back-emfdc motorefficiencymotor
When learning about regenerative braking, I began to ponder as to what the relationship between the number of turns in the coil of a DC motor is to the efficiency of kinetic energy retrieval from the rotating wheel.
In math terms, what is the relationship between N and eta where:
$$\eta = \frac{E_E}{E_K}$$
If you can cite some references as well relating to your answer for further reading, that would be extremely helpful.
Probably you already have it, and just didn't know it. If you are driving a motor with a half-bridge or H-bridge and PWM or similar, you have regenerative braking. Let's consider a half-bridge, since for this analysis we will run the motor in only one direction:
First, let's consider non-regenerative braking. If the bridge output is high (S1 closed, S2 open), the motor will accelerate to full speed. If the bridge is now switched low, the motor won't just gently coast to a stop. It will slam to a stop, as if someone but a brake on it. Why?
A motor can be modelled as a series inductor and voltage source. The motor torque is proportional to current. The voltage source is called back-EMF, and it's proportional to the speed of the motor. This is why a motor draws more current when it is loaded (or worst, stalled): with the speed decreased, the back-EMF is decreased, and it opposes the supply voltage less, resulting in higher current. Let's redraw our schematic with that model, with values as if our motor is spinning at high speed:
This motor is running at full speed. We have a small current to overcome the friction in the motor, and the back-EMF is the supply voltage, less the voltage drop over R1. Not much current flows because the back-EMF cancels most of the supply voltage, so L1 and R1 see only 100mV. Now what happens when we switch the bridge to the low side?
At first, nothing. L1 prevents an immediate change in current. However, this doesn't last long, and very soon (defined by the time constant of \$L_1 / R_1\$, not more than a couple \$ms\$ typically) the back-emf (V1) has reversed the current, and now it's going in the other direction. It's also pretty huge, since now L1 and R1 don't see the small difference of \$V_{B1}-V_1\$ (it was \$100mV\$), but now they see the full 9.9V from V1 alone:
We now have a large current flowing in the opposite direction. Torque is proportional to current, so now instead of applying a gentle clockwise force, just enough to overcome friction, we are applying a hard counterclockwise force, and the mechanical load is rapidly decelerated. As the speed of the motor decreases, so does V1, and consequently so does the current, and the torque with it, until the load is no longer spinning.
Where did the energy go? The kinetic energy of the mechanical load is energy. It can't just disappear, right?
Right. If you look at the circuit again, we have 9.9A flowing through R1. \$P_{R1} = (9.9A)^2 1\Omega = 98.01W\$. The kinetic energy of the load was converted into heat in the motor's winding resistance (and in a practical circuit, also the H-bridge transistors). Some motors will be destroyed by this high power. Others may not. The current generated by the back-EMF is about as strong as the stall current of the motor, so if your motor can run stalled without overheating, it can brake like this all day.
So how do I store the energy, instead of converting it to heat?
Let's look at what's happening a bit after we have started braking, but before we have stopped:
The motor has slowed significantly (back-emf is 1V), and the current has decreased with it. Now what if we switch the bridge to the high side?
Ah ha! We are charging the battery! Of course, if we stay like this very long (again, defined by time constant \$L_1 / R_1\$) then the current direction will reverse, and we will be uncharging our battery, and accelerating our motor, not braking it.
So don't do that. As long as we remain in this state, the current is decreasing. So, we switch back to the other state, with the bridge low, so the back-emf can build the current back up. Then we switch again, and shoot some of it into the battery. Repeat, fast.
If this sounds like what one ordinarily does for PWM motor control, it's because it is. This is why probably you already have it, and just didn't know it.
Once you understand the principle of operation, you can make some simplifications. When a motor is being driven by PWM, the inductance of the motor (L1) works like a flywheel, averaging the voltage you apply to the motor. It's as if you had a real flywheel, and spun it by striking it with a hammer repeatedly. So in this example our supply voltage is 10V. If our PWM duty cycle is 80%, we are effectively driving the motor with 8V (\$80\% \cdot 10V = 8V\$).
Whenever the back-EMF is greater than this voltage, you get regenerative braking. This will happen whenever the PWM duty cycle decreases faster than external forces (friction, for example) will slow the motor. Any resistance in the circuit reduces the energy you can recover from the mechanical load. In the most extreme case where the PWM duty cycle is decreased to 0% and the motor terminals are shorted together, the current is so high that losses reach 100%. (\$P = I^2 R\$)
You can also open all the transistors on the bridge, and the inductor current will die out through the diodes in the bridge. Then neither the back-EMF nor the battery will have a path to drive a current, and the motor will freewheel. Unless of course, some external force accelerates the motor enough to push the back-EMF higher than the supply voltage. A vehicle rolling down a hill is a good example.
In all other cases, you get regenerative braking.
You must consider what you will do with the mechanical energy from the motor. Batteries can absorb energy, but there's a limit to how much, and how fast, that varies on the type of battery. Some power supplies (linear voltage regulators, for example) can't absorb energy at all.
