Yousuf said:
How hard would it be to make sure that a powered USB hub doesn't feed
power back to the PC? I don't know, as I don't know how the design
works, so I'm honestly asking this as a question.
Yousuf Khan
One of the companies that makes USB hub chips, provided a reference
schematic for how to build a hub.
You use a *relay*, to open the path. K1, lower left.
http://replay.waybackmachine.org/20080906233617/http://www.smsc.com/main/tools/usb/usb20h04evb.pdf
When the 5V adapter is plugged into the hub, that energizes
the relay, and breaks the DC path back to the USB host cable.
When the 5V adapter is unplugged, the relay closes, and the
USB host cable is used as a source of +5V.
This method is preferred, from the perspective that usage of
diode steering, even with Schottky diodes, would involve too
much of a voltage drop component, for the voltage drop budget.
The voltage drop budget, was only ever intended to cover
drop due to resistance in the cable. Even a Schottky, when
you pump enough current through it (500mA), the Vfb is
going to be significant.
*******
Firewire solves this problem, by placing an *unregulated* voltage
on the cable. Each device drawing power, has its own regulator ($$$).
Any device can source power, and such devices use a diode to prevent
backflow. It's the fact the bus power is unregulated, that leaves
headroom for backflow solutions. In the example, source#1 could
contribute +12V (typical of a PC). Source#2, could have a +18V
source. With the diodes, the Source#1 diode is reverse biased, and
no current flows into Source#1 from Source#2. The +18V is safely
sent to the peripheral. Since the peripheral can handle any DC voltage
up to the Firewire max, it doesn't care if it sees +12V (when Source#2
is unplugged), or see +18V (when Source#2 happens to be plugged in).
The diodes "pick the winner", and the winner provides the current.
The losers get no backfeed into themselves. USB has no backfeed prevention,
because diodes are a poor solution when the voltages are already regulated
at source.
+12V +18V
Source#1 Source#2 "IEEE1394 cables carry
| | a voltage of +8V to +40V"
--- ---
\ / \ /
===== =====
| |
+-----------+--------- peripheral --- regulator ---- load
|
GND
Note that, for a Firewire 3.5" hard drive enclosure, a buck-boost
switcher would be needed, to make +12V from such a wide range of
input voltages. Which means the power circuit looks pretty ugly.
And having to deal with +40V, probably does not help matters.
That's quite a wide range to have to support.
On USB, the bus power is already regulated, and any degradation
of the regulation means the available device power could go
out of spec (either too low or too high). For example, if you
boosted the source power, you'd think "problem solved". Until
someone used a "zero length cable", there was no drop at all,
and the full voltage appeared across the device. It's pretty
hard to "cheat" and keep all devices happy.
I'm not sure what percentage of USB hubs use no protection,
or throw in a diode solution anyway, or use the SMSC relay example
as their solution. But to me, the SMSC solution is the clear
winner, because I know right away, there won't be an issue with
backfeed.
Could you use a MOSFET to cut the power ? Perhaps. But MOSFETs
need a decent voltage on the gate, to make them work as perfect
switches. You would not want the MOSFET contributing too much
Rds to the power path, when switched on. It would be just
as bad as the diodes, if it did that.
Paul
