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1. 

   x893
   Member

   Why not connect 1.5K resistor directly to pin and set pin to high for USB D+ current ?

   Posted 4 years ago #
2. 

   siy
   Member

   Probably because there is a danger of overvoltage and/or electric discharge. Beside that I see no problems with such an approach (note about bootloader and inversion applies here as well).

   Posted 4 years ago #
3. 

   x893
   Member

   but i see J-Link schematic (on AT91SAM7S chip) and see that thay use simple connect pin to D+ via 1.5K resistor. May be it normal for short line and not accepted for long distane in factory application (joke)?

   Posted 4 years ago #
4. 

   siy
   Member

   By the way: if to use some USB upstream termination IC, which has protection at all lines, direct pin connection might be completely safe.

   Posted 4 years ago #
5. 

   gbulmer
   Moderator

   sly -
   

   ... STF202 USB upstream protection IC and PNP-driver with built-in resistors. Emitter of the transistor is connected to VCC, base to DISC pin and collector to pin 1 of the IC (VBUS). ...

   Don't you lose the protection across VBUS using that circuit?
   (It is still better than the normal USB input which has no protection)

   Did you look at NUF2042XV6?
   It doesn't contain the 1.5K data-line pull-up, but is otherwise very similar.

   Posted 4 years ago #
6. 

   siy
   Member

   Well, protection across VBUS is lost with this approach, but this is minor drawback, as for me. Not all IC's contain it anyway and VBUS is not connected directly to MCU.
   As for other upstream termination/protection IC's, there are a number of them (USBUFxx from ST or STF203, for example). I've used STF202 just because it was simpler to buy in local stores and because it is most convenient for single side board routing :)

   Posted 4 years ago #

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