Tesla Physics

[Krash]

Kudos SucreTease for converting power velocity into torque. Keep in mind that the Power Velocity is average and not adjusted for state of charge. Once that is corrected I think the Tesla numbers will look very understated. Here is how that looks roughly on the empirical data at the time of posting. Check the next Performance Metrics Tracking update for current data...

I believe the published numbers are from Wikipedia. Feel free to recommend corrections.

 

[csanders90D]

While I am looking at the empirical data I am curious as to why the ratio of Current (Amperes) to Power (kW) varies over time...

View attachment 260333

Power (Watts) = Current (Amperes) * Voltage

The battery voltage decreases as current increases, and then will bounce back as current decreases. Guessing at the numbers from your chart, the battery dropped from 385 to 294 volts. There are a variety of factors that affect how much the battery voltage sags, including size of the battery, age, state of charge, temperature, etc.

 

[Mediocrates]

While technically correct, this entire subject covers motion in one dimension, and all cross-products are at right-angles, so scalar values (magnitudes only) are entirely warranted. Also, what you call "colloquial" terms have clear, precise dictionary definitions, and I do not believe that either me or @Krash meant them any other way. Accelerate means to move faster, decelerate means to reduce speed. This is how the terms were used by me and @Krash. Unless @Krash wants to correct me here, the confusion was about concepts, not words.

I recommend reviewing the concept of equivocation.

 

[SucreTease]

I think the following article has good explanations on the complicated relationship behind battery performance and other factors. It shows how complicated the picture is and why Tesla may have made the performance upgrades dependent on battery sensors.

Rising Internal Resistance - Battery University

 

[SucreTease]

Here is a good introduction to electric motors that explain many of the pros and cons of various motor designs. Note that Tesla uses a three-phase A/C induction motor for the Model S/X. The article does not go into the details that will answer some of the more challenging questions that @Krash asks, but will be useful for others who want to understand the basics.

Electric motors and generators

And here is a little more detail on the style of motor that Tesla uses in the S/X:

AC induction motors | How AC motors work

 

[SomeJoe7777]

In the 3rd region (where the power limit ends), the limit is caused by the RPM of the motor becoming such that the back-EMF generated by the motor opposes current flow enough to start limiting power.

In the 2nd region, limitation is maximum power that can be supplied through the inverter and/or motors due to heat limits. In this region, the inverter is controlling the power delivered according to the (simplified) equation:

P = (Eio - Ebemf) * Iio

where P = total power delivered to the motor, Iio = inverter output current, Eio = inverter output voltage, and Ebemf = back-EMF voltage generated in the motor due to RPM. This increases as motor RPM increases.

To regulate power as the car speeds up through region 2, the inverter has to increase voltage output while maintaining current at the maximum level (dictated by battery limits). Once the inverter voltage output increases to the point where it is equal to the battery terminal voltage, then the inverter can no longer maintain power at the maximum level. Inverter output voltage cannot increase past the battery terminal voltage, so you cross into region 3 where the increasing RPM is increasing back-EMF, and that causes power to drop.

To continue to maintain power at the maximum level to higher RPM, you need more available battery terminal voltage. Decreasing battery internal resistance with better chemistry is one way, or a larger battery in a higher voltage configuration is another way. Note that rearranging existing cells into a higher voltage configuration won't do it, as the trade-off would then be a lower current limit.

 

[SucreTease]

In the 3rd region (where the power limit ends), the limit is caused by the RPM of the motor becoming such that the back-EMF generated by the motor opposes current flow enough to start limiting power....

It's great to have you join us here, @SomeJoe7777. I am happy that someone with knowledge can fill in areas where I am ignorant. @Krash will be quite happy, too, that someone can answer some of the questions that keep him up at night.

To regulate power as the car speeds up through region 2, the inverter has to increase voltage output while maintaining current at the maximum level (dictated by battery limits).

To see if I understand all the implications of what you explained, would it be more precise to say, "the inverter has to increase voltage output in order to maintain current at the maximum level"? In other words, the inverter increasing its output voltage is not incidental to the current level, but done because of the need to maintain maximum current, correct?

 

[SomeJoe7777]

To see if I understand all the implications of what you explained, would it be more precise to say, "the inverter has to increase voltage output in order to maintain current at the maximum level"? In other words, the inverter increasing its output voltage is not incidental to the current level, but done because of the need to maintain maximum current, correct?

