Electronics gurus-Powering AA/AAA-cell powered devices w/Li-Ion: Voltage conversions?
I'm surprised to find no Makezine, Instructables, or other guides for modding and powering devices off of Lithium-Ion or Lithium-Polymer battery backs. I want to add DC sockets on modified battery doors on a few gizmos to increase run time and improve operating temperature range. For example, cold weather kills Aalkaline and NiMH/NiCad... I can't afford disposable lithium AAA cells for my helmet cam and I'd like to use it as a CONSTANT DVR/log for documentation and such. In addition, I'd like to add a few other things that take a ton of batteries and convert them to run off the same packs.
I searched for universal Li-Ion packs with selectable voltage outputs and, while many have 5v USB output, they are either all for laptops with 12v+, or only available wholesale from Hong Kong. I figure 5v is a better starting point than most, so I settled on some flexible packs intended to be wrist-band Li-Ion packs.
So, how should I go about stepping down the 5v output to the voltages needed for various devices? The camera takes two batteries in a series, so the ideal output voltage would be between 1.2v*2-to-1.5v*2 (2.4v-to-3v). Another takes 4xAA, but I haven't received it in the mail yeat so I don't know if it's wired in series, parallel, or a parallel series. If in series, so that would be 1.2v*4-to-1.5v*4 (4.8v-to-6v), so I wouldn't have to change anything for that AFAIK.
My electronics skills are very limited. I could solder stuff together if you told me exactly what to buy and how to connect it and I've modded more than a few game consoles, but I really don't know what will work or not. In my fruitless searching for guides, I at least found a few vegetables (ideas!): First is a VRM based on LM388 or 317 voltage requlating transistor. I see kits for $8-$10. I'd have to figure out how to use my meter! Second is to find something like this (whatever it is) that actually supports the needed voltages... because that one doesn't. Third is to run it through three or four diodes for the voltage drop (5v-[.6v*4]=2.4v). Would this cause a current/amperage issue?
Actually, devices like digicams, etc. have a good tolerance range as batteries generally have a higher voltage than nominal when just charged or new non chargeable ones. So most, 2 battery powered devices will work on 3V +/- 1V.
Yup, one or more of those new, low differential, variable or fixed (just switch between several if you need the diff. outputs) voltage reg chips should do you.
.bh.
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Unless you include proper battery management electronics, you will damage the batteries if discharge continues below 3V per cell - not to mention you'll also need a charger designed specifically for them. Li-Ions provide 3.7V per cell, so any batteries not providing a multiple of that already have a built-in regulator - it may be wise to try to find 'plain' batteries, depending on what the electronics in the packs you're looking at really do. It's also recommended to use management circuits with charging multiple cells in series, to insure equal charging - a proper charger will be capable of this (or the packs will have integrated electronics for this).
That UBEC is used to eliminate the need for a receiver battery for electric RC planes/cars/boats - it runs the receiver (normally requires 5-6V) off the electric motor batteries (almost always >6V).
You have three options: diode drop/zener regulator, linear regulator, or buck converter (a form of smps). The first two drop the voltage & dissipate it as heat - efficiency will be Vout/Vin. Any devices requiring considerable current will increase heat dissipation & may require a heatsink in some cases. The buck converter can approach efficiencies of 95%, depending on the design, with little deviation resulting from changes in input/output voltages.
Actually, devices like digicams, etc. have a good tolerance range as batteries generally have a higher voltage than nominal when just charged or new non chargeable ones. So most, 2 battery powered devices will work on 3V +/- 1V.
Yup, one or more of those new, low differential, variable or fixed (just switch between several if you need the diff. outputs) voltage reg chips should do you.
.bh.
You mean the LM3xx chips or whatever that other thing is that I linked to?
Quote:
Originally Posted by PM650
Unless you include proper battery management electronics, you will damage the batteries if discharge continues below 3V per cell - not to mention you'll also need a charger designed specifically for them. Li-Ions provide 3.7V per cell, so any batteries not providing a multiple of that already have a built-in regulator - it may be wise to try to find 'plain' batteries, depending on what the electronics in the packs you're looking at really do. It's also recommended to use management circuits with charging multiple cells in series, to insure equal charging - a proper charger will be capable of this (or the packs will have integrated electronics for this).
Can't I assume that any universal Li-Ion/Li-Po battery pack with a 5v USB port would know that USB-powered devices have no battery regulation circuitry for their external power source and thus there needs to be one built in to the battery pack?
