnetic wire with those thin insulation varnish as the winding. Worked just fine.

Nothing beats building from scratch to see for yourself, or taking things apart. Lol.

I got 2 amazon motors unassembled. The top is the 2nd unassembled unit. After getting it to run. I noticed it won't run if it is in the horizontal positions. This was explained in the thread.

Yes the rotor seems to have some "iron core" block at the sides. Why is it necessary? Can't you create solenoid just by winding wires? The wire size is 0.34mm (0.0139").

The reason i got 2 pcs is so I can re create the rotor to make it so slim that force on current carrying conductor is totally eliminated. I want to see for myself the idea of pole attraction and how it is enough to make the motor run. So what size of wire to wind it and how many turns? And can't it happen without the "iron core" block?

(if maxwell_demons didn't come back, hope others masters in motors can answer. Thanks).

He wrote in message #43 this:

I'm willing to bet, if you could detach that solenoid and stuck a shaft through its center coming from the same direction as the motor shaft, that solenoid would twitch due to the torque of those magenta arrows.

If someone understands what he means. Please reword it. What does it mean to stuck a shaft through its center in the rotor picture above? I want to see it twitch (or whatever it is that can twitch)

 

[jake jot]

Here is the amazon generator. It's much smaller than it looks.

Amazon.com: EUDAX School DIY Dynamo Lantern Educational STEM Building ,Labs Demonstration Motor Activity Teaching Model Hand Cranked Power Electricity DC Electric Generator Physical Science Experiment Education: Toys & Games

View attachment 274330

Comparing the amazon motor and generator. The generator has smaller wires in the winding. I guess they don't want large current in the output to avoid shocking the kids? This is the amazon generator disassembed showing the rotor with thin wires.

Tomorrow I'll try turning the generator into motor by inputting power to the output of the above. I'll start with 1.5v, 3v, 6v. But I think the wires may burn at 6v. Smaller current and there may not be enough force and torque on the conductor carrying wires to move the rotor.

So I'll change the winding wires sizes and turns. Hence I need the cookbook how to design rotor and number of turns (should one count?) and size of winding for the given stator. I'll try up to 12v dc only and see how fast the rotor can rotate. Does this need Maxwell equations? Gimme some tips guys or sample computations. Thanks.

 

[Averagesupernova]

You need to find a way to measure the current drawn by your experimental motors. An assortment of series resistors with a good auto ranging DVM should do the trick.

 

[jake jot]

Comparing the amazon motor and generator. The generator has smaller wires in the winding. I guess they don't want large current in the output to avoid shocking the kids? This is the amazon generator disassembed showing the rotor with thin wires.

View attachment 274433

Tomorrow I'll try turning the generator into motor by inputting power to the output of the above. I'll start with 1.5v, 3v, 6v. But I think the wires may burn at 6v. Smaller current and there may not be enough force and torque on the conductor carrying wires to move the rotor.

So I'll change the winding wires sizes and turns. Hence I need the cookbook how to design rotor and number of turns (should one count?) and size of winding for the given stator. I'll try up to 12v dc only and see how fast the rotor can rotate. Does this need Maxwell equations? Gimme some tips guys or sample computations. Thanks.

I was able to get the amazon generator to run as a motor. I added 1.2v battery one by one. At about 6 volts, the LED lights burned up (do all LED uses the same voltage?). The purpose of the LED lights are just to show directions of the hand crank and current. After adding 6 batteries or 7.2 volts. It still didn't run. It was only when I turned the unit sideways that the rotor can rotate (see above video).

I removed the leads less than 3 seconds as I didn't want the winding to burn up. I want to know the following:

1. Can the winding be just a single conductor? What's the problem with single conductor?

2. What is the formula for numbers of winding and size? Is there a formula or people just do it by trial and error?

3. I'll only remove the windings (destroyed it) and replace with a single conductor if there is theoretical reason the conductor can work too.

4. This proves generator and motor has relationship.

 

[Tom.G]

An LED lights up when the applied voltage is around 2V to 2.5V, and you already measured the current those LEDs draw. So when you crank the generator to light an LED, the generator is putting out around 2.2V at the current you measured.

Now a funny thing about LEDs is that they keep about the same voltage across them. If you apply a higher voltage they will conduct ALOT more current. Eventually they get hot enough to self-destruct internally.