Yes, that is correct. The inverter cannot completely independently regulate voltage and current, they are inter-related by Ohm's law. So yes, in order to maintain current at the maximum permissible level given battery limits, the inverter must increase output voltage.

The inverter works on a pulse-width modulation basis, switching the IGBT devices on and off at high frequency, with the width (more specifically, duty cycle) of those pulses controlling the output voltage. The output section of the inverter contains capacitors and inductors which filter the high frequency square wave produced by the inverter and result in a smooth output.

I don't know the specific design frequency of the Telsa inverter for the switching, but typical PWM switch-mode power supplies use frequencies in the 20-50 kHz range.

Note that this high frequency switching controls the output voltage. The inverter is also slowly varying this output at a much lower frequency to drive the motor. Maximum motor speed is around 18,000 RPM, which in a 2-pole 3-phase induction motor requires 300 Hz. So, we have a 3-phase inverter AC output at anywhere from 0-300 Hz, with voltage controlled by 20 kHz switching that's being filtered.

On top of this, we are driving an induction motor, which means that the inverter drive frequency and the motor rotational speed don't match (motor runs slower during acceleration). The difference in frequency is directly proportional to the torque produced by the motor, and directly affects current. So, the inverter has two axes of control: pulse width at the switching frequency and motor drive frequency. By varying both of these parameters, the inverter can drive the motor to produce the desired torque while limiting other parameters like battery current or total power.

A simplified summary of the decisions made by the drive inverter could be:

• Accelerator pedal is pressed to a certain amount, this is assigned as the torque that the driver desires.
• Based on the actual motor rotational speed at this instant, the necessary drive frequency to deliver that torque is computed and the inverter begins supplying the motor with that drive frequency.
• The new frequency difference between driving frequency and motor rotational speed changes the motor's back-EMF (lowers it), increasing the current flow.
• The capacitor and inductor elements on the inverter output have some stored energy in their magnetic and electric fields, this supplies the current for an instant. After that, the terminal voltage at the inverter output begins to drop due to the increased current flow.
• The inverter increases the PWM width (duty cycle) to increase output voltage to maintain current flow.
• As more current flows, battery terminal voltage sags due to internal resistance, so the inverter will compensate by increasing PWM duty cycle more to maintain output voltage, and therefore current flow.
• The inverter checks total power, total current, and total motor torque constantly, and will limit itself at any point if required.

The inverter for the Model S and Model X, because they have to drive an induction motor, are some of the most complicated, yet best-engineered inverters I've ever seen. About 20 years ago, I worked in an industry where certain motors needed to run at a lower speed than could be done with line frequency, and those motors were equipped with a fixed-frequency inverter to run them at a lower speed. The cabinets were the size of a bookshelf, were only pushing about 50 kW, made a noise like a hoard of locusts, and raised the temperature in that motor room to where you couldn't stay in there very long.

The Model 3 supposedly has a permanent-magnet motor. The inverter to drive that motor is considerably simpler because there is no frequency difference between the supplied frequency and motor speed. Torque is produced by the rotating magnetic field leading the permanent magnet field by a certain angle (called the torque angle), which simplifies calculations and operation.

 

[Krash]

So here is just the empirical quarter mile data that we have. The top two cars are S100Ds. The bottom three are S75Ds. kthoring's car is the only one of the five still corked. The top two and the bottom one are 400v. The two in the middle are 350v. Reported amperage of the three batteries are: BTX6 1850A, BTX5 uncorked to 1300A, BTX8 1500A\1600A might be corked to original 1150A.

Here is what I observe about this third period where the power drops: For the top two BTX6 batteries the 14% SoC difference means lower SoC power starts to drop 3 mph earlier but max power stays the same. For the BTX5 batteries the 29% SoC difference means the lower SoC power starts to drop 4 mph earlier and max power drops 20kW. Max Power difference between these two main battery types is about 75kW and the power drop starts 10 mph earlier for the lower voltage batteries. The BTX8 never experiences any drop.

Note that I have converted from time on the x axis to velocity for this example. Anyone still think Back EMF is causing the decrease in power? Any additional insight as to what causes that drop?

