Why more KV = more power, the Maximum Power Transfer Theorem
Why more KV = more power, the Maximum Power Transfer Theorem
I wrote this up some time ago when I was trying to understand the Maximum Power Transfer Theorem in relationship to sizing a motor and choosing a KV for driving a propeller at high RPM and high load for several top speed quad copters I built.
Some basic electronics knowledge is needed to understand this.
Ohm's law:
Current = Voltage / Resistance
Resistance = Voltage / Current
Voltage = Current * Resistance
Power = Voltage * Current
Kirchoff's current law  which states that all current entering a circuit must be equal to the current leaving it. This is charge conservation.
Motors work in two directions, the convert electricity into motion and motion into electricity. You cannot have one without the other.
When you spin a motor, it produces voltage, this is known as back EMF. Take a 2000KV motor and spin it at 2000 RPM and you produce 1V of back EMF.
Motors have an equivalent resistance (AC has impedance) which is measured in Ohms. In this example we'll say the motor has 0.1 Ohms (100mOhm) of resistance.
If we take a 2000KV motor and supply it with 20V, it will spin to 40,000 RPM at which point it's producing 20V of back EMF and it stops accelerating because 20V input + 20V BEMF = 0V
Using our 20V power supply with infinite current supply capability, we bring the motor up to 10,000 RPM under load.
10,000RPM / 2000KV = 5V of BEMF
This leaves us 15V to power the motor with. The resistance of 0.1 ohms now comes into play. So how much current are we putting into the motor? Using Ohms law we find that 15V / 0.1Ohms = 150A (it's going to get toasty quick). According to Kirchoff's rule, if we put 150A in, we also get 150A out. We know the motor is outputting 5V of BEMF, so now we can calculate.
Output wattage:
150A * 5V BEMF = 750W of shaft output power
We are inputting:
20V * 150A = 3000W of electrical power to the motor.
It is outputting:
5V * 150A = 750W of shaft power available to do work.
Efficiency is output/input:
750W/3000W = 25% efficient.
In the above calculations I list 5V of BEMF and multiply it by the 150A current and call this output shaft power. At first glance this does not appear to be a valid way of of looking at the output power, but you need to remember that the motor is being fed 20V (no PWM involved). The motor is having a load place on it so it is only able to achieve 10k RPM, it is over loaded. That is why it takes 150A of current, very high load. You would not want to do this for any extended length of time. In an EV, this state would most likely only happen upon hard acceleration or high load with the throttle wide open.
The 5V of BEMF is caused by the RPM of the shaft which is performing work on something. That work requires power which it is receiving from the shaft. Power at the shaft is voltage times current. There is 15V of voltage headroom left to add more power to the motor. If the voltage is increased, then the current in the motor will increase as long as it is less than 50% of the 20V bus. Current will stop increasing once the motor reaches RPM = 50% of it's unloaded speed, because after this point, it is unable to add more current to the motor due to the generated BEMF voltage going higher than the voltage source. The current through the wire creates a force proportional to itself as described in Ampere's law / Maxwell equation #4. This is why the output shaft power is equal to the generated BEMF * Input current in these examples.
Equation references with detailed explanations on their derivation.
http://www.maxwellsequations.com/ (click on each equation part for more detail)
http://hyperphysics.phyastr.gsu.edu/hb ... mplaw.html
So the statement I made is that maximum motor power occurs at 50% of the motors unloaded KV for a given bus voltage. Well with 20V input voltage and 2000KV motor we need to be at half of 40,000RPM or 20,000 RPM. 20k RPM occurs at 10V of BEMF.
Math time:
10V BEMF / 0.1 Ohms = 100A of current flowing through the system
100A * 10V BEMF = 1000W of shaft output power
20V * 100A = 2000W of input electrical power
1000W output / 2000W input = 50% efficient
So you might be wondering what about if we have 15V of BEMF because the motor is at 30k RPM.
20V in  15V BEMF = 5V
5V / 0.1 Ohms = 50A of current flowing through the system
50A * 15V BEMF = 750W of shaft output power
20V * 50A = 1000W of electrical input power
750W output / 1000W input = 75% efficient
Notice how we haven't and will not be able to exceed 1000W of output power from the motor and this peak is at 50% of the unloaded KV and is only 50% efficient.
You might notice from the math that if we produce 19.9V BEMF out, the motor is producing 19.9W with 20W input which is 99.5% efficient and at 20V it would be 100% efficient, aka not possible. Motors have several types of losses and I've only dealt with the copper losses in the above examples. The other losses are the iron losses. Iron losses may be calculated (I won't do that now as this is already getting long), but we can find out what all losses are for a given RPM by powering the motor unloaded and noting its current draw. If the motor draws 10A unloaded at 20V supply, the efficiency is not going to be at 100% as the simplified example shows. We won't even be able to get to 40k RPM because of that 10A.
The math changes like this:
20V in  19V BEMF = 1V
1V / 0.1 Ohms = 10A of current flowing through the system
10A of generated current + 10A of loss current = 20A
10A * 19V BEMF = 190W of shaft output power
20V * (10A output + 10A losses) = 400W of electrical input power
190W output / 400W input = 47.5% efficient.