The first stage of destruction is they melt and become almost a short circuit. At this point they no longer produce any light.

The second stage of destruction, if there is enough current available, one of the internal connecting wires will melt, acting like a fuse. Then no current will flow thru them.

From your description, the LEDs are in the first stage of destruction, an almost short circuit.

With the LEDs shorted, little current is available to flow thru the motor. I suspect that's why the motor most doesn't turn when connected to batteries.

Try disconnecting the LEDs by cutting the wire to them then see what the motor does with voltage applied. It just may work with only 1 or 2 batteries! More batteries = higher voltage = faster motor and more current - - - until it overheats and dies.

Have Fun!
Tom

 

[jake jot]

An LED lights up when the applied voltage is around 2V to 2.5V, and you already measured the current those LEDs draw. So when you crank the generator to light an LED, the generator is putting out around 2.2V at the current you measured.

Now a funny thing about LEDs is that they keep about the same voltage across them. If you apply a higher voltage they will conduct ALOT more current. Eventually they get hot enough to self-destruct internally.

The first stage of destruction is they melt and become almost a short circuit. At this point they no longer produce any light.

The second stage of destruction, if there is enough current available, one of the internal connecting wires will melt, acting like a fuse. Then no current will flow thru them.

From your description, the LEDs are in the first stage of destruction, an almost short circuit.

With the LEDs shorted, little current is available to flow thru the motor. I suspect that's why the motor most doesn't turn when connected to batteries.

Try disconnecting the LEDs by cutting the wire to them then see what the motor does with voltage applied. It just may work with only 1 or 2 batteries! More batteries = higher voltage = faster motor and more current - - - until it overheats and dies.

Have Fun!
Tom

The purpose of the leds is to show that turning the hand crank one way or another makes different lights in the led

What do you call this circuit?

I also cut the wire and measured it with multimeter. The leds are now open (not shorted anymore)

Earlier I think your scenario happened where the leds were in short circuit, so the motor didn't run. I'm recharging the batteries to see how many it can run.

I'll keep adding batteries until the rotor spins so fast and the thin wires burn. But before I do this. I ought to know. Why is the motor rotor composed of hundreds of winds, why not just one big conductor? In a transformer, the purpose of multiple windings may be to give relations between primary and secondary. But in a motor, what is the advantage of such windings? Will using just one big wire cause the same magnetic field? Or somehow hundreds of tiny wires can create greater magnetic field than one single whole conductor with the same area? Why so? Thanks!

 

[Tom.G]

What do you call this circuit?

Two LEDs connected in antiparallel.

Why is the motor rotor composed of hundreds of winds, why not just one big conductor?

As a general approximation, the strength of the magnetic field is proportional to the product of current times turns, called Ampere-Turns. You would need big batteries and larger wire connecting them if you want few turns and many amps.

 

[jake jot]

Two LEDs connected in antiparallel.As a general approximation, the strength of the magnetic field is proportional to the product of current times turns, called Ampere-Turns. You would need big batteries and larger wire connecting them if you want few turns and many amps.

Are the formulas for transformer windings the same as the winding in motors? I'm googling about the formulas and I read about inductance. Maybe the single conductor won't work because it would just short the batteries? So all these illustrations in youtube won't work at all because they would short?

Also I tried to determine the force of the rotors using the left hand rule. I can point my right hands better. That's when I read the rule for generator is difference. I read this:

"Fleming's two rules are very similar, but they must not be confused. When a current causes motion (like in a d.c. motor), Fleming's left hand rule applies. When motion causes current (like in a generator), Fleming's right hand rule applies. Since these two phenomena are the exact opposite from each other, it makes sense that direction of current would be opposite from each other. If the rules were not opposite then we could easily make electrically-driven perpetual motion machines. "

Can you give illustration how if the rules were not opposite, we could easily make electrically-driven perpetual motion machines? Just to understand the concept of the oppositeness.

 

[jake jot]

Are the formulas for transformer windings the same as the winding in motors? I'm googling about the formulas and I read about inductance. Maybe the single conductor won't work because it would just short the batteries? So all these illustrations in youtube won't work at all because they would short?