If you don't provide a place for the energy to go, either a battery, or some other load in the circuit, it will go into the power supply decoupling capacitors. If you have enough energy returned from the motor and not enough capacitance, the power supply rail voltage will increase until something breaks.
You must design your circuit so it can't happen. In an electric car, there are complex battery controllers that will apply the conventional brakes if the batteries can't absorb any more of the car's kinetic energy. You can also switch on a power resistor across the supply rails, or design your motor controller to back off on the braking if it gets to be too much.
What happens if we have a motor with no winding resistance, and we have a way to drive it without adding any additional resistance (ideal transistors and wires)? It's more efficient, obviously. But how does the speed of the motor vary with applied voltage and mechanical load? Hint: if you try to change the speed of the motor by increasing or decreasing the mechanical load, what does the back-emf do to the current?
Field Oriented Control (also known as Vector control) of BLDC motors can improve low speed torque and reduce torque ripple at speed. Normally it uses sine wave drive. This works best when matched to motor with sine wave back-emf. However most motors used in multirotor drones are designed for 6 step trapezoid drive.
The following traces show the different back-emfs of a 2 pole coreless ironless BLDC motor and a multipole iron cored motor. The second example might actually perform worse with sine wave drive.
So to get the best out of FOC drive it needs to 'tuned' to the motor. This might explain why the only two FOC ESCs I could find for sale (DJI 1240S/X and EMAX WindTalker) are both intended to be used only with a specific motor.
The vast majority of ESCs used in multirotors don't even use synchronous rectification, so their part-throttle efficiency is already ~5% worse than it could be. Also many of them are not using the latest most powerful processors and high speed drivers, so their switching losses are higher at high frequency (this is more important when using FOC drive, because to produce an accurate waveform requires high frequency PWM).
The only one way to find out what real improvement FOC drive could make is to try it. For a fair comparison you would have to use the same ESC and only change the commutation technique.
Best Answer
The number of turns is irrelevant.
All changing the number of turns does, assuming you also change the wire diameter so that the resulting winding fills the space available, is to change the impedance of the winding, the voltage/current ratio, not the efficiency.
Consider the space filled with two identical windings, running at the same current, voltage, field, dissipation etc. If you connect them in series, it runs at 2V and I, you have twice as many turns as each coil. If you connect them in parallel, it runs at V and 2I, you have double the copper area but the same number of turns. All of the important parameters, field, weight, torque, power lost, cost of the copper, are unaltered, even as you've doubled or halved the number of turns.
That's all to first order of course. There are many small second order things going on. The wires to the motor can be thinner if the motor impedance is higher. Thin wire tends to have a greater part of its cross sectional area occupied by insulation than thick wire, so it's not quite as efficient as medium weight wire. Very thick wire also doesn't pack as well due to its stiffness. A very high voltage motor may well dedicate more space to insulation than a lower voltage motor. There may be other considerations for voltage, for instance availability of 12v batteries, or staying below 40v so you can use 'low voltage' insulation rules.
That's enough about number of turns not affecting efficiency, so what does?
The main loss is \$I^2R\$ loss in the copper. Unfortunately there's little that can be done about the resistivity of copper. Silver is too expensive (and barely any better), cryogenics is very complicated, though superconductor is worth doing if the machine is really, really big (>> 100MW).
High field magnets produce more volts per turn, and more torque per current than low field magnets, so you should use the highest field you can afford. More volts per turn means a higher ratio of 'useful' volts (coupled to the speed of the motor) to 'useless' \$IR\$ drop volts that only generate heat, so better efficiency. A small airgap helps produce a higher field with the same magnets, but requires better machining tolerance and produces more air viscosity losses (aka windage), so it can't be reduced too much.
A high speed motor produces more power than a low speed one at the same output torque, which is another way of saying generates a higher EMF than a slow one. Again, a higher ratio of useful volts to useless volts. However, high speed requires strong materials, and careful balancing, and produces losses that increase rapidly with speed, so it can't be increased too much. The higher rotor frequency means more hysteresis loss in the magnetic materials, and the windage losses increase as the speed cubed, so will rapidly come to dominate at very high speeds.
Because of the way some power and loss terms scale, a big motor, other things being equal, will have lower losses than a small motor. But let's say you want four driven wheels. Is it better to use a small motor per wheel, or a big motor and a mechanical drive train? Almost certainly the former, for lower weight and complexity, even if it's slightly less efficient.
Motor/generator design is a compromise between efficiency and, well, cost really, once you have a motor powerful enough for your specifications. Unfortunately, even the simple things like cost, weight and volume are a multidimensional space you can't simply optimise over. When you add things you can't easily put numbers on, like time to design, maintainability, reliability, the rest of the vehicle efficiency, convenience, use of strategic materials (in strong magnets), there is no one size fits all, no one equation describes all.