Motor performance curves / low duty cycle PWM operation

[foiloco]

I have an unusual application. It is a solar powered tricycle human following robot ( http://solarburro.net/home.html ) which has a limited speed range of about 2 to 4 miles per/hour. I want to use an available bike hub motor to drive this tricycle with a fixed belt reduction drive. I don't want to use a geared motor. I don't like the noise.

I found a hub motor that can be operated with 36 or 48 volts. The manufacture has supplied me with performance data tables.

The data tables show the motor performance at 36 and 48 volts for the motor. The data shows the maximum efficiency of the motor is at 248 rpm for 36 volts and 335 rpm for 48 volts.

Why does the maximum efficiency point increase linearly with applied voltage?

My application needs about 24 watts of power to drive it's wheel on level pavement and 250 watts to climb a paved 20 percent grade at 3 mph.

The smaller my belt reduction ratio the better. The motor will supply the power I need at 36 volt. So I will use a 36 volt battery.

I see that the motor maximum efficiency point increases linearly with voltage. I want to set up my belt reduction drive to operate at the maximum motor efficiency at 3.5 miles per hour with a power output of approximately 30 watts.

My understanding is that under normal operation the analog voltage sent to the motor controller from a throttle assembly is a commanded speed, The motor controller uses PWM to set the output power as required to achieve the commanded speed. So in my case the PWM duty cycle at 3.5 mph on level ground will be about 30/300 or 10 percent. So I am running the motor at the maximum efficiency speed from the performance curve but at a PWM duty cycle of 10 percent.

Argument 1: "My thinking is all wrong and too simple. The ten percent PWM duty cycle makes the voltage equivalent to a steady 3.6 volts. According to the two motor performance curves the motor maximum efficiency will be at 254 x .10 = 25.4 rpm. At 254 rpm the efficiency will be terrible. You've blown it again."

Argument 2: " No you've got it all wrong. When the PWM pulse is applied the motor is seeing 37.5 volts and it is operating at it's maximum efficiency speed so all is good. Between the pulses it's just coasting. there is no problem. You've done everything right. "

I guess that argument 2 is correct. My motor will operate efficiently. I want to hear from others and feel confident that I haven't blown it again. Is Argument 2 correct? If not what should I do? Thanks for your comments.

 

[Lebowski]

basically there's a few rules (ps, you've blown it as your argument 1 is correct)... simplified and not taking mechanical friction losses into account.

- power from the battery equals power delivered by motor plus wasted power
- wasted power goes up with the square of the torque supplied by the motor
- mechanical power is torque times rpm's

So for a certain amount of mechanical power with little losses you want high rpm's and low torque, as this
gives the lowest amount of wasted power (not counting mechanical losses).
The error of your ways lies in chosing a low gearing ratio.

 

[foiloco]

I think you are saying I have made a mistake in selecting this motor. I disagree. I want a high torque at low speed motor because my wheel speed is extremely slow (around 45 rpm), I don't want to listen to a screaming gearbox and I want to reduce the speed to the wheel with one belt and two sprockets - not two/four or three/six. I don't know of a better motor to satisfy my requirements. The motor I selected is reasonably efficient. The efficiency is 83 percent at 248 rpm with 37.5 volts supplied to the motor while outputting 163 watts of mechanical power.

I don't have an answer to my main question yet. To put it another way, What will the motor efficiency be at at 248 rpm with a 37.5 volts, PWM 18 percent duty cycle for about 30 watts motor power output?

Also, if anyone knows of a much better motor for my requirements I would like to hear about it. Thanks.

 

[Miles]

Why does the maximum efficiency point increase linearly with applied voltage?

The motor will operate at maximum efficiency when the active (copper) losses are at parity with the parasitic losses. Increasing the voltage increases the speed, which means higher parasitic losses (not linearly).

 

[liveforphysics]

Also keep in mind, only the current in the windings generate torque. The voltage is fairly irrelevant with respect to efficiency/losses etc.

When you PWM some voltage to a motor, all that matters is the resultant current, which gets filtered through the inductance of the motor to be pretty smooth regardless of how spikey the voltage profile to generate that smooth current may be.