View attachment 274510

Also I tried to determine the force of the rotors using the left hand rule. I can point my right hands better. That's when I read the rule for generator is difference. I read this:

"Fleming's two rules are very similar, but they must not be confused. When a current causes motion (like in a d.c. motor), Fleming's left hand rule applies. When motion causes current (like in a generator), Fleming's right hand rule applies. Since these two phenomena are the exact opposite from each other, it makes sense that direction of current would be opposite from each other. If the rules were not opposite then we could easily make electrically-driven perpetual motion machines. "

Can you give illustration how if the rules were not opposite, we could easily make electrically-driven perpetual motion machines? Just to understand the concept of the oppositeness.

To add. Here is some weird results.

I tried to add all my 8 rechargeable batteries (1.2 x 8 = 9.8v) (all newly charged) waiting for the rotor winding to burn (so I can try winding new one for once in my life). But the motor kept running. I measured the amperage in series to the 8 batteries and it reads 1A average. But when I used the hand crank turning the same unit into generator (original design), the output reads only at most 10mA (with the 2 leds removed). And voltage is only 0.5v! I can't even get it above 1v. How do you explain this?

Is it not if the rotor is a certain given speed. It should output the same voltage or current as when you use the battery of same voltage and current to turn the rotor (at same speed)?The wiring diameter of the rotor is about 0.20mm (0.0078 inches). Again this is how it looks like. How do you know the maximum battery voltage it can take? I no longer have extra batteries to test winding burnout voltage.

 

[Merlin3189]

First, your LEDs are diodes. Each allows the current in one direction only. So depending on the polarity of the supply, either red or green lights
.

It was only when I turned the unit sideways that the rotor can rotate (see above video).
Maybe less friction??
I removed the leads less than 3 seconds as I didn't want the winding to burn up. I want to know the following:

1. Can the winding be just a single conductor? .
What's the problem with single conductor?

2. What is the formula for numbers of winding and size? Is there a formula or people just do it by trial and error?

3. I'll only remove the windings (destroyed it) and replace with a single conductor if there is theoretical reason the conductor can work too.

4. This proves generator and motor has relationship.

1 - Yes, maybe
- You have to put a lot of current through it. If you have 100 turns with 10 mA, that's equivalent to 1 turn with 1000 mA. If it took say 100 mA with 100 turns, then you'd need to put 10 A through a single turn.

2 - You'd need to know the strength of the magnetic field in order to calculate the torque. But the force is proportional to the length of the side parallel to the axis and to the aggregate current in the winding, assuming it's all in the same field (With a fixed size magnet, you don't get more force once your coil is as wide as the magnet.)

3 - Usually we prefer more turns of thinner wire. If you halve the CS area of the wire, have double the number of turns and halve the current, then you get the same force. Resistance is double, current is half, so voltage is the same, but power lost to resistance is less, because it's $I^2R$ or $\frac V R$

4 - Abso-blooming-lutely! The motor IS a generator, even when you are using it as a motor. That's where the "back emf" comes from. That's why the voltage determines the speed - when the generator voltage (plus resistance voltage) equals the battery voltage, the battery can't push more current through, so torque can't increase and it can't accelerate. (A bit over simple, but more or less.)Edit: changed layout as it didn't look ok when posted.

 

[jake jot]

First, your LEDs are diodes. Each allows the current in one direction only. So depending on the polarity of the supply, either red or green lights
.View attachment 274565

1 - Yes, maybe
- You have to put a lot of current through it. If you have 100 turns with 10 mA, that's equivalent to 1 turn with 1000 mA. If it took say 100 mA with 100 turns, then you'd need to put 10 A through a single turn.

2 - You'd need to know the strength of the magnetic field in order to calculate the torque. But the force is proportional to the length of the side parallel to the axis and to the aggregate current in the winding, assuming it's all in the same field (With a fixed size magnet, you don't get more force once your coil is as wide as the magnet.)

3 - Usually we prefer more turns of thinner wire. If you halve the CS area of the wire, have double the number of turns and halve the current, then you get the same force. Resistance is double, current is half, so voltage is the same, but power lost to resistance is less, because it's $I^2R$ or $\frac V R$

4 - Abso-blooming-lutely! The motor IS a generator, even when you are using it as a motor. That's where the "back emf" comes from. That's why the voltage determines the speed - when the generator voltage (plus resistance voltage) equals the battery voltage, the battery can't push more current through, so torque can't increase and it can't accelerate. (A bit over simple, but more or less.)Edit: changed layout as it didn't look ok when posted.

