Author Topic: DC Electric Motors 101  (Read 1348 times)

Offline Allen Smithee

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DC Electric Motors 101
« on: July 21, 2021, 01:36:38 PM »
This question was asked in Gadabout's thread on the superb work he's done on an electric starter for his Seal so I've offered to do a piece on it, which will take a little while. I tend to explain things from first principles because that makes sense to me, but it may not make sense to other people so:

- If I am pitching this at the wrong level please shout. Especially if I'm stating the obvious and talking down to people in an annoyingly patronising manner!
- If something isn't clear please just ask me to clarify further
- If you disagree with something I say please shout. You may well be right and I could learn something!
- I am only going to talk about DC motors. AC motors (synchronous or induction) are a different skillet of cod. We could come to that afterwards (if anyone is still reading) but it's a bigger subject with more maths, so I wouldn't recommended it. But when I say DC motors this includes both normal brushed motors and brushless DC (BLDC) motors because even though they look similar BLDC motors are NOT 3-phase AC motors we will probably talk about that at the end.
- If I have put this thread in the wrong place please feel free to move it

I have to (briefly) start with some very basic stuff about electromagnetism. I have no doubt that you all know this already, but I want to do this brief bit of revision because a heck of a lot of the general misconceptions about electric motors arise because people forget that this is the underlying principle.


Electricity and magnetism are linked. I don't want to go into why or how because I'm no longer current enough to start explaining how magnetism is the inevitable consequence of relativity in electrostatic fields - we don't need to know this! What we DO need to know is what that means in practice. We all know that:

1. If you apply a voltage across a conductor it will cause a current to flow through the conductor
2. If you pass a current through an conductor it produces a magnetic field - the conductor becomes a kind of magnet
3. If you put two magnets near each other they experience a force between them that either attracts or repels
4. If you move a conductor THROUGH a magnetic field (or move a magnetic field past a conductor) it induces a current in the conductor. The faster the movement the bigger the Electro-Motive Force (EMF or "voltage" if you prefer) which induces the current.

The important and basic thing here is that these the are not "either/or" things - they all happen at the same time. So if you take a wire and pass a current through it then it becomes a magnet. If you place the wire in the field of a fixed magnet it experiences a force. If you allow that wire to be MOVED my the force then the external magnetic field also induces a current in the wire. And the important bit is that the current induced by the external field is always in the opposite direction to the current flowing in the wire that produces its magnetic field so in the context of motors we refer to this as the "Back-EMF" because it's a backwards-facing voltage opposing the voltage across the wire. This is a very important concept, so if you don't grasp it to start with please read it agaon until the penny drops. If the penny refuses to drop please ask questions!

The other important concept is what we call "ideal" machines. In science and engineering we have to reconcile the fact that theory is simple, neat and tidy where the real world is often messy and complicated making it very difficult to understand. One of the more common ways of doing this is to start off with a simple, pure theoretical "ideal" that is easy to understand, and when we have that fully sussed out we add in each of the "real" bits separately until the thing we understand looks like the "real" one.

So in the next post I will start talking about an "ideal" electric motor to help understand the basic stuff, and then move on to add the "real" bits.

So if you're sitting comfortably I'll begin...

AS
« Last Edit: July 22, 2021, 06:37:50 AM by Allen Smithee »
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Offline Allen Smithee

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Re: Electric Motors 101
« Reply #1 on: July 21, 2021, 02:39:22 PM »
So now we consider an "ideal" electric motor. This consists of a perfect coil of resistanceless wire wrapped around a core of soft loss-less iron mounted on a shaft with frictionless bearings. The coil has a perfect commutator which connects it to the DC Voltage source with no losses, only connecting the coil when it is at the correct angle.  This assembly sits inside a par of perfect permanent magnets which have lossless magnetic coupling to the coil. You can ignore most of that - it's just a list f the "real" bits we have to add in later. Lets call the moving part a "rotor" and the static bit a "stator". In a brushed motor the "rotor" has the coil while the "stator" is one or more permanent magnets. When we come to brushless motors we'll see why they put the coil in the stator and the permanent magnets in the rotor, but that doesn't really change anything.

So if we connect the power a current flows in the which creates a magnetic field that attracts/repels the field of the permanent magnets producing a torque that rotates the rotor.

As the rotor rotates the fixed magnetic field induces a Back EMF in the coil which is opposite to the voltage from the power supply, reducing the net voltage on the coil and so reducing the current in the wire.

