But this isn't really a battery charger. That's just what you use it for
- it's actually a power supply with USB output connectors. It just has
to give out 5V at as much current as what ever is connected wants.
Ideally it needs the ability to tell what you connect to it how much
current is available so that devices can decide to fast charge if they
want to without fear of overloading it. It's up to the device to protect
it's battery from over charging.
Brian Gregory (in the UK).
To email me please remove all the letter vee from my email address.
It can't be the input because 100-220VAC times the maximum stated
800mA input current is 80 Watts (give or take) input, which would
be about right assuming 50% efficiency.
That is interesting. So, when I plugged in the iPad, it "asked" for
100mA, and then it asked for more, so the charger gave it more?
But, then, why did the cord melt?
I understand that a current "source", which is the charger, is just a
source of current (sort of like being a big tank of water); it's not
going to "push" that current into the current "sink", which is the
iPad at any rate more than what the iPad "asks" for.
So, I'm assuming that the cable is bad (maybe pins are shorted, for
example, between power and ground).
But that still doesn't explain why the device says it's both 40Watts
output and that its maximum output is 6.8Amps at 5VDC.
That's not even close to 40 Watts (it's 15% off).
note that the charger itself doesn't say 40w
100ma is guaranteed without doing anything.
if a device needs more (many do), they need to request more power, and
must do so using less than 100ma.
standard usb ports should be able to provide 500ma. a bus-powered hub,
such as a keyboard, can't supply 500ma to downstream devices (it could
only supply 500-whatever it takes) so it will decline any request,
which is what causes the 'this device uses too much power' warnings.
recent usb ports can source more current because of usb hard drives
that need >500ma to spin up the drive as well as phones and tablets
that want more to charge faster.
it's hard to say without looking at it.
one possibility is a defective cord but there could be other reasons.
if you have a continuity tester, try checking the cable for shorts.
Dave Higton wrote, on Mon, 30 Nov 2015 22:32:11 +0000:
But if you're saying the 40W is actually 40VA, where is the math
that backs that up?
All "my" math (given only the printed numbers), come up with
34 Watts (DC) output with about 80VA (RMS) input.
Nothing is even close to 40W or 40VA.
If you understand inductive or capacitive reactance in AC circuit.
Most load in AC power has inductive reactance meaning voltage leads
current by certain amount depending on how reactive it is. The more
phase angle, less power factor losing power as wasted(non energy
producing) Power factor is expressed as Cosine Phi So when the angle
difference is zero PF factor is 1 which never will be. Less than one,
VA and W=V*I are not equal. I just explained it in layman talk. Draw
a full Sine wave one for voltage and one for current. In this case
voltage wave leads the current wave by few degrees. So both waves does
not super impose. The difference in phase angle expressed in Cosine is
PF. In the case of capacitive reactance current leads voltage. You
should know conjugatory number? +jR is inductive reactance, -jR is
capacitive reactance. AC circuit impedance Z is root mean square of DC
resistive value plus +/-jR Still Ohm's law like in DC cibut reactance is
additional parameter. I hope I made some sense. If I have blackboard
I could show you with graphics.....
+jR is 2*pi*f*L (f in Hertz, L in Henry) -jR is inversed 1/2*pi*f*C
(C in Farad) Opposite of Impedance Z is called Admittance Y=1/Z
If there is step down transformer involved power can incur some loss due
to poor material like heating up the core iron Not only PF is an issue.
Any power supply produce heat(lost energy lowering effciency)
For an example look at the power supply for desk top or laptop PCs.
Good ones have higher efficiency as well as producing clean very
stable outputs. You get what you pay for applies here too, LOL!
Tony Hwang wrote, on Mon, 30 Nov 2015 17:52:11 -0700:
I think the packaging is telling a lie about the 40Watts.
If you look at the device, you see the model of HC363-5U.
If you type "?HC363-5U" into the URL bar, you find the same device
under a different brand name, which lists it clearly as 35 watts!
So, pretty much, I think it's a lie or a "misprint" that the unit
is 40Watts. It's more like 34Watts (35W is reasonably close).
"Hausbell 35W HC363-5U UL Certified (UL No:E310745)
Family-Sized USB Wall Charger Plug Smart Charger,
Single USB Output 2.4A Max,Output total 6.8A Max,
5P-USB Output for Apple and Android Smartphones,Tablets and More(White)"
Do you know which authorities to contact?
Tony Hwang wrote, on Mon, 30 Nov 2015 17:12:02 -0700:
I don't fully understand the power factor but it only applies,
AFAIK, to the AC input voltage time current.
Even then, I think it only applies when you add multiple AC
The voltage and current fluctuate so, even if you pick the
RMS voltage and current of the one AC signal, if you try to
add it to another AC signal with a different phase, then
you have to factor in the power factor.
At least that's how I understood it after reading about it,
but, I don't think the power factor applies here, as you said, b
because we only care about the output power, which is DC.
Danny D. wrote, on Tue, 01 Dec 2015 00:20:57 +0000:
This article says the AC power factor plays a role whenever the
AC current and the AC voltage are not in phase:
I think the whole power factor thing is a red herring because
we're talking output power, which is DC, which has both the
current and voltage in phase.
Only if the 40W is for the input power would the AC power factor
matter, and, if that 40W was for the input, it wouldn't be called
Watts - it would be called VA (volt amperes) anyway.
So, I don't think the AC power factor applies in this question.
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