Jaguars vs Victors

[techhelpbb]

Originally Posted by Matt Krass

I'm a fan of this idea, I didn't mean to accidentally kick off a whole new topic in here to that end I've started a new thread here to continue this discussion.

Back on to Victors and Jaguars, I'd really like the opinion of some of the veteran EEs around here on my idea for a test rig I mentioned earlier. Essentially, I'd be alternating between a Jaguar sourcing current to a Victor and the opposite, using something with a sufficiently intelligent brain to control both and perhaps some kind of active load between them. While doing this, we can monitor all of their vitals (mostly temperature of the FETs in question) and we can do some pretty terrible things to them and see how they handle it.

A bank of FETs with beefy heat sinks combined with some smoothing LC networks off each speed controller would make a nice load I think, how about it?

Matt

A consistent load would be essential or it might poison your experiment.

I'm not sure what you mean by the Jaguar sourcing current to a Victor however.

Would it not be easier to just have a consistent load and a consistent source of power as static factors in your test?

I should think you'll need to monitor the load and source of power quite carefully and if that monitor is digital it'll have to have pretty high bandwidth for transients.

[Matt Krass]

Originally Posted by techhelpbb

A consistent load would be essential or it might poison your experiment.

I'm not sure what you mean by the Jaguar sourcing current to a Victor however.

Would it not be easier to just have a consistent load and a consistent source of power as static factors in your test?

I should think you'll need to monitor the load and source of power quite carefully and if that monitor is digital it'll have to have pretty high bandwidth for transients.

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:

Jaguar M+ (at 12V) ---> Dynamic Load ---> Victor M+ (at 0V)
and
Jaguar M- (at 0V) <--- Dynamic Load <--- Victor M-(at 12V)

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

[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

[Ether]

Originally Posted by techhelpbb

Synchronous rectification on the black Jaguars.
...
I like this description,

I like this one better, see Pages 18* & 19.

The Black Jags use the low side to provide the path for the inductive current during the OFF portion of the PWM cycle.


* there's a typo in the third paragraph in the section at the bottom of Page 18 titled "Switching Scheme". It should read "and Q4 on the low-side"

[techhelpbb]

Originally Posted by Ether

I like this one better, see Pages 18* & 19.

The Black Jags use the low side to provide the path for the inductive current during the OFF portion of the PWM cycle.


* there's a typo in the third paragraph in the section at the bottom of Page 18 titled "Switching Scheme". It should read "and Q4 on the low-side"

Good catch, thanks. Hmm, I wonder why they didn't do this the other way or flip it over for the higher duty cycle.
Wouldn't it make it even more efficient?

Per this:
"During PWM_OFF, assuming a 40 A load, Q2 losses are approximately 40W without synchronous
rectification. This drops to just 4W if synchronous rectification is used (Rds-on = 2.5 mΩ).
Synchronous rectification significantly improves drive-stage efficiency, particularly at lower duty
cycles (50% and less) when the PWM_OFF time is longer that the PWM_ON time."

Besides I would think the Jaguars would tend to spend more time at a higher duty cycle in our application.

[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.

[Ether]

Originally Posted by techhelpbb

Hmm, I wonder why they didn't do this the other way or flip it over for the higher duty cycle.
Wouldn't it make it even more efficient?

Why would that make it even more efficient? All the FETs in the bridge are the same.

[techhelpbb]

Originally Posted by Ether

Why would that make it even more efficient? All the FETs in the bridge are the same.

"During the PWM_ON period, Q1 on the high-side and Q2 on the low-side provide a path that increases current in the motor."

Shouldn't this be Q4, not Q2 (for one thing it's a contradiction to the diagram on the next page, and for another it wouldn't make sense as it is written versus how an H-bridge functions).

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.

[Ether]

Originally Posted by techhelpbb

"During the PWM_ON period, Q1 on the high-side and Q2 on the low-side provide a path that increases current in the motor."

Shouldn't this be Q4, not Q2

Uh, yeah:

Originally Posted by Ether

* there's a typo in the third paragraph in the section at the bottom of Page 18 titled "Switching Scheme". It should read "and Q4 on the low-side"

[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 Ether

Uh, yeah:

Sorry, didn't notice that before when you wrote it. Thanks.

[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.