If the motor has 5A of losses at 20k RPM, then our motor will be a bit less than 50% efficient. It would in fact only be 47.6% efficient. 50% efficiency would happen at 10.5V of BEMF and only output 996W, but 10.5V * 2000KV = 21k RPM, so a bit over half the unloaded KV.
The Maximum Power Transfer Theorem is why going up in KV while keeping the motor size the same produces more power under the same load. Understanding this concept will assist in sizing a motor for a given task. If the goal is peak power, have your motor loaded to half of it's KV. Of course you will most likely be unable to sustain peak power due to thermal limitations.
Some basic electronics knowledge is needed to understand this.
Ohm's law:
Current = Voltage / Resistance
Resistance = Voltage / Current
Voltage = Current * Resistance
Power = Voltage * Current
Kirchoff's current law  which states that all current entering a circuit must be equal to the current leaving it. This is charge conservation.
Motors work in two directions, the convert electricity into motion and motion into electricity. You cannot have one without the other.
When you spin a motor, it produces voltage, this is known as back EMF. Take a 2000KV motor and spin it at 2000 RPM and you produce 1V of back EMF.
Motors have an equivalent resistance (AC has impedance) which is measured in Ohms. In this example we'll say the motor has 0.1 Ohms (100mOhm) of resistance.
If we take a 2000KV motor and supply it with 20V, it will spin to 40,000 RPM at which point it's producing 20V of back EMF and it stops accelerating because 20V input + 20V BEMF = 0V
Using our 20V power supply with infinite current supply capability, we bring the motor up to 10,000 RPM under load.
10,000RPM / 2000KV = 5V of BEMF
This leaves us 15V to power the motor with. The resistance of 0.1 ohms now comes into play. So how much current are we putting into the motor? Using Ohms law we find that 15V / 0.1Ohms = 150A (it's going to get toasty quick). According to Kirchoff's rule, if we put 150A in, we also get 150A out. We know the motor is outputting 5V of BEMF, so now we can calculate.
Output wattage:
150A * 5V BEMF = 750W of shaft output power
We are inputting:
20V * 150A = 3000W of electrical power to the motor.
It is outputting:
5V * 150A = 750W of shaft power available to do work.
Efficiency is output/input:
750W/3000W = 25% efficient.
In the above calculations I list 5V of BEMF and multiply it by the 150A current and call this output shaft power. At first glance this does not appear to be a valid way of of looking at the output power, but you need to remember that the motor is being fed 20V (no PWM involved). The motor is having a load place on it so it is only able to achieve 10k RPM, it is over loaded. That is why it takes 150A of current, very high load. You would not want to do this for any extended length of time. In an EV, this state would most likely only happen upon hard acceleration or high load with the throttle wide open.
The 5V of BEMF is caused by the RPM of the shaft which is performing work on something. That work requires power which it is receiving from the shaft. Power at the shaft is voltage times current. There is 15V of voltage headroom left to add more power to the motor. If the voltage is increased, then the current in the motor will increase as long as it is less than 50% of the 20V bus. Current will stop increasing once the motor reaches RPM = 50% of it's unloaded speed, because after this point, it is unable to add more current to the motor due to the generated BEMF voltage going higher than the voltage source. The current through the wire creates a force proportional to itself as described in Ampere's law / Maxwell equation #4. This is why the output shaft power is equal to the generated BEMF * Input current in these examples.
Equation references with detailed explanations on their derivation.
http://www.maxwellsequations.com/ (click on each equation part for more detail)
http://hyperphysics.phyastr.gsu.edu/hb ... mplaw.html
So the statement I made is that maximum motor power occurs at 50% of the motors unloaded KV for a given bus voltage. Well with 20V input voltage and 2000KV motor we need to be at half of 40,000RPM or 20,000 RPM. 20k RPM occurs at 10V of BEMF.
Math time:
10V BEMF / 0.1 Ohms = 100A of current flowing through the system
100A * 10V BEMF = 1000W of shaft output power
20V * 100A = 2000W of input electrical power
1000W output / 2000W input = 50% efficient
So you might be wondering what about if we have 15V of BEMF because the motor is at 30k RPM.
20V in  15V BEMF = 5V
5V / 0.1 Ohms = 50A of current flowing through the system
50A * 15V BEMF = 750W of shaft output power
20V * 50A = 1000W of electrical input power
750W output / 1000W input = 75% efficient
Notice how we haven't and will not be able to exceed 1000W of output power from the motor and this peak is at 50% of the unloaded KV and is only 50% efficient.
You might notice from the math that if we produce 19.9V BEMF out, the motor is producing 19.9W with 20W input which is 99.5% efficient and at 20V it would be 100% efficient, aka not possible. Motors have several types of losses and I've only dealt with the copper losses in the above examples. The other losses are the iron losses. Iron losses may be calculated (I won't do that now as this is already getting long), but we can find out what all losses are for a given RPM by powering the motor unloaded and noting its current draw. If the motor draws 10A unloaded at 20V supply, the efficiency is not going to be at 100% as the simplified example shows. We won't even be able to get to 40k RPM because of that 10A.
The math changes like this:
20V in  19V BEMF = 1V
1V / 0.1 Ohms = 10A of current flowing through the system
10A of generated current + 10A of loss current = 20A
10A * 19V BEMF = 190W of shaft output power
20V * (10A output + 10A losses) = 400W of electrical input power
190W output / 400W input = 47.5% efficient.
