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Home > Archive > Electrical Engineering > April 2006 > How do I increase lo-current charging power for LiOn batteries?
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How do I increase lo-current charging power for LiOn batteries?
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| Martin Ver Strunk 2006-04-06, 7:21 pm |
| Hello folks,
I'm a DIYer that likes to fabricate solutions to work and home projects.
Unfortunately, electricity is not a strong point for me, and my EE uncle
passed away about 11 years ago.......
I have a lot of portable handheld radios with LiOn batteries that travel
frequently throughout North America and Europe. I currently store/ship them
in small briefcases. The chargers are all 110v un-regulated wall warts, and
in the US I just plug them into a power strip (3-4 at a time) until all
12-15 radios are charged, while overseas I plug them into a basic power
converter (220v-110v), or use a generic adjustable wall wart. In addition
to burning through lots of chargers, the briefcases now contain more
hardware for the charging process than the radios themselves.
I want to fabricate a charging system that utilizes a single, regulated,
universal transformer, then split the output wires to accomodate as many
radios (or charger bases) as possible. Although we use different types of
radios (with different charging requirements/capacities), I'll set up each
system separately so that there's no mixing of the output currents from the
transformers.
Now for my problem(s): None of the transformers that I've looked at in my
research exactly match any of the input/charging requirements of the radios.
I'm not really sure if this is a problem, as I expect that the available
power for multiple chargers will need to be increased from the requirements
of a single charger. In addition, the transformers also give a Wattage
rating, while the wall warts don't mention watts, only Volts and Amps) So,
my question is this: if I know the power requirements for 1 charger, how do
I calculate the requirements for 2,3,4 or more chargers? How will this
affect the batteries when all of the chargers are active simultaneously, or
just 1 of the radios is being charged?
The 5 setups that I've currently identified are this:
a) 6vdc/500mA
b) 6.5vdc/200mA
c) 9vdc/100mA
d) 9vdc/500mA
e) 16vdc/900mA
Any help or guidance is much appreciated. Please reply directly to the
group, as this e-mail is fake.
MvS
| |
|
| Martin Ver Strunk wrote:
> Hello folks,
>
> I'm a DIYer that likes to fabricate solutions to work and home projects.
> Unfortunately, electricity is not a strong point for me, and my EE uncle
> passed away about 11 years ago.......
>
> I have a lot of portable handheld radios with LiOn batteries that travel
> frequently throughout North America and Europe. I currently store/ship them
> in small briefcases. The chargers are all 110v un-regulated wall warts, and
> in the US I just plug them into a power strip (3-4 at a time) until all
> 12-15 radios are charged, while overseas I plug them into a basic power
> converter (220v-110v), or use a generic adjustable wall wart. In addition
> to burning through lots of chargers, the briefcases now contain more
> hardware for the charging process than the radios themselves.
>
> I want to fabricate a charging system that utilizes a single, regulated,
> universal transformer, then split the output wires to accomodate as many
> radios (or charger bases) as possible. Although we use different types of
> radios (with different charging requirements/capacities), I'll set up each
> system separately so that there's no mixing of the output currents from the
> transformers.
>
> Now for my problem(s): None of the transformers that I've looked at in my
> research exactly match any of the input/charging requirements of the radios.
> I'm not really sure if this is a problem, as I expect that the available
> power for multiple chargers will need to be increased from the requirements
> of a single charger. In addition, the transformers also give a Wattage
> rating, while the wall warts don't mention watts, only Volts and Amps) So,
> my question is this: if I know the power requirements for 1 charger, how do
> I calculate the requirements for 2,3,4 or more chargers? How will this
> affect the batteries when all of the chargers are active simultaneously, or
> just 1 of the radios is being charged?
>
> The 5 setups that I've currently identified are this:
> a) 6vdc/500mA
> b) 6.5vdc/200mA
> c) 9vdc/100mA
> d) 9vdc/500mA
> e) 16vdc/900mA
>
> Any help or guidance is much appreciated. Please reply directly to the
> group, as this e-mail is fake.
>
> MvS
>
>
Best to use 2 dc wall wart supplies. One rated at 12 V, 1.5 amps
to provide a, b, c, d and the last one for 16 V 1 amp.