You get some effects/artifacts that are easy to misinterpret for something they aren't when playing with motor simulators, including the illusion that the motor cares what the voltage happens to be (the voltage only needs to be adequate to drive the current you're seeking).

 

[gogo]

If you don't want noise and geared hub motors make more than you care for, but want to use a hubmotor with a small belt reduction, you might try a large diameter Magic Pie hubmotor.

 

[foiloco]

If you don't want noise and geared hub motors make more than you care for, but want to use a hubmotor with a small belt reduction, you might try a large diameter Magic Pie hubmotor.

Thanks for your suggestion. I made me feel a little smarter. That's because I had selected the Smart Pie as the best motor for my application. The smart Pie is the little brother to the Magic Pie. I will put out the 250 watts I need and and weighs less than the Magic Pie. Also it's outside diameter is more than 110mm less than that of the Magic Pie.

 

[foiloco]

Thanks to Miles and liveforphysics for your comments.
I sent myself to 2 hour self instruction on motors. I now understand that torque is proportional to current and back emf is proportional to the motor speed. I understand what no load speed is. Given that I reviewed the performance of the Smart Pie motor (which I plan to use) with 35.8 volts supply. Using the data I estimated the no load speed is 289 rpm. I calculated the resistance of the windings at .925 ohms. At the maximum efficiency point of 248 rpm I calculated the power loss due to coil resistance at 27 watts. The total power loss is 33.34 watts so the other losses amount to 6.34 watts. I estimated what the efficiency would be like for the same speed with a PWM duty cycle of 18.4 percent. The current would be 1.01 amps. Resistance power loss would be .943 watts. What I don't know is how much the 6.34 watts of parasitic power loss would change. I calculated the efficiency assuming it wouldn't change at all. So the total power losses are now 7.37 watts. The efficiency came out at 75 percent. I'm confident this is a worst case. The required power at the motor output is 26.2 watts for my application most of the time (level paved grade). I might need an input power of 26.2/.75 = 34.9 watts. Power loss due to drive motor inefficiencies is about 11.3 watts.

The solar panel should supply a total of 750 watt hours on a sunny day. For 10 hours of operation the energy used by the motor will be about 350 watt hours. Cloudy days and climbing mountains will make things more difficult.

In an ideal world where cost is not a concern I think it would be a good idea to have two or three small motors instead of one big one which is capable of powering the rig up steep hills. Have the second and third motors kick in when required. This would probably increase the efficiency most of the time quite a bit.

This was my favorite article I discovered during my web search. It shares the basics of motors.: Simple Analysis for Brushless DC Motors, Case Study: Razor Scooter Wheel Motor. http://web.mit.edu/first/scooter/motormath.pdf

 

[Miles]

I was a bit puzzled by your concentration on duty cycle and motor speed. For your case (solar burro) the speed is constant. Yes? The torque requirement ratio is as high as 20:1, though. Right?

Parasitic (iron etc.) losses are comprised of: hysteresis losses, which increase linearly with motor speed; eddy current losses, which go up at approximately the square of motor speed; windage losses, which go up as the cube of motor speed; bearing losses etc. which go up linearly. Iron losses also go up with torque output, but not greatly.

Active (copper) losses relate directly to how much torque the motor is producing. Torque is close to directly proportional to phase current but the losses (as heat) go up as the square of the phase current.

 

[foiloco]

I was a bit puzzled by your concentration on duty cycle and motor speed. For your case (solar burro) the speed is constant. Yes? The torque requirement ratio is as high as 20:1, though. Right?

Parasitic (iron etc.) losses are comprised of: hysteresis losses, which increase linearly with motor speed; eddy current losses, which go up at approximately the square of motor speed; windage losses, which go up as the cube of motor speed; bearing losses etc. which go up linearly. Iron losses also go up with torque output, but not greatly.

Active (copper) losses relate directly to how much torque the motor is producing. Torque is close to directly proportional to phase current but the losses (as heat) go up as the square of the phase current.