The bearings are all frictionless, so as long as there is ANY current flowing there will be a torque that accelerates the rotor. It will keep accelerating until it reaches the speed where the Back EMF exactly equals the voltage from the power supply. At this speed there is zero current and zero power but that's OK because our perfect frictionless bearings don't need any to keep the rotor turning. This speed is essentially what's called the "motor characteristic" or "Motor Constant (Kv)" [there is an technical difference but it's not relevant here] which is expressed in RPM per Volt. so if a motor has a Kv of 1,000rpm per volt and you connect it to a 10volt power supply at 10,000rpm the Back EMF would also be 10volts, so it would be expected to turn at 10,000rpm and draw no current.

Now all of that is with no load. If we now use this motor to do actual work on physical load it slows the motor down. This causes the Back EMF to drop, which means the coil now has a voltage across it again and a current flows. This develops power, but only the power that is required by the load we put on it. This is one of the main fundamental differences between electric motors and infernal combustion engines, so it may be worth reading that again until it's clear (and if it isn't please ask for clarification).

With our "ideal" motor the speed returns to the Kv speed (because there is zero resistance in the coil), so the theoretical "ideal motor" runs at a constant speed which is determined by its Motor Constant and the applied voltage, and it draws whatever current it needs to turn the load at that speed.

That sounds like a good place to stop. In the next chapter I'll finally start talking about real motors, and hopefully it will become clear why we had to start with all this theoretical malarky.

AS
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Offline Roger B

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Re: Electric Motors 101
« Reply #2 on: July 21, 2021, 04:49:55 PM »
Looks good to me, however I have some knowledge in this area. Any other views?
Best regards

Roger

Online Kim

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Re: Electric Motors 101
« Reply #3 on: July 21, 2021, 06:15:04 PM »
Very interesting! I've got a background in Electrical Engineering, but I can't say as I know much about electric motors.  And all makes sense to me so far and is quite fascinating.

Thanks for taking the time to write this down, Allen!
Kim

Offline Allen Smithee

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Re: Electric Motors 101
« Reply #4 on: July 21, 2021, 06:45:11 PM »
So, "real" motors, what are the differences and why do we care?

Essentially we can sum these up under three headings and these actually do useful things for us.

1. The coils (windings) of real motors have resistance.
2. The bearings of real motors have friction
3. The magnetic circuit of real motors isn't 100% efficient.

Of these the most important [for us] is (1). If you think back to the "ideal" motor, I said that if we put a load on it then it would slow down and that would reduce the Back EMF so that there was now a voltage across the coil and so it would start drawing current. I also said the coil had zero resistance, so how much current is drawn?

Let's take an example - an ideal 1000rpm/v motor on a 10v battery so turning at 10,000rpm. We put a load on it and slow it down to 9,500rpm so the BackEMF falls to 9.5v and so there is half a volt across the coil. Mr Ohm said:

v=iR, so i=v/R

v=0.5voltrs
R=0 Ohms

Therefore i = (0.5 / 0) = infinity.

In fact ANY drop in Back EMF draws an infinite current in an ideal motor, which is why the RPM is constant. There are paradoxes in this, but it doesn't matter because you can't buy ideal motors. A real motor will have some winding resistance. Typical values for small motors would be 0.01 to 1.0 Ohms, with SPeed 400 size (like that used in the Seal Starter) being around 0.2 Ohms.

So if we repeat the sums using a real resistance we get:

i = v/R = (0.5 / 0.2) = 2.5Amps

Now that's much more like it. So the coil of the motor is now drawing 2.5Amps at 9.5volts, so it is developing 23.75watts of useful power, nearly all of which is going into the load (I'll come to this in a minute).

But we are also putting 2.5Amps through a resistance of 0.2 Ohms, which means we will dissipate some heat:

Power = i2R = (2.5x2.5) x 0.2 = 1.25watts

We call this the "copper losses" = any electrical machine has to do some work against resistance and this appears in the efficiency numbers. It's also one of the two primary "specification limit" characteristics of an electric motor but I just want to get one more bit out of the way before getting to the punch line.