Thanks. I made this video and thinking whether to include it in the amazon review of the product. None mentioned it can be turned into motor.



Many customers are children. And maybe for safety. I'd not post it. So don't make the above public.

Here is the mystery. Putting a multimeter in series to the batteries, the current of the above setup is 1A. But using the hand crank, no matter what rotations, it can't output more than 10mA and 0.5volts. If you will notice in the video, the rotations are not that fast. Well. Even if it's fast when put sideways to avoid friction. The handcrank can make it as fast. Is it possible or the case that in a generator. You need more input power to make it match the voltage and current you would use when using it as motor?

 

[Merlin3189]

First - don't worry about children. Using it as a generator they can't get a significant voltage, so no danger.

Putting a multimeter in series to the batteries, the current of the above setup is 1A. But using the hand crank, no matter what rotations, it can't output more than 10mA and 0.5volts.

The generator emf should depend only on speed.
When using it as a motor, the applied voltage has to match the generator emf plus the Ohmic emf caused by the current passing through the windings (and other bits.)
Looking at the 1 A current with 9.8 V battery, the resistance (windings, connections, battery internal resistance) is less than 9.8 Ohm. How much less depends on the motor speed. Maybe your multimeter could measure the resistance at zero speed?

Looking at the open circuit voltage of 0.5 V at max cranking speed, the short circuit current is 10 mA, which implies a resistance of $\frac {0.5}{0.01) = 50 Ohm$ , which includes the winding, contacts and ammeter.
So, a big discrepancy, for which I have to guess at possible causes.
Is it rotating at the same speed when measuring voltage and when measuring current? There is no load other than friction, when measuring o/c voltage. When measuring s/c current, there will be a load torque added, so it may be going slower and the voltage is really less than the 0.5 V max. But 5 times slower ?
Hand cranking probably gives a varying reading, which will be harder to take accurately.
Is any current still going through the dead LEDs? (But I think you've taken them off.)

As far as I can see, 9.8 Ω must be greater than the winding resistance, so something must be wrong with the generator measurement.

 

[jake jot]

First - don't worry about children. Using it as a generator they can't get a significant voltage, so no danger.

The generator emf should depend only on speed.
When using it as a motor, the applied voltage has to match the generator emf plus the Ohmic emf caused by the current passing through the windings (and other bits.)
Looking at the 1 A current with 9.8 V battery, the resistance (windings, connections, battery internal resistance) is less than 9.8 Ohm. How much less depends on the motor speed. Maybe your multimeter could measure the resistance at zero speed?

Looking at the open circuit voltage of 0.5 V at max cranking speed, the short circuit current is 10 mA, which implies a resistance of $\frac {0.5}{0.01) = 50 Ohm$ , which includes the winding, contacts and ammeter.
So, a big discrepancy, for which I have to guess at possible causes.
Is it rotating at the same speed when measuring voltage and when measuring current? There is no load other than friction, when measuring o/c voltage. When measuring s/c current, there will be a load torque added, so it may be going slower and the voltage is really less than the 0.5 V max. But 5 times slower ?

What is meaning of o/c voltage? overclock? or open circuit voltage? or over charge? Also what is the s/c in "s/c current", shut current? step current?

I measured the resistance of the winding of the amazon generator by directly tapping the leads at the 2 commutators. It is exactly 24 ohms. I put the multimeter in AC instead of DC and tried hand cranking it so fast i don't mind if the rubber band would snap and highest I can get is 1.2 volts and 25mA. If it is set in DC, the voltage highest is 0.7 volts and 20mA. Why does putting multimeter in AC gives higher voltage?

Whereas the voltage (in generator mode) that can even make the rotor rotates by putting it sideways (to avoid friction) should be at least 7 volts dc and it's not rotating as fast. Why couldn't I get it to rotate at 1.2 volts? (i put it sideway to lessen friction from bottom).

