(Please note before reading that I'm a complete electronics noob. I know about ohm's law/etc from highschool but that's about it).

This is a very basic question for the electronics people here: When a ac->dc adapter has a specific voltage and current/wattage written on it, can you substitute the equivalent power supply with batteries and the appropriate resistor?

For example, I'm thinking of getting one of these: http://www.pcengines.ch/alix3d1.htm According to its docs, its power supply should be: Does this mean that I can put 2 9V batteries in a series (18V), with a 21.6 ohm resistor and the appropriate power plug to make a portable power supply? (21.6 ohms because wattage = voltage^2 / resistance = 18^2 / 21.6 = 15 watts). I know they don't make resistors that are exactly 21.6 ohms but I assume I can use a, say, 20 ohm resister, right?

Or am I totally off base here and should I stop before I kill myself via electrocution?

Also, for other devices that have off-voltage (say, 15V) and you only have 9V batteries, how do you "scale down the voltage"? I think I remember something about transformers (or those things with adjacent coils, where the ratio of the length of the coils is the ratio of the voltage difference, but iirc that works only for AC).

You must keep in mind that the batteries will only supply as much power as the device demands. The current draw is what will change, not the voltage, therefore "stepping down the voltage" at idle is not necessary.

You should also keep in mind that simple 9V batteries won't be able to provide that much power for any appreciable length of time.

If you want to use a portable supply I'd suggest looking into something like a lithium ion rechargeable battery pack. It should be quite easy to build and would likely prove to be more useful in the long run.

A current limiting resistor is a good idea, but be sure to include the voltage drop across the resistor in your calculations.

i dint think you'll need a resister.

p = v * i

Let me pull some of my old-Radio Shack knowledge from deep in my brain. Voltage is key, but most electronics are tolerant to a volt here or there, but do NOT go crazy. Your power for battery life will be key...most batteries (AA, AAA) can deliver up to 150ma each, so that can point in the right direction. Issue is with high power electronics (even the "greenest" (anyone remember the Green PC logos in the upper right hand corner on BIOS boots 10 years ago...ha!) the current draw will be something to consider. Lithium ion batteries have been the industry-nod for a couple years now over older-"memory-laden"-technologies like NiCd and NiMH, but they come at an expense. Most laptop replacement batteries end up being the signal for time to go shopping for a new laptop since retail prices on these Li-ion's usually are up to 50% the price of the laptop. In a matter of pure-expense the Shack sells 12v 4Ah batteries (stock number from top of my head 23-289?) that are mainly used in home security panels for power outage backup, through my years at the Shack I've seen these used all over the place for portable packs with a variety of step-down/up electronics. These are heavy as crap though, but are good for say outdoor-booth events etc.... And since the government is spending a trillion here and there, maybe apply for a solar-power computer grant...just give me 10% for the idea ;P

Nope, no resister is required. As stated the board will only draw as much power as it needs. It appears bare board requires 5W and anything attached will be more.

A 9V battery capacity is ~500maH so if the board was really pulling 15W that would be ~800ma and therefore it would only last ~30 minutes. The 12VDC 4Ah battery would last 3.2 hours.

So what is this project all about?

Voltage dividers, regulators, Zenor diodes are various methods to lower a DC voltage.

Oh and please stay away from wall sockets....

I believe google servers use lead-acid batteries ... car batteries as backups ... so yeah it should work. 7-20 V is a wide range, so 12V car battery should work ... has plenty of amps so it should run for a long time,

OK, in reverse order

• "they" do make resistors with all sorts of values, and some a lot closer to 21.6 ohms (but your local hobbyist store doesn't stock them, although you could make up the required value with series/parallel combinations), but that is irrelevant because...
• what is important is to stay in the operating range of the equipment at all times at which the equipment is to operate (and below the absolute maximum allowed voltage at all times). For this, it doesn't seem as if a resistor is required (and even if something like that was required, a voltage regulator may be a more sane choice...but if you are struggling with your understanding of resistors....)
• the reason that the equipment has a fairly wide operating voltage range is that it likely has its own internal voltage regulator. Adding a series resistor isn't really a help, except in that it moves some of the dissipation from the internal regulator to the resistor (but has consequences at higher current draws that may be unnacceptable)

I think the mistake that you were making initially was to calculate stuff about the power as a starting point. What you should have been doing was to regard Volts and Amps as a starting point and regard the power (watts) as something that arose consequentially from volts and amps. That is not to say that watts are unimportant; they are important, but starting from V & I and deriving W is what you should have done, rather than trying to modify W directly with a resistor.

Not transformers, you are correct that is only for AC (although you could convert the DC to AC and convert it back again, please regard this as an advanced subject for the moment).

If the 15V is really 15 +/-2 volts (or 13 to 17 volts), you always have to stay in that range. The battery will have an internal resistance and the open circuit voltage will drop as it discharges, which might mean that this is (slightly) difficult and the battery life on one charge is prematurely reduced by putting a resistor in series. In a case like this, a regulator can be helpful.

If instead the allowable voltage range was 15 +/- 4 volts, then an 18 volt battery could be used (subject to checks on max voltage fully charged, capacity, max current delivery, etc) and it all gets a lot easier.

this thread is related:
http://www.linuxquestions.org/questi...y-room-634019/

As a correction and being informative...from the years in a NOC...your entire telephone company and every reliable Tier 1/2 ISP runs off car batteries...but not the way you think...behind the muxes behind the routers behind the peering points is a room full of car batteries...telecom and data equipment is run off -48VDC...it's a loop of car batteries (well kind of...but wired backwards:? ) and the more batteries your have the longer you last when the local power company has an outage (or shuts you off....as telecom used to go :P )

Hi,
Most high load UPS utilize the 'lead calcium' battery racks for obvious load requirements. The life of a standard lead acid is rather short when you do a deep cycle of the cells. Most UPS are utilized with a monitor system that will take the cells down to a trip point that will protect the cells. Large load systems have dual rail that will allow a switch between racks to prevent damage to the battery system.

Cell maintenance is very important with this type of backup. Therefore the expense of batteries within the rack is rather high if not cycled properly.

Newer battery designs based on composite plates look promising for 'UPS' or even hybrid cell use.