In our list of differences with real motors the second two were friction and magnetic losses. For our purposes w lump these together because they are (within practical limits) constant values in that they change very little with power, speed or load. You won't see any numbers for these in a motor specification, but what you WILL see is a value for "no load current" (usually called "i0"). This number is the current required to overcome friction and magnetic losses, all wrapped up into one handy parameter. Typical values for small motors can be anything from 0.1A to 5A or more, with a typical Speed400 no-load current being around 0.7Amps This number can to some extent  be taken as a quality indicator - the lower the number the better the bearings & magnets and the finer the internal clearances (small gaps bean lower magnetic losses), but it's not the only guide. We tend to call this the "iron losses", but I personally don't like this term because we're also lumping the friction into it.

So if we now go back to the sums. Lets assume our 10,000rpm/v motor has a winding resistance of 0.2 Ohms and a no-load current of 0.5Amps, and we're driving it from a 10volt battery with a load that draws 2.5Amps.

First of all we look at the current and winding resistance. 2.5Amps into 0.2 Ohms gives us copper losses of 0.5v so the windings are seeing 9.5volts.

Then we deduct the no-load current from the total, so the "productive" current is 2.0Amps. So the "net" power coming out of the shaft is:

Power (out) = Volts * Amps = 9.5 * 2 = 19watts

Of course we also know that the *input* power is simply the input voltage times the total current:

Power (in) = Volts * Amps = 10 * 25 = 25watts

So the efficiency is Power (out ) / Power (in) = 19 / 25 = 76%

One of the great features of electric motors is that you don't need a dyno to know the net (output) power - you can work it out directly.

I mentioned a "punchline", which was probably rather over-selling it, but this is an important concept. I showed we can work out the power dissipated against the copper losses:

Power = (total current2) x Winding Resistance = i02 x Rm

This power comes out as heat - it heats up the windings. If the windings can't dissipate the heat they get hotter and hotter until they melt. This is what happens when the magic white smoke escapes from an electric motor. This is the only major constraint on the power a motor can deliver. Within certain limits (in a minute - be patient!) you can whack the voltage up as high as you like, but the CURRENT determines how much heat the windings need to shed.

Why is this important? Well we are all used to defining motors/engines in terms of power ratings. We talk about a 1bhp engine or a 500 watt motor. Well for electric motors this is actually only true at one particular voltage. The true "rating" of the motor is simply the maximum current it can take without melting the windings. As this current limit is constant it means that the torque is constant. increasing the voltage leaves the torque value the same, but increases the revs. There is in fact a second "motor characteristic" called "Kt" or the Torque Constant, which is in units of torque per amp (and if you want to use metric units throughout you find that Kt=1/Kv, but don't worry about that). 

This is getting abstract - let me give a real example:

In the electric RC model world one of the most popular small motors was a Czech motor called the Axi 2820/10 which has been in production for something approaching 20 years now. This is a 1200rpm/v motor which can take around 40Amps for over a minute at a time, which means on a 10cell nicad or 3-cell lipo it delivers well over usable 400Watts - well over half a BHP and more than equivalent to a good 2-stroke 0.25 glow motor.

On 3 cells it turns a 10-6 prop and powers a fast 40" aerobatic model at 400watts. But the same motor used on 2cells turns slower and thus turns a bigger prop. In fact it turns something like a 13-6 prop and will tow up a 100-120" electric soarer in a sprightly manner at around 250Watts. But at the other end of the scale I have seen this same motor used in an electric ducted fan unit on a 12-cell (50volt) lipo producing well over 2.2kW (3bhp) and 2kg of thrust - more than some small turbines! Again, a small practical difference here is that many motors can actually take more current at high voltages because the high revs draw more air through the cooling holes than they get at lower revs. The copper losses are the same at all speeds and the iron losses are more or less constant, so I hope you can see that the higher the voltage the higher the efficiency from the same motor.

In all of these cases the motor is operating within its 40A limit, but the voltage determines how fast it turns and with more voltage you get more revs for the same torque. So the motor does not have a power limit, only a Torque limit which comes from the Current limit. There is only a POWER limit at any specific voltage. And that is the fundamental point I wanted to get across.

As far as I can remember there is only one more "in a minute" thing I said I'd come back to. There is (of course) a second limiting factor to the power, but it's one that we very rarely run into so we usually ignore it. I said we can keep increasing the voltage, but there will be a maximum voltage a motor can take. This is partly a matter of whether the voltage is high enough to start blowing holes in the insulation (very, very unlikely) but mainly that thing about higher voltages mean higher revs. If you keep winding the voltage up you can conceivably get to a speed where the motor can't take the centrifugal forces and bursts. I've seen this happen with early inrunner brushless motors that can shed magnets from the rotor, and I've (once) seen it with a cheap outrunner brushless motor with very agricultural machining in its rotor can that resulted in a split along a classical stress-raiser. But for most general uses this can be ignored.