What I exactly want to know is this principle. Whether if the rotor at motor mode rotates at a certain RPM at a given voltage and current. Whether if the rotor is put in generator mode. It will output the same voltage and current at the same RPM. What do you think?

Hand cranking probably gives a varying reading, which will be harder to take accurately.
Is any current still going through the dead LEDs? (But I think you've taken them off.)

As far as I can see, 9.8 Ω must be greater than the winding resistance, so something must be wrong with the generator measurement.

 

[jake jot]

What is meaning of o/c voltage? overclock? or open circuit voltage? or over charge? Also what is the s/c in "s/c current", shut current? step current?

I measured the resistance of the winding of the amazon generator by directly tapping the leads at the 2 commutators. It is exactly 24 ohms. I put the multimeter in AC instead of DC and tried hand cranking it so fast i don't mind if the rubber band would snap and highest I can get is 1.2 volts and 25mA. If it is set in DC, the voltage highest is 0.7 volts and 20mA. Why does putting multimeter in AC gives higher voltage?

Whereas the voltage (in generator mode) that can even make the rotor rotates by putting it sideways (to avoid friction) should be at least 7 volts dc and it's not rotating as fast. Why couldn't I get it to rotate at 1.2 volts? (i put it sideway to lessen friction from bottom).

What I exactly want to know is this principle. Whether if the rotor at motor mode rotates at a certain RPM at a given voltage and current. Whether if the rotor is put in generator mode. It will output the same voltage and current at the same RPM. What do you think?

I wonder if it has to do with the split commutator. Mine is the 2nd one.

13.7: Electric Generators and Back Emf - Physics LibreTexts

How do you turn a pulsed dc to direct dc? Maybe this was why I was only measuring 1.2 volts at most? (when in motor mode, the voltage that can get the rotate to start rotating must be at least 7 volts?)

 

[jake jot]

I wonder if it has to do with the split commutator. Mine is the 2nd one.

13.7: Electric Generators and Back Emf - Physics LibreTexts

View attachment 274632View attachment 274633

How do you turn a pulsed dc to direct dc? Maybe this was why I was only measuring 1.2 volts at most? (when in motor mode, the voltage that can get the rotate to start rotating must be at least 7 volts?)

Could the reason be the so called back emf? that is.. in motor, you need larger voltage to make it run and not proportional to the voltage in the output when it is in generator mode? I read in article above:

"
The generator output of a motor is the difference between the supply
voltage and the back emf. The back emf is zero when the motor is first turned on, meaning that the coil receives the full driving voltage and the motor draws maximum current when it is on but not turning. As the motor turns faster, the
back emf grows, always opposing the driving emf, and reduces both the
voltage across the coil and the amount of current it draws. This effect is noticeable in many common situations. When a vacuum cleaner, refrigerator
, or washing machine is first turned on, lights in the same circuit dim briefly due to the IR drop produced in feeder lines by the large current drawn by the motor. "

 

[Tom.G]

The OP repeatd this query to me in a private msg. I chose to address it here publically.

Is it not if the rotor is a certain given speed. It should output the same voltage or current as when you use the battery of same voltage and current to turn the rotor (at same speed)?

When measuring as a motor and as a generator they should be measured at the same motor current. Under that condition they should be close to the same. If measured as a generator with little or no load they will be different. The lower voltage and current as generator indicate other, unknown, problems somewhere in the test setup.

Looking at the 1 A current with 9.8 V battery, the resistance (windings, connections, battery internal resistance) is less than 9.8 Ohm. How much less depends on the motor speed. Maybe your multimeter could measure the resistance at zero speed?

As far as I can see, 9.8 Ω must be greater than the winding resistance, so something must be wrong with the generator measurement.

WIth the stated wire size of 0.0078 inches, 9.8 Ohms indicates about 50 feet of wire, the 24 Ohms measured by the OP indicates about 130 to 150 feet on the winding. Bad commutator contact?

Overall, the various measurement inconsistencies point to both an instrumentation problem, such as an average-reading (or maybe peak-reading) meter being used for a pulsed measurement, and probably poor, bouncing, contact between the commutator and the brushes.

I suggest the OP find someone locally that can work on the actual device with him and may have more appropriate equipment and more hands-on experience. Trying to debug convoluted problems remotely with limited information is not very practical.

