# Electric Motor Questions



## hyper24 (Jun 13, 2008)

Lets say you had a 144V motor that put out 20kW.
Now lets say you feed it 72V (half its rating) now this is straight 72V not a 144V signal PWM 50% duty cycle like we do in EV's.

Will the motor produce 10kW? Or will it only do this when its pulsed 144V at 50% duty cycle.

As I thought the efficiency of electric motors goes down as you go down from its rated voltage. Hence why the PWM is used to keep it at its rated voltage and best efficiency.

Then again 144V at 50% duty cycle may be EXACTLY the same as straight 72V?

Just trying to make heads and tails of it all, Not that any of this makes a difference to my EV design but I like to have a better understanding of whats going on, thanks.


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## manic_monkey (Jun 24, 2008)

I too would like to know this. 

In addition, would it also be the same when raising the voltage?


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## Jeremy (Jul 12, 2008)

Generally speaking, motor speed is pretty much a direct function of voltage. There is a constant that motors have that gives you an idea of this, often called kV, which is rpm per volt.

Power is simply torque times rpm, so halving the voltage applied to the motor halves the rpm and halves the power, assuming that the torque remains constant.

There are two ways to get, say, 10hp. One way is to use a motor running at a high voltage and relatively low current, the other is to use a motor running at a low voltage and relatively high current. Generally speaking the high voltage low current motor will be more efficient, as the losses will be lower.

Motor efficiency is a function of the losses in the motor, primarily copper losses due to the electrical resistance of the windings. More current = more resistive losses (and more heat). This also applies to the batteries, wiring and controller - the more current they have to handle the greater will be the resistive losses.

The PWM controller is simply a current/voltage transformer, in effect. It allows you to reduce the motor applied voltage, and so it's speed, by quickly switching the supply on and off. A 50% duty cycle halves the applied voltage to the motor - it's exactly the same as halving the number of series connected batteries in terms of what the motor sees. If you take a look at the waveform across the motor then it doesn't actually look as if it's chopped neatly, as the motor inductance smooths out the current and stops the motor really seeing the individual pulses (assuming that the PWM frequency is fast enough - some older controllers were a bit slow).

There are plenty of other reasons as to why you might choose a particular motor and system voltage, other than pure efficiency. High voltage, high rpm motors can be a pain to connect to a drive train, as the reduction ratios required complicate things. For example, on my motorcycle conversion I opted to use a low'ish voltage (48V) system just to keep the motor speed down and simplify the drive to allow a standard rear sprocket to be used. Had I gone for a faster motor then I'd have had to get a custom rear sprocket made up and make a new chain guard to suit.

Had I been able to then I'd have gone for a higher voltage system, maybe I will once I've ironed the bugs out of the present set-up.

Jeremy


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## ElectriCar (Jun 15, 2008)

Jeremy, you've obviously been working in this area as well as I have. While an AC drive turns the transistors on and off at varying frequencies, the DC controller uses the same transistors to vary the pulse width 0-100% of the cycle time yea? Or do they likewise vary the frequency of the cycles until they're full on at 100% power? Just curious.


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## Jeremy (Jul 12, 2008)

The DC motor controllers usually just vary the pulse width, with the frequency staying pretty constant, at least on the ones I've looked at that's the case (not a big sample though - just two!).

The actual switching frequency for a DC motor controller ideally depends on the motor inductance, if the controller switching frequency is too low for a motor with a low inductance, like a Lynch/Lemco, a Mars brush motor or a Perm, then the current in the motor will have more ripple than is desirable.

Older controllers seem to switch at lower frequencies than some of the newer ones, I think, which makes them a bit smoother and quieter when the motor is running at low speed. As far as I've been able to tell, separately excited motors seem to have a pretty high inductance so can work OK with lower frequency switching. My experience is only with PM motors, both brushed and brushless - I've never played with a separately excited motor, other than change the starter motor on cars.