If the motor has 5A of losses at 20k RPM, then our motor will be a bit less than 50% efficient. It would in fact only be 47.6% efficient. 50% efficiency would happen at 10.5V of BEMF and only output 996W, but 10.5V * 2000KV = 21k RPM, so a bit over half the unloaded KV.
The Maximum Power Transfer Theorem is why going up in KV while keeping the motor size the same produces more power under the same load. Understanding this concept will assist in sizing a motor for a given task. If the goal is peak power, have your motor loaded to half of it's KV. Of course you will most likely be unable to sustain peak power due to thermal limitations.
Last edited by zombiess on Jan 07 2020 1:09pm, edited 3 times in total.
Re: Why more KV = more power, the Maximum Power Transfer Theorem
Assuming no gearing, lower kV is still useful when more torque at lowest possible rpm is required and top speed no issue at all, right?
 spinningmagnets 100 GW
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Re: Why more KV = more power, the Maximum Power Transfer Theorem
I know that there are so many more facets to electric system design than are obvious at first. Thanks for taking the time to type this out. I am still learning.
One of the times I was talking to Luke, it sounded like one of the reasons he preferred a high Kv motor at halfthrottle compared to a lower Kv motor of the same type and size at full throttle (with both resulting in the same speed)...is that the high Kv motor had thicker wire and less resistance.
When copper wire gets warm, it's resistance goes up. If the amount of input watts flowing through it stays the same, then...it will get even hotter just because the wire is now warm. He is also famous for liking a bigger motor than "necessary", plus active motor cooling.
One of the times I was talking to Luke, it sounded like one of the reasons he preferred a high Kv motor at halfthrottle compared to a lower Kv motor of the same type and size at full throttle (with both resulting in the same speed)...is that the high Kv motor had thicker wire and less resistance.
When copper wire gets warm, it's resistance goes up. If the amount of input watts flowing through it stays the same, then...it will get even hotter just because the wire is now warm. He is also famous for liking a bigger motor than "necessary", plus active motor cooling.
 Dauntless 100 GW
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Re: Why more KV = more power, the Maximum Power Transfer Theorem
So this thread
http://endlesssphere.com/forums/viewto ... 4f8cb67e89
Is discussing this hub. https://www.ebikes.ca/shop/electricbic ... 005rc.html#
We're going to have lower performance with hubs, but what is the real story comparing it to a middrive? Should we really want a high gear ratio?
http://endlesssphere.com/forums/viewto ... 4f8cb67e89
Is discussing this hub. https://www.ebikes.ca/shop/electricbic ... 005rc.html#
We're going to have lower performance with hubs, but what is the real story comparing it to a middrive? Should we really want a high gear ratio?
Any sufficiently advanced technology is INDISTINGUISHABLE FROM MAGIC!
 Arthur C. Clarke
 Arthur C. Clarke
Re: Why more KV = more power, the Maximum Power Transfer Theorem
I like higher KV in the mid throttle range as well as it has better throttle response and this is where most riding occurs, in the middle.spinningmagnets wrote: ↑Dec 31 2019 1:49pmOne of the times I was talking to Luke, it sounded like one of the reasons he preferred a high Kv motor at halfthrottle compared to a lower Kv motor of the same type and size at full throttle (with both resulting in the same speed)...is that the high Kv motor had thicker wire and less resistance.
When copper wire gets warm, it's resistance goes up. If the amount of input watts flowing through it stays the same, then...it will get even hotter just because the wire is now warm. He is also famous for liking a bigger motor than "necessary", plus active motor cooling.
A motor with 8KV is slow on 48V, but would be considered a high KV motor on 480V. The design choices are relative to the application.

 Posts: 227
 Joined: Mar 27 2011 8:07pm
 Location: Bay Area, California
Re: Why more KV = more power, the Maximum Power Transfer Theorem
If you only have 1V of headroom over the BEMF, then how are you getting 20A into your 0.1 Ohm motor? Do you mean 18V BEMF?zombiess wrote: ↑Dec 31 2019 1:17pm
....
Notice how we haven't and will not be able to exceed 1000W of output power from the motor and this peak is at 50% of the unloaded KV and is only 50% efficient.
You might notice from the math that if we produce 19.9V BEMF out, the motor is producing 19.9W with 20W input which is 99.5% efficient and at 20V it would be 100% efficient, aka not possible. Motors have several types of losses and I've only dealt with the copper losses in the above examples. The other losses are the iron losses. Iron losses may be calculated (I won't do that now as this is already getting long), but we can find out what all losses are for a given RPM by powering the motor unloaded and noting its current draw. If the motor draws 10A unloaded at 20V supply, the efficiency is not going to be at 100% as the simplified example shows. We won't even be able to get to 40k RPM because of that 10A.
The math changes like this:
20V in  19V BEMF = 1V
1V / 0.1 Ohms = 10A of current flowing through the system
10A of generated current + 10A of loss current = 20A
10A * 19V BEMF = 190W of output power
20V * (10A output + 10A losses) = 400W of input power
190W output / 400W input = 47.5% efficient.
For 19V BEMF, I think you have 0W output power, 190W iron loss, 10W copper loss, and you are 0% efficient.