5v through miniUSB is the only output. I'm sure it isn't limited to 500mA like a host USB port considering that it isn't a USB host. It includes cables to charge various devices that would not have management electronics for their external power source, so I assume it's built into the pack... for example, the original PSP has a Li-Po battery inside and the electronics to charge/manage it, but the official DC charge cable comes from an AC power brick.
These packs obviously have built in charge electronics because they only charge via USB (does that also imply built-in battery management/controller?). They are flexible to conform to my helmet and can be snapped/unsnapped easily. I plan to hide them under a Skull Skin. Is this a sound idea?
Quote:
Originally Posted by Zepper
That UBEC is used to eliminate the need for a receiver battery for electric RC planes/cars/boats - it runs the receiver (normally requires 5-6V) off the electric motor batteries (almost always >6V).
You have three options: diode drop/zener regulator, linear regulator, or buck converter (a form of smps). The first two drop the voltage & dissipate it as heat - efficiency will be Vout/Vin. Any devices requiring considerable current will increase heat dissipation & may require a heatsink in some cases. The buck converter can approach efficiencies of 95%, depending on the design, with little deviation resulting from changes in input/output voltages.
Thanks! I'm not sure I know enough to implement the buck type (output is controlled though a pin?), but it looks to be what I want. Perhaps I can find a guide now that I know what to look for. Then again, there's plenty of airflow on a helmet, so the easier method may just be the diode or VRM path.
Since it charges & discharges through a USB port, it's likely safe to assume it has the proper charge/discharge electronics to prevent damage - if you were to use bare cells, you would need the management electronics I mentioned. As far as usage, I would not put a Li-Po battery in any place that would have the potential for physical damage - if it's going under a helmet you might want to insure the batteries are well padded, but not so much that they run hot (no idea how warm these will run in this application).
The buck converter regulates by adjusting duty cycle of the switching transistor(s), but it determines the duty cycle by comparing a saw-tooth waveform to an error amplifier output, which itself is the output voltage compared to a constant reference voltage. In short, you can change the regulated output with a variable resistor (pot), just as in a linear regulator (see the datasheet for the link I posted - efficiency figures are included for various voltages & currents). The buck converter will consume about the same space as a linear regulator (LM317) solution while being slightly more expensive & efficient, diode(s) will be slightly more compact but voltage range will be in 0.2V steps at best (if schottkys are used).
There are two things to consider. Voltage and current. If the device is running off of 4 AA batteries with a typical 2000mah capacity per battery, then it is likely expecting about 4.8volts @ 2200mah or a total of 4.8 x 2.2 amps = 10.56 watts total power. NiMh batteries would be around 4.4 volts @ 2000mah = 8.8 watts total.
The packs you linked are 7.4V @ 1500mah packs regulated down to 5v. We can't say what they are using for regulation so we don't know what the loss is. I would guess the output is probably 10 watts total. About the same as 4 AA batteries.
The good part is the packs take care of the charging which is one of the biggest problems for end users working with LiPo batteries . The 5vdc could just be connected in place of the 4AA and the device should work fine. To get less voltage all you need to add is something to drop the 5vdc down to whatever you need. If it is a device that only used 2 AAA batteries then that is 3v @ 2000mah or 6 watts power. Buy 3 silicon diodes, 1N4004 are fine or the RS ones below, and put them in series between the + and the device and that will provide the needed drop , you can heatshrink it into the wiring if you want.
If using the diodes above you only need 2, each drops 1 volt.
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Last edited by Modelworks; 11-03-2009 at 05:57 PM.
Since it charges & discharges through a USB port, it's likely safe to assume it has the proper charge/discharge electronics to prevent damage - if you were to use bare cells, you would need the management electronics I mentioned. As far as usage, I would not put a Li-Po battery in any place that would have the potential for physical damage - if it's going under a helmet you might want to insure the batteries are well padded, but not so much that they run hot (no idea how warm these will run in this application).
The main reason for a flexible battery pack is to conform to the curved outer shell of the helmet. I will be slipping them under an elastic helmet skin. My current Bluetooth kit is Li-ion/Li-Po and the control unit itself attaches to the rest (mic boom, antennas, internal speakers, etc) externally.