 

[jake jot]

The OP repeatd this query to me in a private msg. I chose to address it here publically.When measuring as a motor and as a generator they should be measured at the same motor current. Under that condition they should be close to the same. If measured as a generator with little or no load they will be different. The lower voltage and current as generator indicate other, unknown, problems somewhere in the test setup.WIth the stated wire size of 0.0078 inches, 9.8 Ohms indicates about 50 feet of wire, the 24 Ohms measured by the OP indicates about 130 to 150 feet on the winding. Bad commutator contact?

Overall, the various measurement inconsistencies point to both an instrumentation problem, such as an average-reading (or maybe peak-reading) meter being used for a pulsed measurement, and probably poor, bouncing, contact between the commutator and the brushes.

I suggest the OP find someone locally that can work on the actual device with him and may have more appropriate equipment and more hands-on experience. Trying to debug convoluted problems remotely with limited information is not very practical.

Please see message #97 and #98. Is it not caused by back EMF? Meaning you need to use larger voltage when used as generator to counter any back EMF?

 

[jake jot]

The OP repeatd this query to me in a private msg. I chose to address it here publically.When measuring as a motor and as a generator they should be measured at the same motor current. Under that condition they should be close to the same. If measured as a generator with little or no load they will be different. The lower voltage and current as generator indicate other, unknown, problems somewhere in the test setup.WIth the stated wire size of 0.0078 inches, 9.8 Ohms indicates about 50 feet of wire, the 24 Ohms measured by the OP indicates about 130 to 150 feet on the winding. Bad commutator contact?

Overall, the various measurement inconsistencies point to both an instrumentation problem, such as an average-reading (or maybe peak-reading) meter being used for a pulsed measurement, and probably poor, bouncing, contact between the commutator and the brushes.

I suggest the OP find someone locally that can work on the actual device with him and may have more appropriate equipment and more hands-on experience. Trying to debug convoluted problems remotely with limited information is not very practical.

Reading more about Back EMF. I think it's not the cause of the motor setup of the amazon generator requiring more voltage and current. Isn't it. Maybe it's the commutator contacts. I used a load that should light up if 3v or more and it still could never light up.

It's not good idea to bring the amazon generator for kid into a university and ask the professor when I wasn't enrolled in the school. And technician at the washing machine motor rewinding shop may not want to look at my little amazon motor/generator or the power company.

Anyway I just want to understand the basic concepts and I think it's all enough for me.

One last question I need to know to conclude the thread is the following.

In pole attraction motor like brushless motors or the amazon motor where the solenoid with iron core creates north-south poles. So the purpose of commutator is to make sure it is north or south at one side so the rotor continues to rotate. I understand this concept well. And the purpose of the current is to produce the poles electromagnetically and make it switch or maintain north, south via commutator. I need to know the following thought experiment to make sure I understand the basic concept.

Supposed you had a pole attraction motor where the rotor has one end a monopole that switches north and south, and the other end also a monopole that switches north and south opposite to the other. Will the motor runs continuously without any current? So it will keep rotating because of Newton's 1`st law which states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it?

Kindly answer the above and not say it is bad example. Because this whole thread is to understand the above thought experiment. Thank you.

 

[Merlin3189]

Sorry, just saving typing.
o/c = open circuit. ie. negligible current, only voltage. Voltmeter has very high resistance, 20 kΩ - 10 MΩ
s/c = short circuit. ie negligible voltage, only current. Ammeter has very low resistance.

"I measured the resistance of the winding of the amazon generator by directly tapping the leads at the 2 commutators. It is exactly 24 ohms."
So this should be correct and your other measurements must have faults.

"Why does putting multimeter in AC gives higher voltage?"
Measuring DC is easy: there is only one value.
AC is more confusing: you could measure the peak value, the RMS value (which is equivalent to DC value in resistive circuit), or average value (average modulus, otherwise it would be zero for pure AC!) It gets even more difficult when the voltage is a mixture of AC and DC. What you see depends on your meter. Many meters give an average, but modern electronic meters can do RMS (and maybe peak as well.)
But, for all that, I can't see how you get the values you do.

"What I exactly want to know is this principle. Whether if the rotor at motor mode rotates at a certain RPM at a given voltage and current. Whether if the rotor is put in generator mode. It will output the same voltage and current at the same RPM. What do you think?"

