batteries are rated in AH (Amp Hours), or mAh (Milli Amp Hours). there are 1000 mAh in 1 Ah. amp hours are exactly how it sounds, a 1 Ah battery can source 1 amp for 1 hour. it is just a ratio: it can source 2 amps, but only for .5 hours. if your only pulling .5 amps, it will last for 2 hours. just for reference, a 12v car battery (lead acid, 15kg) is about 50 ah. the 12v batteries we use for robotics (gelcell, 7kg) are 17ah. a NiMh AA battery is ~1300mAh, but only 1.3v
Ok, so I have two NiMh AA batteries rated at 1800mah with 1.2v(each) and a NiMh at 150mah with 9v. Add them together and I get 3750mah = about 3ah. So it would last about 3 hours, thats a lot mroe than I thought.
wait wait wait, NO when you hook batteries up in series (to get 12v i assume) you cannot add current (because you are adding voltage) in series, each battery must contribute an equal amount of current. this means that the capacity of the whole array is the capacity of the smallest battery. you could only take 150mah out of it before the 9v battery died, meaning its resistance goes to infinity and the circuit is broken. if you were to hook 10 NiNH AA's together in series (to get ~12v, when they are fully charged they are 1.2v) you would have a TOTAL capacity of 1800 mAh. keyword: total so your system would last about half an hour.
Ok so what about when I hook them up in.....My mind just whent blank. But you know the other way to hook something up. What about that.
parallel is the other way to hook them up... in parallel, current is added, but voltage is not. if you try to hook a 9v battery and a 1.2v battery up in parallel, the 9v battery will charge the 1.2v battery until its dead, then your array will output a grand total of 1.2v. even 10 1.2v batteries in parallel will output only 1.2v this is why there are specific power saving laptop parts, and why people don’t build their own CD players its not as easy as it looks
Dang...And I just spent all my time working on it, LOL. Well I guess I am going to walmart to buy a cheap one. Ehh here are some pictures any who. The drive. I liked how it stood on its edge. ANd all that polishing work. SHINY. In real life it does look real cool (except the front). Any one got any ideas for it. (there is NO room in my case)
Oh wait I just got an idea, what if when they are in parrallel I put a diode in front of each adding the voltage and Ah. Code: ---<--diode----------<---diode---------<--diode--<- | | | | CD 9 1.2 1.2 Player Volt Volt Volt | | | | ---<--diode----------<---diode---------<--diode--<-
That wouldn't work at all You are putting diodes in pointing in opposite directions, which would cause no electricity to flow. I don't know where you get your idea's from, but simply put, there is no way (that I know of) to add both voltage and capacity of batteries. Why don't you just buy a single 12v battery? It's much simpler. Also, don't just pull 12v directly from a car battery for anything from a computer. The voltage can vary from 7-15 volts approx. _C
no... i assume that you meant to put the diodes on the bottom facing the other way, as you have it now, no current will flow (diodes allow current to flow only in one direction) so if they are facing the other way, each diode will drop the voltage .7v before any current will flow. this will eliminate the 1.2v batteries and making the whole thing put out 7.6 v with 150mAH. even if there is no voltage drop (hypothetical) they are still in parallel... thus still only adding the current capacity, and not the voltage. what would happen, if your circuit did function, would be that the load would see 9v for 150mah, and only when the 9v battery was dead would it see the 1.2v current. the potential of the 9v battery would not allow the 1.2v current to flow. the ONLY way that you can make 12v out of dissimilar batteries is to have a separate voltage multiplier circuit for each voltage of battery and then have the outputs of the multipliers in parallel. the problem with a voltage multiplier is that you divide the current by whatever you multiply the voltage by, because nothing is free…. and of course there is loss too. Nice try though
Ahh foiled again. *looks up voltage multipliers* EDIT: And yes the diodes were ment to go the other way on the bottom.
Ahh I got another Idea. Ok so what if I did some thing like this. Code: 12V---------(Diode?)->------12V | | | | VOltage Multiplier | = 12 volt at 1800Ah. | | | | 1.2V--------(diode?)->------1800Ah
alright foxx as much fun as your having... lets get one thing straight. V=I/R where V is voltage, I is current, and R is resistance. that formula means for any fixed load, the I/V ratio is equal to a constant. so you can never increase I without decreasing V, and you can never increase V without decreasing I. so, lets say this is a perfect world, without any loss related to component design or resistance. (READ: this could never happen) lets take your 2 AA and 1 9v batteries and see what we can do with them. first lets up convert your 9v 150mah battery: 9/12 = .75 so 75% of 150mah = 112mah @ 12v then also up convert two AA NiMh 1800mah batteries: 1.2 / 12 = .10 so 10% of 1800mah = 180mah x 2 = 360 mah @ 12v and add the 2 together to get a grand total of 472 MaH @ 12v remember, this is purely theoretical, in reality you would never get such a good conversion rate, there would be losses to heat, ect... but anyway... you have your 472 MaH at 12v. that will run your 2a (a generous assumption) CD player with controller for.... drum roll... 15 minutes get the picture? im telling you, this is a
Oh sorry, And maybey add some diodes between the two red slanted arrows or no diodes at all? EDIT: Just read your second post, Well as a modder/enigneer/inventer/ect I feel it is my right to try anything that pops up. EDIT2: Could also be connected through serial? EDIT3: After looking at my diagram and reading what you put, it is gunna be pritty much imposible.
no problem, thanks for the revision. (start theoretical point) this is the same prob with using a 9v battery and 1.2 volts at the same time... you cant get the current from 1.2v until the 12v is gone, and in this case as soon as the 12v is gone the 1.2v battery is dead and there is no 1800Mah current. (/end theoretical point) (start reality check)if you really were to hook up a battery in parallel with a voltage multiplier also powered by the battery, you would in effect short the battery out through a RC occlator... bad idea(/end reality check) and actualy i enjoy teaching, so ask away, it doesnt bother me at all... voltage multipliers cant be placed in serial with the devices that their being powered by, it has to do with the way they work... yep thats the idea. why dont you find a nice place on the shelf for that CD player, maybe someday youll find a computer that needs an extra one
The best and one of the only practical ways to do it, is to use a ~3.6V Li-Ion or NiMH pack or whatever. And run this to a 3.3V LDO regulator for 3.3V for your hardware. This would mean using a CF card or something that takes a 3.3V power supply. Alternativly, 4.8V from NiMH cells should run a notebook HDD. Maxim/DS have charging IC's for 3.6V LiIon's that can be powered from a USB port.
The shadow: Oh I think I have already found another idea. I was thinking of making a small sound sistem scince my old boom box doesnt support my cd-rws. IT would not work in my pc cause I coverd the thing in tape (to keap it from rusting)
