# Why does a DC motor have a restricted speed?

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• leafy
In summary: Well, my main purpose is to see if we can attain an EMF greater than the supply voltage. When the magnet pass by the coil, let's say it's supper fast that exceed the supply voltage, we aren't apply voltage to the coil because if we do, the magnet slow down in transfering energy to the supply. So we wait until the magnet about to reverse direction to turn on the coil because at that instant, the magnet induce very low EMF and the supply voltage can overcome that.
leafy
TL;DR Summary
Here I have a rotating magnet attached to a disk and a power coil that turn on and off to provide power to turn the disk. The voltage waveform as the magnet pass the coil is as shown. If I avoid turning power on at peek voltage and turning power on at low voltage (at center of the wave form). Can I have unlimited speed increased in theory?

My understanding is that the main limit to motor speed is the structural integrity of the motor itself. Vibrations and structural failure from the centripetal acceleration are what limit the speed

I read that the factor that limit motor speed is back EMF. I might have to go to an electrician for this one.

Dale
leafy said:
I read that the factor that limit motor speed is back EMF. I might have to go to an electrician for this one.
Back EMF increases with motor speed so the current reduces and, consequently the torque. Eventually the friction losses just balance the motor speed - that is if it hasn't jumped off the table by then - which sets an upper limit.

A DC motor is the same as a DC generator. When you connect it to a battery, the maximum speed as a motor, will be the same as you must turn it to generate that voltage, and so start to charge that battery.

As I drew in the voltage waveform, when the magnet change direction from incoming to outgoing, the back EMF is so small that we can inject current into the coil. Of course the wave form get bigger as speed increase, so we have to shorten the ON time of the coil. But we can always inject current no matter how fast the magnet go.

leafy said:
As I drew in the voltage waveform, when the magnet change direction from incoming to outgoing, the back EMF is so small that we can inject current into the coil. Of course the wave form get bigger as speed increase, so we have to shorten the ON time of the coil. But we can always inject current no matter how fast the magnet go.
The coil is an inductor and it will become increasingly difficult to change the current through it as the frequency of your 'injections' increases. You will eventually choke the current down to negligible levels, preventing any further increase in the speed of the magnet.

DaveE
Drakkith said:
The coil is an inductor and it will become increasingly difficult to change the current through it as the frequency of your 'injections' increases. You will eventually choke the current down to negligible levels, preventing any further increase in the speed of the magnet.
Yes, I guess we have to assume the coil can be on instantly without being choke. What I'm trying to say is can we attain EMF greater than the supply voltage.

leafy said:
Yes, I guess we have to assume the coil can be on instantly without being choke.
I really don't think you can assume this if you want an accurate answer.

leafy said:
As I drew in the voltage waveform, when the magnet change direction from incoming to outgoing, the back EMF is so small that we can inject current into the coil.
You don't 'inject current'. You apply a voltage and current is drawn through the device. How much current depends on the impedance of your device, the applied voltage, and frequency. I only bring this up to emphasize that just saying something like 'inject current' is handwaving away a lot of potentially important concepts. What's happening in the coil based on your input might be critical to understanding this.

leafy said:
What I'm trying to say is can we attain EMF greater than the supply voltage.
I'm not an expert in this area, but I have my doubts. May I ask what you think is going on during the time when you aren't applying a voltage to the coil but the magnet is? Is there current in the coil?

DaveE, phinds and sysprog
Drakkith said:
I really don't think you can assume this if you want an accurate answer.You don't 'inject current'. You apply a voltage and current is drawn through the device. How much current depends on the impedance of your device, the applied voltage, and frequency. I only bring this up to emphasize that just saying something like 'inject current' is handwaving away a lot of potentially important concepts. What's happening in the coil based on your input might be critical to understanding this.I'm not an expert in this area, but I have my doubts. May I ask what you think is going on during the time when you aren't applying a voltage to the coil but the magnet is? Is there current in the coil?
Well, my main purpose is to see if we can attain an EMF greater than the supply voltage. When the magnet pass by the coil, let's say it's supper fast that exceed the supply voltage, we aren't apply voltage to the coil because if we do, the magnet slow down in transfering energy to the supply. So we wait until the magnet about to reverse direction to turn on the coil because at that instant, the magnet induce very low EMF and the supply voltage can overcome that. Picking the right time to turn on the coil would further increase magnet speed. Of course when the coil is OFF, no current flow.

leafy said:
Well, my main purpose is to see if we can attain an EMF greater than the supply voltage.
If you mechanically spin the motor faster than the normal motor speed it will become a generator and produce a higher voltage than normal.

The torque produced by a DC motor is proportional to the current that is flowing in the armature. The commutator inside the motor should smooth out the armature rotor current so it should not fall to zero.