This is a model circuit for a DC motor. There is a source of emf, whose value depends on the speed of rotation and the motor constant - a function of the stator magnetic field and rotor windings.
At a given speed the emf is determined, kω. The current is then determined by the internal resistance R and the external load resistance. The voltage you measure at the output terminals or across the load will be less than Bemf by the voltage across the internalresistance R. If there is no load and no current, then you measure the Bemf.

When you drive it as a motor, the current is determined by the resistance, the applied voltage and the speed that it rotates - which will depend on the mechanical load and friction.

Say, at 1000 rpm the emf is 5 V (and R is 24 Ω as you measured) then attaching a 26 Ω load will give a current of $\frac {5 V}{24 +26 Ω } = 0.1 A$ with $0.1 A \times 26 Ω = 2.6 V$ across the load.

To drive it as a motor you will need to apply more than 5 V to get that same 1000 rpm, because you need some current to give the torque to overcome friction.
If you apply 5 V, then $\frac {5V}{24 Ω } = 208 mA$ will flow (neglecting battery resistance.)
If the torque that generates is enogh to overcome friction, then the rotor will accelerate.
That will cause some Bemf which will reduce the current. For eg. if the rotor got up to 900 rpm, the Bemf will be 4.5 V, so the current will drop to $\frac {5-4.5 V}{24 Ω } = 21 mA$
If that is just enough to overcome friction at that speed, then that's the speed it will run at. If it isn't, then it won't ever reach that speed.

If for eg. it needs about 25 mA to have the torque needed to overcome friction, that requires a voltage across R of $0.025 A \times 24 Ω = 0.6 V$
So with 5 V battery the Bemf can rise to 4.4 V , leaving 0.6 V across R. But Bemf = 4.4 V means it is rotating at $1000 \times \frac {4.4}{5} = 880 rpm$

If you now add a mechanical load, that requires extra torque, which requires extra current.
Say eg. the extra load requires a current of 100 mA. Now the total required current is 125 mA to give the torque needed for the load and friction.
That requires a voltage across R of $0.125 A \times 24 Ω = 3 V$
So now we can afford only 2 V of Bemf and the motor must slow to $1000 \times \frac{2}{5) = 400 rpm$

If we wanted the motor to run at 1000 rpm with that load, then we need our 125 mA current to provide the correct torque and enough voltage to overcome the 5 V Bemf. So with 125 mA needing 3 V across R, we need a supply of 8 V.

That's probably laboured the point a bit, but I hope it shows the nature of the model and the source of the values.

 

[jake jot]

Sorry, just saving typing.
o/c = open circuit. ie. negligible current, only voltage. Voltmeter has very high resistance, 20 kΩ - 10 MΩ
s/c = short circuit. ie negligible voltage, only current. Ammeter has very low resistance.

"I measured the resistance of the winding of the amazon generator by directly tapping the leads at the 2 commutators. It is exactly 24 ohms."
So this should be correct and your other measurements must have faults.

"Why does putting multimeter in AC gives higher voltage?"
Measuring DC is easy: there is only one value.
AC is more confusing: you could measure the peak value, the RMS value (which is equivalent to DC value in resistive circuit), or average value (average modulus, otherwise it would be zero for pure AC!) It gets even more difficult when the voltage is a mixture of AC and DC. What you see depends on your meter. Many meters give an average, but modern electronic meters can do RMS (and maybe peak as well.)
But, for all that, I can't see how you get the values you do.

"What I exactly want to know is this principle. Whether if the rotor at motor mode rotates at a certain RPM at a given voltage and current. Whether if the rotor is put in generator mode. It will output the same voltage and current at the same RPM. What do you think?"

View attachment 274676This is a model circuit for a DC motor. There is a source of emf, whose value depends on the speed of rotation and the motor constant - a function of the stator magnetic field and rotor windings.
At a given speed the emf is determined, kω. The current is then determined by the internal resistance R and the external load resistance. The voltage you measure at the output terminals or across the load will be less than Bemf by the voltage across the internalresistance R. If there is no load and no current, then you measure the Bemf.

When you drive it as a motor, the current is determined by the resistance, the applied voltage and the speed that it rotates - which will depend on the mechanical load and friction.