Ok, that's been me in "hosepipe" mode. On the off-chance that anyone's still reading, are there any actual questions?!

 :)

AS
« Last Edit: July 29, 2021, 11:31:40 PM by Allen Smithee »
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Online crueby

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Re: Electric Motors 101
« Reply #5 on: July 21, 2021, 07:40:28 PM »
Great stuff Allen!  I remember back when working on printer R&D learning how the electric motors would change efficiency and power as they got hot during extended use.
 :popcorn:

Offline Charles Lamont

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Re: Electric Motors 101
« Reply #6 on: July 21, 2021, 09:00:43 PM »
Allen, as one of those guilty of making the enquiry, thank you very much for your most thorough exposition, which must have taken quite some time to put together. I have read all three posts in one sitting, and now need to digest that, and to go back over post 3 a couple more times before I will know if I need to ask questions.

You mention in part 2 the motor only delivering the power 'required' by the load. This perhaps needs a little sidebar clarification. The load may well be speed dependent, as, for example, with a centrifugal pump, in which the power absorbed is theoretically proportional to the cube of the speed. So the motor and load will settle to a speed at which the motor power output curve crosses the load input power curve. I don't see that this differs from an IC engine, except in the typical shape of the power curve.   

Offline Allen Smithee

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Re: Electric Motors 101
« Reply #7 on: July 21, 2021, 10:18:36 PM »
Firstly remember that all of this is strictly DC motors. There are similarities in some respects with some kinds of AC motors, but it's a much more complex subject so don't try to apply what I've said to an AC motor*. I also think it probably only applies to sensorless brushless brushless motors - I think sensored ones behave differently (that's something tatt only occurred to me this evening!).

Right, that's out of the way - to the question:

Quote
You mention in part 2 the motor only delivering the power 'required' by the load. This perhaps needs a little sidebar clarification. The load may well be speed dependent, as, for example, with a centrifugal pump, in which the power absorbed is theoretically proportional to the cube of the speed. So the motor and load will settle to a speed at which the motor power output curve crosses the load input power curve. I don't see that this differs from an IC engine, except in the typical shape of the power curve.

How can I describe this? OK, let's go back to the "ideal" 1,000rpm/v motor. If you run this from a 10v battery on a 10-6 prop it will turn it at 10,000rpm and draw something like 50Amps. If you take this prop off and replace it with an 8-6 prop it will also turn it at 10,000rpm, but it will only draw about 25A. Now a "real" 1,000rpm/v motor won't be quite the same in that it will turn the 8-6 prop a little faster than it turns the 10-6 prop - but not A LOT faster. Its main response to the reduced load is to draw less current. Even if you shaft-run a DC motor it will never exceed the characteristic speed for that voltage(in fact this is one empirical method of getting a rough estimate of a motor's Kv value) and it will only ever develop enough torque to drive the load that is applied to it.

This is where it differs from an IC engine. The IC engine at a given throttle/mixture setting will develop a given amount of torque and then increas its speed until the load actually gets big enough to adsorb it**. If you shaft-run an IC engine (unthrottled) it will run up to a huge speed until either the friction holds it back, or the gas flow can't keep up. Of course in reality it usually doesn't get there because the big end bursts and sends a rod through the block long before it hits the friction limit!

The IC engine has a power curve because the torque drops with speed as the friction increases, the gasflow fails to fully charge the cylinders and the combustion speed becomes too slow to heat the charge in time to push the pistons down. The electric motor only has a small increase in friction with RPM. All the other forces act at what are effectively infinite speeds.

Not sure if that answered the question, so please ask again if it doesn't.

AS

* if that's what you want to understand buy yourself a copy of Hughes Electrical Technology and start working your way through it - it was one of my undergrad textbooks and while I still have it I wouldn't claim to be a master of its content these days!

** this is a very simplistic description of the IC engine which ignores tuning effects of intake/exhaust devices and gas flow limitations,  but it's reasonable for this purpose
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Offline dieselpilot

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Re: Electric Motors 101
« Reply #8 on: July 21, 2021, 11:29:30 PM »
And if you double the voltage with a given prop current increases by 4!. This is where you get into trouble. Overloaded, the motor pulls as much current as it takes to try to achieve the RPM. With nothing limiting current, a poorly chosen load or voltage results in enough current to destroy motors, controls, wiring, and current sources.

