Modeling motor control

[Kevin Sevcik]

I've now deciphered the current sampling strategy employed by the Jaguars. It's a pretty simple scheme, they probably just wanted a robust fast solution over the broad range of motor inductances, etc. Anyways, the current, bus voltage, analog input, and temperature are sampled every PWM cycle, with an offset to put the sample in the middle of the PWM pulse (15.625 kHz). Voltage, analog, and temperature are used as-is, no filtering whatsoever.

The voltage returned over CAN is, indeed, the instantaneous bus voltage scaled by the percent command.

Current, on the other hand, is put through a 16x super-sampling, 8-point moving average filter. That is, 16 consecutive samples are put into a bucket, and when the bucket is full, it's averaged with the last 7 buckets, and that's your new current value till the next bucket's full.

So you end up with something like a 976Hz update rate for the current, but a bandwidth of something like 122Hz or less. Of the instantaneous value in the middle of the PWM pulse, lest we forget. This is the value reported back over CAN, and also used for current control mode. Interestingly, the instantaneous value is used for I2T overcurrent protection, which is set for 23A over 46A for 2 seconds (ie. 69A for 2s, or 78A for 1s, etc.).

No, I don't know why the overload specs differ from the specs in the data sheet. Either they fudged the DS numbers down for safety, or their conversion functions are wrong by that amount, which seems unlikely.

This is all assuming the FRC firmware is identical in these regards to the reference design firmware on Luminary Micro's site, of course. I'm not going to decompile the FRC firmware unless lives are at stake of something.

[Ether]

Originally Posted by Kevin Sevcik

I've now deciphered the current sampling strategy employed by the Jaguars...

current... [is] sampled every PWM cycle, with an offset to put the sample in the middle of the PWM pulse...

Current ... is put through a 16x super-sampling, 8-point moving average filter.

Nice work! Thank you for this information!

So the Jag firmware is doing both: digital filtering and phase-locking the sample to the bridge switching.

So you end up with something like a 976Hz update rate for the current, but a bandwidth of something like 122Hz or less.

Well, the 976Hz makes sense, since the PID in the Jags is updated at about the same rate.

Of the instantaneous value in the middle of the PWM pulse, lest we forget.

Right. Now the question now becomes: what is the relationship between this metric, RMS, and algebraic average.

[EricVanWyk]

Marshall -

To account for motor heating, consider re-ordering your tests pseudo-randomly. This will spread the time dependent effects across your sample space. Hopefully, this will result in a curve that makes more sense (but will likely be "wider").


Calculating effective motor voltage from Vbus * Duty Cycle is actually a pretty good method as long as the motor current is continuous (doesn't go to zero), and your measurement of Vbus is good enough.

Measuring Vbus half way through the pulse is a reasonably standard method. Assume that the applied voltage waveform looks roughly like a trapezoid: It starts at Vbus nominal when the switch starts, linearly(ish) droops due to increasing IR loss, and then goes to zero when the switch stops. The measured Vbus looks just like this, but with some ugly ringing around the transitions. By measuring halfway through, we kill two birds:
1) We are far away from the ringing.
2) We effectively take the average of the Vbus across the cycle (if it droops linearly).
Pretty sweet, eh? Simple way to get a darn good estimation. If you think about 2) a bit, you'll see why there usually isn't a filter in place - it would mess up the reading! If there is one, it usually has a very small time constant compared to the pulse width.

Continuous current is absolutely key to using effective motor voltage calculations - if the current stops, the underlying assumptions fall apart entirely. This is "The Big Difference"(TM) between Jaguars and Victors, and is why they have such different responses.
In a continuous current control mode, the effective motor voltage is (nearly) equivalent to what would happen with a pure DC source of the same voltage.

The reason for this gap is that when current is stops flowing, the motor is effectively floating with no applied voltage. If current is always flowing, the motor controller is always in control of the applied voltage.

Continuous current occurs whenever the off period is shorter than the amount of time it takes the current to decay to zero. Mathematically, this puts a relationship between switch frequency (on and off periods, actually) and minimum average current. That is to say that for a given frequency and duty cycle, there is a minimum average current to establish continuous current control. For a Jaguar, this is a few amps. For a Victor, this is a few hundred.

[Ether]

Look at the difference in current waveforms for a CIM with locked rotor driven at 5% duty with a Jag vs a Victor.

This is a simulation using R=0.090 ohm & L = 200uH.


At 15KHz, the current varies from ~7.4amps to 7.7amps. The RMS is 7.6amps and the algebraic average is 7.6amps.

At 150Hz, the current varies from ~1.8amps to 18.7amps. The RMS is 8.4amps and the algebraic average is 6.7amps.

Since average motor torque is created by algebraic average current, and motor heating is created by RMS current, the 150Hz waveform produces less average torque and more motor heating.

[kamocat]

Thank you for the straightforward answers.

Originally Posted by Ether

To a very good approximation, yes, as long as we're talking about the same waveform of 5A in each case.

How would I measure it in a way that any waveform with the same current would create the same torque? You said algebraic average?

[Ether]

Originally Posted by kamocat

Thank you for the straightforward answers.

How would I measure it in a way that any waveform with the same current would create the same torque? You said algebraic average?

That's a fair question but the answer has a lot of ifs, ands, and buts.

If the waveform is nice and flat like a 5% 15Khz CIM curve then any sample taken with be both the algebraic average and the RMS, approximately.