Jeremy


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## Nate (Jul 10, 2008)

*I would say a series wound motor that can produce 20kW on 144V would produce closer to 5kW on 72V, 1/4 the output rather than 1/2 the output. This is based on pure DC figures not PWM. If you cut the voltage by half you would cut the current by almost half also. When you figure your wattage (volts x amps) you get 1/4 the wattage. True that your motor RPM would drop which would lower your CEMF and allow the current to rise a little.*

*If you have a motor on 144V at 50% PWM your output would be closer to 10kW because your motor amps are higher than the straight 72VDC motor amps.*

*Most DC PWM controllers are fixed frequency and vary the duty cycle, some older Curtis controllers would lower their frequency when they got warm.*


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## Jeremy (Jul 12, 2008)

The thing is that the current doesn't necessarily halve when you halve the voltage, as motors aren't a purely resistive load and don't follow Ohms Law particularly well. A good example is to look at the winding resistance of a motor and then just apply Ohms Law and see what current you get. The answer is the motor stall current, usually an extremely high value (it's up around a couple of thousand amps or more for my motor just on 48V). Clearly something else controls the motor current when it's spinning, otherwise it'd draw this current all the time.

The current through the motor is primarily determined by the difference between the back EMF (generated by the motor as a consequence of it's rotation - think of it as a generator as well as a motor), the supplied voltage and the circuit total resistance. What this means in practice is that as you slow the motor down, by applying a torque to it, the current increases. The reason for the current increase is because, by slowing the motor, you've reduced the back EMF, so have increased the difference between it and the supply voltage. The circuit resistance remains the same, so the current increases (this is a bit of an over simplification, but hopefully explains the basics of what makes things happen the way they do).

Motors also have a torque "constant", which is pretty much independent of the applied voltage, as it's a function of the available magnetising force and physical size of the motor. As the magnetising force is a function of the number of turns in the windings and the current flowing through them, it follows that torque is directly proportional to current (square of the current for a series motor).

This means that it's possible to get the same torque from a motor over a wide range of supply voltages (provided the controller does it's job). PWM controllers are, as I mentioned earlier, effectively voltage/current transformers - the motor current can exceed the battery current at less than full power.

Jeremy


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## ElectriCar (Jun 15, 2008)

Jeremy you lost me at the last paragraph. And that motor explanation cleared up how the motor maintains it speed, or tries to when a load is applied. It's never been explained like that to me, or at least it never sunk in.


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## Jeremy (Jul 12, 2008)

Sorry about that, perhaps I should have tried a bit harder...... 

Let's assume that the controller is running at 50% throttle, so is giving a 50% duty cycle (on and off periods are the same) and that the supply voltage is 72V. The motor will "see" an average voltage of 1/2 the supply, due to the PWM, so it will run at a no load speed that's equivalent to that if it were fed with 36V.

When it's loaded, with a torque, it doesn't really "see" the difference between its back EMF and the notional 36V though, as it's being fed with 72V pulses. The controller can actually push far more current through the motor during the pulse "on" time as the difference in voltage between the back EMF and the pulse voltage is very much greater. The net result is complex, because of the effect of motor inductance limiting the current rise time, but essentially the motor current can exceed the battery current for duty cycles less than 100%. Thus it's effectively converting volts to amps (this is a bit of a simplistic way to describe it, though - apologies to the purists............).

Here's an example (completely mythical numbers): The motor circuit has a resistance of 1 ohm. The supply voltage is 72 volts. The motor back EMF at a nominal working torque is 30 volts.

With 72V applied directly to the motor the motor draws a current of: (72 - 30) / 1 = 42A

With 36V applied directly to the motor the motor draws a current of: (36 - 30) / 1 = 6A

For our 50% duty cycle PWM controller, running on 72V, the motor sees 42A pulses of current, but only for half the time. This gives an average motor current of 21A, a lot more than if it were connected directly to a 36V supply.

I hope this makes it a bit clearer!

Jeremy


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## ElectriCar (Jun 15, 2008)

Are you talking series wound motors only? Though the pulses are instant on and off, the motor is still turning and generating counter emf correct? Isn't that how we get regeneration? Or is the reason we can't regen on series motors the same reason this happens? Did that make any sense?