Matching the velocity constant and voltage to the load is vital.

Offline gadabout

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Re: Electric Motors 101
« Reply #9 on: July 22, 2021, 01:49:46 AM »
AS,
Thanks am enjoying and learning!
Maybe the title could be changed to “DC Electric Motors 101” ?
Regards
Mark

Offline Allen Smithee

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Re: Electric Motors 101
« Reply #10 on: July 22, 2021, 09:39:16 AM »
Quote
You mention in part 2 the motor only delivering the power 'required' by the load. This perhaps needs a little sidebar clarification. The load may well be speed dependent, as, for example, with a centrifugal pump, in which the power absorbed is theoretically proportional to the cube of the speed. So the motor and load will settle to a speed at which the motor power output curve crosses the load input power curve. I don't see that this differs from an IC engine, except in the typical shape of the power curve.

Not sure if that answered the question, so please ask again if it doesn't.

I was right - I hadn't answered the question!

With something like the centrifugal pump you mention the DC electric motor would turn at its characteristic speed (with the practical variations as discussed above), or at least it would attempt to. If turning the pump at that speed required more current than its windings could handle it would get hot and burn out in the process. This is where it differs from (say) an IC engine which would simply accelerate until (as you say) the engine's torque curve and the pump's torque demand curve intersect and then it would run at that speed.

It works the other way as well - if turning the pump at the motor's characteristic speed only loaded it to a quarter of its current capacity it would simply run at that speed and draw the small amount of current it needed. If a centrifugal pump was turned by an IC engine it would keep accelerating until the load matched the power output. If the pump running at its desired speed presented a load that was a quarter of the engine's power output the engine would keep accelerating until the pump was running fast enough to absorb the power even if this speed was beyond the safe RPM limit of the motor and/or the pump. SO the IC engine needs some kind of speed regulator or throttle where the electric motor is (in this respect) inherently self-regulating.

Is that a clearer answer?

I would just add one other thing. The characteristic speed of an electric motor is determined by many things including the number if turns in the coil and the "strength" of the permanent magnets. The higher this strength the lower the Kv. The permanent magnets will lose their strength if they are overheated - each material has a particular temperature (called its "Curie Point") at which the magnetic properties will fail, and the magnetic properties will start to degrade as this temperature is approached. So if an electric motor is allowed to overheat and that temperature is allowed to spread to the permanent magnets the magnets will degrade, which increases the Kv. Increasing the Kv reduces the Back EMF in the coil, which increases the current, which in turn increases the heat generated by the copper losses. So if an electric motor overheats there is a "runaway" effect that leads to rapid destruction, burn-out and meltdown boring holes down to the core of the earth and ending all life on the planet*.

AS

* I may have made that bit up
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Offline Allen Smithee

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Re: Electric Motors 101
« Reply #11 on: July 22, 2021, 09:40:56 AM »
AS,
Thanks am enjoying and learning!
Maybe the title could be changed to “DC Electric Motors 101” ?
Regards
Mark

I tried, but it only changed the title of the first post so perhaps a Mod could do it?

AS
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Offline Jo

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DC Electric Motors 101
« Reply #12 on: July 22, 2021, 11:55:23 AM »
The thread title has changed Pete  ;)  But for the on going posts to take on the new title it is the last post that needs changing so people reply with the corrected title.

Jo
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Online MJM460

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Re: DC Electric Motors 101
« Reply #13 on: July 22, 2021, 12:06:03 PM »
Hi Allen, thank you for a well written, logical and informative primer on dc motors.  My learning in this area has been informal and piecemeal, and no work involvement.  You have put it together beautifully and filled in many gaps in my knowledge.

I am looking forward to the next chapter.

(And perhaps you will add a little on three phase motors at the end.)

MJM460

The more I learn, the more I find that I still have to learn!

Offline Allen Smithee

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Re: DC Electric Motors 101
« Reply #14 on: July 22, 2021, 12:17:30 PM »
I'm not going to do anything on AC motors (single or 3-phase) because it gets very mathematical very quickly, and because I haven't rally done anything with them since university (back in the early Palaeozoic Era). So I'm not sure what you want next? I could do something about brushed vs brushless motors if you like

AS
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