USB charger

[3Good5U]

I'd like to make an USB charger. I can break out some USB port from old PC parts. I am thinking of just connecting 3x rechargable AA batteries (4.5V) but a USB port is usually 5V, I assume it's not a problem to go under 5V ? Should I place a voltage regulator as well ? (P.S. I have no experience with a voltage regulator, if the voltage is below the specified voltage, will it go up to that voltage or only limit to that voltage ?)
Thanks in advance !

Axel

[kxenos]

You will need a circuit that's called step-up or boost converter (2Euros in ebay) for this or have 4 batteries and use an LDO linear regulator but there are other issues with this. Linear or LDO regulators used like this will typically give the input voltage minus about 0.3 - 0.5V but it will not be regulated, meaning it will not be steady: As your batteries discharge the voltage will drop.

[Simon]

This is something you can buy off the shelf or are you trying to learn ? If you want to learn then basically your learning about SMPS or linear regulators, you might be best doing that in a less challenging design first like trying to convert 12V to 5V

[3Good5U]

I am an electromechanics student so I guess this is good practice. I might as well buy these chargers as they only cost around €15 

[Simon]

Yes, build yourself a car adapter first, it will be a bit easier as you have more voltage head room. Although I can't see why your phone would have a problem with 4.5-4.75V

[Spikee]

Microchip has some nice low cost easy to use charger ic's. Like the MCP73831 (lion/lipo's).
http://www.microchip.com/wwwproducts/Devices.aspx?dDocName=en024903

[Rick Law]

I'd like to make an USB charger. I can break out some USB port from old PC parts. I am thinking of just connecting 3x rechargable AA batteries (4.5V) but a USB port is usually 5V, I assume it's not a problem to go under 5V ? Should I place a voltage regulator as well ? (P.S. I have no experience with a voltage regulator, if the voltage is below the specified voltage, will it go up to that voltage or only limit to that voltage ?)
Thanks in advance !

Axel

From how you worded this, sounds like you want to use three AA rechargeable to power a charger that charges perhaps Li Ion cells.

3 AA rechargeable would be 3.6V nominal.  NiMH is only 1.2V per cell.  You actually have anywhere from >5V to below 3V.  Most of the capacity will be delivered at or around 3.6V for good cells like Eneloop.  Other cells may deliver most of the capacity below 4.6V.

So, with 3AA NiMH, you better use a charger setup that operate down to 3.0V since most of your capacity would be below 4V.  A booster is a good option too but it is not efficient.  You would loose 40% of your power to boosting, but you would have steady 5V 750mA-ish for (if your cells are good) around 30-45 minutes, or even an hour if your cells are really good. 

I made a 3AA "power supply and charger" after an extended storm power outage.  Using a Chinese 2577 booster board with 3xAA Eneloop.  It powers 4xlow power LED without booster, or G4-5050smd 12V-1W lantern, or a USB plug so I can adjust the 2577 to get 5v to charge my phone on emergency.  (Crazy set up, since it is for the "last stand" stage of an emergency when even my my car is low on gas and my 12V SLA's are all flat,.  For such rare use, I went for throwing together something cheap rather than something nicer.)

I can get days of 4xlow power led, or 2 hours of 12V-1W light which even with boost went down to 9V in 2 hours; or adjust the 2577 to 5v and power my charger.  My charger is at about 500mA.  If I recall, it was around 1 to 1.5 hours before the 3xAA drops too low to power the booster.

You need to test endurance for your batteries for exact time of how long you can run the charger.

With 2000mAH Eneloop, the math would be total power @5V@60% boost efficiency = 0.6*2000*3.6/5 =864mAH @5V total;
With prior tests, I know Eneloop does delivery at least 80% of it power at around nominal voltage or above, so 80% is 691mA for an hour.  The rest is gravy.

Good luck with yours.

Rick

[DanielS]

Yes, build yourself a car adapter first, it will be a bit easier as you have more voltage head room. Although I can't see why your phone would have a problem with 4.5-4.75V

It could certainly be a problem if the power management IC on the phone/tablet/whatever interprets low voltage on the USB power lines as excessive voltage drop due to current draw and tries backing off to let USB voltage recover. I would not be surprised if 4.75V caused many devices to charge much slower than they should.

[Rick Law]

Yes, build yourself a car adapter first, it will be a bit easier as you have more voltage head room. Although I can't see why your phone would have a problem with 4.5-4.75V

It could certainly be a problem if the power management IC on the phone/tablet/whatever interprets low voltage on the USB power lines as excessive voltage drop due to current draw and tries backing off to let USB voltage recover. I would not be surprised if 4.75V caused many devices to charge much slower than they should.

Yeah, in some ways, that Vdrop as cutoff to current increase is a pain in the buds.

If he is using 3AA to charge tablet, might as well forget it.  It is not worth the time.  500mA for an hour for a 8000mAH battery is not worth the trouble.  With a phone however, 500mA for an hour would half-charge most phone batteries.  Trying to draw 1A off AA battery is almost silly.  Like in the desert, it gets very hot, and goes dry quick.

[kxenos]

You would loose 40% of your power to boosting

Boost converter efficiencies can exceed 90%. Why 60%??

[sacherjj]

Do some reading about a "Minty Boost".   A fun little single AA USB charger and decoding the resistor divider charge currents of Apple. 

[Rick Law]

You would loose 40% of your power to boosting

Boost converter efficiencies can exceed 90%. Why 60%??

Well, typically the most favorable conditions are used.  So, to know the exact lost, you would need to hook it up in your typical environment and measure power-in verses power-out.

Most important factor to remember: The lower the input voltage, the more the lost.  Laws of Physics, can't get around that.

Rick