Quote:
Originally Posted by PM650
The buck converter regulates by adjusting duty cycle of the switching transistor(s), but it determines the duty cycle by comparing a saw-tooth waveform to an error amplifier output, which itself is the output voltage compared to a constant reference voltage. In short, you can change the regulated output with a variable resistor (pot), just as in a linear regulator (see the datasheet for the link I posted - efficiency figures are included for various voltages & currents). The buck converter will consume about the same space as a linear regulator (LM317) solution while being slightly more expensive & efficient, diode(s) will be slightly more compact but voltage range will be in 0.2V steps at best (if schottkys are used).
Thanks.
Quote:
Originally Posted by Modelworks
There are two things to consider. Voltage and current. If the device is running off of 4 AA batteries with a typical 2000mah capacity per battery, then it is likely expecting about 4.8volts @ 2200mah or a total of 4.8 x 2.2 amps = 10.56 watts total power. NiMh batteries would be around 4.4 volts @ 2000mah = 8.8 watts total.
The packs you linked are 7.4V @ 1500mah packs regulated down to 5v. We can't say what they are using for regulation so we don't know what the loss is. I would guess the output is probably 10 watts total. About the same as 4 AA batteries.
The good part is the packs take care of the charging which is one of the biggest problems for end users working with LiPo batteries . The 5vdc could just be connected in place of the 4AA and the device should work fine. To get less voltage all you need to add is something to drop the 5vdc down to whatever you need. If it is a device that only used 2 AAA batteries then that is 3v @ 2000mah or 6 watts power. Buy 3 silicon diodes, 1N4004 are fine or the RS ones below, and put them in series between the + and the device and that will provide the needed drop , you can heatshrink it into the wiring if you want.
If using the diodes above you only need 2, each drops 1 volt.
I see a lot of complaints about those diodes having different specs than advertised and I don't see the 1v drop specified... unless that's what the "1000-PIV" means, in which case, people say that it's really 700-800 PIV (whatever that is). If they are the right ones after all, I'll go ahead and buy them. Thanks!
Oh, and what will happen if there isn't enough or is too much current? I've plugged many 500mA USB devices into my 1A official iPhone 3G USB AC charger without issue. Come to think of it, it's the recalled one without the colored dot so I'd better go exchange it. I expect that because this isn't a USB host device that is can provide at least 800mA-1,000mA like many USB devices require even though it is officially out-of-spec.
I see a lot of complaints about those diodes having different specs than advertised and I don't see the 1v drop specified... unless that's what the "1000-PIV" means, in which case, people say that it's really 700-800 PIV (whatever that is). If they are the right ones after all, I'll go ahead and buy them. Thanks!
Diodes have a couple different ratings. the PIV is peak inverse voltage. Diodes act like a one way gate for electricity. The PIV voltage is the amount of voltage before the diode becomes a two way gate. You are nowhere near 1000 volts so no worry
The other ratings are' If ' which shows the current , those are 1.5A , so plenty for devices at 3v that would normally run of AAA batteries. The last is Vf or energy lost going forward for normal use. That varies from diode to diode but has to be within 10% spec . Those diodes are rated a Vf of 1.1V so roughly between 1 - 1.2 volts, way within the specs for what you need.
Quote:
Oh, and what will happen if there isn't enough or is too much current? I've plugged many 500mA USB devices into my 1A official iPhone 3G USB AC charger without issue. Come to think of it, it's the recalled one without the colored dot so I'd better go exchange it. I expect that because this isn't a USB host device that is can provide at least 800mA-1,000mA like many USB devices require even though it is officially out-of-spec.
If not enough the device will not work, can't ever have too much current, exception is devices like LED light. A circuit will only consume what it can use.
Seeing that the diodes are rated at 1.5A if you want to protect from a fault or shorted wiring it would be a good idea to put a fuse right after the battery leads. That way if something shorts or breaks you don't short the battery pack, which could cause it to overheat. I would use this for a fuse holder and put a 1.5A fuse in it: http://www.radioshack.com/product/in...ductId=2102784
__________________
"Education is what remains after one has forgotten everything he learned in school."
"If we knew what it was we were doing, it would not be called research, would it?"
A. Einstein
Electronics gurus-Powering AA/AAA-cell powered devices w/Li-Ion: Voltage conversions?
Then again, there's plenty of airflow on a helmet, so the easier method may just be the diode or VRM path.
I expect that because this isn't a USB host device that is can provide at least 800mA-1,000mA like many USB devices require even though it is officially out-of-spec.