 

[SomeJoe7777]

We can view this region 3 phenomenon in a different way by analyzing the torque equation and the inverter response.

Let's briefly review the way an induction motor works. The electrical power is supplied via 3 phases to the stator, which is fixed and does not rotate. However, the sinusoidal AC waveform applied to the 3 stator phases causes a rotating magnetic field to appear inside the stator. The speed of rotation is given by:

where N=the speed of the magnetic field rotation in RPM, f = the applied AC waveform frequency in Hz, and P = the number of poles (pairs of windings) per phase in the stator of the motor.

This speed is referred to as the "synchronous speed". Take a normal AC motor in an industrial application, such as a pump. With 2 poles and a 60 Hz line frequency, the synchronous speed is 3600 RPM.

The rotor of the induction motor lies within this rotating magnetic field, and contains shorted windings. Like a transformer, the rotating magnetic field from the stator induces a current to flow in the rotor windings (hence the name induction motor). The current flow in the rotor causes the rotor to create its own magnetic field. The interaction between the stator's rotating magnetic field and the rotor's magnetic field now creates a torque that causes the rotor to rotate.

The important thing to realize is that the rotor cannot turn at synchronous speed. If it does, the net difference in speed between the stator's rotating magnetic field and the rotor will be zero, and no current can be induced in the rotor. This causes the magnetic field of the rotor to fall to nothing, and therefore no torque can be produced.

Thus to produce torque, the rotor always must rotate slower than synchronous speed. The difference between the two speeds is referred to as slip and is given by the equation:

​

s is the slip (near 0 when the motor is unloaded, 1 when the motor is at standstill), Ns is the synchronous speed, and Nr is the rotor speed.

A running induction motor developing torque must always have some slip. Torque is directly proportional to slip, and the more slip, the more torque.


Now, let's explain why the Model S hits region 3, where the power begins to decrease. Let's start with the torque equation for an induction motor:

T is total torque
ns is the synchronous speed of the stator's rotating magnetic field
s is the slip
E2 is the rotor's induced voltage. This is proportional to the stator's voltage E1 which is applied by the inverter.
R2 is the rotor winding's resistance (fixed).
X2 is the rotor winding's inductive reactance (fixed).

Now remember that the inverter can vary both the applied frequency (and therefore the synchronous speed, Ns, and thus also the slip, s) and the applied voltage E1 (which then directly affects E2).

Let's assume a Model S is accelerating at maximum capability through region 2, where the inverter is keeping power limited to the maximum that the inverter elements can handle. Let's hold the inverter parameters constant, and observe what happens in the torque equation if the car speeds up a small amount:

​

As the car speed up, rotor speed goes up, which reduces the slip, s. The net result is that torque would go down. The inverter has to react to this to maintain torque. The inverter can do two things: Increase applied frequency and/or increase applied voltage. At first, it might seem that we can completely counteract the increase in rotor speed by just increasing applied frequency. This will bring the slip back up. Unfortunately, there is another effect in that increasing frequency also increases synchronous speed, and the Ns parameter in the denominator means that it will bring torque down:

​

Thus, the inverter can't completely counteract the torque decrease by only raising frequency. It will have to increase applied voltage as well:

​

Now the torque can be brought back up along with the current, and our power can remain constant.

Note that the inverter is taking battery voltage and dividing it. Some of that voltage is dropped across the IGBT's in the inverter, the rest is delivered to the motor (E1). With what just happened above, the division of the voltage has changed: Less of the battery's voltage is being dropped across the IGBT's, and more is being delivered to the motor. This process continues throughout region 2.

Eventually, we hit a point where nearly all of the battery's voltage is being delivered to the motor, and virtually none of it is being dropped across the IGBT's. At this point, E1 (and therefore E2) can no longer be increased by the inverter. We enter region 3.

Now, the only thing the inverter can do is increase frequency. We result in the same equations as above:

​

The vehicle speeds up, the slip decreases, and therefore the torque decreases. The inverter will try to compensate by increasing applied frequency, but that can't fully stem the torque decrease.

​

With lower torque, current drops. With constant voltage and less current, we have less power.