You will need a greater voltage to push a greater current through a DC motor. Toy motors, (with only two poles), have a short period, twice per turn when minimum current is flowing. There is no advantage in having current flow then, as the commutators are changing over and that is when the magnets are at a useless angle to the rotor.

DaveE, sysprog and Drakkith
Tell me if this circuit possible. The switch close and open to spin the magnet. The magnet attain enough speed to charge two batteries in series.

leafy said:
Tell me if this circuit possible. The switch close and open to spin the magnet. The magnet attain enough speed to charge two batteries in series.View attachment 294684
To me it looks like a battery-powered motor-based heater.

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hutchphd
leafy said:
Well, my main purpose is to see if we can attain an EMF greater than the supply voltage
Yes you can and before the advent of active electronics motor-generators were used to shift voltages and interconvert DC and AC power.

Of course you cannot extract more energy than you supply.

sysprog
hutchphd said:
Yes you can and before the advent of active electronics motor-generators were used to shift voltages and interconvert DC and AC power.

Of course you cannot extract more energy than you supply.
We had MGs at Amdahl Corp for the plug-to-plug compatible (in re peripherals) IBM mainframe 'clones'. The gererator was perched atop the motor, both inside a lcabinet about the size of a commercial refrigerator. The connectors were all under the raised floor in the more-cooled air. The Amdahl boxes used ECL (emitter coupled logic) chips and required 400hz, so we ran the motor using 60hz and tapped off at 400hz.

The first time I saw one of the 'frequency converters' (about 3' deep x 6' tall x 15' long) in '87, one of the FEs, with another one standing next to him (Field Engineers ##-## they were exclusively hardware guys ##-## I was a Sytems Engineer (SE} ##-## mostly systems software, but we had to be hardware-competent) opened the cabinet ##-## I saw row upon row of arrayed and ganged electrolytic capacitors, each the size of a big oatmeal carton, with 1' x 1##\frac 3 8## slabs of metal joining them, and he asked me "what do you think about that?" and I said immediately "don't touch?" and he said "that's the right answer", and closed the cabinet door.

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hutchphd
leafy said:
my main purpose is to see if we can attain an EMF greater than the supply voltage
Why not just use a step-up transformer?

sophiecentaur said:
Back EMF increases with motor speed so the current reduces and, consequently the torque. Eventually the friction losses just balance the motor speed - that is if it hasn't jumped off the table by then - which sets an upper limit.
This reads like nonsense in the context of a Permanent Magnet Motor. The PMM is the equivalent of a Shunt Wound Motor , which has the same field at all speeds.
However, I learned the above in the context of Series Wound Motors which have an electromagnetic field magnet in series with the armature. This was long before serious PMMS were available. In these, the field reduces as the current reduces and so does the back EMF. This means that the back EMF will not reach the supply volts until the rotation speed is infinite (or limited by resistance and friction).

leafy said:
Tell me if this circuit possible. The switch close and open to spin the magnet. The magnet attain enough speed to charge two batteries in series.
Again, not an expert, but I don't think two dead batteries are supposed to be in parallel with the source and the input circuit of the generator. The current is going to flow out of the source and straight into the batteries until they are the same voltage as the source. The magnet here is pointless. It will just spin up until it reaches its maximum no-load speed and has little influence on the circuit. It certainly won't charge anything, as this is the input side of your generator, not the output.

hutchphd said:
Yes you can and before the advent of active electronics motor-generators were used to shift voltages and interconvert DC and AC power.
Motor-generators are still in regular use today. We had several in the Air Force that were used to convert mains 60 Hz power to 400 Hz power for use in testing our cruise missiles.

sysprog
leafy said:
Tell me if this circuit possible. The switch close and open to spin the magnet. The magnet attain enough speed to charge two batteries in series.View attachment 294684
OK, it's not great for a bunch of reasons, but I get what you're trying to do. The motor is completely irrelevant, or at least not optimal, for generating high voltages. Maybe in 1910, not in 2010.

Now days, this would be done with a switching power supply in a boost or flyback configuration. In essence, everything you think the motor is providing can be done with a simple inductor, which will be smaller, cheaper, more reliable, easier to design with, and will work better.

Next step: google "boost converter" or "flyback converter".

DaveE said:
OK, it's not great for a bunch of reasons, but I get what you're trying to do.
I'm still not. First it was about the generator speed, then getting a larger back EMF than input EMF, now it's about a generator design to charge some batteries that appears to defy circuit logic.
@leafy With respect, it would be immensely helpful if you were clear about what you are really trying to accomplish here.

doc538, DaveE and sysprog
Base on the replies I think the answer is yes. Let's assume that everything is ideal. The current going into the magnet - current induced by magnet = 0 for steady state. Let's assume all batteries are 12Vs.

