Jaguars vs Victors

[techhelpbb]

Originally Posted by Matt Krass

The configuration I had in mind was a Jaguar running forward, bringing it's load node up high to +12V, and the Victor running reverse, bringing it's load node down to 0V, so that current flows:



In this configuration, both active legs see a stable constant current load, so both get a complete work out. Halfway through the test I'd flip the directions and work the other legs just as hard. Since the loss in the FETs are resistive and related to current, we'd only need constant current, not constant power, so the declining battery voltage is not necessarily a problem.

If my description still doesn't make sense, I can draw it up really quick and see if that makes more sense. The idea is the load current passes through both legs of both bridges while the FETs are monitored.

The problem with a consistent load is that resistors heat up and change resistance, and other components have similar temperature dependence. However I have not (yet) done the math to see if this variance is enough to warrant the trouble of a dynamic load. Perhaps it isn't necessary, though I agree I'd not want to use a digital controller to regulate the load, merely to program an analog feedback loop to regulate it due to bandwidth concerns.

Matt

I see what you're thinking but be careful, I'm thinking the Jaguars don't entirely work like that in the off part of the cycle.

[Matt Krass]

Originally Posted by techhelpbb

I see what you're thinking but be careful, I'm thinking the Jaguars don't entirely work like that in the off part of the cycle.

Simple solution, run both at full blast and let the dynamic load regulate the current (this actually makes sense for a bunch of different simplicity reasons, but it also deprives us of the results of switching loss/downtime for the FETs which will very likely affect the results)

I'm curious, what effect are you expecting to see during the off cycle part of things that would be a problem?

Also, this idea is really not fully fleshed out, it's still on the back of the napkin, so I'm expecting people to poke lots of holes in it, so they can hopefully be filled in

Matt

[techhelpbb]

Originally Posted by Matt Krass

I'm curious, what effect are you expecting to see during the off cycle part of things that would be a problem?

Synchronous rectification on the black Jaguars.
http://www.ti.com/lit/an/spma033a/spma033a.pdf

I like this description, so with respect to ST Microelectronics I quote it here from this datasheet http://www.datasheetcatalog.org/data...onics/7048.pdf page 8.
(This is a courtesy to the audience that may not understand what this is. I'm quoting this because I like the description in this datasheet. The datasheet does not pertain to any part in our systems.)

A motor is an inductive load. When driven in PWM mode, motor current is switched on and off at the
25kHz frequency. When the MOS is switched off, current can not instantaneously drop to zero, a so-called
”free-wheel” current arises in the same direction than the power current. A path for this current must be
provided, otherwise high voltage could arise and destroy the component. The classical way to handle this
situation is to connect a diode in an anti-parallel configuration regarding to the MOS, so that current can
continue to flow through this diode, and finally vanishes by the means of ohmic dissipation, mainly in the
diode due to its 0.8V direct voltage. For high currents, dissipation can be an important issue (eg: 10A x
0.8V makes 8 W!). Furthermore, high speed diodes have to be used, and are expensive.

A more efficient way to handle this problem is to use the high side MOS as a synchronous rectifier. In this
mode, the upper MOS is switched ON when the lower one is switched OFF, and carries the free-wheel
current with much lower ohmic dissipation. Advantages are : one expensive component less (the fast
power diode), and more reliability due to the lower dissipation level.

However, we have to take care not to drive the two MOS simultaneously. To avoid transient problems
when the MOSare switched, a deadtime is inserted between the opening of one MOS, and the closing of
the other one. In the TD340 device, the deadtime is fixed to about 2.5 microseconds. This value is the time
between the commands of the gate drivers, not the deadtime between the actual MOS states because of
the rising and falling times of the gate voltages (due to capacitance), and the MOS characteristics. The
actual value of the deadtime for a typical configuration is about 1.5 microseconds.

