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We are having a debate on the implication of powering this circuit. The image is only a snipped of the complete hardware- but the debate arises from providing power using a PIN header connected to the +5v bus located somewhere else.

I suggested not to do this and rather connect to the usb +5/gnd pins as the +5v should be used for supply of external circuits and not really to power the entire hardware.

Will the filtering caps have any relevance if we put +5v on the +5V0 Rail instead of VBUS from the USB connector; like originally designed?

Will this circuit actually have any benefit at all?

Would powering the circuit using the TP1/2 be better than directly powering +5V bus?


It is not powered from a PC but a USB charger rated at 5v 1A - The device peaks 500ma~800-ma depending on what its doing. People want to connect 2A or higher to the 5v0 pin because they cannot find a USB charger with that sort of output. I am trying to establish what can happen if the +5 is connected after this circuit instead of how it was desinged.. from the USB port side.

  • 2
    \$\begingroup\$ Related discussion that is the source of technical inquiry. This might be interesting to those that want to see more about the discussion going on, but not required reading or very pertinent here. \$\endgroup\$
    – Kortuk
    Aug 16, 2012 at 15:00
  • 1
    \$\begingroup\$ To clarify a bit, are you replacing the USB power supply with a new one at the GPIO pins, or adding another power supply while the USB supply is still powered? The two answers do not make the same assumption. \$\endgroup\$
    – W5VO
    Aug 16, 2012 at 15:11
  • \$\begingroup\$ The inrush current for the design shown exceeds the USB spec. You are allowed to put a maximum of 10 uF across the USB 5V supply. \$\endgroup\$
    – markrages
    Aug 16, 2012 at 16:34
  • \$\begingroup\$ Why does this not mention the Raspberry Pi? \$\endgroup\$ Aug 16, 2012 at 18:07
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    \$\begingroup\$ @AlexChamberlain I think the user is asking in reference to the design of this specific circuit as best they can, no the overall product it is a part of. \$\endgroup\$
    – Kortuk
    Aug 16, 2012 at 18:57

1 Answer 1


If you apply power at +5V0 and you have some sort of an overvoltage condition (if the source loses control, or is inadvertently swapped for a 12V supply, etc), the current that will be drawn when the TVS clamps will not pass through the 1.1A fuse. The TVS may survive forever, or for a while, or for a few milliseconds. Once it gives up there's no line of protection left to prevent your downstream components and capacitors.

If you apply power at VBUS and get an overvoltage condition, the TVS clamp current will pass through the fuse. If the TVS, fuse, and source power supply are rated 'correctly', the TVS should be able to draw enough current (and the power supply should be able to provide enough current) to clear the fuse and safely isolate the supply from the load.

Properly placed caps should be close to where they're needed. Small decoupling caps should be near the ICs that need decoupling. Your power connection should 'see' the \$220 \mu F\$ cap before 'seeing' the load in terms of PCB geometry, but this really depends on how the board is laid out.

  • \$\begingroup\$ The three caps that are in the circuit. Will they actually do anything usefull? I thought the current going left to right will be maintained "smoothed" but if powered from another source(other side of pcb via an exposed pin direct to 5v0 rail) these caps would be useless? or not? \$\endgroup\$
    – Piotr Kula
    Aug 16, 2012 at 13:09
  • \$\begingroup\$ No, there is no directionality in the supply rail of the schematic, though some parts of the implementation could be unsuitable as a feed point. There is of course a significance to being on the opposite side of the polyfuse from the load. \$\endgroup\$ Aug 16, 2012 at 13:35
  • \$\begingroup\$ Directionality is irrelevant on a net, unless the PCB designer takes special care to set up a certain flow in the actual copper. That's why some designers make sub-nets connected with zero ohm resistors, to force the PCB designers to group certain parts together (the resistors can be removed once the parts are routed.) \$\endgroup\$ Aug 16, 2012 at 18:52

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