I think you should be careful with your usage of the Kv term. I think it should say, "If the goal is peak power, have your motor loaded to half of its max rpm (Kv*bus voltage)."If the motor has 5A of losses at 20k RPM, then our motor will be a bit less than 50% efficient. It would in fact only be 47.6% efficient. 50% efficiency would happen at 10.5V of BEMF and only output 996W, but 10.5V * 2000KV = 21k RPM, so a bit over half the unloaded KV.
The Maximum Power Transfer Theorem is why going up in KV while keeping the motor size the same produces more power. Understanding this concept will assist in sizing a motor for a given task. If the goal is peak power, have your motor loaded to half of it's KV.
Re: Why more KV = more power, the Maximum Power Transfer Theorem
But but but he stutters.Output wattage:
150A * 5V = 750W of output power
We are inputting:
20V * 150A = 3000W of power to the motor.
It is outputting:
5V * 150A = 750W of work.
Efficiency is output/input:
750W/3000W = 25% efficient.
@zombiess:
BEMF is the voltage resisting the input voltage, not the voltage doing mechanical work.. In your calc it’s all backwards or the motor would output max power at max back emf.. that is at noload where efficiency is zero.
If your calcs above would be correct then higher kV motor would always have the upper hand. But with high kV and resulting higher rpm for a given voltage eventually the eddy losses become so high that motor burns or motor explodes from the centrifugal forces created. So there’s an optimum balance to be found.
Motor winding and kV has to be matched to controller current and voltage for highest output
And motors are not run at short circuit so the I=U/R calc to get current is not useful  it only tells what the absolute maximum current can be.
Take my moto setup as an example with 5mohm phase resistance that’s run at 72V. That’s 14400A short circuit current at zero rpm. Enough to kill both motor and controller if this current could be sustained long enough for calculating output. That’s incorrect you might say, as max power is at half noload rpm and BEMF then lowers effective voltage and short circuit current to half? Well then short circuit current is only 36/0.005 which is 7200A.. I don’t have a 259kW motor.
Post is simplified to the point of being wildly misleading.. kV is not a performance figure.
Last edited by larsb on Jan 05 2020 4:32am, edited 1 time in total.
Ride on!
Re: Why more KV = more power, the Maximum Power Transfer Theorem
>> Assuming no gearing, lower kV is still useful when more torque at lowest possible rpm is required and top speed no issue at all, right?
Would be greatly appreciated if this could be answered, I realize a bit prosaic for such highlevel discussion
But only if you are **sure** that you know, get so many disagreements based in theory only. . .
Would be greatly appreciated if this could be answered, I realize a bit prosaic for such highlevel discussion
But only if you are **sure** that you know, get so many disagreements based in theory only. . .
Re: Why more KV = more power, the Maximum Power Transfer Theorem
There is no such thing as a ”high torque motor” in the sense that it always creates more torque with low kV than high kV with same copper fill and it’s driven with enough current to match kV difference.
This is valid when controller is powerful enough but all controllers i’ve had limit current at low rpm so that the high kV motor can create lower torque than the low kV as a result from this.
How significant it is isn’t possible to say generally as it depends on load, controller, winding etc. Controller manufacturers should state stall current, then we’d know.
The apt9660 controller has 28% set as "motor lock" current which i think represents stall current (this is locked setting) so it only puts out 168A instead of 600A at stall. That means that for the QS 3000w mid motor you can only get 27Nm torque instead of 56Nm at zero rpm. If motor had half the kV this would create almost double torque from the start with this motor/controller combination. I don’t know at what RPM current is scaled up, if it’s at 10, 50, or 100 rpm
This is valid when controller is powerful enough but all controllers i’ve had limit current at low rpm so that the high kV motor can create lower torque than the low kV as a result from this.
How significant it is isn’t possible to say generally as it depends on load, controller, winding etc. Controller manufacturers should state stall current, then we’d know.
The apt9660 controller has 28% set as "motor lock" current which i think represents stall current (this is locked setting) so it only puts out 168A instead of 600A at stall. That means that for the QS 3000w mid motor you can only get 27Nm torque instead of 56Nm at zero rpm. If motor had half the kV this would create almost double torque from the start with this motor/controller combination. I don’t know at what RPM current is scaled up, if it’s at 10, 50, or 100 rpm
Ride on!
Re: Why more KV = more power, the Maximum Power Transfer Theorem
Yes, it is useful. With caveats.
To explain why let's go back to first principles. Torque is caused by the interaction of magnetic fields. In any normal magnetic structure, this is given by the B field. The stronger the B field, the more torque you get.
B field is induced in a magnetic structure (like stator windings) by current, specifically ampturns. Ampere's Law describes how much current produces how much magnetic field. One of the interesting results of this is that you can get more field two ways  by increasing the current through a single turn, or by increasing the number of turns. However, when you increase the number the number of turns, inductance (the tendency of the structure to resist changes in current) also goes up.
So let's say you have a perfect power supply and inverter. It will give you thousands of amps, thousands of volts, or any combination thereof  and in any waveform you want  without complaint. In that case the number of turns isn't too important (as long as you don't melt or overvolt the windings.) If there are a low number of turns, the current can go skyhigh to get whatever torque you want.
