Hey guys, this is my first time on this site, so I'm sorry if this post is in the wrong topic.
My question is;

If I wire 4 AA batteries in series I get;
6 Volts

If I wire 4 AA batteries in parallel I get;
1.5 Volts

But I need to figure out how many amps each of these circuits would produce. I've done about an hours worth of digging and I couldn't find anything except for people talking about Ah (amps/hour).
Thank you so much in advanced for all answers. Thanks

It's dependent on the batteries and what you have connected, since they're not ideal voltage sources.

You could find the internal resistance by experiment with, say, a decade resistor box in series with the batteries as a voltage source in order to find the Thevenin resistance (R_th) of the source as a whole. You could also measure V_oc (open circuit voltage) and I_sc (short-circuit current) to get R_th via Ohm's Law (well, Thevenin's Theorem, but it's still I = V / R).

Wikipedia notes that typical AA alkaline batteries can safely deliver about 700 mA without heating up.

This sounds like the sort of thing that one would see on on basic Circuits homework.. what's the reason for this query?

Ok well here's what I'm doing.

I want to take 4 AA batteries and most likely wire them in series to bring me to 6 Volts. I want to power a USB device, but in order to do that I believe I need 5V +/- 0.25V. So I was thinking of adding a 1n4001 diode to the DC output of the batteries so the voltage would drop to 5.3V. From there I was thinking of adding a resistor after that to bring it down to 5 Volts? I thought a 2 ohm 2 watt resistor would be okay, but I want to make sure I know exactly what I'm doing.

A resistor wouldn't work as you don't know the load resistance ahead of time (you'd effectively be forming a voltage divider, and the more current your USB device would draw the lower its effective load resistance would be and so the lower its supply voltage would be).

I reckon your best bet would be to use a 5V linear regulator as this will produce a clean and precise 5V supply with a minimal part count (usually just the voltage regulator and two or three capacitors). The problem is that regular voltage regulators have a drop-out voltage (i.e. difference between output voltage and minimum input voltage) of around 2V, so you would need to find a part with a drop-out voltage that is smaller than 1V (these are typically referred to as "low drop-out" regulators).



An alternative would be to use four rechargeable batteries as these supply 1.2V each rather than 1.5V.

[bah, Ben beat me to it. Ah well]
A resistor in series is definitely not what you want, then.

If you look at it (assuming the batteries are ideal voltage sources capable of sourcing infinite current), you're probably going to end up pulling huge amounts of current through the resistor (5 volts across it yields a huge 2.5A, for example), which is amazingly wasteful in power. Even more troublesome, the voltage drop in that large resistor is dependent on what the rest of the circuit is doing.

There are a couple of better ways to do this better.
First, you could just leave it at 5.3V as coming out of your initial diode. A typical rule of thumb for DC circuits (particularly around computers) is plus or minus 10% tolerance, so 5.3V is pretty safe for USB.

A 5V Zener diode rather than that plain diode will allow you to get closer to 5V (spot on within the diode's tolerances), but cuts out when input voltage drops below 5V, which may or may not be useful (it protects you from undervoltage when the batteries are somewhat depleted..).

Third, a cheap linear regulator such as the LM7805 will also clamp the output voltage at 5V, but it has a small voltage drop on the input (similar to a diode) and so has a very small band in this case where the output voltage will actually be 5V before it drops out of spec.

Fourth, similar to your current solution, you could use a voltage divider with a 5/6 ratio. With a voltage divider, though, to account for drop as the batteries age, I might prefer a ratio more like 9/12 to keep output a bit higher as input drops out of spec.

A 5.1V Zener diode would be my choice, personally, especially considering in my experience you need at least a 1.5V gap for something like a linear regulator to be able to work properly.

hardwarebook.info claims the max Voltage for USB is 5.25 Vdc, so 6V with 0.7 V through a diode would be about right. A linear regulator might be a little better, as you could set it up to put out 4.75, and then the batteries would last longer. More complex than a single diode, thought.

Dave

For the linear voltage regulators. I start with 6V, I could use the low voltage dropout 5V regulator and be fine, but what about those 3 capacitors, are they necessary? If they are, is there a formula I can use to calculate the correct uF I need for a 6V input? In the video he has a 12 or a 16V input, can't remember, and he used a 10uF capacitor, I doubt I'd have to use the same.

Do you have a particular objection to the 5.1V zener diode option, Mrmcman? I'm inclined to think that might be the easiest alternative.

Kerm, I have no experience with this, zener diode; I'm not sure if radio shack would have them, and if they did, how would I put it in my circuit?

edit: This has been revised so that it doesn't confuse anyone.
(At the first post)
You can draw as many amps from a battery as you want, just know that the ESR of the battery will cause it to heat up. I got from wikipedia that the ESR of a AA is about 0.9 Ohms. That means that if you draw 2 amps from a AA (P=I squared R) that the energy dissipated in watts would be about 4 watts. That's a lot of heat, and I am SURE the ESR was higher than that. Anyway, if you put batteries in parallel, you cut the ESR just like if you put resistors in parallel. That's how you should calculate it. (Kerm mentions that this is in an IDEAL world, so please leave room for error. also there are physical limits to this as well, involving saturation of the battery.)

Also, keep in mind the discharge curve of a battery. The more you current you draw, it dies about twice as fast. (actually, superlinearly as Kerm would put it. just know that it dies REALLY fast the more you push the envelope. (the one you aren't pushing anyway cuz ur current is so low))

(at the last post)
What I would do is use the 7805 voltage regulator, seeing as how he's a beginner. A zener diode offers no protection, a 7805 can take being wired backwards, short circuits, AND cuts off if the heat is too high.

