How do I increase lo-current charging power for LiOn batteries?

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

Reply to
Martin Ver Strunk
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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:
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+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

Reply to
ehsjr

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

Reply to
Martin Ver Strunk

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

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.

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

Reply to
ehsjr

Thanks Ed, that looks like the way to do it. MVS

inconsequential,

Reply to
Martin Ver Strunk

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