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USB Hosts Hubs Transceivers

New Contributor

Hello,

I gone through the application note for power switch implementation.

I noticed that the example image having active high enable power switch. So the PWR# outputs are pull-up by 100K

If i'm using active low enable power switch, i have to use pull-down resistor as per the application note statement.

I have formulated the below table as per my understanding. I would like to the expected behavior for Case3 and Case4

For example,

if am using active low enable power switch and set the PWR#=0. What will be the behavior of the downstream port.

  1. Is there any specific reason to use EN=1 power switch when PWR#=0.
  2. Is it advisable to use EN#=0 power switch when PWR#=0

    

USB Hub PWR# OutputPower switch Enable inputRemarks
Case10 (default)1 (EN)Power switch with active HIGH enable input to be used
Case21 (PWR_PIN_POL=1)0 (EN#)Power switch with active LOW enable input to be used
Case30 (default)0 (EN#)
Case41 (PWR_PIN_POL=1)1 (EN)

Awaiting your response eagerly.

Regards,

Malathi T

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1 Solution
Moderator
Moderator

Hi Malathi,

Please note that the power switch in the image is for an active low input on the pin so pull up resistors are recommended which would keep the pins high unless driven low by the hub. Similarly, for active high input power switch, you would have to connect pull down resistors which would keep the pins low unless driven high by the hub.

I couldn't understand the first and second cases you mentioned, could you please explain? The hub uses the PWR pins to enable or disable the power switch based on the overcurrent condition so please set the polarity of the hub to that required by the power switch. So, in cases 3 and 4, the hub will drive the pin low or high based on the PWR_PIN_POL for enabling the switch and thus provide VBUS on the downstream ports.

Best Regards,
Sananya

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11 Replies
Moderator
Moderator

Hi Malathi,

Please note that the power switch in the image is for an active low input on the pin so pull up resistors are recommended which would keep the pins high unless driven low by the hub. Similarly, for active high input power switch, you would have to connect pull down resistors which would keep the pins low unless driven high by the hub.

I couldn't understand the first and second cases you mentioned, could you please explain? The hub uses the PWR pins to enable or disable the power switch based on the overcurrent condition so please set the polarity of the hub to that required by the power switch. So, in cases 3 and 4, the hub will drive the pin low or high based on the PWR_PIN_POL for enabling the switch and thus provide VBUS on the downstream ports.

Best Regards,
Sananya

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New Contributor

Hi Saranya,

Thanks for the response.

Yes, I noted just now.

Please Ignore the table logic.

Regards,

Malathi T

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New Contributor

Hi Saranya,

Can you confirm about the gang mode implementation?.

Please refer the below image and provide your feedback

gang mode.PNG

Reagrds,

Malathi T

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Moderator
Moderator

Hi Malathi,

Yes, this implementation can be done but please ensure that the current limits for the power switches 1 and 2 include the current consumption of both the ports.

Best Regards,
Sananya

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New Contributor

Hi Saranya,

Thanks for the response.

Regards,

Malathi T

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New Contributor

Hi Saranya,

If OUTA & OUTB of power switch is NOT shorted together in the above block diagram, will it work in gang mode?

Regards,

Malathi T

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Moderator
Moderator

Hi Malathi,

If they are not shorted together, only OUTA will be considered for PWR#1 enable/disable and only the first port will be controlled for overcurrent events.

Best Regards,
Sananya

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New Contributor

Hi Saranya,

As per your statement, if the design will be done in below configuration, Por1 only will have overcurrent detection. So the below figure will work like individual mode NOT in gang mode. Correct me if my understanding is wrong.

If the design will work in Gang mode all the out of power switch signals should be shorted right?

(or) All the down stream ports are shorted to OUTA for gang mode if we are using single port power switch

pastedImage_0.png

Regards,

Malathi T

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Moderator
Moderator

Hi Malathi,

Thanks for attaching the figure, I will clarify the previous statement. The EN pins and FLAG pins from the power switches should be shorted together to PWR#1, OVR#1 pins of the hub respectively in Gang mode as done in the figure. But the outputs can be separated and connected to the respective port only. If you are using a single power switch with combined current limit, then the same output OUTA should be connected for all the ports.

Best Regards,
Sananya

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New Contributor

Hi Saranya,

You mean that the below type of connection also will work as Gang mode.

If i am using a power switch as per application note. It capable to deliver 500mA current for each channel. Then the down stream current limit cannot be set into 100mA per each channel. In this case how it will work as Gang mode. Could you please explain me in detail.

gang.png

Regards,

Malathi T

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Moderator
Moderator

Hi Malathi,

Yes, this should work as long as you set the current limit for each port as per bus-powered or self-powered mode. This would generate the overcurrent event when any port exceeds the limit and cause the hub to turn off power to both the ports. Please note that if the hub is self-powered, each port can draw upto 500mA as well. You can refer to the usage of the power switch in the CY4607 DVK which supports both bus-powered and self-powered modes in Individual mode. The GANG mode will be different from the individual mode since it turns off the power for all the ports whenever any port triggers the overcurrent event through only OVR#1, PWR#1 pins.

Best Regards,

Sananya

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