Please keep in mind that people often use the body diodes of these MOSFETs as the anti-parallel high speed Schottky. So these diodes get hot and that heats up the MOSFET transistor as a whole. That's the advantage of this feature in the black Jaguars as apposed to the previous generation.

Course it might be trouble if other things are injecting power in the bridge. Usually motors can't return more power than you put into them unless something turns them faster than the speed control (it may be more voltage than you put into them with less current, but in the sense of power it must be less than you put in unless something adds mechanical energy and makes the motor into a generator). I may be misunderstanding but from your original proposal it would seem that a black Jaguar might dissipate the energy in the load from the Victor....that would be quite a bit more energy than I think would be normal even for a motor that had extra mechanical energy being added to it's rotation (at least in the scope of what we can build with these parts).

[Matt Krass]

Originally Posted by techhelpbb

Synchronous rectification on the black Jaguars.

Please keep in mind that people often use the body diodes of these MOSFETs as the anti-parallel high speed Schottky. So these diodes get hot and that heats up the MOSFET transistor as a whole.

I was aware of the Jaguars synchronous behavior, but I thought it wouldn't be a problem because, if I'm looking at my sketch correctly, when the off cycle occurs the terminals of the Jaguar reverse polarity and match those of the Victor, at roughly (but never at exactly the same time) that the Victors low side FETs disengage. Perhaps it just oversight on my part, but I don't believe this would be a problem, since the terminals would either be open at the Victor, or connected to the same node as the corresponding Jaguar terminals, not flowing any current?

It's entirely possible I'm missing something right in front of my face, am I?

Also, since the Victors don't have synchronous rectification (to the best of my knowledge) and only switch the low side (again, to the best of my knowledge) I suppose the body diode on the bottom FETs get's a work out when the Victor is in it's off cycle and the Jaguar isn't, I assume this is where your concern about heating comes in?

Matt

[techhelpbb]

Originally Posted by Matt Krass

I was aware of the Jaguars synchronous behavior, but I thought it wouldn't be a problem because, if I'm looking at my sketch correctly...
Matt

Sorry, I'm not following. It might just be me, but I think I should take you up on your offer to provide that illustration of what you propose.

[Matt Krass]

Originally Posted by techhelpbb

Sorry, I'm not following. It might just be me, but I think I should take you up on your offer to provide that illustration of what you propose.

Excuse the crudity of this model, it's not to scale...

So, with the Jaguar in forward, and the Victor in reverse, in the on period for both, Node A will be at +12V, and Node C will be at 0V, flowing some current through dynamic load R1, and B would be at 0V and D at +12V, so current is flowing the opposite way through R2.

When the PWM goes off, let's assume the Victor stays on longer for the sake of argument, so C stays at 0V, D stays at +12V, and A also goes to 0V, and B goes to +12V... No current flows.

Ok, so what obvious thing did I miss?

[techhelpbb]

Originally Posted by Matt Krass

Excuse the crudity of this model, it's not to scale...

So, with the Jaguar in forward, and the Victor in reverse, in the on period for both, Node A will be at +12V, and Node C will be at 0V, flowing some current through dynamic load R1, and B would be at 0V and D at +12V, so current is flowing the opposite way through R2.

When the PWM goes off, let's assume the Victor stays on longer for the sake of argument, so C stays at 0V, D stays at +12V, and A also goes to 0V, and B goes to +12V... No current flows.

Ok, so what obvious thing did I miss?

Now I see what you intended. Thanks.

Well for one thing there's the current sensing resistor in the Jaguar which is not in your diagram.
If you look at Ether's link of the manual it's in the schematic on page 23 pretty much dead center.

[Matt Krass]

Originally Posted by techhelpbb

Now I see what you intended. Thanks.

Well for one thing there's the current sensing resistor in the Jaguar which is not in your diagram.

I left out things like that and the control electronics for brevity, I was more referring to any obvious reasons that my scheme wouldn't work that I overlooked.