If there are a large number of turns, the current can be much lower, due to that ampturn part of the formula for flux. But since the voltage is what "forces" current into the windings, and the larger inductance (caused by more turns) opposes a rapid rampup of that current, you need a higher voltage to force enough current through the windings to reach (or maintain) a given speed, since higher speeds = shorter phase ontimes = shorter time to ramp current up.*
So with a perfect supply, you don't have to worry too much about number of turns  your power supply will just adapt.
However, most of us don't have ideal batteries or inverters. A typical setup might have a battery that can source 52 volts, and a battery+inverter limit on phase current of 30 amps. This means if there are too many turns on the wind, there may be too much inductance to force enough current through the winding at high speed. At low speed, it's not an issue because there is more time to ramp up the current due to long "on" times for each phase.
If you have too few turns, then your battery may not be able to supply the high current needed to get the torque you want for climbing (for example.) On the other hand, at higher speeds, the lower inductance means that you won't have any problems getting enough current through the windings.
So for real world combinations of battery and inverter, lower Kv ratings will give you more torque at the bottom end, at the expense of speed at the top end.
(*  Yes, I am purposely simplifying here and neglecting backEMF for now. BackEMF also sets a limit on speed based on voltage. As a motor spins up, it starts generating voltage on its own, and that opposes the voltage coming in. Once you hit a motor's base speed, you cannot accelerate past that (due to back EMF) with normal commutation. More turns = lower base speed. That also is a factor in voltage/winding selection.)
bill von
Re: Why more KV = more power, the Maximum Power Transfer Theorem
I'm glad others have responded to this thread as it's been bugging me since it was posted. I've reread it many times and even started but then deleted a reply because I'm not sure if I'm not following it correctly or there is a big oversight.
Unless I'm missing something, motor efficiency is being muddled up with impedance matching:
Apart from peak motor power typically being at (approx!) 50% of no load speed and corresponding efficiency being also 50%, I think the relationship to maximum possible power transfer in an impedancematched system also being 50% is coincidental.
I accept zombiess has stated it's a simplified analysis, neglecting core losses. I also accept the basic premise of matching motor KV to your system voltage to give a useful powerband, but I fear larsb is correct in saying the post is oversimplified...
* It might just be a typo but something else that didn't parse was how input power is 150A x 20V = 3000W, yet output can never exceed 1000W?
Unless I'm missing something, motor efficiency is being muddled up with impedance matching:
I don't believe motor efficiency is related like this, unless "efficiency" means "effectiveness" of power transfer to the motor coils. Describing BEMF as "motor output" and using it calculate motor efficiency is at best confusing terminology.
Apart from peak motor power typically being at (approx!) 50% of no load speed and corresponding efficiency being also 50%, I think the relationship to maximum possible power transfer in an impedancematched system also being 50% is coincidental.
I accept zombiess has stated it's a simplified analysis, neglecting core losses. I also accept the basic premise of matching motor KV to your system voltage to give a useful powerband, but I fear larsb is correct in saying the post is oversimplified...
* It might just be a typo but something else that didn't parse was how input power is 150A x 20V = 3000W, yet output can never exceed 1000W?
Re: Why more KV = more power, the Maximum Power Transfer Theorem
I like threads like this, think about Volts vs KV for top speed, cruising speed and my limited amps vs low end torque.
My thinking was to match my Volts/KV to my cruising for efficiency (efficiency is great). Now it seems set it up for better torque response at cruising speed which seems to be at 50% not 100%. feels like we are selecting camshafts.
Edit: After this post went into Grinn's motor simulator and added my two motors (6KV Leafmotor, 10KV Edge) in to compare them at their respected winding. https://www.ebikes.ca/tools/simulator.h ... heel_b=26i
Used my setup to see what I would gain/lose from changing them out and having a higher KV.
What is more power? Torque or speed! I would need to stay above 35 KPH to get any benefit from more KV but lose below it.
My thinking was to match my Volts/KV to my cruising for efficiency (efficiency is great). Now it seems set it up for better torque response at cruising speed which seems to be at 50% not 100%. feels like we are selecting camshafts.
Edit: After this post went into Grinn's motor simulator and added my two motors (6KV Leafmotor, 10KV Edge) in to compare them at their respected winding. https://www.ebikes.ca/tools/simulator.h ... heel_b=26i
Used my setup to see what I would gain/lose from changing them out and having a higher KV.
What is more power? Torque or speed! I would need to stay above 35 KPH to get any benefit from more KV but lose below it.
Last edited by ZeroEm on Jan 07 2020 1:00pm, edited 1 time in total.
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Re: Why more KV = more power, the Maximum Power Transfer Theorem
What I'm trying to illustrate is that when the motor is run at 19V, while unloaded and draws 10A, this 10A is not doing useful work, only 1V is available to convert into useful work.thepronghorn wrote: ↑Jan 04 2020 4:17pm
If you only have 1V of headroom over the BEMF, then how are you getting 20A into your 0.1 Ohm motor? Do you mean 18V BEMF?
For 19V BEMF, I think you have 0W output power, 190W iron loss, 10W copper loss, and you are 0% efficient.
