Re: Failed attempt to explain JAG linearity
 

I beleive that when IFI talked about the low duty cycle control they were talking mainly about the minimum duty cycle. The minimum duty cycle of the Victor 883 was 10%, while the minimum duty cycle of the Victor 884 is 3%.

Here is a visual indication of how the jaguar and victor 884 do with their minimum duty cycles: http://forums.usfirst.org/showpost.p...08&postcount=4

Re: Failed attempt to explain JAG linearity
 

OK,
I think I am seeing this your way now. The low side FET D is always "on" according to Scott so that when the PWM goes low, the high side FET A opens. Current developed by the decaying inductance now flows through the FET D across the common (negative terminal) and forward biases the diode in FET C. I think then, that the expectation is that neither diodes at A or B would conduct although conditions might allow that to happen for a short period of time. Does the discharge path then have significantly less resistance than the charging path and would that need to be compensated for in the calculations? That is until the discharge falls to CEMF less the voltage drop across the diode and FET on resistance.

Re: Failed attempt to explain JAG linearity
 

Thank you Mr. Ross,
Yes, that is a real world confirmation of the jags linearity and controlability. Now back to Chris and AL. Is the PWM frequency the main factor between the 2 controllers or is the switching method also a factor? Why did IFI drop the frequency if there are so many benifits to a higher fequency? Did Luminary MIcros give us linearity and controlability but also introduce added stress that may have caused a number of failures that accurred?

Re: Failed attempt to explain JAG linearity
 

You were a tough fish to reel in, but welcome aboard.


good question.. my models have not assumed any difference in R for charge and discharge. We know this is a simplification since in addition to the motor resistance the charge path includes the full system installation losses (battery, power panel, wires etc) and the discharge only includes some motor wires and some internal JAG losses. These could be estimated and one would expect the discharge time constant to be longer than the charge time constant (maybe 30% to 40% higher). I suspect this would make the Victor look a little more linear. If you post some typical numbers, I could add these effects in easily. How about updating your ppt slide to show all the losses and your estimate of what they might be. This can be added to the new ppt I just posted.

Re: Failed attempt to explain JAG linearity
 

Ok..now that Al is onboard I updated the ppt to represent the JAG / Victor and added the Brake/Coast configurations and the ABCD notation. See JAG/Victor Hbridge.5.25.ppt zip attach. We can all contribute to this as a document and capture stuff as we go.

If you compare the high side and low side OFF circuits there is no difference in how one would model them. There is still only one diode in the return path around the motor. The charge paths are identical. So I would say the switching method is not a factor.


EDIT: There was typo in Brake mode title. Should be "JAG BRAKE CONFIGURATION C and D FET's ON". Fixed and reposted rev1 file.

Re: Failed attempt to explain JAG linearity
 

Whoa, not on board yet, just in the same lake.
In your power point, the FET groups B and C are turned on for direction control. The CEMF is the opposite polarity of that so current is reversed from what is drawn when in the "brake" mode. The effect should be the same for either controller, just a difference in which FET groups are in effect. In thinking the brake mode through, at least initially, the CEMF of the motor could be high enough to force enough current through the path to actually turn on one of the diodes.
At this point, the Rdson of the FET groups should be brought into the discussion. The Jaguar uses a FET with lower Rdson and is equivalent of 1.8 mohms per group. The Victor is 4 mohm per group. However, the specs are tested under different conditions and I suspect that they are somewhat closer than you would expect. The Victor is speced for pulse widths less than 400 micro seconds @<2% duty cycle, while the Jaguar FET is a little less obvious as to minimum pulse width and duty cycle.
As to series resistance in the initial run phase (charge the inductance), typical wiring techniques would give you the same series resistance up to the controllers and from controller to motor. I usually look at worst case analysis so 11 mohm for the battery, 4 mohm for connectors, breakers, terminals, etc., and 6 mohm for wire using #10. Add in the 0.5 mohm current sense resistor for the Jaguar as well. So for Victor total series resistance could be as high as 29 mohms and for the Jaguar and 25.5 mohms for the Jag.
Those are small differences so the only thing left is the actually switching frequency, which we both have proved results in lower the current for the Jag compared to the Victor due to the L/R factor and charge current of the motor inductance. Realistically, the lower current, would translate to lower speed and therefor, lower CEMF for the Jag. We need to get true duty cycle vs speed data for each controller to the same motor. I have no availability in the near future for experiment. Analog TV shutdown is on the horizon and several weeks of transmitter duty are in my future.