I think in this scheme, the sychronous behaviour of the Jaguar is not a problem. I also suspect some inductive and capacitive components on each ESC might be wise to smooth out the output so the load is more stable, especially given the (wildly) different switching frequencies involved.

EDIT: I'm just curious, could you describe what you thought I meant?

Matt

[techhelpbb]

Okay, but the current sensing resistor in this case is limiting the current to the Victor as well. So you can't fully test the Victor's power range like that. Plus I don't think they have the same on-state resistance so the Jaguar might drop more voltage. Might not matter for what you plan.

I thought you planned on powering the same load with both speed controls, because you started off in the singular tense in the first proposition post. Clearly, however, you intend to use 2 loads and that's just fine.

The only thing is that the terrible things (as you put it before) involved here are not really inductive or motors?

I sort of ask because while I understand the heat concerns entirely, obviously there are quite a few other concerns in the comparison.

[Matt Krass]

Originally Posted by techhelpbb

Okay, but the current sensing resistor in this case is limiting the current to the Victor as well. So you can't fully test the Victor's power range like that. Plus I don't think they have the same on-state resistance so the Jaguar might drop more voltage. Might not matter for what you plan.

I thought you planned on powering the same load with both speed controls, because you started off in the singular tense in the first proposition post. Clearly, however, you intend to use 2 loads and that's just fine.

The only thing is that the terrible things (as you put it before) involved here are not really inductive or motors?

I sort of ask because while I understand the heat concerns entirely, obviously there are quite a few other concerns in the comparison.

I fully expect they don't have the same on-resistance, that is why I want to go for constant current loads, so we can measure the performance of the FETs, getting an empirical comparison of the 3 FET leg of the Victor vs the 2 FET leg of the Jag. If we measure the voltage drop and temperature across the FETs on each side we can see which heats up faster, which loses more power to heat, etc.

As long as the power source doesn't run out of volts and the load can compensate, the current sense resistor should be canceled out by the adaptive load.

Matt

[techhelpbb]

(Back to what I was communicating with Ether about...)

I wrote:

On what I just proposed I'm not sure but it would seem to me that dumping that energy into the low side of the bridge runs the risk to mess with the lower side reference 'ground'. Since the high side would have to be driven higher than the supply to saturate the MOSFET anyway, because the high side is N-Channel MOSFETs, wouldn't it less risky to dump that energy to the side that is already able to exceed the supply rail which might shift down anyway? I mean might not a shift in the lower reference cause the lower MOSFET to not be entirely saturated?

Hmmm, might not matter, guess it would depend on just how not ideal the lower reference 'ground' really is.

The high side current limit resistor is probably why they used the low side as the current going back up through the high side would heat that resistor. So that resistor makes the high side's path back to the battery positive less than ideal as long as it exists (even if the transistor remained fully saturated while it did that).

Trade off I guess.

[Ether]

Originally Posted by techhelpbb

The high side current limit resistor is probably why they used the low side as the current going back up through the high side would heat that resistor. So that resistor makes the high side's path back to the battery positive less than ideal as long as it exists (even if the transistor remained fully saturated while it did that).

Trade off I guess.

The current sense resistor would not be in the path of the inductive current if the high side were used for shunting

The inductive current would flow in a loop: through the motor, through one pair of high-side FETs, then through the other pair of high-side FETs back to the motor.

[dsirovica]

Guys,

if you use a Jag to drive a Victor you will get into a pickle. The two PWMs are free running and will produce all sorts of beating products. It will be very confusing to figure out whats going on. May be fun though!

What we have is really pretty simple. For the Jag we have decent documentation for the Vic we don't so we are guessing a bit for the Vics.

For the Jag, during the HIGH PWM cycle, we have:
Battery,
Current sensing Resistor,
Two Mosfets in parallel,
the Load (Motor),
Two more Mosfets in parallel.

During the LOW PWM cycle, we have:
the Motor that is shorted via two sets (in series) of two parallel Mosfets.