Your statement "If the goal is peak power, have your motor loaded to half of its max rpm (Kv*bus voltage)." is exactly what caused me to write this up. I've made that statement several times and have had almost no one agree with it. It's a slippery concept grasp and it is an important one to understand.thepronghorn wrote:ZombieSS wrote: If the motor has 5A of losses at 20k RPM, then our motor will be a bit less than 50% efficient. It would in fact only be 47.6% efficient. 50% efficiency would happen at 10.5V of BEMF and only output 996W, but 10.5V * 2000KV = 21k RPM, so a bit over half the unloaded KV.
The Maximum Power Transfer Theorem is why going up in KV while keeping the motor size the same produces more power. Understanding this concept will assist in sizing a motor for a given task. If the goal is peak power, have your motor loaded to half of it's KV.
I think you should be careful with your usage of the Kv term. I think it should say, "If the goal is peak power, have your motor loaded to half of its max rpm (Kv*bus voltage)."
Re: Why more KV = more power, the Maximum Power Transfer Theorem
Clarification, I'm using a theoretical motor and simplified math to demonstrate a concept, the numbers may not be realistic, detailed knowledge of the types of losses and/or mechanical limits are not needed for the concept, but are mentioned to demonstrate limits.larsb wrote: ↑Jan 04 2020 4:35pm
But but but he stutters.
@zombiess:
BEMF is the voltage resisting the input voltage, not the voltage doing mechanical work.. In your calc it’s all backwards or the motor would output max power at max back emf.. that is at noload where efficiency is zero.
If your calcs above would be correct then higher kV motor would always have the upper hand. But with high kV and resulting higher rpm for a given voltage eventually the eddy losses become so high that motor burns or motor explodes from the centrifugal forces created. So there’s an optimum balance to be found.
Re: Why more KV = more power, the Maximum Power Transfer Theorem
I don't think I referred to BEMF as motor output anywhere, but if I did please point it out so I can change the wording. When I'm speaking of motor output, I'm talking about shaft power.Punx0r wrote: ↑Jan 06 2020 6:00amUnless I'm missing something, motor efficiency is being muddled up with impedance matching:
I don't believe motor efficiency is related like this, unless "efficiency" means "effectiveness" of power transfer to the motor coils. Describing BEMF as "motor output" and using it calculate motor efficiency is at best confusing terminology.
Apart from peak motor power typically being at (approx!) 50% of no load speed and corresponding efficiency being also 50%, I think the relationship to maximum possible power transfer in an impedancematched system also being 50% is coincidental.
It's not coincidental, this theorem applies to electrical systems.
It's purposefully over simplified. If you already have a decent amount of motor knowledge you can add all kinds of exceptions to show limitations in an ideal concept vs practical application.I accept zombiess has stated it's a simplified analysis, neglecting core losses. I also accept the basic premise of matching motor KV to your system voltage to give a useful powerband, but I fear larsb is correct in saying the post is oversimplified...
* It might just be a typo but something else that didn't parse was how input power is 150A x 20V = 3000W, yet output can never exceed 1000W?
I'm not sure where you are seeing a 1000W limit on that 150A*20V=3000W.
Re: Why more KV = more power, the Maximum Power Transfer Theorem
BTW, I know my title was clickbait I just wanted to talk nerd with other nerds for a bit, test out the explanation of the relationship between KV and and Power.
Re: Why more KV = more power, the Maximum Power Transfer Theorem
Wow most excellent 'splaining goin' on there.
The key parts of my use case that drives the truth of the "popular misconception being debunked" are
no gearing, need for super strong torque at low rpm, and
Not giving a fig about inefficiency there, nor any desire for speed, even 1520mph tops is fine
The key parts of my use case that drives the truth of the "popular misconception being debunked" are
no gearing, need for super strong torque at low rpm, and
Not giving a fig about inefficiency there, nor any desire for speed, even 1520mph tops is fine
billvon wrote: you need a higher voltage to force enough current through the windings to reach (or maintain) a given speed, since higher speeds = shorter phase ontimes = shorter time to ramp current up.*
…
there may be too much inductance to force enough current through the winding at high speed. At low speed, it's not an issue because there is more time to ramp up the current due to long "on" times for each phase.
…
So for real world combinations of battery and inverter, lower Kv ratings will give you more torque at the bottom end, at the expense of speed at the top end.
Re: Why more KV = more power, the Maximum Power Transfer Theorem
Here:
You've multiplied the applied current by the BEMF to arrive at output power.If we take a 2000KV motor and..Using our 20V power supply with infinite current supply capability, we bring the motor up to 10,000 RPM under load.
10,000RPM / 2000KV = 5V of BEMF
This leaves us 15V to power the motor with. The resistance of 0.1 ohms now comes into play. So how much current are we putting into the motor? Using Ohms law we find that 15V / 0.1Ohms = 150A (it's going to get toasty quick). According to Kirchoff's rule, if we put 150A in, we also get 150A out. We know the motor is outputting 5V of BEMF, so now we can calculate.
Output wattage:
150A * 5V = 750W of output power
Here:I'm not sure where you are seeing a 1000W limit on that 150A*20V=3000W.
