input 0.18A @ 220V output 1A @ 5V

I read these spec from an AC-to-DC adapter for a 4-port USB hub.

Does it sound weird? I meant the current...

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On Wed, 14 Dec 2011 18:29:04 +0800, Man-wai Chang rearranged some
electrons to say:

I read these spec from an AC-to-DC adapter for a 4-port USB hub.

Does it sound weird? I meant the current...

Other than it being extremely inefficient, what are you concerned about?

I read these spec from an AC-to-DC adapter for a 4-port USB hub.
Does it sound weird? I meant the current...

Other than it being extremely inefficient, what are you concerned about?

I understand by definition it looks OK.

But how do u explain in electron level how 0.18A become 1A?

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Man-wai Chang wrote:
I read these spec from an AC-to-DC adapter for a 4-port USB hub.
Does it sound weird? I meant the current...

Other than it being extremely inefficient, what are you concerned about?

I understand by definition it looks OK.

But how do u explain in electron level how 0.18A become 1A?


If you ignore the efficiency issue, "power is conserved".

If you want 5V at 1A on the output, and your source is 20V,
you'd expect 20V at 0.25A on the input side. The product of
V*I should be constant.

The power supply is not an ideal "transformation device". It is
not 100% efficient. So instead of putting in 20V at 0.25A
to make 5V at 1A, you might need 20V at 0.30A. The excess is
dissipated as heat, in things like power transistors, transformers
or diodes.

So when you use a very high voltage like 220V, you expect a
very low current number to go with it (the product of the two,
to give you 5 watts total).

*******

That still doesn't explain the above numbers though.
0.18A is too high.

First of all, what it said on the adapter was probably
something like this:

100-220V 0.18A

In other words, a "universal" adapter, capable of running
at two different line voltages. Simply change adapter
cords, to connect to a different kind of wall socket.

The 0.18A value is quoted at the *lowest* applied voltage, which
is 100V in this example. 100V * 0.18A = 18W.

Now, compare to the output of 5W. This is terrible efficiency.
And as a practical check, feel the outside of the adapter.
Does it feel like 13W is wasted as heat, ever ? Probably
not. It probably feels like it is running cooler than that.

Could we account for it a bit, by assuming a 0.7 power factor.
That probably wouldn't account for all the difference between
18W input and 5W output.

I can't adequately explain it. I've seen things like this
before, but have never investigated any further for an answer.
Obviously, they've measured some current, or they've attempted
to estimate the worst case current flow value. That's about
all I can suggest, it's an estimate that just never happens
in practice.

Maybe someone else has an idea where those imaginative
numbers come from.

Paul

The 0.18A value is quoted at the *lowest* applied voltage, which
is 100V in this example. 100V * 0.18A = 18W.

Now, compare to the output of 5W. This is terrible efficiency.
And as a practical check, feel the outside of the adapter.
Does it feel like 13W is wasted as heat, ever ? Probably
not. It probably feels like it is running cooler than that.

It's not hot. But I have never pushed it to its limits.

It's actually the power supply for a 4-port USB hub. I don't know why
only a 1A power supply was bundled with it when you should need 500mA
times 4.

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Man-wai Chang wrote:
The 0.18A value is quoted at the *lowest* applied voltage, which
is 100V in this example. 100V * 0.18A = 18W.

Now, compare to the output of 5W. This is terrible efficiency.
And as a practical check, feel the outside of the adapter.
Does it feel like 13W is wasted as heat, ever ? Probably
not. It probably feels like it is running cooler than that.

It's not hot. But I have never pushed it to its limits.

It's actually the power supply for a 4-port USB hub. I don't know why
only a 1A power supply was bundled with it when you should need 500mA
times 4.


Yes, but how many customers would notice that ?

If the adapter is overloaded, it will shut off.

Paul

On Dec 14, 12:36 pm, Man-wai Chang wrote:

But I have never pushed it to its limits.

It's actually the power supply for a 4-port USB hub. I don't know why
only a 1A power supply was bundled with it when you should need 500mA
times 4.

