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The Overcurrent detection in the CY7C65630 datasheet has the following description:
After detecting an overcurrent condition, hub reports overcurrent condition to the host and disables the PWR # output to the external power device.
Does this "disable" go from Low to High?
Or, go from Low to Hi-z?
Also, is a pull-up resistor required in addition to the inverter? If yes, what is the resistance value? I don't understand when this pin output Hi-z state.
Since the opposite connection destination of the PWR is active HIGH, I would like to know if it works properly by inserting the inverter.
Thanks,
Tetsuo
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USB Hosts Hubs Transceivers
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Hello Tetsuo,
Apologies for the delay in response.
If the destination of PWR pin is active high, the polarity of PWR pin can be changed from default active low to active high from the Blaster utility.
The PWR pin has an internal pull-up, so there is no need of externally pulling it up.
Regards,
Mallika
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Hello Tetsuo,
Apologies for the delay in response.
If the destination of PWR pin is active high, the polarity of PWR pin can be changed from default active low to active high from the Blaster utility.
The PWR pin has an internal pull-up, so there is no need of externally pulling it up.
Regards,
Mallika
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I want to change the polarity of the PWR output through the inverter. Because I don't want to use EEPROM. Therefore, I want to invert the polarity of active LOW through the inverter.
Does "Disable PWR output" in the data sheet mean LOW → HIGH? If so, the polarity can be reversed only with the inverter. If the states that the PWR can take are LOW and Hi-z (not High), I can't invert it with an inverter.
Is the value taken by "Disable" Hi-z or High?
Thanks,
Tetsuo
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If the polarity of the PWR pin is active low, it is low when the port is active. When power is disabled for a port, the PWR pin will be in a high state till power is enabled.
Regards,
Mallika
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Mallika-san,
Thank you for your reply.
My understanding is that the PWR output takes only two values, LOW or HIGH (not Hi-z). So it is possible to invert by inserting an inverter. Is it correct?
Thanks,
Tetsuo
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Yes, your understanding is correct.
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Mallika-san,
Thank you for your help.
Tetsuo