Re: Failed attempt to explain JAG linearity
 

.
Ha..just meant that the OFF return now has one diode..I know you aren't on board with anything else yet.

There is a typo in the title: Should be "JAG BRAKE CONFIGURATION B and D FET's ON". Otherwise the slide is ok.

I don't see how since both ends of the motor are shorted thru B and D "on" FETs.

Your calculations agree with the 30mhoms I assumed in rerunning the sim with different R_on and R_off. R_on and R_off are the total effective resistances during the PWM ON and OFF phases. I checked the speed changes with R_on = .12 ohms and R_off = .09 (basically just the motor resistance). I saw essentially no differences for the Victor and slight differences for the JAG vs running the cases with R_on=R_off = .12 ohms. So this is a secondary effect.

We have two data sets to work with. The Beach Bots provided us with 884 CIM 01 responses and we have several JAG linearity plots posted which were supposedly for a CIM. The motor variations and installations would be slightly different but at least we have some data. Hmmmm, if we could only get Joe Ross to run some more data comparing the two.

Re: Failed attempt to explain JAG linearity
 

Chris,
I am thinking out loud that a motor in a drive on a 150lb FRC robot might be able to produce significant current in "brake" mode that might just bias one of the diodes for a period of time.
I am looking at your slide 9 and the CEMF would cause a current to flow opposite of the direction shown in slide 12. No real effect on result.
Can you link to a Jaguar table showing the same duty cycle and motor speed for a CIM as documented in the Beach Bot data?

Re: Failed attempt to explain JAG linearity
 

Sorry guys...going a bit mental.

Should be "JAG BRAKE CONFIGURATION B and D FET's ON". The fix is ok in rev1.

AL; re slide 9 and 12 . I don't understand what you are saying here. As I see it, there is no relationship between slide 9 and 12. Slide 12 "brake mode" stands alone. The current will flow in either direction and is dictated by Vemf sign. I have shown it for only one direction of course. Because B and D FET's are on I don't see how there can be any diodes in the picture.

I couldn't find the links...I saw one recently but don't remember where.
However, I did find a JAG curve that I think I downloaded from the FRC 2009 beta site. There was some discussion there and this was my first look at a JAG response. It also had a nice Victor curve. See attached .xls

Re: Failed attempt to explain JAG linearity
 

Chris,
Slide 9 shows current flow for the motor spinning in one direction. If slide 12 is showing the current flow due to CEMF for a motor running in the same direction in brake mode, then it should have the current flow in slide 12 reversed.

The attachment sheet tells the tale doesn't it? If a team was designing for the CIM motor to be in the more efficient part of it's curve or around max power, then the area between 3000 & 4000 RPM is the place to look. These curves are as we expected and do represent what is likely the interaction of the motor inductance, switching frequency and CEMF. The Jaguar is no doubt more linear in the region of interest but both controllers are linear in the area above max power and through the area of greatest efficiency above 4000 RPM.

Does the Luminary Micro title on page 2 indicate this came from Luminary?

Re: Failed attempt to explain JAG linearity
 

Yes. Here's the full thread on the FIRST forums: http://forums.usfirst.org/showthread.php?t=10182

Re: Failed attempt to explain JAG linearity
 

I see what is bothering you. I really didn't intend slide 12 to be sequential to any of the previous slides. But if it was ..the current in slide 12 must match the current in slide 9 or any other prior slide at initial switch over. The voltage across L will force this since the current cannot change instaneously. The current will eventually change sign and head for Vemf/R which is what you are looking for.

Thanks Joe for finding that link.. I knew I'd seen it recently.

Re: Failed attempt to explain JAG linearity
 

I woke up thinking about your excel http://www.chiefdelphi.com/forums/sh...9&postcount=20
and how you have implied (subtly) that this supports the case for lower JAG currents and hence lower Vemf. I cannot support this implication because it is involves comparing apples to oranges. The excel results simply state that if you run a 12 v pulse into an LR ckt you will drive less current if the pulse is 100 times shorter.