If the waveform is an exaggerated sawtooth like a 5% 150Hz CIM curve, then you'd have to sample the instantaneous current and run it through a low-pass filter of some sort. You'd have to make sure your samples captured a reasonable picture of the entire waveform and were set up in such a way that aliasing didn't distort the results. OR, I think a better way would be to hardware-filter the raw analog signal (the amplified voltage across the shunt resistor) before it goes to the ADC. I think a filter could be designed that would give the algebraic average.

[Al Skierkiewicz]

Originally Posted by EricVanWyk

That is to say that for a given frequency and duty cycle, there is a minimum average current to establish continuous current control. For a Jaguar, this is a few amps. For a Victor, this is a few hundred.

Erik, can you explain this?

Ether, in the graphs you posted for a locked rotor CIM it appears that the Victor has a distinctive current inductance charge/discharge curve while the Jaguar does not. The inductance charge is much more pronounced in the Jag due it's higher frequency. The charge current rise is very long compared to the pulse width at 15kHz. Using the values R=.090 and L=200uH, the time constant is 2.2msec or full current will only be achieved through the motor in 11 msec.
At 150 Hz and 5% duty cycle that is an ON time pulse width of .33msec so the current will reach 18.35 amps by the end of the pulse for a Victor. At 15kHz and 5% duty cycle that is an ON time pulse width of 3.3usec and the current will only reach 0.199 amps by the end of the pulse for the Jaguar. Neither controller provides a discharge path during the OFF period. It is this phenomena that gives the Jag it's apparent linear response.

[Ether]

Originally Posted by Al Skierkiewicz

Neither controller provides a discharge path during the OFF period. It is this phenomena that gives the Jag it's apparent linear response.

The FETs in both the Tan and the Black Jags have a zener diode to provide a path for the current to keep flowing through the motor during the OFF portion of the cycle.

http://www.chiefdelphi.com/forums/sh...30&postcount=9

The Jag's linear response is due to its higher switching frequency, which creates an effectively flatline current waveform whose amplitude (which is the average since the wave is flat) varies linearly with the voltage duty cycle.

The Vic's slow switching frequency allows to current to decay to zero during the off portion. The average value of this sawtooth waveform does not vary linearly with the voltage duty cycle, so neither does the motor torque.

[Al Skierkiewicz]

The FETs in both Jags and Victors have a forward voltage drop around 1 volt. Therefore to make these conduct (when OFF) would require that the motor be generating at least 2 volts more than the power supply.

[Kevin Sevcik]

Actually, you're both wrong. Sort of. The control scheme for the motor only pulses the high side FET for a direction, and leaves the low side FET on. So one low side FET remains conducting during the entire PWM cycle, and you only have the forward drop from the opposing low side FET when the current cycles back through in that direction. And since we're talking about the motor inductance driving the current in the off phase, the voltage is going to increase to compensate for that 1V drop anyways.

[Ether]

Originally Posted by Kevin Sevcik

Actually, you're both wrong. Sort of. The control scheme for the motor only pulses the high side FET for a direction, and leaves the low side FET on.

Just to be clear: I didn't say that the Jag turns the low side FET off for pulsing. I just showed what the current path would look like if it were switching the low side... and I asked if anyone knew, definitively, what the switching method is:

See attached images. For simplicity, I showed path through only one FET of each pair, and only for + currents. Green path is during ON portion. Blue path is during OFF portion (for high-side switching). Red path is during OFF portion (for low-side switching).

I don't know if the Jags use high-side, or low-side, or some other switching method. Does anyone know, definitively, what switching method the Jags use? I've heard many different claims.

How do you know that the Jag switches only the high side? Did you discover this during your analysis of the firmware, that you posted earlier? Or did you find it in some document somewhere?

So one low side FET pair remains conducting during the entire PWM cycle, and you only have the forward drop from the opposing low side FET pair's zener diodes when the current cycles back through in that direction.

I added the bold stuff for clarity. You also have the opposing back emf if the motor is spinning. See attached PDF.

And since we're talking about the motor inductance driving the current in the off phase, the voltage is going to increase to compensate for that 1V drop anyways.

Exactly. See attached PDF.

[Al Skierkiewicz]

The only knowns are the Victors turn off both high and low side FET groups except when "brake" mode is enabled and throttle is zero. Luminary reps (a while ago) posted that the tan Jags only turn off the high side FETs while leaving the low side FETs turned on for direction. "Brake" operation is the same. I have no known source for operation on the black jags but assume there was no change.
While the Jags only turn off one FET group, they also have the current sense resistor (.0005 ohm for tan and .001 for black) in series with the discharge path to battery. Again, this path to the battery only exists while the voltage generated in the motor can forward bias the body diode(s).

[Ether]

Originally Posted by Al Skierkiewicz

While the Jags only turn off one FET group, they also have the current sense resistor (.0005 ohm for tan and .001 for black) in series with the discharge path to battery. Again, this path to the battery only exists while the voltage generated in the motor can forward bias the body diode(s).

The current we are discussing in this thread is the current that flows through the motor during the off portion of the duty cycle. This current does not flow through the current sense resistor. It does not flow to the battery.

[Kevin Sevcik]

Hey all. Just to add to the confusion and all, the Black Jags seem to operate somewhat differently than the Tans. I know, I was surprised too. Anyways, the reference design of the Black Jags is also high-side switching, BUT it switches the high and low side FETS on the pulsed leg.

Ex: at +25% command, M- low is ON, M+ high is ON 25% of the cycle, M+ low is ON 75% of the cycle.

So on the Black Jags, you don't need to worry about any diode drops in the off phase, save for the vanishingly small slice of dead time to prevent shoot through.

Isn't this so much more fun than rumor mongering?

[Ether]

Originally Posted by Kevin Sevcik

Isn't this so much more fun than rumor mongering?

Yes. You are the man !