You might ask, well, can't we just increase slip even further by raising applied frequency and get more torque? At this point, no you can't. There is a balancing act for increasing the applied frequency. At low values of slip, increasing the applied frequency will indeed increase the torque, but past a certain amount, the ns term in the torque equation becomes dominant, and increasing applied frequency past that point will result in a torque decrease, not an increase. The inverter will keep the applied frequency at the exact point throughout region 3 such that the torque is maximal.

 

[SucreTease]

View attachment 261671Anyone still think Back EMF is causing the decrease in power? Any additional insight as to what causes that drop?

The four "Maxwell's Equations" describe all known behavior of macroscopic electric and magnetic phenomena. One of the fundamentals of electrodynamics is that a a time-varying magnetic field will produce an electric field proportional to the time derivative of the magnetic field. This is expressed in one of Maxwell's Equations as:

(1) ∇ × E = -∂B/∂t

This equation describes the spatial variation of an electric field (on the left) that is produced by a time-varying magnetic field (on the right). The real key, here, is that pesky minus sign, which we shall return to momentarily.

This spatially varying induced electric field produces results in voltage differences across space which, in turn, produce a voltage difference across an electric circuit, thus causing current to flow. Another one of Maxwell's equations describes how, reciprocally, an electric current (or a time-changing electric field) induces a magnetic field:

(2) ∇ × B = ₒ(J + ₒ∂E/∂t)

This equation describes the spatial variation of the magnetic field (on the left) that is produced by a current J or time varying electric field (on the right).

So, in essence, what happens inside the motor is that an AC current flows though the windings of the stator of the motor. This induces a time and spatially varying magnetic field within the motor (per equation 2). This, in turn, produces a time and spatially varying electric field (per equation 1), thus inducing within the windings of the rotor a current flow. Per equation 2, this current flow in the rotor produces its own magnetic field. The magnetic field of the rotor is repelled and attracted to different areas of the magnetic field produced by the stator and the rotor turns. By appropriate modulation of the current flow through the windings in the stator, we can cause the rotor to keep moving in a desired direction.

Now, back to equations 1 and 2. We note that there is a reciprocal relationship between them, in that a changing magnetic field produces a electric field (and, therefore, a current in an electric circuit), and a current will produce a magnetic field (running us through both equations).

That pesky minus sign means that if the original magnetic field is changed (i.e. due to the AC current causing it, or due to the motion of the rotor through the spatially varying magnetic field), an electric field will be induced in opposite polarity to the one already driving the induced current in the rotor. A similar effect is going on within the windings of the stator because of the changing magnetic fields caused by the current flow in the rotor. In each case, the voltage (EMF, electro-motive force) induced by the time-varying magnetic field is in the opposite direction to the voltage driving the current through the windings. This "back-EMF" thus reduces the net voltage applied to the pole windings, reducing the current that can be pushed through them. The faster the magnetic field changes (i.e. the faster the rotor spins), the stronger the back-EMF until it equals, and cancels, the externally applied voltage, and the net current flow through the windings approaches zero.

I hope this helps.

 

[john pane]

This is all very interesting. We have seen that Tesla has been installing various battery configurations. There are 400V and 350V batteries, and they vary in the number of cells in parallel (which I naively assume is related to maximum current they can deliver). I realize there are other variables, like the drive unit and vehicle weight. But I would be very curious to see, holding other factors constant, the above theory translated into theoretical acceleration curves for the various battery configurations. To my knowledge, there are at least the following:
400V, 86P (cells in parallel)
400V, 74P
350V, 80P
350V, 74P

 

[Krash]

Thank you SomeJoe7777 and SucreTease. I think I get the basics. It will take me a bit to get through the math.

If I understand correctly the batteries are DC and the inverter is converting into AC for the three phase motor. Does it not matter what the battery pack (DC) voltage is then in determining when Back EMF happens? Is it only the overall current of the battery that determines Back EMF and the higher voltage batteries just happen to have higher current which is why the higher voltage batteries appear to deal with Back EMF better? I understand that current and voltage are related.

What would the proper units be to measure the loss of power over time over time due to EMF, say worst case from 65 mph to 155mph over 30 seconds starting at 0 to at most 200W lost? Watt Seconds or Joules? 100W average over 30 seconds would be 3000 Joules?

