Question about connecting cell phone charger to transformer

T

Ted S.

Guest
I realize you may have to make a guess here, but this is my question...

IF I wanted to take my cell phone car charger and plug it in to a 12 volt
transformer instead of the car battery, what amp rating to you think the
transformer would have to be? I know it would have to be an output of 12V
DC and know it would need to put out sufficient current but don't know what
is enough for my Lucky Goldstar V111 Verizon cell phone.

Any guesses? Any way to tell? I could call LG, but I'd probably spend all
day trying to get an answer like that.

Thanks.

Ted
 
In article <bjdh3j016tt@enews1.newsguy.com>,
tswirsky@SPAMNOT.screaminet.com says...
| IF I wanted to take my cell phone car charger and plug it in to a 12 volt
| transformer instead of the car battery, what amp rating to you think the
| transformer would have to be? I know it would have to be an output of 12V
| DC and know it would need to put out sufficient current but don't know what
| is enough for my Lucky Goldstar V111 Verizon cell phone.
|
|

12 to 14.2 DC would be fine. Check the mA hour rating marked on the
battery, time how long it takes to charge it from flat to fully
charged.

Divide the hours into the mA hour rating, then multiply that figure by
3 (to allow a bit of spare capacity). That last figure should be an
adequate mA rating guesstimate for your PSU.

--
Regards,
Harry (M1BYT)...

Remove the 'NOSPAM' in my email address to reply.

Free Amateur Radio Courses:-
http://www.ukradioamateur.org
 
Yikes. I need an example to make sure I understand.

So it my 1200 ma hr battery takes 4 hours to fully charge, I get 300, then
multiply by 3 to get 900. So you say I should use a 12V DC power supply
that outputs at least 900 ma. Right???

Um, I assume no damage will be done if my PS is rated at higher than that,
correct?

Also, what happens if it is much lower, say 100 ma? My guess is that all
will work fine with no damage (do you notice that is a big concern :) )
although it will take much longer to charge the battery. Am I correct on
this?

Thanks for all the help!

Ted





"Harry Bloomfield" <harry.m1bytNOSPAM@tiscali.co.uk> wrote in message
news:MPG.19c46f127a88bf7d98993d@news.individual.net...
In article <bjdh3j016tt@enews1.newsguy.com>,
tswirsky@SPAMNOT.screaminet.com says...
| IF I wanted to take my cell phone car charger and plug it in to a 12
volt
| transformer instead of the car battery, what amp rating to you think
the
| transformer would have to be? I know it would have to be an output of
12V
| DC and know it would need to put out sufficient current but don't know
what
| is enough for my Lucky Goldstar V111 Verizon cell phone.
|
|

12 to 14.2 DC would be fine. Check the mA hour rating marked on the
battery, time how long it takes to charge it from flat to fully
charged.

Divide the hours into the mA hour rating, then multiply that figure by
3 (to allow a bit of spare capacity). That last figure should be an
adequate mA rating guesstimate for your PSU.

--
Regards,
Harry (M1BYT)...

Remove the 'NOSPAM' in my email address to reply.

Free Amateur Radio Courses:-
http://www.ukradioamateur.org
Click to expand...
 
Darn. That's a good idea. Check the output on the wall charger (if it has
one) and use at least that ma rating? I ask again, am I correct that there
will be no damage if my PS is rated for a higher ma value?

On a different subject, how does one tell if an output is AC or DC? I have
a multitester. Is it enough to set it on DC Volts and see if switching the
leads makes it go from positive to negative? This would mean DC, right?
Can this hurt the tester? Is there another way? There must be.

And you are correct, I was sloppy with my wording when I called the wall
plug DC power supply a 'transformer'. My knowledge of electronics is basic
and rusty so I really appreciate the help you guys give.

Ted




"Peter Bennett" <peterbb@interchange.ubc.ca> wrote in message
news:uvqklvog841cvehchc0rt81rcapd3pic8r@news.supernews.com...
On Sat, 6 Sep 2003 16:45:45 -0400, "Ted S."
tswirsky@SPAMNOT.screaminet.com> wrote:

I realize you may have to make a guess here, but this is my question...

IF I wanted to take my cell phone car charger and plug it in to a 12 volt
transformer instead of the car battery, what amp rating to you think the
transformer would have to be? I know it would have to be an output of
12V
DC and know it would need to put out sufficient current but don't know
what
is enough for my Lucky Goldstar V111 Verizon cell phone.

Any guesses? Any way to tell? I could call LG, but I'd probably spend
all
day trying to get an answer like that.

Thanks.

Ted


The charger for my LG TM520 says it wants 12 volts DC at 0.3 amp (3.6
watts), and the wall-wart supplied with it claims to output the same.
Like most wall plug supplies, this one is unregulated - it outputs
about 17 volts with no load.

