55pilot:
Internet surely is confusing, -sigh-
It sure is
Ok, so are you saying that putting rechargeable batts in a flashlight that requires 2 AA is scary?? Assuming they are both fully charged (or equally charged)
Not really, because you take them out and charge them together and put them back in together. If one has slightly higher leakage than the other, then over time it will get discharged more and charged less (depending on the charger). Eventually one may end up completely discharged while the other has remaining charge. Depending on the battery, that itself is enough to destroy the battery, however further use can cause the battery to be charged in reverse as the other cell(s) are used, leading to failure.
This is one of the reasons why lead acid batteries are routinely charged beyond the level at which they accept more charge, it acts to level the charge in all of the cells ensuring that a single weak cell does not prematurely fail.
The same thing happens in large battery banks such as those used in electric vehicles, where a very clever battery monitoring circuit ensures that no cell is ever allowed to over charge or discharge. I believe smaller versions of this are used in the electric model community for handling LiPo fast charging.
What's a 'ripple voltage', ESR? Please explain to someone that English isn't his native
A capacitor (or a battery) cannot be charged or discharged at an infinite rate. One thing that prevents this is the amount of resistance within the capacitor or battery. It acts exactly the same as having a perfect device with a resistor in series with it. Thus this internal resistance is called Equivalent Series Resistance (ESR)
Because of this resistance current flowing into the device will raise it's voltage due to IxR (in addition to any charging effect) and discharge will reduce the voltage by IxR in addition to any discharging effect.
So one effect of high ESR is that the voltage rises and falls more than it would otherwise for a given current.
The more important effect though is the (I^2)*R heating effect. This is one of the things that causes batteries to heat up as they are charges and discharged.
Ripple current is the current that flows in and out of a capacitor as it attempts to filter the peaks and troughs from it. Low ESR is important for this, and especially important where these currents are high. For switchmode power supplies it is even more important because ESR tends to rise with frequency (actually caused by inductance in the capacitor) and because high frequency noise on power rails tends to interfere with circuit operation more than lower frequencies (which have lower rise and fall times).
Then again, what type of batt is one of your choice for higher voltage?
It all depends on the load, the weight and size restrictions, charging constraints, etc, etc. There is no one perfect battery.
Steve:
I'm sorry if I mentioned anything confusing, I was saying a normal disposable 9V where you can get in store for $1
Yep
But the rechargeable one, I meant for my alternative, the 3AAs
Yep, remember that rechargeable cells often have a lower energy density than non-rechargeable cells of the same size. Obviously with the benefit of being able to be recharged.
And my purpose is, I want to power up things and not having to find a plug every time I need to charge my devices, phones, mp3s, etc. which from what I've searched, some has the capability of drawing as much as 500mA so I'm working with 500mA for the worst case.
Personally, I'd look at a 9V to 12V power supply using AA cells and a well smoothed buck regulator (on the basis that the 5V rail in most PCs comes from something similar and the devices should be able to handle it). But do you have the experience to do this?
At a minimum you need to take a part and follow the manufacturers recommendations *exactly*, and especially as regards to the printed circuit design and required capacitors, and diode, and inductor.., ok, everything. Some of these look a bit like 7805's (except for a few more legs) but you can't be anywhere near as cavalier with them.
I am a little more generous than 55p. I'd agree that you should not attempt to *design* a SMPS, but there are some out there that with guidance (principally from the manufacturer's datasheet) you should be able to use an existing good and simple design.
I guess the question then becomes, why do you need a LED illuminated when current is being drawn? It doesn't tell you how much capacity is being used, or how much capacity remains.
You mean, linear regulator looses difference as heat
whereas for buck regulator, it preserves P=IV as much as possible so voltage drop = current boost?
Yep, that's it. And voltage increase --> current decrease.
for perfect efficiency, Vin*Iin = Vout*Iout
but nothing is perfect. Depending on the application you may be looking at efficiencies from 75% to 98% (higher = harder and more expensive and larger etc, etc.)
So with an efficiency E (say 0.75) Vin*Iin*E = Vout*Iout