Why wouldn't Tesla simply switch to advanced capacitors in or near the inverters to cover this gap? I am not sure what that amount of energy is in Farads but isn't the cost in the thousands of dollars rather than tens of thousands? I have a feeling that when I do the math it will be obvious why this won't work since the new inverters and old perform almost identically in this regard, unless the new inverters are somehow still corked.

It sounds like Back EMF by itself doesn't generate any extra heat.

Thanks again for the detailed explanation.

 

[SucreTease]

Thank you SomeJoe7777 and SucreTease. I think I get the basics. It will take me a bit to get through the math.

I think it is awesome that we had @SomeJoe7777 join us here. My theoretical knowledge only goes so far, and he seems to have great knowledge of motor design and performance (applied physics).

If I understand correctly the batteries are DC and the inverter is converting into AC for the three phase motor.

That is correct.

Does it not matter what the battery pack (DC) voltage is then in determining when Back EMF happens? Is it only the overall current of the battery that determines Back EMF and the higher voltage batteries just happen to have higher current which is why the higher voltage batteries appear to deal with Back EMF better? I understand that current and voltage are related.

The inverter can only do its job if the necessary resources are available to it. So, yes, it does matter what the input voltage is (current will follow from the input voltage--it is not independent of it). The current from the battery is not directly related to back-EMF (the inverter creates its own time-varying voltage and current levels out of the resources provided by the battery). It is the magnetic field generated by the induced current in the spinning rotor which is causing the back-EMF in the stator. The faster it spins, the stronger the back-EMF. Also, the more current flowing through the stator, the stronger will be its magnetic fields, and, in turn, the stronger the current they induce in the rotor, and the stronger the magnetic fields generated by the rotor, and the larger the back-EMF, in a vicious cycle).

What would the proper units be to measure the loss of power over time over time due to EMF, say worst case from 65 mph to 155mph over 30 seconds starting at 0 to at most 200W lost? Watt Seconds or Joules? 100W average over 30 seconds would be 3000 Joules?

Power is the time-rate of change of energy delivered (first time-derivative of energy). It has units of energy/time. E.g. Joules/s = Watt.

To get the total energy delivered over a period of time, you would integrate the power over the time. For a constant power level, that is just the power times the time. W*s = (J/s)*s = J, but most of the time, in applications using batteries, it is expressed as W*hr or kW*hr.

But that is not what you are wanting to do. As you said, you want to measure a change of power over time, which is the rate of power loss (drop). A time-rate of change in power is measured as change of power over change in time (first time-derivative of power, or second time-derivative of energy). Thus, it has units of power/time, i.e. W/s or kW/s ("velocity of power" as it were). You could write it as J/s² also, but no one ever does that.

Why wouldn't Tesla simply switch to advanced capacitors in or near the inverters to cover this gap? I am not sure what that amount of energy is in Farads but isn't the cost in the thousands of dollars rather than tens of thousands? I have a feeling that when I do the math it will be obvious why this won't work since the new inverters and old perform almost identically in this regard, unless the new inverters are somehow still corked.

This is in the realm of motor design, so outside of my expertise.

It sounds like Back EMF by itself doesn't generate any extra heat.

That is correct. A voltage, by itself, does not cause energy loss (e.g. heat). What actually causes energy loss is current flowing through a medium with electrical resistance. This is the electrical analogy of friction. The rate of power loss is equal to I²R, where I is the current (measured in amps), and R is the resistance (measured in ohms). This power loss is converted to heat. As you can see, this increases by the square of the current (just as aerodynamic friction increases with the square of the speed), and is one of the two reasons why you get worse efficiency when you drive faster (or accelerate faster, pulling more current).

 

[Krash]

Interesting carbon wrapped copper performance revealed in the new plaid motors 10 June...

Goodbye back EMF. I assume the common leftmost slope is for simplification since the fixed max torque setting is not the same for those motors, Also because those motors have different 0-60 times which are mostly determined before the cars reach the different power limits.

Knowing the plaid has two (plus a front motor), the slope could even be lower for an individual motor.

Here are a few questions about that carbon wrapped technology...
1. Is the variable cost per motor expensive, or like the mega presses is it just expensive to setup?
2. Related to the above, does/would it benefit the PM motor in front?
3. Would/will it benefit the 3/Y motors as well?
4. Does the carbon wrap change the motors’ efficiency?
5. Anyone think the quarter and half mile times will be the really amazing change?