Thank you for your comment. Now I have a much better idea of the what makes up the parasitic losses. I think a lot of the time I will be trying to stretch my stride out long and walk 4 miles per hour. I don't have much experience in walking without a backpack and on pavement so this will be new for me. I studied the efficient range of the Smart Pie. It's max efficiency point is 248 rpm. It's efficiency is greater than 70 percent from 191 to 273 rpm. I decided to set 3.5 mph at 232 rpm. This means I have 5.125:1 belt reduction between the motor and the 26" diameter wheel. At 4 mph the rpm will be 265. On level paved surface my speed shouldn't vary too much in an absolute sense. Uphill sections will slow me down a fair amount and downhills will speed me up a bit. In a percent point sense my speed will vary quite a bit.

I estimated the torque for level ground to be 39 inch pounds at the wheel. For an 8 percent grade the torque will increase to 247 inch pounds at the wheel. For an uncommon 20 percent grade it will increase to 546 inch pounds. So I would say the normal torque range would be about 247/39 = 6.3. I'm not sure what the back drive torque of the motor will be. Perhaps I will be able to use some steady regenerative braking on steeper down grades to charge the battery a bit.

 

[Miles]

You need some idea of what the parasitic losses of the motor are. The current drawn at no load, for the nominal voltage, is a start.

Obviously, if the parasitic losses are 30Watts, efficiency isn't going to be that great when your traction power requirement is 30Watts....

 

[foiloco]

You need some idea of what the parasitic losses of the motor are. The current drawn at no load, for the nominal voltage, is a start.

Obviously, if the parasitic losses are 30Watts, efficiency isn't going to be that great when your traction power requirement is 30Watts....

Based on my calculations using the motor performance data I was satisfied that the efficiency of the motor run at 248 rpm and 36 volts and about 18 percent PWM duty cycle will be no less than 75 percent and is probably higher. I find that to be acceptable.

 

[liveforphysics]

When seeking an optimization for efficiency, you're seeking a balancing of losses.

Adding more motors adds more losses, while having the potential to reduce copper loss from splitting the torque demand between the two motors. However, if you're on the side of the balance where all non-copper based losses are greater than copper losses, by adding another motor you only reduce efficiency (from the increase in non-copper-losses).

If you're at a point at which copper loss exceeds the non-copper losses (core/eddy/hysteresis/windage/increased mass etc), then adding another motor to reduce copper losses MAY improve efficiency.

I suspect you may find at some extreme slope it's more efficient to have a second motor, but for your application you want to optimize for the conditions it will be spending the most time in, not the outlying case steepest hill climb.

 

[major]

The electric motor efficiency is best represented by a map based on load and speed, not just RPM alone as you seem intent on doing. See this chart (or map):



Notice the constant power curves overlaid. So once you have determined the power required for your load, you adjust your gear ratio to position your operating point (torque and RPM) in the zone with the highest efficiency. The "highest efficiency" will seldom be a single RPM value as you appear to believe.

Notice how efficiency stays relatively high for large area and drops off at light loads (low torque) or low RPM.

Obviously this map is for a larger motor. The source is: http://in-wheel.com/technology/optimization/

And as others have mentioned, you should not dwell on the PWM duty cycle. It is just the method the motor controller uses to adjust the motor voltage.

 

[foiloco]

The electric motor efficiency is best represented by a map based on load and speed, not just RPM alone as you seem intent on doing. See this chart (or map):



Notice the constant power curves overlaid. So once you have determined the power required for your load, you adjust your gear ratio to position your operating point (torque and RPM) in the zone with the highest efficiency. The "highest efficiency" will seldom be a single RPM value as you appear to believe.

Notice how efficiency stays relatively high for large area and drops off at light loads (low torque) or low RPM.

Obviously this map is for a larger motor. The source is: http://in-wheel.com/technology/optimization/

And as others have mentioned, you should not dwell on the PWM duty cycle. It is just the method the motor controller uses to adjust the motor voltage.

Thanks for your input. I have made a decision to not incorporate the complexity have a variable ratio transmission and associated micro controller software in my application. I think that maintaining simplicity in this regard is sensible. Plus I don't have an unlimited budget. I have set up my output belt reduction ratio so the motor operates is at maximum efficiency for the most common operation point which is level operation on pavement.

I don't think I am dwelling on PWM duty cycle. I am aware of it and now I have a basic understanding of how the motor efficiency changes as the motor current is set by changing the PWM duty cycle.

