Tesla car - mathematical calculations - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: July 12, 2019, 03:41:57 pm
content: I was wondering if we measure DC current which goes to the inverter of Tesla's car. And we know the magnetic flux, can we then calculate the torque on the crankshaft of the induction motor? By T = psi*I?

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: July 12, 2019, 03:58:58 pm
content: Not a motor expert. But surely you would have to determine some proportional constants to account for losses?IIRC there's Kv, Km, Kt?

author: MagicSmoker
author_rank: Super Contributor
timestamp: July 12, 2019, 06:09:34 pm
content: Quote from: nForce on July 12, 2019, 03:41:57 pmNope. While DC input current to the inverter will go up as torque from the motor goes up, this assumes the speed is constant; DC input current will also go up at a constant torque output if speed is increasing.This is because the inverter effectively acts as a buck converter when the motor is operating as a motor (and a boost converter when the motor is regenerating).CurrentTorque is most directly correlated with phase current, but even then you need to know the slip percentage and phase angle displacement with the integrated voltage waveform to calculate it.Determining torque of a PM or other kind of synchronous AC motor is far easier - at least when operated below field-weakening speed - as then it is directly proportional to phase current (neglecting some effects from saturation, etc.).EDIT - corrected striked out word

author: max_torque
author_rank: Super Contributor
timestamp: July 12, 2019, 06:54:00 pm
content: If you measure the DC voltage and DC current that flows into (and out of) the inverters DClink, you know the power that is being used by the inverter and motor.If you know the speed of the electric motor, then you can estimate the torque of that motor, because power = speed x torque.I say "estimate" because you don't know either the inverter or motor efficiencies, however if you estimated a 5% loss for the inverter and a 5% loss for the motor you wouldn't be a mile off........

author: nForce
author_rank: Frequent Contributor
timestamp: July 14, 2019, 05:42:28 pm
content: Thanks guys.I have one question if we measure the DC current to the inverter, should then the 3-phase currents if we sum them up be this DC current, minus the losses in the inverter?

author: MagicSmoker
author_rank: Super Contributor
timestamp: July 14, 2019, 07:37:05 pm
content: Quote from: nForce on July 14, 2019, 05:42:28 pmNope again for an induction motor (less wrong for a synchronous motor). This is because some of the phase current driving an induction motor is used to induce the field in the rotor, and some of it is reactive current that is inversely proportional to power factor (generally speaking, power factor goes up with higher loading) and which merely sloshes back and forth between the winding inductance and the DC link capacitance (causing losses in the semiconductors but not doing anything useful).In synchronous motors there may or may not be reactive current - it is possible to make this type of motor appear either inductive or capacitive by slightly altering the phase angle of the current with respect to the actual rotor angle and/or varying the field intensity (for wound field motors) - but otherwise the DC input current to the inverter is roughly equal to the RMS current in any *one* phase multiplied by 3^0.5. That is, if the DC input current is 173A then the expected current in any one phase should be approximately 100A.EDIT - flipped the currents in my example... doh!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 14, 2019, 07:45:01 pm
content: No, or not necessarily.Reactive (AC) power can be cycled in and out of the inverters, pushing current through the motor without necessarily spinning it, and only incurring resistive and switching losses, without delivering mechanical power.The most fundamental case of this is your butt. Your butt, when sitting, delivers a force into the chair (and vice versa), but no work is being done, because there is no motion induced by that force. The act of sitting down, or standing up, transfers energy from that system -- reactive power -- but that [reactive power] is separate from the average work (real power) done in the system (which in this case, is a small amount of heat loss due to your butt squishing into the cushion).An AC motor delivering torque at zero RPM, is also a case of this.One could argue still further, and perhaps less usefully -- that the time scale of this reactive power should be even longer. See, the motor can deliver torque into nonzero RPM, and therefore real power; but if we get that power right back during regenerative braking, it's not reallyrealpower, is it? It's just power we've taken out of the battery pack, then put back in. Just like reactive power delivered at the motor's stator frequency*, or like reactive power delivered at the inverter's switching frequency (mainly in the PWM filter choke, which may be the winding inductance).*Which, speaking of, these are synchronous motors, aren't they? For these, the stator frequency equals the rotor frequency, and therefore no AC power is delivered at zero RPM, and no reactive power either. This analysis really isn't all that useful then -- it's not invalid per se, but it's not useful if the reactive power happens to be zero anyway. It does however apply to induction motors, where a low frequency is applied to the stator to deliver torque at zero RPM.Regardless, the key insights are:- Torque is a rotating force; a force can exist without motion and therefore without  power. The force still stores some energy, which is released when the force is removed. This can be considered a reactance.- Reactive power is not average [real, DC] power, so we can't necessarily expect the average DC current draw to reflect it. We can expect some AC ripple to reflect it, but this may be filtered at the inverter -- reactive power can be cycled between different reactive elements, namely capacitors and inductors in an electronic filter -- so we cannot necessarily get the full picture just by measuring current at the battery pack, even if we have a higher bandwidth current probe than a clampmeter.Another gotcha is, even if we know flux (say because we know the driven frequency if any, and the inverter output voltage), we don't know the turns ratio without taking apart the motor, and therefore we don't know the -- in essence, electrical gearing, of the system.But if we know all those things -- as the power train designers do -- we very well can, and do, make use of that, that current and torque are related.Tim

author: MagicSmoker
author_rank: Super Contributor
timestamp: July 14, 2019, 07:47:10 pm
content: Quote from: T3sl4co1l on July 14, 2019, 07:45:01 pmThe Model 3 uses a PM synchronous motor, but the other models use induction motors.

author: IanB
author_rank: Super Contributor
timestamp: July 14, 2019, 08:18:36 pm
content: Quote from: nForce on July 12, 2019, 03:41:57 pmI would say it's easier to work back through the mechanical drive train from external observations. For example, if you know the weight of the car and the time taken to accelerate from 0 to 60 mph (or 0 to 100 kph) then you can work out the torque at the axles. From there, knowing gear ratios, you can determine the torque and power at the motors.Here's an example of the kind of calculation:https://www.eevblog.com/forum/chat/some-musings-on-the-traction-current-of-an-electric-train/msg114506/#msg114506

author: GeorgeOfTheJungle
author_rank: Super Contributor
timestamp: July 14, 2019, 09:57:23 pm
content: Quote from: nForce on July 12, 2019, 03:41:57 pmPower out = Power In - lossesPower out = torque * rpmPower In = Iin * Vin=>torque= Pout / rpm = (Pin - losses) / rpm =((Iin*Vin) - losses) / rpm