So from the input we have IV = current of 12V x 12V = Power input
From the output (charging) we have IV = same current x 24V = Power output

DaveE
Why would you think the currents would be the same? I mean other than magic free energy fairies

sysprog
leafy said:
Base on the replies I think the answer is yes.
I don't know why you think this. I mean, it's possible to construct such a circuit. The question is whether it's useful or not. If you're trying to charge the batteries using your generator, then no, your circuit makes no sense. You're just building a generator that has no load on it, and thus will just freely spin while the voltage source charges the batteries a bit every time the switch is closed.

leafy said:
From the output (charging) we have IV = same current x 24V = Power output
What output? That entire circuit is the input to the generator. You have no output.

The current input has to be equal to current output have steady state speed. If there is only input current, the wheel would keep speeding up. And when we draw output current, that is a load. Charging battery is a load.

leafy said:
The current input has to be equal to current output have steady state speed. If there is only input current, the wheel would keep speeding up. And when we draw output current, that is a load. Charging battery is a load.
That load is not on the generator, it is on the power source. The total current through the power source is the sum of the current through the bottom branch (two batteries) plus the generator input. The first of those never changes assuming ideal components (they are just resistive devices that dissipate energy). The bottom branch doesn't care about the top branch. You could cut the wires to the generator and nothing would happen to the bottom branch.

Trace the current through each branch. You'll notice that the current through the bottom branch never goes through the top and vice versa. The only thing changing in your circuit is the voltage across the top branch and its associated current as the generator gets up to speed. At first the voltage is 12 V and the current is maxed out (let's assume a small amount of resistance so we don't have infinite current). As the generator spins up the back EMF increases, which could be represented by a voltage source placed in series with the top branch and in opposition to the source voltage. This reduces the voltage across the top branch, which reduces the current. The back EMF increases as the generator spins up until the back EMF is equal to the source voltage and a steady state is reached. Since 'ideal' components would mean an infinite spin rate, let's assume a small amount of resistance. At this point the voltage across the top is very small since the back EMF is almost equal to the source voltage.

But nothing has happened to the bottom. It always has the source voltage, 12V in this case, across it regardless of what the generator is doing.

sysprog
Hm... you might have read the diode wrong. The bottom 24V has diode that only allow current for charging direction . The bottom 24V cannot be the supply. When the switch close, current is draw from the 12V through the coil, speeding up the wheel. Then the switch open, it'll close again on the next turn. Notice the 12V cannot be charge because the diode only allow current out.

leafy said:
The current input has to be equal to current output have steady state speed.
First, for clarity you are calling the current through the one battery the input and the current through the two batteries the output, right?

If so then what you say is wrong. Steady state current thorough the motor is not zero, and as long as the motor current is not zero then the input and output currents will not be equal.

sysprog
Dale said:
If so then what you say is wrong. Steady state current thorough the motor is not zero, and as long as the motor current is not zero then the input and output currents will not be equal.
I do not get what you mean. How is it not zero?

leafy said:
I do not get what you mean. How is it not zero?
Motors require current to keep them running. Is that not obvious?

Dale said:
Motors require current to keep them running. Is that not obvious?
Unload motor require current to overcome friction. Loaded motor require current for the load. We assumed ideal, no friction, so the only requirement is current for the load.

leafy said:
Unload motor require current to overcome friction. Loaded motor require current for the load. We assumed ideal, no friction, so the only requirement is current for the load.
Even without friction there are still electromagnetic losses, but OK let’s imagine all of that is somehow recaptured. So now you have a motor that you rev up and then just let it spin. If there is no current through it then it does nothing but spin. What is the point of having it there?

Dale said:
Even without friction there are still electromagnetic losses, but OK let’s imagine all of that is somehow recaptured. So now you have a motor that you rev up and then just let it spin. If there is no current through it then it does nothing but spin. What is the point of having it there?
But there is current through it. Each time the switch close, current flow to speed up the wheel. Each time current flow to the 24V, curren flow slow down the wheel. Each cycle, there is a speed up and slow down. The net is zero current with is steady state.

I know that you think, or want, this problem to be about motors. But, you are missing some fundamental understanding about energy storage and transfer with magnetics. I strongly suggest you set aside the motor part and study a boost converter. It is the simplest circuit configuration that does this sort of thing. Understand the voltage and current waveforms in this simpler case first. Then move on to a DC motor, if you want.

I think solid state is harder.

leafy said:
But there is current through it.
If there is current through the motor then the current through the output is not equal to the current through the input. Also, if you are switching back and forth there is no reason that the “forth” current would equal the “back” current.

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