I'll be honest I find your post hard to follow and I'm not sure if it's because there are mistakes or if it's correct but I can’t understand it. Or maybe it’s right but for the wrong reasons. For example, maximum power transfer theorem is about matching load to source and is generally unrelated to efficiency of the load. Kirchoff's current law simply says current in = current out, but doesn't tell you what comes out of a motor shaft. Motor speed/torque/power/efficiency curves are what they are but seem distinct to both these concepts. Unless the underlying physics of all this is more joined up than I realised. OR if you simplify a motor to the point where it’s just an coil and you assume a straight 1:1 conversion to mechanical power, then it’s all valid and what you say is all correct. I don’t know.Notice how we haven't and will not be able to exceed 1000W of output power from the motor and this peak is at 50% of the unloaded KV and is only 50% efficient.
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Re: Why more KV = more power, the Maximum Power Transfer Theorem
Hello. Me again. I know I said I didn't want to get involved in a discussion about motor kV, but at this very moment I have a few minutes to spare. My knowledge is nowhere near as complete as some other members who have already contributed to such threads, but I think I can help fill in the blanks.john61ct wrote: ↑Jan 06 2020 11:15pmWow most excellent 'splaining goin' on there.
The key parts of my use case that drives the truth of the "popular misconception being debunked" are
no gearing, need for super strong torque at low rpm, and
Not giving a fig about inefficiency there, nor any desire for speed, even 1520mph tops is finebillvon wrote: you need a higher voltage to force enough current through the windings to reach (or maintain) a given speed, since higher speeds = shorter phase ontimes = shorter time to ramp current up.*
…
there may be too much inductance to force enough current through the winding at high speed. At low speed, it's not an issue because there is more time to ramp up the current due to long "on" times for each phase.
…
So for real world combinations of battery and inverter, lower Kv ratings will give you more torque at the bottom end, at the expense of speed at the top end.
Motor kV is indeed a very useful parameter for defining the performance envelope of an EV power system, but no more than any other parameter available to us.
What motor kV does not dictate (assuming the same copper fill) is the performance limit of a specific motor. Two different windings are capable of exactly the same performance, they just take a different combination of voltage and current (totalling the same wattage) to get there.
I've linked to the thread previously, but there was one specific post by Luke (liveforphysics) that just made the whole subject 'click' for me:
viewtopic.php?t=64907#p974320
“Lets say we have some motor, and it's wound with 2 pieces of say 14awg wire in parallel, and it's wound with 4 turns around each tooth. If we take this same motor, and rewind it to have 8 turns of a single pieces of 14awg (identical copper fill %) or rewind it with 2 turns of 4 pieces of 14awg in parallel (identical copper fill).
So, we have 3 winding options for this motor now, each one yielding a substantial difference in kV, but do any have a performance difference?
We know the 8turn needs half the amount of current to make a given amount of torque than the 4 turn, yet the 8 turn is 50% of the cross section of wire AND twice as long, giving it a 4x increase in phase resistance. This 4x resistance increase perfectly balances the I^2*R difference, both make precisely the same amount of loss per unit of torque produced, both are capable of exactly the same continuous power, continuous torque, etc. Now let's look at the 2 turn, it needs 2x the phase current to generate the same torque as the 4 turn, yet, it's resistance dropped by 4x, because the wire is now twice the cross section and only half as long. It is also capable of exactly the same continuous torque, continuous power, same loss per unit of torque produced, etc.”
Keep reading the above post a few times and hopefully you'll get the same “eureka” moment.
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Re: Why more KV = more power, the Maximum Power Transfer Theorem
No I had absorbed all that and pretty sure I grokked it.danielrlee wrote:Keep reading the above post a few times and hopefully you'll get the same “eureka” moment.
I am in no way saying changing the kV winding does anything for performance **overall**.
What I am clear on, is that **in this use case**
 DD hub only, no gearing
 speed past 20mph is irrelevant
 heavy weight, say 450+ lbs
standing starts & long steep hills, so highest possible torque needed at well under 10mph, even 36mph would be worth sacrificing efficiency
heat being the limiting factor
trying to avoid a 30lb motor
** then** the lowvolts / highamps context means the kV winding makes a significant difference.
That is ALL I am saying, only within that use case, which does not apply to 99.9% of bikers out there.
And not overstating its importance, not like reducing the wheel size for example, but definitely fighting the myth that it's "always" insignificant.
, so please don't assume some straw man to just keep repeating the same arguments.
billvon wrote: you need a higher voltage to force enough current through the windings to reach (or maintain) a given speed, since higher speeds = shorter phase ontimes = shorter time to ramp current up.
…
there may be too much inductance to force enough current through the winding at high speed. At low speed, it's not an issue because there is more time to ramp up the current due to long "on" times for each phase.
…
So for real world combinations of battery and inverter, lower Kv ratings will give you more torque at the bottom end, at the expense of speed at the top end.
 danielrlee 10 kW
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Re: Why more KV = more power, the Maximum Power Transfer Theorem
I must be missing the point you are trying to make.john61ct wrote: ↑Jan 07 2020 8:42amNo I had absorbed all that and pretty sure I grokked it.danielrlee wrote:Keep reading the above post a few times and hopefully you'll get the same “eureka” moment.
I am in no way saying changing the kV winding does anything for performance **overall**.