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## Jeremy (Jul 12, 2008)

It applies to any DC motor, series and PM motors behave similarly when run from PWM controllers.

You're right, regeneration is produced by the motor working as a generator. The back EMF is the voltage that the motor would produce if run at that speed as a generator and is always lower than the supply voltage needed to make the motor run at that speed.

Regeneration relies on the voltage being generated by the motor being greater than the supply voltage. To make this happen you need to use a controller that actively switches both sides, instead of just switching the "ground" side and using a flywheel diode. This converts the controller/motor combination into a boost type switch mode supply, in effect, allowing voltages greater than the motor normal back EMF to be generated. The topic's a bit complex, but there are some reasonable web sites that cover this stuff. I've just Googled up this student paper that might help: http://eprints.usq.edu.au/501/1/DeanTHOMPSON-2005.pdf

Jeremy


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## major (Apr 4, 2008)

Jeremy said:


> Here's an example (completely mythical numbers): The motor circuit has a resistance of 1 ohm. The supply voltage is 72 volts. The motor back EMF at a nominal working torque is 30 volts.
> 
> With 72V applied directly to the motor the motor draws a current of: (72 - 30) / 1 = 42A
> 
> ...


Hi Jeremy,

You have come to an invalid conclusion here. A motor behaves the same, practically speaking, with a 36 volt source or a 72 volt, 50% PWM, even under load.

I think you have picked a poor example. For one thing, assuming you mean the same load torque when you say "at a nominal working torque", then the motor current would have to be the same for both cases, 72 and 36 volts. But the 1 ohm motor cannot draw 42 amps at 36 volts, because that is beyond stall. So the example falls apart.

And if you were working with a suitable example and went from 72 volts to 36 volts, the Eg (back EMF) would be drastically lower, not stay the same at 30 because the RPM would be a lot lower.

Regards,

major


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## Jeremy (Jul 12, 2008)

I'll admit the numbers were a poor choice, I chose the daft value of 1 ohm to make the maths easy, maybe that confused things............... 

The sums still stack up though and I didn't imply that the 42A current could be driven through a 1 ohm load by a 36V supply.

For the 72V case, with a 30V back EMF, there is 42V to drive current through the motor. With 1 ohm motor circuit resistance the peak current is (72V - 30V) / 1 ohm = 42 amps.

For the 36V case, with the same 30V back EMF, then there is only 6V to drive current through the motor. (36V - 30V) / 1 ohm = 6 amps.

The motor torque has to be much greater for the high current case, as torque is proportional to current (or the square of the current for a series motor). The important point is that the motor RPM has to be the same for both cases. To achieve this the motor must be loaded with a an appropriate nominal (not the same) load torque for each case, to keep the motor at the same rpm. Without a torque load the motor rpm would be different for the two cases, so changing the back EMF.

Jeremy


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## major (Apr 4, 2008)

Jeremy said:


> To achieve this the motor must be loaded with a an appropriate nominal (not the same) load torque for each case, to keep the motor at the same rpm. Without a torque load the motor rpm would be different for the two cases, so changing the back EMF.
> 
> Jeremy


Well Jeremy,

You have me baffled as to how this then relates to hyper24's original post. He said a 144V motor was running at 20 KW. Then asked if that motor run at 72V could do 10 KW and if that would be the same as the motor running at 50% PWM on a 144V battery. Answer is yes. With some qualifiers like ventilation. But basically yes. It could run at 72V, or 144V on 50% PWM, at the same torque, same current and half voltage (compared to 144V, 100%) and would run at half speed and half power, 10 KW. Wouldn't you agree?

Regards,

major


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## Jeremy (Jul 12, 2008)

I guess we've run into a bit of thread drift, as I was very specifically addressing the post later in the thread requesting clarification about the voltage/current transformation effect of a PWM controller, not the original posters question that I addressed in post #3.

Halving the voltage at the motor pretty much halves the motor rpm, so halves the power, assuming that the current, and hence torque, remains the same. I thought I'd been clearer in making this point than I had, sorry.

Jeremy


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