Say, at 1000 rpm the emf is 5 V (and R is 24 Ω as you measured) then attaching a 26 Ω load will give a current of $\frac {5 V}{24 +26 Ω } = 0.1 A$ with $0.1 A \times 26 Ω = 2.6 V$ across the load.

To drive it as a motor you will need to apply more than 5 V to get that same 1000 rpm, because you need some current to give the torque to overcome friction.
If you apply 5 V, then $\frac {5V}{24 Ω } = 208 mA$ will flow (neglecting battery resistance.)
If the torque that generates is enogh to overcome friction, then the rotor will accelerate.
That will cause some Bemf which will reduce the current. For eg. if the rotor got up to 900 rpm, the Bemf will be 4.5 V, so the current will drop to $\frac {5-4.5 V}{24 Ω } = 21 mA$
If that is just enough to overcome friction at that speed, then that's the speed it will run at. If it isn't, then it won't ever reach that speed.

If for eg. it needs about 25 mA to have the torque needed to overcome friction, that requires a voltage across R of $0.025 A \times 24 Ω = 0.6 V$
So with 5 V battery the Bemf can rise to 4.4 V , leaving 0.6 V across R. But Bemf = 4.4 V means it is rotating at $1000 \times \frac {4.4}{5} = 880 rpm$

If you now add a mechanical load, that requires extra torque, which requires extra current.
Say eg. the extra load requires a current of 100 mA. Now the total required current is 125 mA to give the torque needed for the load and friction.
That requires a voltage across R of $0.125 A \times 24 Ω = 3 V$
So now we can afford only 2 V of Bemf and the motor must slow to $1000 \times \frac{2}{5) = 400 rpm$

If we wanted the motor to run at 1000 rpm with that load, then we need our 125 mA current to provide the correct torque and enough voltage to overcome the 5 V Bemf. So with 125 mA needing 3 V across R, we need a supply of 8 V.

That's probably laboured the point a bit, but I hope it shows the nature of the model and the source of the values.

Thanks for the details.

One full hand crank revolution at bottom is equal to 5 rotor full turn. How do you compute the RPM based on a normal hand effort to turn it? (anyway I measured 2 revolution per sec tops when I turned the hand crank). So in one minute there are 2 x 60 x 5 = 600 RPM or make it 500 RPM since I can't maintain full 2 revolution per sec in one minute continuously.)

The winding resistance is 24 ohms, and it measured 1 A when connected to 9.8v battery in motor mode. What should be the output in generator mode based on these data? Assuming there is no load in either the generator and motor mode (except the meter and batteries).

 

[Merlin3189]

V=IR so 9.8 V = 1 A x 24 Ω means something is not right.
It should need 24 V to drive 1 A through 24 Ω resistance, even when it's not moving. When it moves, it should require even more voltage.

As someone else said, it's hard to guess which bit is wrong when you're not there to see exactly what's going on. My guess would be to distrust the resistance reading. Have you checked the meter shows zero when you short the test leads together?
Otherwise, the 1 A has to be too high.

 

[jake jot]

V=IR so 9.8 V = 1 A x 24 Ω means something is not right.
It should need 24 V to drive 1 A through 24 Ω resistance, even when it's not moving. When it moves, it should require even more voltage.

As someone else said, it's hard to guess which bit is wrong when you're not there to see exactly what's going on. My guess would be to distrust the resistance reading. Have you checked the meter shows zero when you short the test leads together?
Otherwise, the 1 A has to be too high.

I'm recharging the 8 rechargeable batteries.

But in other loads, 8 of these can make 1A, isn't it?

I'll try to use other multi tester. I need to get to the bottom of it. Because after determining why it gives such. I'd put the amazon motor and generator in my cabinet and I'd not touch them again for my whole life. So this is one time to figure it out.

Also the product at amazon is now available. It's the only hand crank available for about 7 billion humans. So maybe you can try to purchase one and measure it using your multimeter? For science. For education. For humanity.

Amazon.com: EUDAX School DIY Dynamo Lantern Educational STEM Building ,Labs Demonstration Motor Activity Teaching Model Hand Cranked Power Electricity DC Electric Generator Physical Science Experiment Education: Toys & Games