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 = i

^{2}R = (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 "i

_{0}"). 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 current

^{2}) x Winding Resistance = i

_{0}^{2} x R

_{m}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