(omit this paragraph if you are easily confused by my ranting) In terms of the dropoff voltage, who cares? If it drops below about 5.7 volts, the voltage out will drop linearly with the battery. If the battery is at 5.6, the voltage out of the 7805 typically (in my experience) will be 4.9. If the battery is at 5v, the output of the 7805 usually is around 4.3 volts. He'll just have to change his battery a little more often because the dropout wastes 0.7 volts of the usable battery life.

If it was me, I would use a 9v battery because they are a lot smaller anyway, and you won't have to worry about dropout voltages at all because the battery is dead long before it gets to 5.7 volts. Just don't draw much more than an Amp and you'll be fine. He said he was powering a USB device, and that's NEVER more than an amp (and I'm being generous here Kerm ) NEVER more than an amp. You'll blow a fuse in your computer if you draw more than that. Go with the 9v MrMcMan.

If it was myself, I would use a switching regulator with the proper inductor, capacitors, and resistors. Those will give the most efficient regulation possible (more efficient than linear regulators OR zener diodes). Just make sure the capacitor is big enough to smooth out the ripple for your load . (Kerm likes to use zeners though. )

It's a nice thought, but unfortunately, the reality of batteries doesn't quite match the math. There's a limit at which you simply can't pull any more current out of the batteries, even disregarding safety.

That has nothing to do with the discharge curve. To be more accurate, in fact, the remaining capacity will decrease at a superlinear rate the more current you pull, so if it lasts A minutes at B current, at 2B current it will last fewer than (even significantly fewer than) A/2 minutes.

Because if he's just trying to power a USB device with some AA batteries, a zener diode does a great job of giving him roughly 5V, is tiny and cheap and simple, and assuming he's not a total retard, the reverse-polarity protection is not a big deal.

I don't understand what you're saying here. How can the battery be producing two different voltages at the same charge level? Are you quoting loaded/unloaded potential? If so, that totally depends what the load is.

For what it's worth, the USB spec prevents devices from drawing more than 500mA from the host for a powered hub, and 100mA for an unpowered hub (although some are generous and will provide slightly more).

And I would do the Zener diode solution, going for simplicity for a solid-state load and fairly dependable supply.

I went with your approach and used a 5.1 volt zener diode. But while I was at radio shack I noticed their 4 aa battery packs weren't wired and I've never wired a battery pack before, so I went ahead and bought a couple 9v battery snaps with the wires already on them. Here's how I wired my circuit;

From the + side of the 9v I went to the anode side of the zener, then from the cathode side of the same zener to the red wire of the small USB output. I took the black wire from the same small USB output and ran that back to the black wire on the 9v button snap.

My phone takes 5volts to charge at 700ma, how come this isn't working? Any ideas?
P.s I soldered the wires together correctly so its gotta be something else.

Thanks everyone for all your help so far you guys are awesome!!!!

P.s.s I read online some radio shack battery holders have been known to be wired backwards, ie red to ground and black to hot... I switched the wires to make sure and it melted my solder.... yeahhh radiomshack had it correct haha

Nope, not quite. You need a loading resistor and the Zener parallel (but against) the flow. It will dissipate voltage above 5.1V:

How might I calculate how much resistance I need?

A back-of-the-envelope calculation says that with a 9V battery and a max current draw of 700mA, tossing in another 10% for 770mA for safety, you'd need a 3 Watt, 5 ohm resistor and a 3.8 Watt 5.1V Zener diode. That's quite a bit on the high side, so going with a 6V supply instead, you need only a 1 Watt resistor, roughly, but at around 1 ohm, a ridiculous value. I'm starting to think that a Zener regulator may not actually be appropriate for this application; I didn't realize you needed such high current. A linear/switching regulator might be better for such a high-current application.

You're not pulling 700 mA out of USB, that's for sure. That's probably the max current rating for the AC adaptor you're using.

A 9V battery is good for ~ 600 mAh. In other words, you can pull 600 mA for 1 hour, 100 mA for 6 hours etc. They are pretty useless for this type of thing, as they are expensive for the capacity you get.

AA's are good for 2500 mAh, so there's much more capacity there. plus the batteries are cheaper.


Take a look at this kit

http://www.adafruit.com/products/14

it looks like what you want to do, with a PCB and components.

2500mAh at 1.5V compared with 600mAh at 9V, though, so 3.75 Watt-hours for a AA and 5.4 Watt-hours for a 9V battery. Of course, put the four AAs together and you have 15 Watt-hours.

From a Watt hour point of view, a 9V isn't so bad. If you're powering something @ 5V then you need a switching regulator to get any sort of efficiency going 9V to 5V, otherwise you're dropping all your power in a linear regulator. To pull 100 mA at 5V from a 9V with a linear means you're dumping 100 mA*4V = 400mW minimum into heating up the linear regulator. With a Switcher, you'll pull ~ 70 mA from a 9V depending on switcher efficiency. Even then, you'll only get 5 to 6 hours of run time.

With a 4AA cell stack, you start out with 4*1.57 = 6.28 V (Alkaline cells) and so a linear regulator, assuming 6.28V in and 5V out, dumps 128mW into heating up a linear regulator, and your runtime is 5 times as long, with a much simpler, lower cost design.

Definitely agreed on all the points you made there, rfdave. Either way, linear or switching regular, it sounds like a regulator of sorts is necessary for something as relatively high-current as the OP is seeking to deal with.