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Re: Jaguars vs Victors
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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. |
Re: Jaguars vs Victors
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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 |
Re: Jaguars vs Victors
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Re: Jaguars vs Victors
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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 |
Re: Jaguars vs Victors
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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.) Quote:
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). |
Re: Jaguars vs Victors
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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 |
Re: Jaguars vs Victors
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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" |
Re: Jaguars vs Victors
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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. |
Re: Jaguars vs Victors
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Re: Jaguars vs Victors
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Re: Jaguars vs Victors
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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. |
Re: Jaguars vs Victors
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Re: Jaguars vs Victors
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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? :) |
Re: Jaguars vs Victors
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Re: Jaguars vs Victors
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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. |
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