I'd therefore suggest you need a 12 volt DC supply capable of 0.3 amps
or more (but look for labels on your charger, in case it is
different).

Note that a transformer will output AC, and you need DC (however, we
do often get a little sloppy and call the wall plug DC power supplies
"transformers").



--
Peter Bennett, VE7CEI
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
Click to expand...
 
So it my 1200 ma hr battery takes 4 hours to fully charge, I get 300, then
multiply by 3 to get 900. So you say I should use a 12V DC power supply
that outputs at least 900 ma. Right???

Um, I assume no damage will be done if my PS is rated at higher than that,
correct?

Also, what happens if it is much lower, say 100 ma? My guess is that all
will work fine with no damage (do you notice that is a big concern :) )
although it will take much longer to charge the battery. Am I correct on
this?
Click to expand...
Your supply will probably die and certainly overheat if your phone tries to
take more than the supply can give. No damage at all will occur if the
supply can supply more than the charger needs. This assumes that you are
using the charger that you used to plug into the cigar lighter of your car.
If you intend to do it without the car charger then the problem is a lot
more complex and I wouldn't recommend it particularly if you have LiIon
batteries in your phone.
 
Thanks! You all are sure nice to lend your knowledge to others. (I do too,
but not in electronics :) )

Firstly, yes, I'd absolutely be using the car charger.

Secondly, I just noticed something else. My battery says 3.7 volts.
(doesn't tell amps) The house charger I use says it supplies 5.2 VDC..
BUT, the terminals from it attach to a different place than does the plug
for the car charger and the other type of house charger which I do not have.
(With the house charger, I have to set the phone in a base to charge and
cannot use it while charging. The car charger can talk and charge at the
same time.)

So, this kills the other idea of looking at the output of my wall charger
(5.2V) since it connects to a different place on the phone.

Well, I do have some DC power supplies from 12 to 14V that are from .6 to
2A. Do you think I could try the lower ones and just keep watching
(feeling) the PSU to see if it is getting too warm? It would take a few
minutes to overheat, wouldn't it?

Any informative web sites I could read (and understand) on this topic?

Thanks again for all the help. If I do decide to try this, you've all be
really helpful.

Ted




"Mjolinor" <mjolinor@hotmail.com> wrote in message
news:EbB6b.204$Oc4.48103@newsfep2-gui.server.ntli.net...
So it my 1200 ma hr battery takes 4 hours to fully charge, I get 300,
then
multiply by 3 to get 900. So you say I should use a 12V DC power supply
that outputs at least 900 ma. Right???

Um, I assume no damage will be done if my PS is rated at higher than
that,
correct?

Also, what happens if it is much lower, say 100 ma? My guess is that
all
will work fine with no damage (do you notice that is a big concern :) )
although it will take much longer to charge the battery. Am I correct
on
this?

Your supply will probably die and certainly overheat if your phone tries
to
take more than the supply can give. No damage at all will occur if the
supply can supply more than the charger needs. This assumes that you are
using the charger that you used to plug into the cigar lighter of your
car.
If you intend to do it without the car charger then the problem is a lot
more complex and I wouldn't recommend it particularly if you have LiIon
batteries in your phone.
Click to expand...
 
In article <bjg23502mls@enews4.newsguy.com>,
tswirsky@SPAMNOT.screaminet.com says...
| Well, I do have some DC power supplies from 12 to 14V that are from .6 to
| 2A. Do you think I could try the lower ones and just keep watching
| (feeling) the PSU to see if it is getting too warm? It would take a few
| minutes to overheat, wouldn't it?
|
|

Start with the heavier current PSU's and work down from there. Make
sure the battery is is need of a charge each time and check the
temperature of the PSU case after 15 minutes. They should get
appreciatively warm to the touch (guess 50 deg C internal), but not
hot.

A vehicle power supply will normally vary between 11.5v and 14.2v, so
it should be perfectly happy with any of your described supplies.

--
Regards,
Harry (M1BYT)...

Remove the 'NOSPAM' in my email address to reply.

Free Amateur Radio Courses:-
http://www.ukradioamateur.org
 
I realize you may have to make a guess here, but this is my question...

IF I wanted to take my cell phone car charger and plug it in to a 12 volt
transformer instead of the car battery, what amp rating to you think the
transformer would have to be? I know it would have to be an output of 12V
DC and know it would need to put out sufficient current but don't know
what
is enough for my Lucky Goldstar V111 Verizon cell phone.

Any guesses? Any way to tell? I could call LG, but I'd probably spend
all
day trying to get an answer like that.
Click to expand...
The way to tell is to measure it when you have it plugged into the car, the
battery is flattish and it is making a phone call, that should be maximum
current requirement. If you can make it a weak signal as well then that
would be more accurate. Multiply that figure by 3 or 4 and use that as a
minimum.
 
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