Voltage, High or Low: The Maths

Blackboard with mathematics sketches - vector illustration

More, with Maths, from our buddy Biff:

If you just increase the voltage of the system you only get more power if you can handle the higher speed of the motor. (True you will get more power without increasing Maximum RPM in the field weakening range, but that is not relevant to the discussion to follow, you get the same conclusion.)
If you adjust the number of turns in the motor, and increase the current capability of the controller you end up with the same result as increasing voltage – more power (if you can spin the motor faster) without added heat in the motor compared to just going higher voltage and spinning faster too.

I know most people have short attention spans, so everything below is just the math and physics to back up the statement I just made, so you can believe me and stop reading now, or continue reading and see the explanation.

– Motor configuration 1 = 10 turns, 100A continuous, 50Nm continuous, 10,000 max RPM
– Motor configuration 2 = 5 turns, ? A continuous, ?Nm continuous, ? max RPM now lets figure out the ?’s

Copper loss pretty much determines the continuous torque of a motor = I^2 * R .  Resistance of the phase leads = cross section area * length * resistivity of copper.  You already have filled the motor with copper, so halving the turns gives you 2x the cross section per turn and about 1/2 the length, resistivity of copper stays the same, so 1/4 the resistance. Put that into the equation and set the copper loss (power being turned into heat) of both motor configurations to be the same to determine the new motor configuration’s continuous current for the same heat production.

100^2*R = I^2 * R/4 => I = sqrt(100^2 * 4) = 200A

Torque is proportional to current * the number of turns (this is because magnetic field strength = number of turns * current).  For a given motor geometry the torque can be determined by T = C * Amps * Turns and C remains constant.

So with the first motor we know that 50 = C * 10 * 100 which gives C = 50/1000.  The second motor we get ? = C * 5 * 200, substitute C = 50 / 1000 and you get ? = 50/1000*1000 = 50

Electrically, Max RPM is defined by the BEMF and the system voltage.BEMF is related to the number of turns (again because of the magnetic field strength being proportional to the number of turns), if you 1/2 the number of turns, you essentially get 1/2 the BEMF / RPM, so you double the maximum RPM (provided the motor can handle that speed) for the same voltage.

Now we have a pretty good idea how motor 2 is going to perform.

Motor configuration 1 = 10 turns, 100A continuous, 50Nm continuous, 10,000 max RPM
Motor configuration 2 = 5 turns, 200 A continuous, 50Nm continuous, 20,000 max RPM

…so if you can handle the higher speed, and you can find a controller with 2x the current, you could get up to 2x the power out of the same motor using the same voltage.

With higher voltage it is sort of easier to derive, you just need to consider the BEMF/ RPM, and as stated above, max RPM is defined by BEMF and system voltage, so if you double the voltage to can spin the motor about 2x faster to get about 2x the power. with the same current.

So there you have it, double current and rewind the motor with 1/2 the turns, or double voltage, the motor doesn’t care, you will get the same result assuming you can allow the motor to spin faster. now the question is what is easier to design, doubling your system voltage , or doubling your current, and that really is determined by the tools and components you have available.

OK, I’ll be the first to admit, I’ve read this through a few times and am still trying to absorb it…  but that’s what we call “learning”, is it not?

Thanks again, Biff!


One response to “Voltage, High or Low: The Maths

  1. While I agree with the basic premises of the post it’s really much more complex in real life. Voltage means pricier FETs more complex gate drive circuitry, dV/dT issues, layout considerations, etc. Add to that the expense of high voltage connectors battery chargers and the fact that switching losses generally increase with voltage and the current vs voltage argument gets muddier. The good news is that I^2 usually trumps a lot of these things. Cost must always be a consideration in engineering. Voltage is a great way to get power density, but it comes at the expense of system cost and complexity.


Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s