zombiess
10 MW
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.maxwells-equations.com/ (click on each equation part for more detail)
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/amplaw.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.maxwells-equations.com/ (click on each equation part for more detail)
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/amplaw.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.