 

[Miles]

I don't think I am dwelling on PWM duty cycle. I am aware of it and now I have a basic understanding of how the motor efficiency changes as the motor current is set by changing the PWM duty cycle.

The motor current is set by the load on the motor. The duty cycle has some indirect control over this but, in your case, with a relatively constant speed but wide variation in torque, it's not that relevant. The association of efficiency with motor speed causes no end of confusion.............

 

[foiloco]

When seeking an optimization for efficiency, you're seeking a balancing of losses.

Adding more motors adds more losses, while having the potential to reduce copper loss from splitting the torque demand between the two motors. However, if you're on the side of the balance where all non-copper based losses are greater than copper losses, by adding another motor you only reduce efficiency (from the increase in non-copper-losses).

If you're at a point at which copper loss exceeds the non-copper losses (core/eddy/hysteresis/windage/increased mass etc), then adding another motor to reduce copper losses MAY improve efficiency.

I suspect you may find at some extreme slope it's more efficient to have a second motor, but for your application you want to optimize for the conditions it will be spending the most time in, not the outlying case steepest hill climb.

I was just making a guess. I do think since the mechanical power required is about 25 watts almost all of the time and under 60 watts most of the time and that having a small motor that would put out 0 to 60 watts would probably be a good idea efficiency wise. I am aware that there are many places with 20 percent grade hills. Like San Francisco, Vermont, New Hampshire and so on. So I am making a mistake using a 300 watt motor instead of using two motors efficiency wise. I suppose I could have two motors and have a freewheel or a clutch on the second "not used often" motor. I think I am probably better off using one motor and keeping hardware and software simple and the system cost lower.

 

[foiloco]

I don't think I am dwelling on PWM duty cycle. I am aware of it and now I have a basic understanding of how the motor efficiency changes as the motor current is set by changing the PWM duty cycle.

The motor current is set by the load on the motor. The duty cycle has some indirect control over this but, in your case, with a relatively constant speed but wide variation in torque, it's not that relevant. The association of efficiency with motor speed causes no end of confusion.............

Thanks for your comment. I think I now have a much better understand how the motor works. I know that I want the motor to operate at it's maximum efficiency for a condition which is around 3.5 miles per hour driving a 26 inch bicycle wheel and about 28 watts motor output power. I only have performance data which was collected at 35.8 volts. I will use a 36 volt battery. I have the attached the motor data. It is also a given that I want to be able to climb a 20 percent grade at 2.5 miles per hour now and then. I would like for a single pass belt drive ratio to be practical.
Perhaps I should run a bunch of new data points for the motor. I will need to make a dynamometer. it will cost more money. I really dread the thought of that.
My solution was : use a 5.125:1 belt reduction.
I invite anyone to determine what the belt drive ratio should be for my application and tell me why.

 

[foiloco]

I fogot to attach the data sheet file. My apologies. here it is.

 

[Miles]

So, parasitic losses at 36V, no load, are about 25 Watts. It looks like your cruising efficiency with a 5:1 reduction will be about 60%. (see row 4).

39 pound inches is about 4.5Nm or 4500mNm. Divide that by 5 to get the motor torque.

 

[major]

Thanks for your input. I have made a decision to not incorporate the complexity have a variable ratio transmission and associated micro controller software in my application.

I was not inferring the use of a multiple ratio drive or micro controller (beyond the motor controller).

I invite anyone to determine what the belt drive ratio should be for my application and tell me why.

The efficiency map is used to determine the best single ratio to accomplish your objectives; ie. high efficiency at normal load and occasional hill climbing capability.

 

[liveforphysics]

I would avoid the added losses of additional power transfer stages entirely. Direct drive is the pinnacle of efficiency.

 

[Miles]

I would avoid the added losses of additional power transfer stages entirely. Direct drive is the pinnacle of efficiency.

Yes, I haven't even included the losses on the belt drive. Allowing for that, and controller losses, cruising efficiency would be around 50%.

 

[Miles]

...

 

[foiloco]

I would avoid the added losses of additional power transfer stages entirely. Direct drive is the pinnacle of efficiency.

Thanks for the great idea. I love it. You need to follow through. But please tell me what commercially available gearless hub motor will meet the requirements of my application.