The Mosfets are driven hard-ON and hard-OFF at 15KHz by the driver chip.

All the other components are “in the noise” - that is they matter very little and can be ignored for our purposes here. It is fair to assume that these other components are engineered properly, and are not limiting the behavior we observe.

Based on the previous discussion in this thread it seems that the practical limitation we have is thermal overheat due to high amperage. The Jags have software that cuts out way before meltdown (perhaps too soon, sacrificing maximum power), while the Victors can burn in the field but as a result deliver more power than the Jags. This gives Victors an apparent power advantage, and it seems many teams prefer them for that. The sad thing is that the Jags actually could deliver more power than the Vics if allowed.

I've been playing with a simulator – it is really cool and answers many of the questions we posed here. I would advise all to play with this, see:
http://www.linear.com/designtools/software/
It is very noble for LT to provide this for free – its a great educational and practical tool.

LT doesn't have the exact MOSFETS we have, but you can get close, or design your own.
I entered a simplified model with just one Mosfet. I don't have a Motor model, so I assumed a Stalled-Motor equivalent of an Inductor and series Resistor. A good approximation for a CIM is L=200uH, R-80mOhms (please advise if you have better data on this). See:
http://i.imgur.com/1cNXB.png

You can see the Current through the Inductor is ~60A. In reality we were getting 55A see real picture of a PWM driven Jag and stalled CIM Search for the "Voltage vs. PercentVbus” thread for more motor pics:
http://imgur.com/0IjKf

To give more details on the simulation parameters if you want to play with it (it takes just minutes to do this):
V2 is 12VDC. R3 is the internal Battery resistance (ours measured about 20mOhms)
L1 is 200uH, and R2 is 80mOhms these numbers seem to match a stalled CIM based on some measurements we ran in the aforementioned thread. It would be great if someone has the SPICE models for FIRST motors!
I just used a Diode D1 for the flyback shunt. You can see this would dissipate 30W if the Jags didn't use the Mosfets for this! As the current through D1 is 60A during the LOW PWM.
V1 is the PWM:12V, Period=75us (for 15 Khz), I gave it a rise and fall time of 2.5us each. It runs for 200 cycles or 15ms.
R1 is an arbitrary value. You can see that the Mosfet's parasitic capacitance is spiked through this R and that is where the Capacitive loss is converted to heat – that stays in the Mosfet Driver chip – I will let you all run this to see the trace of the current through R1 – it is cool!
The Mosfet I picked is IRFH5004, this has an RDS of 2.2mOhms and I doubt this simulator takes heating into account so RDS probably stays at 2.2mOhms. In our reality we are heating these puppies driving RDS to about 150% of typical value.

The main things still missing are the forced-air cooling equations for the Jag&Vic (ie. What thermal C/W number should we use) and all the documentation that we don't have for the Vics.

And for the advanced student, here is the Fast Fourier Transform (FFT) of this PWM signal on V1 – it is mind boggling what LT Spice can do !!!
http://i.imgur.com/7Fjh2.png
You can see the main component at 15KHz, and the square wave's harmonics at 3X intervals.
This is much simpler and hugely more powerful than the Fortran punched card program I used for very primitive Fourier Analysis in my college days :-)

[techhelpbb]

Originally Posted by Ether

The current sense resistor would not be in the path of the inductive current if the high side were used for shunting

The inductive current would flow in a loop: through the motor, through one pair of high-side FETs, then through the other pair of high-side FETs back to the motor.

This is very true, but I would think only after the delay when the other half of the high side turns on.

( I hate to put this below a post like the one directly above, dsirovica that was a very good post.)

[Ether]

Originally Posted by techhelpbb

This is very true, but I would think only after the delay when the other half of the high side turns on.

Are you talking about the 2 microsecond shoot-through delay? During that time, the intrinsic diode of the non-conducting high-side FET pair would complete the current loop. The inductive current doesn't flow through the sense resistor, because there is no path to ground.