What I am clear on, is that **in this use case**
 DD hub only, no gearing
 speed past 20mph is irrelevant
 heavy weight, say 450+ lbs
standing starts & long steep hills, so highest possible torque needed at well under 10mph, even 36mph would be worth sacrificing efficiency
heat being the limiting factor
trying to avoid a 30lb motor
** then** the lowvolts / highamps context means the kV winding makes a significant difference.
That is ALL I am saying, only within that use case, which does not apply to 99.9% of bikers out there.
And not overstating its importance, not like reducing the wheel size for example, but definitely fighting the myth that it's "always" insignificant.
, so please don't assume some straw man to just keep repeating the same arguments.
billvon wrote: you need a higher voltage to force enough current through the windings to reach (or maintain) a given speed, since higher speeds = shorter phase ontimes = shorter time to ramp current up.
…
there may be too much inductance to force enough current through the winding at high speed. At low speed, it's not an issue because there is more time to ramp up the current due to long "on" times for each phase.
…
So for real world combinations of battery and inverter, lower Kv ratings will give you more torque at the bottom end, at the expense of speed at the top end.
You don't need any specific winding to achieve what you have described. You could use a higher kV winding with a proportionally lower voltage and higher current. You'd end up with the same torque, the same top speed and the same efficiency.
A certain winding might be better suited in a specific usage scenario where other parameters have been constrained to a fixed value, but it doesn't mean that winding is any more capable at one thing or the other.
Last edited by danielrlee on Jan 07 2020 9:05am, edited 4 times in total.
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Re: Why more KV = more power, the Maximum Power Transfer Theorem
The original premise of 50% being the peak power point is generally recognized. Look at a typical motor graph:
But 50% of what? Changing the kV changes the voltage and current at which 50% happens but it is still 50%. Most motors will burn up if run at this peak power point for very long and nobody really wants to run the motor at 50% efficiency.
But 50% of what? Changing the kV changes the voltage and current at which 50% happens but it is still 50%. Most motors will burn up if run at this peak power point for very long and nobody really wants to run the motor at 50% efficiency.
"One test is worth a thousand opinions"
Re: Why more KV = more power, the Maximum Power Transfer Theorem
Max power at half max rpm is just an effect of the torquespeed curve shape. I think it’s pretty well explained here:
http://lancet.mit.edu/motors/motors3.html
Also there’s a lot of fine data on the ebikes.ca motor simulator covering some motor physics here: https://www.ebikes.ca/tools/simulator.html
http://lancet.mit.edu/motors/motors3.html
Also there’s a lot of fine data on the ebikes.ca motor simulator covering some motor physics here: https://www.ebikes.ca/tools/simulator.html
Last edited by larsb on Jan 07 2020 1:36pm, edited 2 times in total.
Ride on!
Re: Why more KV = more power, the Maximum Power Transfer Theorem
I appreciate you taking the time to post and sort this out. It points out places I can improve on the explanation. I've updated my original post and changed around a bit of the wording. I specified output power as shaft output power to make it more clear that it is mechanical in nature. I'm trying to separate the mechanical/electrical because a motor needs to be viewed as both a motor and a generator.Punx0r wrote: ↑Jan 07 2020 4:26amI'll be honest I find your post hard to follow and I'm not sure if it's because there are mistakes or if it's correct but I can’t understand it. Or maybe it’s right but for the wrong reasons. For example, maximum power transfer theorem is about matching load to source and is generally unrelated to efficiency of the load. Kirchoff's current law simply says current in = current out, but doesn't tell you what comes out of a motor shaft. Motor speed/torque/power/efficiency curves are what they are but seem distinct to both these concepts. Unless the underlying physics of all this is more joined up than I realised. OR if you simplify a motor to the point where it’s just an coil and you assume a straight 1:1 conversion to mechanical power, then it’s all valid and what you say is all correct. I don’t know.
If you think of a motor as a generator, then what you have connected to the terminals is the load. If you have a voltage source connected to the motor, then the load is connected to the shaft. That is until things start moving, then you get both.
I've added the following additional text to the post to try and clarify more.
In the above calculations I list 5V of BEMF and multiply it by the 150A current and call this output shaft power. At first glance this does not appear to be a valid way of of looking at the output power, but you need to remember that the motor is being fed 20V (no PWM involved). The motor is having a load place on it so it is only able to achieve 10k RPM, it is over loaded. That is why it takes 150A of current, very high load. You would not want to do this for any extended length of time. In an EV, this state would most likely only happen upon hard acceleration or high load with the throttle wide open.
The 5V of BEMF is caused by the RPM of the shaft which is performing work on something. That work requires power which it is receiving from the shaft. Power at the shaft is voltage times current. There is 15V of voltage headroom left to add more power to the motor. If the voltage is increased, then the current in the motor will increase as long as it is less than 50% of the 20V bus. Current will stop increasing once the motor reaches RPM = 50% of it's unloaded speed, because after this point, it is unable to add more current to the motor due to the generated BEMF voltage going higher than the voltage source. The current through the wire creates a force proportional to itself as described in Ampere's law / Maxwell equation #4. This is why the output shaft power is equal to the generated BEMF * Input current in these examples.
Equation references with detailed explanations on their derivation.
http://www.maxwellsequations.com/ (click on each equation part for more detail)
http://hyperphysics.phyastr.gsu.edu/hb ... mplaw.html