Motor theory: freewheel RPM vs RPM under load

Hello all,

I have a reasonable understanding of electronics and I'm slowing improving my knowledge of the electrical side of these vehicles. One thing is still confusing me. As a motor builds speed, the back EMF increases until it matches the applied voltage - at which point the motor has reached peak RPM. I understand that the back EMF is proportional to RPM, it isn't related to applied load. If there is no load on the motor (the back wheel is in the air), then the peak RPM is higher than if the motor is under load (pushing the rider/bike along).

What is the electrical/magnetic mechanism that causes this. More work is being done, but how does the friction force translate into electrical reluctance^W impedance?

With thanks, Alex

It's because when the wheel is in the air, the motor will spin fast enough until the back emf is so great, it can't make the wheel spin any faster under it's own (small) load. If you were to run a motor not in a wheel, it would probably spin marginally faster than one laced into a wheel no load, because there is less weight, and the wheel isn't creating wind resistance.

Now it gets even more amplified when that load becomes rolling resistance, and wind resistance.

I probably didn't explain that all that well, and i'm sorry for that.

At first it doesn't appear like anything is changing, except the motor magically knows to increase current somehow as you apply load.

This question you ask shows that you're a smart guy, you do deep thinking, and I welcome you to our forum.

The answer to your question is all in the position of the magnet relative to the electrical pole. As load increases a bit, the position of the magnet angle vs the pole slides back a few degrees. As you increase load again, that position of the magnet is further retarded.

With no magnets going past, the stator is still pulses an electromagnetic field, but it all collapses back on itself, and it essentially makes a big heating element with an inline inductor. Put the magnets in the picture, and position them at the correct time to have a positive magnet sweeping by right as a negative field is building at the tip of the pole, and the fields attract to each other as it sweeps past, this attraction, combined with the magnet moving towards the pole. With out the magnet moving, no work is done, and the pole field energy is just distorted by the magnet, it creates a force, but with no change in distance, no work is done, and you still just have an inductively delayed heating element. With the magnet moving, the force of the fields attracting combined with the motion of the magnet gives us the force x distance needed to accomplish the conversion of magnetic fields into work. As you shift that magnet timing towards retarded (by applying load), you cause the field off the pole to grow longer before interacting with the retarded magnet position, this field growing to meet the more retarded magnet is what causes the increase in current. On a sensored motor, you can also increase current directly by advancing hall sensor placement, or doing it electronically. By causing the coils to fire sooner, you can offset the 0 timing KV saturation point, and cause a motor to spin faster, at the cost of additional current (from needing to grow the field larger to hit the more retarded magnet.)

I hope this helps answer your question. If it doesn't, I know there are a bunch of other folks here who can provide a better explanation, or correct any errors I may have made above, as I just worked all night, and it's way past time for me to sleep. lol

Best Wishes, and welcome to Endless Sphere,
-Luke

Short and sweet answer-

When a motor gets a load applied, the back EMF decreases in relation to the applied voltage. This "asks" for more amperage to overcome the load.

johnrobholmes said:

Short and sweet answer-

When a motor gets a load applied, the back EMF decreases in relation to the applied voltage. This "asks" for more amperage to overcome the load.

Click to expand...

And does your motor ask politely like a little orphan boy;

Please sir, may I have some more?

…or perhaps with Eddie Murphy attitude?

Give me some more of that juice fool!

-R

More like Eddie Murphy. It don't care if the batteries like it, it is gonna take what it wants!

That was a great answer live-etc. If I can nearly understand it, it's well explained indeed. I now understand the why of what I knew to be true, stall your motor and it makes it hot.

Hi guys,

I'm new here, but this one I can explain short and sweet as it is important to understand with electric motorcycles.

As you apply a current load to a battery, the potential difference will drop by up to 25%. This doesn't mean the battery is being damaged. There is a member of this forum that had a ducati with dual agni motors named Steve Jozzer. His battery voltage goes from over 100v to under 80v when he accelerates hard in his youtube video. I notice similar drops. The second the load is removed the voltage goes back to normal.

You are familiar with this in your house. When the air conditioner kicks on, you can notice the lights dim slightly, because the voltage is dropping because there is a draw. And the power company can be considered to have a very powerful battery suppling your current As soon as it cuts off, the lights get brighter.

So free wheeling you will see a RPM that you will never actually see with a load on the motor (air friction at near top speed is a incredible load) If you have a 72v system (about 78-80v actual voltage) under load you may drop to 60-66v reducing the RPM potential of the motor and hence your top speed.

The more Amp hours you have, the less the voltage will drop. If you double your battery pack in parallel, (keeping the pack voltage the same but doubling the amp hours) you will increase your top speed, and more than double the range as the peukert effect on the batteries will be less by splitting the draw in half.

liveforphysics said:

At first it doesn't appear like anything is changing, except the motor magically knows to increase current somehow as you apply load.

This question you ask shows that you're a smart guy, you do deep thinking, and I welcome you to our forum.

Click to expand...

Thank you for the in depth answer, I'm it that went right over my head. I think I need to understand better how real motors function rather than the noddy one we built in physics class all those years ago, then come back to your answer.

In the mean time, a hopefully simpler question. In the world of bikes and scooters there are:
- Brushed motors, which are simple to control
- Brushless DC Motors which are (I believe) more efficient, require a more complex 3-phase PWM controller and give the option of regenerative breaking.

In the world of 4 cars and trucks there are also brushed DC motors, but the high end motors that require 3-phase controllers and give the option of regenerative breaking are called AC motors. Are these 2 names for the same type of motor, or are there differences? Is an AC motor controller doing more than PWM to mange the current to the motor?

The "AC" motor is really the same as the DC brushless, if we are talking about permanent magnet designs. Many times AC will describe an induction motor which is a different beast altogether.

moreati said:

In the mean time, a hopefully simpler question. In the world of bikes and scooters there are:
- Brushed motors, which are simple to control
- Brushless DC Motors which are (I believe) more efficient, require a more complex 3-phase PWM controller and give the option of regenerative breaking.

Click to expand...

You can also do regenerative braking with brushed motors.

Okay, stop me if this sounds bogus. I've been brushing up on circuits/electromagnetism, and I think it's simpler than I was expecting. Considering a brushed DC motor

• Given no load, on an ideal motor, back EMF is proportional to RPM until back EMF matches driving voltage and no net current flows
• Real motors aren't ideal, bearings have friction which cause an opposing torque proportional to RPM.
• Load (wind resistance, rolling resistance) adds a greater opposing torque.
• From Lorentz equation, {F = I * L cross B}, driving force on the armature is proportional to Current, winding length and field strength (of the permanent magnet). Driving torque = armature force * armature radius.
• Windings and magnets are fixed for a given motor, so maximum driving torque is determined by maximum current.
• Maximum RPM is reached when driving torque + opposing torque = 0

So under load maximum RPM is determined by current limits. Because to generate torque to overcome the load torque more current is needed and any given battery/controller can only deliver so much current. Am I making sense?

Alex

Yes, the speed will drop until the current flow/torque balances the load applied - equilibrium state.