Modeling motor control

[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).

[Kevin Sevcik]

Originally Posted by Ether

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?

Yup. Again, with the caveat that I looked at the reference design source code for the Tan Jag. YMMV with the FRC firmware, though I'd be terribly surprised if they changed much, as the reference design looks very good and entirely functional for FIRST purposes.

[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 !

[Ether]

Originally Posted by Kevin Sevcik

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

Did you mean M+ low ?

[Kevin Sevcik]

Originally Posted by Ether

Did you mean M+ low ?

Erm.. Typo? What typo? But yes, thanks for the catch.

[kamocat]

This is very noticeable in motor control when you're slowing down, BTW.
If you go from 100% voltage to 10% voltage on a Tan Jag, the motor just coasts down. If you do the same on a Black Jag, the motor jumps down in speed.

[Ether]

Originally Posted by kamocat

This is very noticeable in motor control when you're slowing down, BTW.
If you go from 100% voltage to 10% voltage on a Tan Jag, the motor just coasts down. If you do the same on a Black Jag, the motor jumps down in speed.

Was the jumper set for "coast" on the Black Jag when you ran this test ?

[kamocat]

Remember this thread?
http://www.chiefdelphi.com/forums/sh...733#post973009

EDIT:
I believe it did happen both in brake and coast mode on the Black Jaguar. It was one of the things I discovered when I was first playing with the BDC-COMM.

EDIT2:
Yes, this happens in coast mode. I just checked, and I keep my jumpers in coast position so I don't have the trouble with Jaguars failing from power generated by the motors when the robot is off.
I was experiencing this locked anti-phase when I was testing yesterday.