One must run both cases for the same time to compare. Let us pick the Victor period as the comparison time. At 100% duty cycle both the JAG and Victor will drive to the same current in that time period. You know this.

At anything less than 100% duty, you must now look at what happens during the off phase of the cycle to draw any conclusions re the average current over the period. The average current is the only thing that enters into the duty vs speed graphs we are looking at. In fact, you must look at the steady state average current that allows Vemf to drift into the picture as the current pulses are accumulated. The steady state criteria has not been defined yet by either of us but we would like it to be at least 5 time constants of a loaded motor.

So my good friend, the captain should not let you in the boat based upon this evidence. You must stay in the lake until you provide the analytical justification for extending your excel to steady state or do the brute force pulse accumulation with a simulation. Or accept the rationale and sim results I have put forth (I pray you will). Or just remain a wet skeptic ... that is fine also. I suspect you will favor the last option:)

Note: I have conviently defined "boat" as holding only passengers who have tried to explain the steady state. This way I can stay dry and have some beers on the poop deck. Others are welcome if they understand and accept anyones explaination. Skeptics are allowed but only as guests so we can still have a beer together.

I like you as a skeptic lurking in the water as a shark (a kind and gentle shark) waiting to feed on weak theories trying to get on board. It makes us all stronger.

Re: Failed attempt to explain JAG linearity
 

Chris,
The spreadsheet from Luminary supports lower currents and therefore lower speeds at less than full throttle for the Jaguar as expected. If the motor is turning slower, the EMF will be lower as well. The two graphs show that motor speed for both controllers approach specified free speed for the CIM at max PWM input. Luminary has also done the work I was hoping to be able to perform at some future date. They compared their controller with a resistive load and an inductive load (the CIM motor) although I question the resistive load "output voltage" and how it was measured. A simple RC filter would give that result. I would like to see the two compared for identical input PWM to make the graphs perfect but it is easy to see the differences. In either case, a mechanical solution and/or a software solution would be able to make either controller perform similar. A final note, we need to remind our readers that the Jaguar never reaches 100% duty cycle as it needs the pulse to keep the bootstrap capacitor charged. Without it the output FETs would not be turned on. This does not play into our discussion except at full throttle.
As a midwesterner and therefore confined to freshwater, the shark analogy is a stretch for me. Let's just say I am the unknown in the dark and muddy waters under the boat.
Chris, no need to edit. I received your request as a moderator.

Re: Failed attempt to explain JAG linearity
 

Al,
I'm glad you are still lurking in this thread. I didn't want it to end on my last quirky post.

Anyway, I was thinking that the Victor 120hz could be made linear by simply changing the switching sequence. If the normal charge phase had A and D fets on, then the off phase as it stands now would just be D on. What if instead they went to the brake mode in the off phase by also turning on B fet. This would eliminate the diode return and the circuit would then behave linearly just like the JAG. I suppose the reason they don't do that is for fear that both A and B fets might be on during the switching transient and cause a high current near short in the AB fet path. It seems that one could delay the turn on of the B fet (say 10 usec or whatever is necessary). Then most of the current could be captured in the off phase and keep the inductor from running dry. Any thoughts on this.

Post note:
I think I answered my own question: The resistive power loss would be much higher in the Victor without the diode. The current would drive to -Vemf/R on each off cycle so there is a large power loss during this phase now. I modified the LABVIEW sim to calculate avg_pwr just to check:
duty = .3 JAG pwr =. 737 watts with and with out diode
duty = .3 Vic pwr = 3.2 watts with diode and a whopping 121 watts without diode.


So for the trial switching of the Victor without the diode the power loss is ridiculous but the transfer function would be linear.

The data shows that for the normal case the resistive loss of the Jag is less than the Victor.
If the power loss of the diode is included, the Jag total power loss would be closer to the Victor since the diode is conducting during the PWM off time. The diode loss adds about 2.7v*1.2amps*(1-duty)=2.3 watts extra or 3.04 watts total.