But, as you haven't, then the onus remains essentially yours. . .both
for having chosen an external powered port for selecting a particular
USB device(s) to premise reason, if then at what point reason is
exceeded by demand or expectation;- It's either of two prongs: a) when
you've got an error report in the operating system, or not, that b) a
fault condition occurs when exceeding specifications (2nd & 4th
passages respectively from wiki). . .

A bus-powered hub is a hub that draws all its power from the host
computer's USB interface. It does not need a separate power
connection. However, many devices require more power than this method
can provide, and will not work in this type of hub.

USB current (related to power) is allocated in units of 100 mA up to a
maximum total of 500 mA per port. Therefore a compliant bus powered
hub can have no more than four downstream ports and cannot offer more
than four 100 mA units of current in total to downstream devices
(since the hub needs one unit for itself). If a device requires more
units of current than the port it is plugged into can supply, the
operating system usually reports this to the user.

In contrast a self-powered hub is one that takes its power from an
external power supply unit and can therefore provide full power (up to
500 mA) to every port. Many hubs can operate as either bus powered or
self powered hubs.

However, there are many non-compliant hubs on the market which
announce themselves to the host as self-powered despite really being
bus-powered. Equally there are plenty of non-compliant devices that
use more than 100 mA without announcing this fact (or indeed sometimes
without identifying themselves as USB devices at all). These hubs and
devices do allow more flexibility in the use of power (in particular
many devices use far less than 100 mA and many USB ports can supply
more than 500 mA before going into overload shut-off) but they are
likely to make power problems harder to diagnose.

Some self-powered hubs do not supply enough power to drive a 500mA
load on every port. For example, many seven port hubs have a 1A power
supply, when in fact seven ports could draw a maximum of 7 x 0.5 =
3.5A, plus power for the hub itself. Designers assume the user will
most likely connect many low power devices and only one or two
requiring a full 500mA.

Man-wai Chang wrote:

I read these spec from an AC-to-DC adapter for a 4-port USB hub.

Does it sound weird? I meant the current...

15W in (120VAC/sqrt2 = 85Vrms; 85Vrms * 0.18A = 15W)
5W out (5V + 1A = 5W)
10W loss (15W - 5W = 10W; transformer dissipates 10W)
33% efficiency (very poor)

What did you think was weird? That you got less power out then went in?
That's how this universe works (lookup "entropy"). Despite entropy (due
to inefficiency), there is generally a conservation of energy, so you
get [mostly] out what you put in. Ohm's Law can be rewritten as R = V*I
(power = voltage times amperes). If voltage goes down then current has
to go up for power to remain the same. Due to inefficiency (dissipated
as heat) in transferring power, the power output will be less. Read up
on Ohm's Law:

http://en.wikipedia.org/wiki/Ohms_law

Is your concern that this unit is highly inefficient? So maybe it's a
super cheapy design. Is your concern that there is only 1A available
for all 4 ports instead of .5A for each one for a total of 2A available
overall? Well, that would be poor design but there's lots of hardware
poorly designed.

Is this a secret 4-port USB hub? If not, why no URL link to it so
others can see what you are asking about?

In another of your posts, you said "explain in electron level how 0.18A
become 1A?". What? You've never heard of [step-down] transformers or
not a clue how they work? If so, start he

http://en.wikipedia.org/wiki/Transformer
http://www.energyquest.ca.gov/how_it...ansformer.html
http://sound.westhost.com/xfmr.htm
http://www.youtube.com/watch?v=ZjwzpoCiF8A
http://www.youtube.com/watch?v=GMePE7NZcxw

But, as you haven't, then the onus remains essentially yours. . .both
for having chosen an external powered port for selecting a particular
USB device(s) to premise reason, if then at what point reason is
exceeded by demand or expectation;- It's either of two prongs: a) when
you've got an error report in the operating system, or not, that b) a
fault condition occurs when exceeding specifications (2nd& 4th
passages respectively from wiki). . .

The power supply was hidden in the package. How could I find out? By the
price alone?

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Thanks. I will do my homework first.

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