5V output circuit

Your homework assignment would probably have specified the output current required of this circuit. amongst other things (ripple, noise, step response to name but a few).

Cheers

PeteS

look for DC-DC converters. you can get them already made in mods..

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

I'm making a USB charger so I'm not to sure about all the deets All I know is that it should output 5V or else the device will blow..

How much current do you need to deliver?

No Idea, however much USB needs??? Think it's either 100 or 500mA

EDIT: device (iPod) needs 500mA of power 100mA is not enough to power it.

A correct USB port will source 500 mA.

Don

This is more complex than it seems. Any device that requires more than

100mA, is meant to 'ask' the system if it can have the extra. Most USB ports _will_ supply 500mA, without such a request, but quite a few will give 'overcurrent' errors, if a device tries to draw this, without first getting the approval...

Best Wishes

So the theory is there but what will the circuit diagram look like??

OK, you want 5 V at up to 500 mA from two AA cells. That will run down the AA cells fairly quickly, but not absurdly quickly... I don't know what the battery life will be.

Since you're stepping up the voltage, this has to be a switching regulator, probably with an inductor (of carefully specified type) in the circuit. That is not nearly as easy as stepping the voltage down.

See this data sheet:

And this:

Energizer Alkaline AA lists 2850mAh nominal capacity at C/100, that is, at

25mA.

But 5V at 500mA is 2.5W; assume 70% efficiency from a boost converter home-made by a beginner, so the batteries need to supply about 3.6W total. Each battery supplies half of that power, so the draw is 1.8W/1.5V = 1.2A when the batteries are charged, going up to 1.8W/0.8V = 2.2A at the nominally-discharged point.

From the curves at

one can see that drawing around 2A is much higher than the battery is rated for, and substantially reduces its capacity. Extrapolating from those curves I'd guess the actual capacity at that rate is somewhere around 500mAh, meaning the batteries will last about 15 minutes. Probably not enough for a charge.

...

I wonder if he really needs 500 mA. What will this be powering?

I want to power my iPod Video, USB 2.0 device, technically it should use 500mA, maybe not. I've seen other designs on the internet using

2AA's powering the iPod for at least 3hours.

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Well, there's an easy way to find out. Get some 5V power supply and a mA meter.

petrus bitbyter

lol, First I need to MAKE a 5V power source???

Then it definitely does not consume 500 mA all the time.

The Maxim chip that I mentioned earlier is probably the way to go. There are not any large differences in efficiency between different chips of that general type.

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Not necessarily but that's what I should do if I hadn't one already. You can try to borrow one or use it at a school lab. You can even buy a 5V wallwart or a 9-12V type and connect a 5V regulator to it. But one thing is for sure: If you want to make something with nothing you will end up with nothing.

petrus bitbyter

The maxim chip seems great but maybe a little complex, I got my ideas from here:

And would now like to expand on the idea, I dont want to buy the kit because I want to marvel at my own design, I have a basic idea but now need to put th eplan into action. How woul 4 AA's 6V with a regulator compare with the 3V step up, the 6V would be much easier.