Jaguars vs Victors

[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.

[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"

[techhelpbb]

Originally Posted by Ether

Uh, yeah:

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