The 12 volt supply can feed 4 regulators: 2 7809's for
the 9 volts you need, 1 7806 for the 6 volts. For the
6.5, use an LM317 with R1 at 240 ohms and R2 at 1K.
That will give you 6.458 volts - close enough to
6.5. If you want it exactly 6.5 volts, put an 8 ohm
resistor in series with the 1K. See the datasheet:
http://cache.national.com/ds/LM/LM117.pdf
The circuit is on page 1
------
+12 -+--Vin|LM7809|Vout---> 9V
| ------
| |
| Gnd
|
| ------
+--Vin|LM7809|Vout---> 9V
| ------
| |
| Gnd
|
| ------
+--Vin|LM7806|Vout---> 6V
| ------
| |
| Gnd
|
| ------
+--Vin|LM317 |Vout---+---> 6.5V
------ |
| [240]
| |
+---[1K]---+
|
Gnd
None of the above is a *charger*, per se. They are all
supplies that match the numbers you posted, except that
they will provide more than the currents you wrote, if
the device tries to draw more. You'll need to put the
regulators on a heat sink.
Ed
| |
| Martin Ver Strunk 2006-04-10, 9:21 pm |
| Ed,
I went to the site and poked around a little.....WAY over my head.....
Ideally, I'd like to use generic, otc products that I can just splice the
output wires from. I just don't have any knowledge about what is required
to accomplish this, given the information I have (from below). If I want to
power three, 9vdc (500mA) radios, can I just assume that I need a 27vdc
(1.5A) power converter?
My understanding is that the available number of Amps is inconsequential,
the device(s) will only draw what it needs, with the remainder in reserve.
I just don't understand the relationship between Amps, Volts and Watts when
adding additional devices to the output side of the AC/DC converter. Is it
a simple linear equation of the power; device1 + device2 + device3 +
deviceX.....?
TIA,
Marty
"ehsjr" <ehsjr@bellatlantic.net> wrote in message
news:UXmZf.3389$v9.680@trndny01...
> Martin Ver Strunk wrote:
them[color=darkred]
and[color=darkred]
addition[color=darkred]
of[color=darkred]
each[color=darkred]
the[color=darkred]
my[color=darkred]
radios.[color=darkred]
requirements[color=darkred]
So,[color=darkred]
do[color=darkred]
or[color=darkred]
>
> Best to use 2 dc wall wart supplies. One rated at 12 V, 1.5 amps
> to provide a, b, c, d and the last one for 16 V 1 amp.
>
> The 12 volt supply can feed 4 regulators: 2 7809's for
> the 9 volts you need, 1 7806 for the 6 volts. For the
> 6.5, use an LM317 with R1 at 240 ohms and R2 at 1K.
> That will give you 6.458 volts - close enough to
> 6.5. If you want it exactly 6.5 volts, put an 8 ohm
> resistor in series with the 1K. See the datasheet:
> http://cache.national.com/ds/LM/LM117.pdf
> The circuit is on page 1
>
> ------
> +12 -+--Vin|LM7809|Vout---> 9V
> | ------
> | |
> | Gnd
> |
> | ------
> +--Vin|LM7809|Vout---> 9V
> | ------
> | |
> | Gnd
> |
> | ------
> +--Vin|LM7806|Vout---> 6V
> | ------
> | |
> | Gnd
> |
> | ------
> +--Vin|LM317 |Vout---+---> 6.5V
> ------ |
> | [240]
> | |
> +---[1K]---+
> |
> Gnd
>
>
> None of the above is a *charger*, per se. They are all
> supplies that match the numbers you posted, except that
> they will provide more than the currents you wrote, if
> the device tries to draw more. You'll need to put the
> regulators on a heat sink.
>
> Ed
| |
| ehsjr 2006-04-11, 10:21 pm |
| Martin Ver Strunk wrote:
> Ed,
>
> I went to the site and poked around a little.....WAY over my head.....
>
> Ideally, I'd like to use generic, otc products that I can just splice the
> output wires from. I just don't have any knowledge about what is required
> to accomplish this, given the information I have (from below). If I want to
> power three, 9vdc (500mA) radios, can I just assume that I need a 27vdc
> (1.5A) power converter?
No, you don't want that. To *power* 3 9vdc 500 mA radios
from one source, get a 12 VDC 1.5 amp (or higher) wall wart.
Then add one LM7809 wired like this for each radio:
-----
| o |
|_____|
/______/|
|LM7809||
|______|/
| | |
+12 ------+ | +----- +9V to radio
| | |
- | -
| | | | | C1 = .33 uf 50 volt capacitor
C1 |_| | |_|C2 C2 = .1 uf 50 volt capacitor
| | |
Gnd ------+--+--+------ Gnd to radio
>
> My understanding is that the available number of Amps is inconsequential,
> the device(s) will only draw what it needs, with the remainder in reserve.
With the above circuit, that is true, provided the number
of amps is high enough. When you have a regulated voltage
that matches the voltage of the equipment, the source current
capability can be anything equal to or higher than what the
equipment needs.
> I just don't understand the relationship between Amps, Volts and Watts when
> adding additional devices to the output side of the AC/DC converter. Is it
> a simple linear equation of the power; device1 + device2 + device3 +
> deviceX.....?
Yes, in a parallel circuit for a given voltage, you can just
add the amperage that each device uses to find the total amperage
required from the supply. The supply can be capable of higher
amperage, but that does not matter, as long as the supply provides
the specified voltage. Your radios must be on a parallel circuit.
A series circuit would not work, as all 3 radios would need to
be turned on in order for any of them to work. (There are other
reasons too, but no need to go into that.) With three 9 volt radios
at 500 mA each, it adds to 1500 mA (or 1.5 amps). You cannot add
the voltages in this computation of a parallel circuit.
If it was a series circuit for something like 3 light bulbs in
an xmas tree light string, then you can add the voltages to get
27, but you must not add the current. Three 9 volt, 500 mA
bulbs in series would need a supply of 27 volts, and 500 mA
would be enough. And, as long as the supply provided 27 volts,
it would not matter if it was capable of supplying hundreds of
amps - the bulbs would use only half an amp (500 mA).
Ed
>
> TIA,
>
> Marty
>
>
>
| |
| Martin Ver Strunk 2006-04-13, 7:21 pm |
| Thanks Ed, that looks like the way to do it.
MVS
"ehsjr" <ehsjr@bellatlantic.net> wrote in message
news:enY_f.4258$ee6.2191@trndny01...[color=darkred]
> Martin Ver Strunk wrote:
the[color=darkred]
required[color=darkred]
want to[color=darkred]
>
> No, you don't want that. To *power* 3 9vdc 500 mA radios
> from one source, get a 12 VDC 1.5 amp (or higher) wall wart.
> Then add one LM7809 wired like this for each radio:
>
> -----
> | o |
> |_____|
> /______/|
> |LM7809||
> |______|/
> | | |
> +12 ------+ | +----- +9V to radio
> | | |
> - | -
> | | | | | C1 = .33 uf 50 volt capacitor
> C1 |_| | |_|C2 C2 = .1 uf 50 volt capacitor
> | | |
> Gnd ------+--+--+------ Gnd to radio
>
>
inconsequential,[color=darkred]
reserve.[color=darkred]
>
> With the above circuit, that is true, provided the number
> of amps is high enough. When you have a regulated voltage
> that matches the voltage of the equipment, the source current
> capability can be anything equal to or higher than what the
> equipment needs.
>
when[color=darkred]
it[color=darkred]
>
> Yes, in a parallel circuit for a given voltage, you can just
> add the amperage that each device uses to find the total amperage
> required from the supply. The supply can be capable of higher
> amperage, but that does not matter, as long as the supply provides
> the specified voltage. Your radios must be on a parallel circuit.
> A series circuit would not work, as all 3 radios would need to
> be turned on in order for any of them to work. (There are other
> reasons too, but no need to go into that.) With three 9 volt radios
> at 500 mA each, it adds to 1500 mA (or 1.5 amps). You cannot add
> the voltages in this computation of a parallel circuit.
>
> If it was a series circuit for something like 3 light bulbs in
> an xmas tree light string, then you can add the voltages to get
> 27, but you must not add the current. Three 9 volt, 500 mA
> bulbs in series would need a supply of 27 volts, and 500 mA
> would be enough. And, as long as the supply provided 27 volts,
> it would not matter if it was capable of supplying hundreds of
> amps - the bulbs would use only half an amp (500 mA).
>
> Ed
>
>
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